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XXXIV OM - I - Problem 1
$ A $ tosses a coin $ n $ times, $ B $ tosses it $ n+1 $ times. What is the probability that $ B $ will get more heads than $ A $? | Consider the following events. $ Z_0 $ - the event that player $ B $ has flipped more heads than player $ A $, $ Z_r $ - the event that $ B $ has flipped more tails than player $ A $. These events are equally probable: if the coin is fair, it does not matter which side is called heads and which is called tails. Thus, $ P(Z_0) = P(Z_r) $. Now let $ a_0 $, $ a_r $, $ b_0 $, $ b_r $ be the number of heads and tails flipped by $ A $ and the number of heads and tails flipped by $ B $, respectively. According to the conditions of the problem, $ a_0+ a_r = n $, $ b_0+b_r=n+1 $. Therefore, by subtracting the equations, $ (b_0-a_0)+ (b_r-a_r) = 1 $. From this equation, it follows that one and only one of the numbers $ b_0-a_0 $ and $ b_r-a_r $ is positive, meaning that exactly one of the events $ Z_0 $ and $ Z_r $ must occur. In other words, these events are mutually exclusive and cover all possibilities. Therefore, $ P(Z_0)=P(Z_r)= 1/2 $. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LI OM - II - Problem 4
Point $ I $ is the center of the circle inscribed in triangle $ ABC $, where $ AB \neq AC $. Lines $ BI $ and $ CI $ intersect sides $ AC $ and $ AB $ at points $ D $ and $ E $, respectively. Determine all possible measures of angle $ BAC $ for which the equality $ DI = EI $ can hold. | We will show that the only value taken by angle $ BAC $ is $ 60^\circ $.
By the Law of Sines applied to triangles $ ADI $ and $ AEI $, we obtain $ \sin \measuredangle AEI = \sin \measuredangle ADI $. Hence,
om51_2r_img_6.jpg
First, suppose that the equality $ \measuredangle AEI = \measuredangle ADI $ holds (Fig. 1). Then also $ \measuredangle AIE = \measuredangle AID $, which means that triangles $ AEI $ and $ ADI $ are congruent (angle-side-angle criterion). Therefore, $ AD = AE $. This proves that triangles $ ADB $ and $ AEC $ are also congruent (angle-side-angle criterion). Hence, we obtain $ AB = AC $, which contradicts the assumptions made in the problem statement.
om51_2r_img_7.jpg
The remaining case to consider is when $ \measuredangle AEI + \measuredangle ADI = 180^\circ $ (Fig. 2). Then points $ A $, $ E $, $ I $, $ D $ lie on a single circle. Therefore,
From this, we obtain
which means $ \measuredangle BAC = 60^\circ $.
To complete the solution, it remains to show that there exists a triangle $ ABC $ in which $ AB \neq AC $, $ \measuredangle BAC = 60^\circ $, and $ DI = EI $. We will show more: in any triangle $ ABC $ with $ \measuredangle BAC = 60^\circ $, the equality $ DI = EI $ holds.
If $ \measuredangle BAC = 60^\circ $, then
Therefore, a circle can be circumscribed around quadrilateral $ AEID $. Since $ AI $ is the angle bisector of $ \angle EAD $, then $ DI = EI $. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXIV OM - II - Problem 6
For a given number $ n $, let $ p_n $ denote the probability that when a pair of integers $ k, m $ satisfying the conditions $ 0 \leq k \leq m \leq 2^n $ is chosen at random (each pair is equally likely), the number $ \binom{m}{k} $ is even. Calculate $ \lim_{n\to \infty} p_n $. | om34_2r_img_10.jpg
The diagram in Figure 10 shows Pascal's triangle written modulo $2$, meaning it has zeros and ones in the places where the usual Pascal's triangle has even and odd numbers, respectively. Just like in the usual Pascal's triangle, each element here is the sum of the elements directly above it, according to the addition table modulo 2: $0+0=0$, $0+1=1$, $1+0=1$, $1+1 = 0$. The drawn horizontal lines divide this triangle into layers:
- layer $W_0$ consisting of row number $0$,
- layer $W_1$ consisting of row number $1$,
- layer $W_2$ consisting of rows number $2$ and $3$,
- layer $W_3$ consisting of rows number $4$, $5$, $6$, $7$,
- $\ldots$
- layer $W_j$ consisting of rows with numbers from $2^{j-1}$ to $2^j-1$
- $\ldots$
We will prove by induction that in any layer $W_j$ there are two disjoint triangles, each identical to the triangle formed by the layers $W_0, W_1, \ldots, W_{j-1}$, and that these two triangles are separated by an inverted triangle composed entirely of zeros; the top row of the layer consists of extreme ones and zeros, and the bottom row of the layer consists of all ones.
For $j = 1, 2, 3$, this is visible from the diagrams (the diagonal lines define the division into the triangles mentioned). Assume the above relationships are true for some $j$ and consider the layer $W_{j+1}$. By the induction hypothesis, the last row of layer $W_j$ has the form $111\ldots 111$, and thus, according to the addition rule given above, the first row of layer $W_{j+1}$ has the form $100\ldots 001$. The extreme ones of this row give rise to two new copies of Pascal's triangle, and between them are zeros. Layer $W_{j+1}$ has as many rows as the layers $W_0, W_1, \ldots, W_j$ taken together, so at the base of each of these two triangles at the bottom of layer $W_{j+1}$ will be the same pattern of digits as in the last row of layer $W_j$, i.e., a row of ones. They will fill the last row of layer $W_{j+1}$. This completes the proof of the inductive thesis.
The number $p_n$ defined in the problem is the ratio of the number of zeros in the rows numbered from $0$ to $2^n$ to the number of all elements in these rows. These rows cover the layers $W_0, W_1, \ldots, W_n$ and the first row of layer $W_{n+1}$. From the relationship proved above, it follows that in the sum of layers $W_0, \ldots, W_j$ there are three times as many ones as in the sum of layers $W_0, \ldots, W_{j-1}\ (j = 1, 2, 3, \ldots)$, and since there is one one in layer $W_0$, there are $3^n$ ones in the sum of layers $W_0, \ldots, W_n$. Adding to this the two extreme ones of the next row, we see that there are $a_n = 3^n + 2$ ones in the considered fragment of Pascal's triangle. The total number of elements in this fragment is $b_n = 1 + 2 + 3 + \ldots + (2^n + 1) = (2^n + 1)(2^n + 2) / 2$, and thus there are $b_n - a_n$ zeros. The probability in question is
\[
\frac{b_n - a_n}{b_n} = \frac{(2^n + 1)(2^n + 2) / 2 - (3^n + 2)}{(2^n + 1)(2^n + 2) / 2}
\]
Hence,
\[
\lim_{n \to \infty} p_n = 1
\] | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Problem 8
Given a natural number $ n \geq 2 $ and an $ n $-element set $ S $. Determine the smallest natural number $ k $ for which there exist subsets $ A_1, A_2, \ldots, A_k $ of the set $ S $ with the following property:
for any two distinct elements $ a, b \in S $, there exists a number $ j \in \{1, 2, \ldots, k\} $ such that the set $ A_j \cap \{a, b\} $ is a singleton. | We will show that $ k = [\log_2(n-1) +1] $, i.e., $ 2^{k-1} < n \leq 2^k $.
Given the number $ k $ defined as above, we construct sets $ A_j $ as follows: we number the elements of set $ S $ with $ k $-digit binary numbers (leading zeros are allowed). We thus have $ 2^k \geq n $ numbers available. Then, we take $ A_i $ to be the set of elements of $ S $ whose number has a 1 in the $ i $-th position. Any two different elements $ a $ and $ b $ of set $ S $ are then assigned different numbers, which differ, say, in the $ j $-th position. In this case, exactly one of the elements $ a $, $ b $ belongs to the set $ A_j $.
The number $ k $ found in this way is the smallest. Suppose, for instance, that there exist sets $ A_1, A_2, \ldots, A_k $ satisfying the conditions of the problem for $ 2^k < n $. We assign to each element $ x $ of set $ S $ a sequence of $ k $ numbers $ (c_1, c_2, \ldots, c_k) $ according to the following rule:
Since $ 2^k < n $, there exist two different elements $ a, b \in S $ that are assigned the same sequence of numbers. This means that element $ a $ belongs to exactly the same sets among $ A_1, A_2, \ldots, A_k $ as element $ b $. | [\log_2(n-1)+1] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XLVIII OM - II - Problem 3
Given a set of $ n $ points ($ n \geq 2 $), no three of which are collinear. We color all segments with endpoints in this set such that any two segments sharing a common endpoint have different colors. Determine the smallest number of colors for which such a coloring exists. | Let's assume that the given points are vertices of some $n$-gon (if $n = 2$, of course, one color is enough). All segments of the same color determine disjoint pairs of vertices of the polygon. The maximum number of disjoint two-element subsets of an $n$-element set is $[n/2]$. Therefore, the number of colors is not less than
We will show that this number of colors is sufficient to meet the desired condition.
When $n$ is an odd number, we take the set of vertices of an $n$-sided regular polygon and paint the side and all diagonals parallel to it with the same color; segments not parallel are painted with different colors. We used $n$ colors.
When $n$ is an even number, we take the set of vertices of an $(n-1)$-sided regular polygon $V$ and one more point $P_0$. We color the sides and diagonals of the polygon $V$ in the manner described above, using $n - 1$ colors. For each family of parallel segments of the same color $k_i$, there remains one vertex $P_i$ of the polygon $V$ that is not an endpoint of any of these segments. We paint the segment $P_0P_i$ with color $k_i$. We used $n - 1$ colors, meeting the required condition. | [n/2] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - III - Problem 6
Determine all pairs of integers $ a $, $ b $, for which there exists a polynomial $ P(x) $ with integer coefficients, such that the product $ (x^2 + ax + b)\cdot P(x) $ is a polynomial of the form
where each of the numbers $ c_0,c_1,\dots ,c_{n-1} $ is equal to 1 or -1. | If the quadratic polynomial $x^2 + ax + b$ is a divisor of some polynomial $Q(x)$ with integer coefficients, then of course the constant term $b$ is a divisor of the constant term of the polynomial $Q(x)$. Therefore, in the considered situation, $b=\pm 1$.
No polynomial $Q(x)$ of the form $x^n \pm x^{n-1} \pm x^{n-2} \pm \ldots \pm x \pm 1$ has real roots with an absolute value greater than or equal to 2. Suppose, for instance, that a number $r$ with $|r| \geq 2$ satisfies the equation $Q(r) = 0$. Then
skipping some steps, we conclude that $|r| < 2$. This is a contradiction.
On the other hand, every root of the quadratic polynomial $T(x) = x^2 + ax + b$ is also a root of the polynomial $T(x) \cdot P(x)$.
Therefore, the quadratic polynomial $T(x)$ has no roots in the set $(-\infty, -2] \cup [2, +\infty)$.
Thus, $T(-2) > 0$ and $T(2) > 0$, which means $4 - 2a + b > 0$ and $4 + 2a + b > 0$.
If $b = 1$, then we have $5 > 2a$ and $5 > -2a$, which implies that $a$ is one of the numbers $-2, -1, 0, 1, 2$. If, however, $b = -1$, then $3 > 2a$ and $3 > -2a$, and the cases to consider are when $a$ is one of the numbers $-1, 0, 1$.
Direct verification shows that all the obtained pairs $(a, b)$ satisfy the conditions of the problem:
Answer: The conditions of the problem are satisfied by the following eight pairs $(a, b)$: (-2,1), (-1,-1), (-1,1), (0,-1), (0,1), (1,-1), (1,1), (2,1). | (-2,1),(-1,-1),(-1,1),(0,-1),(0,1),(1,-1),(1,1),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXVIII - I - Task 3
Let $ a $ and $ b $ be natural numbers. A rectangle with sides of length $ a $ and $ b $ has been divided by lines parallel to the sides into unit squares. Through the interiors of how many squares does the diagonal of the rectangle pass? | Consider a rectangle $ABCD$ with base $\overline{AB}$ of length $a$ and height $\overline{BC}$ of length $b$. Let's consider its diagonal $\overline{AC}$ connecting the lower left vertex $A$ with the upper right vertex $C$ (Fig. 4).
First, assume that the numbers $a$ and $b$ are relatively prime. If a vertex $E$ of a unit square contained within the rectangle $ABCD$ lies on $\overline{AC}$, then we have $\displaystyle \frac{a}{b} = \frac{AB}{BC} = \frac{AF}{EF}$, which means $a \cdot EF = b \cdot AF$, where $F$ is the projection of point $E$ onto line $AB$. Since $a$ and $b$ are relatively prime and $EF$ and $AF$ are natural numbers, it follows from the last equation that $a$ is a divisor of $AF$. This is impossible because $AF < AB = a$.
The obtained contradiction proves that the diagonal $\overline{AC}$ does not pass through any vertex of a unit square contained within the rectangle $ABCD$. Therefore, it defines a sequence of unit squares whose interiors it consecutively intersects. Each subsequent square in the sequence shares a side with the previous one and is located to the right or above it. By following the path defined by the squares of this sequence, we perform $a - 1$ moves to the right and $b - 1$ moves upwards. In total, we have $a + b - 2$ moves, meaning the path consists of $a + b - 1$ squares, i.e., the diagonal $\overline{AC}$ passes through the interiors of $a + b - 1$ unit squares.
If, however, the numbers $a$ and $b$ are not relatively prime and their greatest common divisor is $d$, then let $a = da'$ and $b = db'$. In this case, the numbers $a'$ and $b'$ are relatively prime. The diagonal of a rectangle with sides of length $a'$ and $b'$ consists of $d$ equal segments, each of which is the diagonal of a rectangle with sides of length $a'$ and $b'$. Based on the already considered case, each such segment of the diagonal passes through the interiors of $a' + b' - 1$ unit squares, and therefore the entire diagonal passes through the interiors of $d(a' + b' - 1)$ unit squares. | \gcd(,b) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - I - Problem 4
The school organized three trips for its 300 students. The same number of students participated in each trip. Each student went on at least one trip, but half of the participants in the first trip, one-third of the participants in the second trip, and one-fourth of the participants in the third trip only went on one trip.
How many students went on each trip? How many participants in the first trip also participated in the second, and how many of them also participated in the third trip? | Let $ x $ denote the number of participants in each trip, and $ y $, $ z $, $ u $, $ w $ denote, respectively, the numbers of students who went: a) only on the first and second trip, b) only on the first and third trip, c) only on the second and third trip, d) on all three trips.
The total number of students equals the sum of: 1) the number of participants in only one trip, i.e., the number $ \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13}{12}x $, 2) the number of participants in only two trips, and 3) the number of participants in all three trips, thus
The number of participants in the first trip is the sum of the numbers: 1) students who went only on the first trip, 2) students who went on the first and second or first and third trip, and 3) students who went on all three trips, thus
Similarly, for the number of participants in the second and third trips, we obtain the equations
We need to find non-negative integer solutions to the system of equations (1) - (4). Suppose such a solution is the set of numbers $ x $, $ y $, $ z $, $ u $, $ w $. From equation (1), it follows that $ \frac{13}{12} x $ is an integer, so $ x $ is divisible by $ 12 $. Let $ x = 12t $; equations (1) - (4) take the form
Adding equations (5) and (8) side by side, we get the equation
Since $ y \geq 0 $, it follows from (9) that $ t \leq \frac{300}{22}=13.6\ldots $, thus
Adding (6) and (7) side by side and subtracting (8), we get
Since $ w \geq 0 $, it follows from equation (11) that $ y \leq \frac{5}{2} t $, from which, using equation (9), we conclude that $ 22t + \frac{5}{2}t \geq 300 $, thus
$ t \geq \frac{600}{49} = 12.2\ldots, $ hence
From (10) and (11), it follows that
From equations (9), (11), (6), and (7), we obtain sequentially $ y = 14 $, $ w = 37 $, $ z = 27 $, $ u = 53 $. It is easy to verify that the found numbers satisfy the system of equations (1) - (4), thus they constitute the only solution to the problem. | x=120,y=14,z=27,u=53,w=37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LIV OM - III - Task 3
Determine all polynomials $ W $ with integer coefficients that satisfy the following condition: for every natural number $ n $, the number $ 2^n-1 $ is divisible by $ W(n) $. | Let $ W(x) = a_0 + a_1x +\ldots + a_rx^r $ be a polynomial with integer coefficients, satisfying the given condition. Suppose that for some natural number $ k $, the value $ W(k) $ is different from $ 1 $ and $ -1 $; it therefore has a prime divisor $ p $. Let $ m = k + p $. The difference $ W(m) - W(k) $ is the sum of differences of the form $ a_j(m^j - k^j) $, which are numbers divisible by $ m - k $, i.e., by $ p $. Therefore, $ W(m) $ is divisible by $ p $.
From the condition of the problem, it follows that $ p $ is a divisor of the numbers $ 2^k - 1 $ and $ 2^m - 1 $; it is therefore an odd prime divisor of the number $ 2^m - 2^k = 2^k (2^p - 1) $, i.e., it is a divisor of the number $ 2^p - 1 $. This, however, is not possible, because
the values of the binomial coefficient $ {p \choose i} $ are for $ i = 1,\ldots,p- 1 $ numbers divisible by $ p $.
The contradiction proves that $ W(n) = \pm 1 $ for every natural number $ n $. Hence, it follows that $ W $ is a polynomial identically equal to $ 1 $ or identically equal to $ -1 $. Of course, each of these two polynomials satisfies the postulated condition. | W(x)=1orW(x)=-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
VI OM - II - Task 3
What should be the angle at the vertex of an isosceles triangle so that a triangle can be constructed with sides equal to the height, base, and one of the remaining sides of this isosceles triangle? | We will adopt the notations indicated in Fig. 9. A triangle with sides equal to $a$, $c$, $h$ can be constructed if and only if the following inequalities are satisfied:
Since in triangle $ADC$ we have $a > h$, $\frac{c}{2} + h > a$, the first two of the above inequalities always hold, so the necessary and sufficient condition for the existence of a triangle with sides $a$, $c$, $h$ is the inequality
From triangle $ADC$ we have $h = a \cos \frac{x}{2}$, $\frac{c}{2} = a \sin \frac{x}{2}$; substituting into inequality (1) gives
or
and since $\frac{x}{4} < 90^\circ$, the required condition takes the form
or
Approximately, $4 \arctan -\frac{1}{2} \approx 106^\circ$ (with a slight deficit). | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXIX OM - II - Problem 4
From the vertices of a regular $2n$-gon, 3 different points are chosen randomly. Let $p_n$ be the probability that the triangle with vertices at the chosen points is acute. Calculate $\lim_{n\to \infty} p_n$.
Note. We assume that all choices of three different points are equally probable. | Let $ A_1, A_2, \ldots , A_{2n} $ be the consecutive vertices of a regular $ 2n $-gon, and $ O $ - the center of the circle circumscribed around this polygon. We will investigate how many triangles $ A_1A_iA_j $, where $ i < j $, have the angle $ \measuredangle A_iA_1A_j $ not acute.
Since the inscribed angle in a circle has a measure that is half the measure of the central angle subtended by the same arc, therefore
Thus, $ \measuredangle A_iA_1A_j $ is not acute if and only if $ j - i \geq n $. We will calculate the number of such pairs $ (i, j) $, such that
From these inequalities, it follows that $ i \leq j - n \leq 2n - n = n $. For each $ i $ satisfying $ 2 \leq i \leq n $, there are $ n - i + 1 $ values for $ j $ satisfying (1), namely $ j = i + n, i + n + 1, \ldots, 2n $. Therefore, the number of such pairs is
We have thus proved that the number of non-acute triangles $ A_1A_iA_j $, in which the angle $ \measuredangle A_iA_1A_j $ is not acute, is equal to $ \frac{n(n-1)}{2} $. Since in a triangle at most one angle is not acute, the number of all non-acute triangles $ A_iA_jA_k $ is equal to
The number of all triangles $ A_iA_jA_k $ is equal to the number of three-element subsets of the $ 2n $-element set $ \{A_1, A_2, \ldots , A_{2n} \} $, that is, the number
Therefore,
Thus, $ \displaystyle \lim_{n \to \infty} p_n = 1 - \frac{3}{4} = \frac{1}{4} $.
om29_2r_img_12.jpg
Note. The triangle $ A_1A_iA_j $, where $ 1 < i < j \leq 2n $, is acute if and only if $ i \leq n $, $ j > n $, and $ j - i < n $. By associating the triangle $ A_1A_iA_j $ with the point $ (i,j) $, we see that the acute triangles correspond to lattice points located in the shaded triangle (Fig. 12), and any triangles correspond to lattice points belonging to the triangle $ AOB $. The ratio of the number of lattice points in the shaded triangle to the number of all lattice points in the triangle $ AOB $ approaches the ratio of the areas of these triangles as $ n \to \infty $. This ratio of areas is equal to $ \frac{1}{4} $. | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
IX OM - II - Task 2
Six equal disks are placed on a plane in such a way that their centers lie at the vertices of a regular hexagon with a side equal to the diameter of the disks. How many rotations will a seventh disk of the same size make while rolling externally on the same plane along the disks until it returns to its initial position? | Let circle $K$ with center $O$ and radius $r$ roll without slipping on a circle with center $S$ and radius $R$ (Fig. 16). The rolling without slipping means that different points of one circle successively coincide with different points of the other circle, and in this correspondence, the length of the arc between two points of one circle equals the length of the arc between the corresponding points of the other circle.
Consider two positions $K$ and $K_1$ of the moving circle, corresponding to centers $O$ and $O_1$ and points of tangency with the fixed circle $A$ and $B_1$; let $B$ be the point on the circumference of circle $K$ that, in the new position, passes to the point of tangency $B_1$, and let $\alpha$ and $\beta$ denote the radian measures of angles $ASB_1$ and $AOB$. Since the length of arc $AB_1$ of the fixed circle equals the length of arc $AB$ of the moving circle, we have $R \alpha = r \beta$. The radius $OA$ of the moving circle will take the new position $O_1A_1$, with $\measuredangle A_1O_1B_1 = \measuredangle AOB = \beta$. Draw the radius $O_1A_0$ parallel to $OA$; we get $\measuredangle A_0O_1B_1 = \measuredangle ASB_1 = \alpha$. The angle $A_0O_1A_1 = \alpha + \beta$ is equal to the angle through which circle $K$ has rotated when moving from position $K$ to position $K_1$. If $R = r$, then $\beta = \alpha$, and the angle of rotation is then $2 \alpha$.
Once the above is established, it is easy to answer the posed question.
The moving disk $K$ rolls successively on six given disks; the transition from one to another occurs at positions where disk $K$ is simultaneously tangent to two adjacent fixed disks. Let $K_1$ and $K_2$ be the positions where disk $K$ is tangent to fixed disk $N$ and to one of the adjacent disks (Fig. 17). Then $\measuredangle O_1SO_2 = 120^\circ$, so when moving from position $K_1$ to position $K_2$, the disk rotates by $240^\circ$. The disk will return to its initial position after $6$ such rotations, having rotated a total of $6 \cdot 240^\circ = 4 \cdot 360^\circ$, i.e., it will complete $4$ full rotations.
Note. We can more generally consider the rolling of a disk with radius $r$ on any curve $C$ (Fig. 18). When the point of tangency $A$ traverses an arc $AB_1$ of length $l$ on the curve, the angle through which the disk rotates is $\alpha + \beta$, where $\alpha$ is the angle between the lines $OA$ and $O_1B_1$, i.e., the angle between the normals to the curve $C$ at points $A$ and $B$, and $\beta = \frac{l}{r}$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXVI OM - III - Problem 1
Determine the largest number $ k $ such that for every natural number $ n $ there are at least $ k $ natural numbers greater than $ n $, less than $ n+17 $, and coprime with the product $ n(n+17) $. | We will first prove that for every natural number $n$, there exists at least one natural number between $n$ and $n+17$ that is coprime with $n(n+17)$.
In the case where $n$ is an even number, the required property is satisfied by the number $n+1$. Of course, the numbers $n$ and $n+1$ are coprime. If a number $d > 1$ were a common divisor of $n+1$ and $n+17$, then it would divide the difference $(n+17)-(n+1) = 16$, and thus would be an even number. However, since $n+1$ is an odd number, such a $d$ does not exist. Therefore, the numbers $n+1$ and $n(n+17)$ are coprime.
In the case where $n$ is an odd number, the required property is satisfied by the number $n + 16$. Indeed, the numbers $n+16$ and $n+17$ are coprime. Similarly, as above, we conclude that $n$ and $n+16$ are also coprime. Therefore, $n+16$ and $n(n+17)$ are coprime.
In this way, we have shown that $k \geq 1$.
Consider the number $n = 16!$. The numbers $n+2, n+3, \ldots, n+16$ are not coprime with $n(n+17)$, because for $j = 2, 3, \ldots, 16$, the numbers $n+j$ and $n$ are divisible by $j$. Only the number $n+1$ is coprime with $n(n+17)$. From this example, it follows that $k \leq 1$. Therefore, $k = 1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LVIII OM - I - Problem 1
Problem 1.
Solve in real numbers $ x $, $ y $, $ z $ the system of equations | Subtracting the second equation of the system from the first, we get
Similarly, by subtracting the third equation from the first, we obtain the equation $ 0=(x-z)(x+z-2y+5) $. Therefore, the system of equations given in the problem is equivalent to the system
The product of two numbers is equal to zero if and only if at least one of the factors is equal to zero. Therefore, system (1) is equivalent to the following alternative of four systems:
In the first of the systems (2), the second and third equations lead to the conclusion that $ x=y=z $. Using this relationship in the first equation of this system, we obtain the quadratic equation $ 3x^2+5x=2 $, which has two solutions: $ x=-2 $ and $ x={1\over 3} $. Thus, the considered system of equations has two solutions: $ (-2,-2,-2) $ and $ ({1\over 3},{1\over 3},{1\over 3}) $.
Now let's examine the second system (2). The second equation of this system means that $ x=y $; combining this with the third equation, we see that $ z=-x+2y-5=x-5 $. Therefore, the first equation of the system takes the form
$$2=x^2+2yz+5x=x^2+2x(x-5)+5x=3x^2-5x.$$
The quadratic equation $ 3x^2-5x=2 $ has solutions $ x=2 $ and $ x=-{1\over 3} $. This gives two solutions of the system: $ (2,2,-3) $ and $ (-{1\over 3},-{1\over 3},-{16\over 3}) $.
Similarly, we solve the third system (2), obtaining the solutions $ (2,-3,2) $ and $ (-{1\over 3},-{16\over 3},-{1\over 3}) $.
Finally, let's move on to the last system (2). Combining the second and third equations, we see that $ y-2z=z-2y $, which means $ y=z $. Therefore, the second equation takes the form $ y=x+5 $, and the first can be written as
$$2=x^2+2yz+5x=(y-5)^2+2y^2+5(y-5)=3y^2-5y,$$
which, after solving the quadratic equation, gives $ y=2 $ and $ y=-{1\over 3} $. As a result, we get two solutions: $ (-3,2,2) $ and $ (-{16\over 3},-{1\over 3},-{1\over 3}) $.
Finally, it remains to verify that each of the 8 triples indeed represents a solution to the system of equations given in the problem.
Answer: The solutions $ (x, y, z) $ of the given system of equations are: | (x,y,z)=(-2,-2,-2),(\frac{1}{3},\frac{1}{3},\frac{1}{3}),(2,2,-3),(-\frac{1}{3},-\frac{1}{3},-\frac{16}{3}),(2,-3,2),(-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - I - Problem 6
Factor the quadratic polynomial into real factors
where $ p $ and $ q $ are real numbers satisfying the inequality
Please note that the mathematical expressions and symbols are kept as they are, only the text has been translated. | From the inequality $ p^2 - 4q < 0 $, it follows that the quadratic polynomial $ y^2 + py + q $ has no real roots; therefore, the quartic polynomial $ x^4 + px^2 + q $ also has no real roots. If the number $ x $ were a real root of the second polynomial, then the number $ y = x^2 $ would be a real root of the first polynomial. Therefore, the desired factorization can only have the form
Comparing the coefficients of the same powers of $ x $ on both sides of the above equality, we obtain the equations
\[ (1) \qquad \alpha + \gamma = 0, \]
\[ (2) \qquad \beta + \delta + \alpha \gamma = p, \]
\[ (3) \qquad \alpha \delta + \beta \gamma = 0, \]
\[ (4) \qquad \beta \delta = q. \]
From equation (1) we have $ \gamma = - \alpha $, substituting into equation (3) gives
a) If $ \alpha = 0 $, then from equations (2) and (4) it follows that
This system of equations has no real solutions, because $ p^2 - 4q < 0 $.
b) $ \delta = \beta $; from equation (4) we then obtain
The above values of $ \beta $ and $ \delta $ are real, since from the inequality $ p^2 - 4q < 0 $ it follows that $ 4q > p^2 $, hence $ q > 0 $.
According to equation (2) $ \alpha \gamma = p - \beta - \delta $, so to determine $ \alpha $ and $ \gamma $ we have the system of equations
Therefore
From the inequality $ 4q > p^2 $ it follows that $ 2 \sqrt{q} > |p| \geq p $, hence $ 2 \sqrt{q} - p > 0 $; Since $ 4q - p^2 = (2 \sqrt{q} - p) (2 \sqrt{q} + p) > 0 $, then also $ 2 \sqrt{q} + p > 0 $.
To obtain a real value for $ \alpha $, we need to take the upper sign in the above formula for $ \alpha^2 $. We obtain the following real values of the coefficients $ \alpha $, $ \beta $, $ \gamma $, $ \delta $:
The desired factorization is expressed by the formula: | x^4+px^2+(x^2+\alphax+\beta)(x^2-\alphax+\beta) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Zadanie 5
Wyznaczyć najmniejszą liczbę naturalną $ n $, dla której liczba $ n^2-n+11 $ jest iloczynem czterech liczb pierwszych (niekoniecznie różnych).
|
Niech $ f(x) = x^2-x+11 $. Wartości przyjmowane przez funkcję $ f $ dla argumentów całkowitych są liczbami całkowitymi niepodzielnymi przez $ 2 $, $ 3 $, $ 5 $, $ 7 $. Przekonujemy się o tym badając reszty z dzielenia $ n $ i $ f(n) $ przez te cztery początkowe liczby pierwsze:
\begin{tabular}{lllll}
&\multicolumn{4}{l}{Reszty z dzielenia:}\\
&przez 2&przez 3&przez 5&przez 7\\
$ n $&0 1&0 1 2 &0 1 2 3 4&0 1 2 3 4 5 6\\
$ f(n) $&1 1&2 2 1&1 1 3 2 3& 4 4 6 3 2 3 6
\end{tabular}
Zatem dowolna liczba $ N $ będąca wartością $ f $ dla argumentu naturalnego i spełniająca podany w zadaniu warunek musi mieć postać $ N = p_1p_2p_3p_4 $, gdzie czynniki $ p_i $ są liczbami pierwszymi $ \geq 11 $.
Najmniejsza z takich liczb $ N= 11^4 $ prowadzi do równania kwadratowego $ x^2-x+11=11^4 $ o pierwiastkach niewymiernych. Ale już druga z kolei $ N = 11^3 \cdot 13 $ jest równa wartości $ f(132) $. Funkcja $ f $ jest ściśle rosnąca w przedziale $ \langle 1/2; \infty) $ wobec czego znaleziona minimalna możliwa wartość $ N $ wyznacza minimalną możliwą wartość $ n $. Stąd odpowiedź: $ n = 132 $.
| 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXV OM - III - Task 2
Salmon swimming upstream must overcome two waterfalls. The probability that a salmon will overcome the first waterfall in a given attempt is $ p > 0 $, and the probability of overcoming the second waterfall in a given attempt is $ q > 0 $. We assume that successive attempts to overcome the waterfalls are independent. Calculate the probability of the event that the salmon will not overcome the first waterfall in $ n $ attempts, given that in $ n $ attempts it will not overcome both waterfalls. | Let $ A_n $ be the event that the salmon does not overcome the first waterfall in $ n $ attempts, and $ B_n $ - the event that the salmon does not overcome both waterfalls in $ n $ attempts. Since the probability that the salmon does not overcome the first waterfall in one attempt is $ 1 - p $, and the attempts are independent, therefore
Event $ B_n $ will occur when either the salmon does not overcome the first waterfall in $ n $ attempts, or it overcomes the first waterfall in the $ k $-th attempt $ (1 \leq k \leq n) $ and does not overcome the second waterfall in the remaining $ n-k $ attempts. Therefore,
If $ p = q $, then formula (2) takes the form
If $ q = 1 $, then
In this case, the salmon would overcome the second waterfall in the first attempt. Therefore, event $ B_n $ will occur when the salmon does not overcome the first waterfall in $ n $ attempts or overcomes it only in the $ n $-th attempt.
Formula (4) follows from formula (2) if we assume that $ 0^0 = 1 $.
Finally, if $ p \ne q $ and $ q < 1 $, then using the formula for the sum of the terms of a geometric sequence, we transform (2) as follows:
We thus have in this case
If event $ A_n $ occurs, then event $ B_n $ will certainly occur. Therefore, $ A_n \cap B_n = A_n $. From the formula for conditional probability, we obtain
If $ n = 1 $, then from the conditions of the problem, $ p(A_n) = 1 - p $ and $ p(B_n) = 1 $. Therefore, from (6) we obtain that $ p(A_n \mid B_n) = 1 - p $. Henceforth, we will assume that $ n \geq 2 $.
Notice that if $ p \ne q $, then $ p(B_n) \ne 0 $. Indeed, if it were, then from formula (5) we would obtain $ p(1 - q)^n = q(1 - p)^n $. From this,
If, for example, $ p < q $, then $ 1 - p > 1 - q $ and equality (7) cannot hold. Similarly, we prove that (7) does not hold if $ p > q $.
If $ p = q $ and $ p(B_n) = 0 $, then from formula (3) it follows that $ p = 1 $. Therefore, in the case where $ p = q = 1 $ (and only in this case) the conditional probability $ p(A_n \mid B_n) $ does not exist. We calculate it in the remaining cases.
If $ p = q < 1 $, then from (1), (3), and (6) we obtain
If $ p < q = 1 $, then from (1), (4), and (6) we obtain
If $ p \ne q < 1 $, then from (1), (5), and (6) we obtain
Note.
We will examine in each case the number $ g = \lim_{n \to \infty} p(A_n \mid B_n) $.
If $ p = q < 1 $, then from formula (8) it follows that $ g = 0 $.
If $ p < q = 1 $, then from formula (9) it follows that $ g = 1 - p = 1 - \frac{p}{1} $.
If $ p < q < 1 $, then $ 1 - p > 1 - q $ and $ \lim_{n \to \infty} \left( \frac{1-p}{1-q} \right)^n = \infty $. Therefore, from (10) it follows that $ g = \frac{p-q}{-q} = 1 - \frac{p}{q} $.
If $ q < p \leq 1 $, then $ 1 - p < 1 - q $ and $ \lim_{n \to \infty} \left( \frac{1-p}{1-q} \right)^n = 0 $. Therefore, from (10) we obtain that $ g = 0 $.
Thus, in all cases, $ g = \max \left(0, 1 - \frac{p}{q} \right) $. | \max(0,1-\frac{p}{q}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - I - Task 7
In an $ n $-person association, there are $ 2n-1 $ committees (any non-empty set of association members
forms a committee). A chairperson must be selected in each committee. The following condition must be met: If
committee $ C $ is the union $ C = A\cup B $ of two committees $ A $ and $ B $, then the chairperson of committee $ C $ is also
the chairperson of at least one of the committees $ A $, $ B $. Determine the number of possible selections of chairpersons. | Answer: n!.
We will establish a one-to-one correspondence between the possible choices of committee chairs
and the ways of assigning members of the association the numbers $1, 2, \dots, n$ (where different members
are assigned different numbers). There are, of course, as many of the latter as there are permutations of an $n$-element set,
which is $n!$.
Suppose, then, that in each committee a chair has been chosen in accordance with the assumptions of the problem.
We assign the members of the association numbers inductively as follows:
The number 1 is assigned to the chair of the committee that includes all members of the association.
The number 2 is assigned to the chair of the committee that includes all members of the association except the member with number 1.
The number 3 is assigned to the chair of the committee that includes all members of the association except those with numbers 1 and 2, and so on.
We will show that with this method of assigning numbers, the following condition is satisfied:
(*) The chair of any committee is the one among its members who has been assigned the smallest number.
Indeed, consider any committee $A$ and let $k$ be the smallest of the numbers assigned to the members
of this committee. Let $B$ be the committee consisting of all members of the association who have been assigned numbers
not less than $k$. From the method of assigning numbers, it follows that the chair of committee $B$ is the member
with number $k$. Moreover, committee $B$ is the union of two committees: $A$ and $B \setminus A$. Since the member with number
$k$ does not belong to committee $B \setminus A$, according to the conditions of the problem, he must be the chair of committee $A$.
We have thus defined a mapping that assigns a way of assigning the members of the association the numbers $1, 2, \dots, n$ to a choice of committee chairs. To complete the solution of the problem, we will now define the inverse mapping.
Namely, if the members of the association have been assigned the numbers $1, 2, \dots, n$, then in each committee
we choose as chair the member who has the smallest number (condition (*) will then be satisfied).
If committee $C$ is the union $C = A \cup B$ of two committees $A, B$ and the member with number $k$ is the chair
of committee $C$, then he belongs to one of the committees $A, B$, in which he is the member with the lowest number — and thus is
its chair. We see, then, that the condition given in the problem statement is satisfied. Therefore, if
we start with the choice of committee chairs, assign numbers to the members of the association according to
the previously defined rule, and then reconstruct the committee chairs from these numbers, we
will obtain the original choice of chairs. Similarly, if we assign numbers to the members of the association, choose the committee chairs
based on these numbers, and then assign the members numbers according to the described inductive procedure, the
numbers will be identical to the original numbers. Thus, the defined mappings are mutually
inverse, which completes the solution. | n! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XVII OM - I - Problem 4
On a plane, a circle and a point $ M $ are given. Find points $ A $ and $ B $ on the circle such that the segment $ AB $ has a given length $ d $, and the angle $ AMB $ is equal to a given angle $ \alpha $. | Suppose $ A $ and $ B $ are points on a given circle satisfying the conditions $ AB = d $, $ \measuredangle AMB = \alpha $ (Fig. 5).
Let us choose any chord $ CD $ of length $ d $ in the given circle. Rotate the triangle $ AMB $ around the center $ O $ of the given circle by the angle $ HOK $, where $ H $ and $ K $ denote the midpoints of the chords $ AB $ and $ CD $. After the rotation, point $ A $ will be at point $ C $, and point $ B $ at point $ D $ or vice versa; point $ M $ will move to some point $ N $. Point $ N $ a) lies on the circle with center $ O $ and radius $ OM $, b) lies on such an arc of the circle whose endpoints are points $ C $ and $ D $ and which contains the inscribed angle $ \alpha $. Conversely, if we determine a point $ N $ satisfying the above conditions a) and b), then by rotating the triangle $ CND $ around $ O $ by the angle $ NOM $, we obtain the triangle $ AMB $, whose base is the desired segment. Since there are two arcs with endpoints $ C $, $ D $ containing the inscribed angle $ \alpha $, the number of solutions will be $ 4 $, $ 3 $, $ 2 $, $ 1 $, or $ 0 $, depending on the length of the segment $ OM $. | 4,3,2,1,或0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXII - I - Problem 10
Determine all functions $ f $ mapping the set of all rational numbers $ \mathbb{Q} $ to itself that satisfy the following conditions:
a) $ f(1)=2 $,
b) $ f(xy) = f(x)f(y)-f(x+y)+1 $ for $ x, y \in \mathbb{Q} $. | If the function $ f $ satisfies conditions a) and b), then the function $ g: \mathbb{Q} \to \mathbb{Q} $ defined by the formula $ g(x) = f(x)- 1 $ satisfies the conditions
a') $ g(1) = 1 $
and $ g(xy)+1 = (g(x)+1)(g(y)+1)-g(x+y)-1+1 $, $ g(xy)+1 = g(x)g(y)+g(x)+g(y)+1-g(x+y) $, i.e.
b') $ g(xy)+g(x+y) = g(x)g(y)+g(x)+g(y) $.
Substituting $ y = 1 $ in the last equality, we get
thus, by a')
Hence
Substituting $ x = 0 $ and $ x = -1 $ in the last equality, we will determine, remembering a')
By obvious inductive reasoning, it follows from this and (*), that
for any integer $ n $, and
for an integer $ m $ and a rational $ x $.
Now let $ m $ be an integer, and $ n $ a natural number.
To determine $ g\left( \frac{m}{n} \right) $, substitute $ x = m $, $ y = \frac{1}{n} $ in equality b).
Since $ g(m)= m $ and $ g\left( m+\frac{1}{n} \right)=m+g \left( \frac{1}{n} \right) $, we have
If $ m = n $, then $ 1 = g(1) = n \cdot g \left( \frac{1}{n} \right) $, from which it follows that
Therefore
This means that if the function $ g $ satisfies conditions a') and b'), then $ g(x) = x $ for every rational $ x $. Therefore, if the function $ f $ satisfies conditions a) and b), then $ f(x) = x+ 1 $. On the other hand, the function $ f $ defined by this formula obviously satisfies conditions a) and b). The only solution to the problem is therefore the function $ f: \mathbb{Q} \to \mathbb{Q} $, defined by the formula $ f(x) = x+1 $. | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XII OM - II - Task 4
Find the last four digits of the number $ 5^{5555} $. | \spos{1} We will calculate a few consecutive powers of the number $ 5 $ starting from $ 5^4 $:
It turned out that $ 5^8 $ has the same last four digits as $ 5^4 $, and therefore the same applies to the numbers $ 5^9 $ and $ 5^5 $, etc., i.e., starting from $ 5^4 $, two powers of the number $ 5 $, whose exponents differ by a multiple of $ 4 $, have the same last four digits. The number $ 5^{5555} = 5^{4\cdot 1388+3} $ therefore has the same last $ 4 $ digits as the number $ 5^7 $, i.e., the digits $ 8125 $. | 8125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLVII OM - III - Problem 1
Determine all pairs $ (n,r) $, where $ n $ is a positive integer and $ r $ is a real number, for which the polynomial $ (x + 1)^n - r $ is divisible by the polynomial $ 2x^2 + 2x + 1 $. | Let $ Q_n(x) $ and $ R_n(x) $ denote the quotient and remainder, respectively, when the polynomial $ (x + 1)^n $ is divided by $ 2x^2 + 2x + 1 $. The pair $ (n,r) $ is one of the sought pairs if and only if $ R_n(x) $ is a constant polynomial, identically equal to $ r $.
For $ n = 1,2,3,4 $ we have:
Therefore,
For every integer $ n \geq 0 $, the following equality holds:
Hence, $ R_{n+4}(x) = -\frac{1}{4}R_n(x) $. From this and the formulas (1), we obtain by induction the following equalities (for $ k = 1,2,3,\ldots $):
As we can see, $ R_n(x) $ is a constant polynomial only for $ n = 4k $; its constant value is then $ ( -\frac{1}{4})^k $.
Conclusion: the sought pairs $ (n,r) $ have the form $ (4k,(-\frac{1}{4})^k) $, where $ k $ is a positive integer. | (4k,(-\frac{1}{4})^k) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - II - Zadanie 4
Wyznaczyć wszystkie pary liczb rzeczywistych $ a, b $, dla których wielomiany $ x^4 + 2ax^2 + 4bx + a^2 $ i $ x^3 + ax - b $ mają dwa różne wspólne pierwiastki rzeczywiste.
|
Oznaczmy rozważane wielomiany przez $ P $ i $ Q $:
Załóżmy, że wielomiany $ P $ i $ Q $ mają wspólne pierwiastki rzeczywiste $ x_1 $, $ x_2 $ ($ x_1\neq x_2 $). Liczby $ x_1 $ $ x_2 $ są wówczas także pierwiastkami wielomianu
Stąd $ a \neq 0 $ i $ \Delta = 9b^2--4a^3 > 0 $.
Zgodnie z wzorami Viete'a
Ponieważ $ Q(x_1) = Q(x_2) = 0 $, a $ x_1 \neq x_2 $, zachodzą równości
Zatem $ b = 0 $ i $ 0 < \Delta = -4a^3 $, czyli $ a < 0 $.
Także i odwrotnie: jeśli spełnione są warunki $ a < 0 = b $, to wielomiany $ P $ i $ Q $ przybierają postać
i liczby $ \sqrt{-a} $, $ -\sqrt{-a} $ są ich wspólnymi pierwiastkami.
Rozwiązaniem zadania są więc wszystkie pary postaci $ (a, 0) $, gdzie $ a < 0 $.
| (,0),where0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - II - Problem 1
In how many ways can $ k $ fields of a chessboard $ n \times n $ ($ k \leq n $) be chosen so that no two of the selected fields lie in the same row or column? | First, one can choose $ k $ rows in which the selected fields will lie. This can be done in $ \displaystyle \binom{n}{k} $ ways. Then, in each of these rows, one must sequentially choose one of $ n $ fields, one of $ n - 1 $ fields lying in the remaining columns, etc. - one of $ n - k + 1 $ fields. Therefore, the total number of ways is $ \displaystyle \binom{n}{k} n(n - 1) \ldots (n - k + 1) $. | \binom{n}{k}n(n-1)\ldots(n-k+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXIII OM - III - Problem 1
Indicate such a way of arranging $ n $ girls and $ n $ boys around a round table so that the number $ d_n - c_n $ is maximized, where $ d_n $ is the number of girls sitting between two boys, and $ c_n $ is the number of boys sitting between two girls. | Consider any arrangement of $ n $ girls and $ n $ boys. Let's call a group of $ k $ girls (for $ k > 1 $) any sequence of $ k $ girls sitting next to each other, if the first and the $ k $-th girl in this sequence are seated next to a boy at the table. Let $ D_n $ be the number of groups of girls in a given arrangement. Similarly, we define a group of $ k $ boys. Let $ C_n $ be the number of groups of boys. Since every group of girls is adjacent to a group of boys or a single boy, and every group of boys is adjacent to a group of girls or a single girl, we have
\[
D_n = C_n
\]
If $ D_n = 0 $, then each girl is adjacent to two boys, which implies that each boy is adjacent to two girls, so $ C_n = 0 $ and $ d_n - c_n = 0 $. If, however, $ D_n \geq 1 $, then from the equality (*), we get
\[
d_n - c_n = \left[ \frac{n}{2} \right] - 1
\]
On the other hand, $ C_n $ is the number of groups of boys, and each group contains at least two boys, so $ C_n \leq \frac{n}{2} $. Moreover, $ C_n $ is an integer, so $ C_n \leq \left[ \frac{n}{2} \right] $. Therefore,
\[
d_n - c_n \leq \left[ \frac{n}{2} \right] - 1
\]
for any arrangement of $ n $ girls and $ n $ boys.
We will now show an arrangement where $ d_n - c_n = \left[ \frac{n}{2} \right] - 1 $. If $ n = 2k $, we seat two boys, one girl, two boys, one girl, and so on, finally the last two boys and the remaining girls. In this arrangement, $ k - 1 $ girls have boys on both sides, while no boy has two female neighbors.
If $ n = 2k + 1 $, we seat similarly: two boys, one girl, two boys, one girl, and so on, finally the last three boys and the remaining girls. In this situation, $ k - 1 $ girls sit between two boys, and no boy has two female neighbors.
Therefore, the maximum value of $ d_n - c_n $ is $ \left[ \frac{n}{2} \right] - 1 $. | [\frac{n}{2}]-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
III OM - I - Task 4
a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Determine four mutually parallel planes passing through points $ A $, $ B $, $ C $, $ D $, respectively, so that the distances between adjacent parallel planes are equal. | a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let this line be, for example, line $b$. In this case, points $A$ and $C$ are equidistant from line $b$ and lie on opposite sides of it; therefore, line $b$ intersects segment $AC$ at its midpoint $M$. From the fact that points $A$, $B$, $C$ do not lie on a straight line, it follows that point $M$ is different from point $B$.
Hence the construction: we draw line $b$ through point $B$ and through the midpoint $M$ of segment $AC$, and then we draw through points $A$ and $C$ lines $a$ and $c$ parallel to line $b$ (Fig. 12). The parallel lines $a$, $b$, $c$ determined in this way solve the problem, since points $A$ and $C$, and thus lines $a$ and $c$, are equidistant from line $b$ and lie on opposite sides of this line.
We found the above solution assuming that among the sought lines $a$, $b$, $c$, the line $b$ lies between lines $a$ and $c$; since the "inner" line can equally well be $a$ or $c$, the problem has three solutions.
b) Suppose that the planes $\alpha$, $\beta$, $\gamma$, $\delta$, passing through points $A$, $B$, $C$, $D$ respectively and being mutually parallel, satisfy the condition of the problem, that is, the distances between adjacent planes are equal. Let these planes be in the order $\alpha$, $\beta$, $\gamma$, $\delta$. We mean by this that plane $\beta$ is equidistant from planes $\alpha$ and $\gamma$, and plane $\gamma$ is equidistant from planes $\beta$ and $\delta$.
In this case, points $A$ and $C$ are equidistant from plane $\beta$ and lie on opposite sides of it, so plane $\beta$ passes through the midpoint $M$ of segment $AC$. Similarly, plane $\gamma$ passes through the midpoint $N$ of segment $BD$. From the fact that points $A$, $B$, $C$, $D$ do not lie in a plane, it follows that point $M$ is different from point $B$, and point $N$ is different from point $C$.
From this, we derive the following construction. We connect point $B$ with the midpoint $M$ of segment $AC$, and point $C$ with the midpoint $N$ of segment $BD$ (Figure 13 shows a parallel projection of the figure). Lines $BM$ and $CN$ are skew; if they lay in the same plane, then points $A$, $B$, $C$, $D$ would lie in the same plane, contrary to the assumption. We know from stereometry that through two skew lines $BM$ and $CN$ one can draw two and only two parallel planes $\beta$ and $\gamma$.
To do this, we draw through point $M$ a line $m$ parallel to line $CN$, and through point $N$ a line $n$ parallel to line $BM$; plane $\beta$ is then determined by lines $m$ and $BM$, and plane $\gamma$ by lines $n$ and $CN$.
Finally, we draw through points $A$ and $D$ planes $\alpha$ and $\delta$ parallel to planes $\beta$ and $\gamma$; we can determine them, as indicated in Figure 13, by drawing through each of points $A$ and $D$ lines parallel to lines $BM$ and $CN$.
The planes $\alpha$, $\beta$, $\gamma$, $\delta$ determined in this way solve the problem, since points $A$ and $C$, and thus planes $\alpha$ and $\gamma$, are equidistant from plane $\beta$ and lie on opposite sides of this plane - and similarly, planes $\beta$ and $\delta$ are equidistant from plane $\gamma$ and lie on opposite sides of this plane.
We found the above solution assuming that the sought planes are in the order $\alpha$, $\beta$, $\gamma$, $\delta$. For other orders of the sought planes, we will find other solutions in the same way. The number of all possible orders, or permutations of the letters $\alpha$, $\beta$, $\gamma$, $\delta$, is $4!$, i.e., $24$. However, note that two "reverse" permutations, such as $\alpha$, $\beta$, $\gamma$, $\delta$ and $\delta$, $\gamma$, $\beta$, $\alpha$, give the same solution. Therefore, the problem has $\frac{24}{2} = 12$ solutions corresponding to the permutations. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXIII OM - I - Problem 6
Determine for which digits $ a $ the decimal representation of a number $ \frac{n(n+1)}{2} $ ($ n\in \mathbb{N} $) consists entirely of the digit $ a $. | Of course, $ a $ cannot be zero. If the decimal representation of the number
$ \displaystyle t_n = \frac{1}{2} n (n + 1) $ consists of $ k $ ones, where $ k \geq 2 $, then $ 9 t_n = 10^k - 1 $. From this, after simple transformations, we obtain $ (3n + 1) \cdot (3n + 2) = 2 \cdot 10^k = 2^{k+1} \cdot 5^k $. Since the consecutive natural numbers $ 3n + 1 $ and $ 3n + 2 $ are relatively prime and $ 2^{k+1} \leq 2^{2k} = 4^k < 5^k $, it follows that $ 3n + 1 = 2^{k+1} $ and $ 3n + 2 = 5^k $. Subtracting these equations side by side, we get $ 1 = 5^k - 2^{k+1} > 4^k - 2^{k+1} = 2^k (2^k - 2) \geq 8 $, since $ k \geq 2 $. The obtained contradiction proves that $ a $ cannot be one.
Notice that $ 8 t_n + 1 = (2n + 1)^2 $. If the decimal representation of the number $ t_n $ ends with the digits $ 2 $, $ 4 $, $ 7 $, or $ 9 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 7 $ or $ 3 $. However, the last digit of the square of any natural number can only be $ 0, 1, 4, 5, 6 $, or $ 9 $. Therefore, $ a \ne 2, 4, 7, 9 $.
If the decimal representation of the number $ t_n $ ends with the digits $ 33 $ or $ 88 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 65 $ or $ 05 $. However, the decimal representation of the square of any natural number $ m $ does not end with the digits $ 65 $ or $ 05 $. Such a number $ m $ would be odd and divisible by $ 5 $, i.e., $ m = 10k + 5 $. Therefore, the number $ m^2 = 100k^2 + 100k + 25 $ would give a remainder of $ 25 $ when divided by $ 100 $, meaning the decimal representation of the number $ m^2 $ would end with the digits $ 25 $. Therefore, $ a \ne 3, 8 $.
However, $ a $ can be equal to $ 5 $ or $ 6 $. We have $ t_{10} = 55 $, $ t_{11} = 66 $, $ r_{36} = 666 $.
Note. It has been proven that among the numbers $ t_n $ for $ n > 3 $, only the numbers $ t_{10} $, $ t_{11} $, and $ t_{36} $ have a decimal representation composed of the same digits. See David W. Ballew and Ronald C. Weger in Notices Amer. Math. Soc. 19 (1972), p. A - 511, abstract 72 T - A 152. | 5,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXVI - I - Problem 11
A ship is moving at a constant speed along a straight line with an east-west direction. Every $ T $ minutes, the direction of movement is randomly chosen: with probability $ p $, the ship moves in the eastern direction for the next $ T $ minutes, and with probability $ q= 1-p $, it moves in the western direction. At a point outside the line, there is a submarine whose task is to torpedo the ship. The travel time of the torpedo from the point of firing to any point on the ship's track is $ 2T $. The captain of the submarine knows the value of $ p $ and aims to maximize the probability of hitting the ship. How should $ p $ be chosen to minimize the probability of the ship being torpedoed? | Let's assume that a submarine fires a torpedo at the moment when a lottery is taking place on the ship. In the case of firing the torpedo at a different time, the reasoning proceeds similarly. We can also assume, without loss of generality, that $ p \geq q $, i.e., $ \displaystyle p \geq \frac{1}{2} $. After $ T $ minutes from the moment the torpedo is fired, the next lottery takes place. The results of these two lotteries can be as follows: east-east, east-west, west-east, west-west with probabilities $ p^2 $, $ pq $, $ qp $, $ q^2 $, respectively. In the first case, after $ 2T $ minutes from the moment the torpedo is fired, the ship will be located at point $ A $ to the east of the initial point $ O $, in the second and third cases - at point $ O $, and in the last case - at point $ B $ to the west of $ O $. The probability of the event that the ship will be at point $ A $, $ O $, $ B $ at that time is therefore $ p^2 $, $ 2pq $, $ q^2 $, respectively.
Since $ p \geq q $, then $ p^2 \geq q^2 $. The inequality $ p^2 > 2pq $ holds if and only if $ p > 2q $. Therefore, if $ p > 2q $, i.e., $ \displaystyle p > \frac{2}{3} $, then the largest of these three numbers is $ p^2 $ and thus the torpedo should be fired in the direction of point $ A $. If, however, $ 2q > p \geq q $, i.e., $ \frac{2}{3} > p \geq \frac{1}{2} $, then the number $ 2pq $ is the largest and the torpedo should be fired in the direction of point $ O $. For $ \displaystyle p = \frac{2}{3} $, we have $ p^2 = 2pq > q^2 $ and thus the torpedo can be fired in the direction of point $ A $ or $ O $ with the same probability of hitting. The probability of hitting $ f(p) $ is therefore
The function $ f(p) $ is increasing in the interval $ \displaystyle \langle \frac{2}{3};\ 1 \rangle $ and decreasing in the interval $ \displaystyle \langle \frac{1}{2};\ \frac{2}{3} \rangle $. Therefore, the smallest probability of hitting is at $ \displaystyle p = \frac{2}{3} $. It is $ \displaystyle \frac{4}{9} $. | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LI OM - I - Task 5
Determine all pairs $ (a,b) $ of natural numbers for which the numbers $ a^3 + 6ab + 1 $ and $ b^3 + 6ab + 1 $ are cubes of natural numbers. | Let $ a $ and $ b $ be numbers satisfying the conditions of the problem. Without loss of generality, we can assume that $ a \leq b $. Then
Since the number $ b^3+6ab + 1 $ is a cube of an integer, the following equality must hold
Transforming the above relationship equivalently, we obtain step by step
It remains to determine for which pairs $ (a, b) $ satisfying condition (2) the number
is a cube of an integer. By virtue of relation (2), the above number is equal to $ a^3 + 12a^2 - 6a + 1 $. From the inequality
it follows that if the number $ a^3 + 12a^2 - 6a + 1 $ is a cube of an integer, then it is equal to $ (a+ 1)^3 $, $ (a + 2)^3 $, or $ (a + 3)^3 $. Therefore, we need to consider three cases:
(a) $ a^3 + 12a^2 - 6a + 1 = (a + 1)^3 $
In this case, we obtain $ 9a^2 - 9a = 0 $, from which $ a = 0 $ or $ a = 1 $.
(b) $ a^3 + 12a^2 - 6a + 1 = (a + 2)^3 $
Then we have $ 6a^2 - 18a - 7 = 0 $. The obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is.
(c) $ a^3 + 12a^2 - 6a + 1 = (a + 3)^3 $
Then we get $ 3a^2 - 33a - 26 = 0 $. Similarly to case (b), the obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is equal to $ 0 $.
Ultimately, we have found only one pair $ (a, b) = (1, 1) $ satisfying the conditions of the problem. | (1,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXV OM - I - Problem 9
Three events satisfy the conditions:
a) their probabilities are equal,
b) any two of them are independent,
c) they do not occur simultaneously.
Determine the maximum value of the probability of each of these events. | om35_1r_img_4.jpg
Suppose that $ A $, $ B $, $ C $ are these events, each with a probability of $ p $. Given the assumption of independence between any two of the considered events, the probability of each intersection $ A \cap B $, $ B \cap C $, $ C \cap A $ is $ p^2 $. Furthermore, $ A \cap B \cap C \ne 0 $. Therefore,
Since of course $ P (A \cup B \cup C) \geq P (A \cap B) + P (B \cap C) + P (C \cap A) = 3p^2 $, it follows that $ 3p-3p^2 \geq 3p^2 $, which implies that $ p \leq \frac{1}{2} $.
The value $ p = \frac{1}{2} $ is assumed, for example, by the following events. When flipping a coin twice, we take $ A $ to be the event that a tail appears on the first flip, $ B $ - that a head appears on the second flip, $ C $ - that the same result appears on both flips. Clearly, $ P(A) = P(B) = P(C) = \frac{1}{2} $, the events $ A $, $ B $, $ C $ are pairwise independent and cannot occur simultaneously.
Therefore, the maximum value of the considered probability is $ \frac{1}{2} $. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
VII OM - I - Task 3
In a square $ABCD$ with area $S$, vertex $A$ is connected to the midpoint of side $BG$, vertex $B$ is connected to the midpoint of side $CD$, vertex $C$ is connected to the midpoint of side $DA$, and vertex $D$ is connected to the midpoint of side $AB$. Calculate the area of the part of the square that contains its center. | The solution to the problem is illustrated in Fig. 1, where $ M $, $ N $, $ P $, $ Q $ denote the midpoints of the sides, and $ O $ the center of the square; through the vertices $ A $, $ B $, $ C $, $ D $ lines parallel to the lines $ DQ $, $ AM $, $ BN $, $ CP $ have been drawn. If the figure is rotated by $ 90^\circ $ around the point $ O $, point $ A $ will fall on point $ B $, point $ Q $ on point $ M $, etc., and the figure will overlap with itself. From this, it is easy to conclude that the figure consists of a grid made up of 9 equal squares. According to the Pythagorean theorem, the area of square $ ABCD $ equals the sum of the areas of the squares constructed on the legs $ AK $ and $ BK $ of the right triangle $ AKB $. Denoting the sought area by the letter $ x $, we therefore have | \frac | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLVIII OM - III - Problem 2
Find all triples of real numbers $ x $, $ y $, $ z $ satisfying the system of equations | Suppose that the numbers $ x $, $ y $, $ z $ satisfy the given system; from the second equation it follows that
Further, we have
which, by the first equation of the system, gives:
There is also the inequality
that is,
By property (1), we can multiply inequalities (3) and (4) side by side:
According to the second equation of the system, inequality (5) must be an equality. This means that either both multiplied inequalities (3) and (4) are equalities, or the expressions on both sides of (4) are equal to zero. In the first case, the equality sign in relations (2) and (3) leads to the conclusion that $ x = y = z = \pm 1/3 $. In the second case, the products $ zy $, $ yz $, $ zx $ are equal to zero, which means that two of the numbers $ z $, $ y $, $ z $ are equal to zero; the third must be equal to $ \pm 1/\sqrt{3} $, in accordance with the first equation of the system. This gives eight triples $ (x, y, z) $ that satisfy the given system of equations:
where $ \varepsilon = \pm 1 $. | (x,y,z)=(\1/3,\1/3,\1/3)(x,y,z)=(\1/\sqrt{3},0,0),(0,\1/\sqrt{3},0),(0,0,\1/\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 3
In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $. | Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since
$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $ CP=CQ $. Points $C$, $Q$, $E$, $P$ lie on the same circle, from which $ \measuredangle CED =\measuredangle CEP =\measuredangle CQP = 45^\circ $. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXX OM - I - Task 8
Among the cones inscribed in the sphere $ B $, the cone $ S_1 $ was chosen such that the sphere $ K_1 $ inscribed in the cone $ S_1 $ has the maximum volume. Then, a cone $ S_2 $ of maximum volume was inscribed in the sphere $ B $, and a sphere $ K_2 $ was inscribed in the cone $ S_2 $. Determine which is greater: the sum of the volumes of $ S_1 $ and $ K_1 $, or the sum of the volumes of $ S_2 $ and $ K_2 $. | Let $ S $ be any cone inscribed in a given sphere $ B $, and $ K $ - the sphere inscribed in the cone $ S $. The axial section of the cone $ S $ is an isosceles triangle $ ACD $, where $ A $ is the vertex of the cone (Fig. 7).
om30_1r_img_7.jpg
Let $ O $ be the center of the sphere $ B $, and $ Q $ - the center of the sphere $ K $. Then $ O $ is the center of the circle circumscribed around the triangle $ ACD $, and $ Q $ - the center of the circle inscribed in this triangle. Let $ P $ be the midpoint of the segment $ \overline{CD} $, and $ M $ and $ N $ respectively - the orthogonal projections of the points $ Q $ and $ O $ onto the line $ AD $.
Let $ \alpha = \measuredangle PAD $, $ R = AO $. We have the following relationships:
Therefore, the volume $ V_s $ of the cone $ S $ is equal to
or
where we have taken $ a = \sin \alpha $.
We then have $ MQ = QP $. Thus $ AP \sin \alpha = QP \sin \alpha + AQ \sin \alpha = MQ \sin \alpha + MQ $ and therefore
The volume $ V_K $ of the sphere $ K $ is then equal to
Therefore, the sum $ V = V(a) $ of the volumes of the cone $ S $ and the sphere $ K $ is, by (1) and (2):
From the conditions of the problem, it follows that $ 0 < \alpha < \frac{\pi}{2} $. Therefore, the number $ a = \sin \alpha $ belongs to the interval $ (0,1) $.
In this interval, the function $ f(a) = a (1 - a^2) $ attains its maximum value at $ a = \frac{\sqrt{3}}{3} $, since this is the only zero of the derivative of $ f $ in this interval, and $ f(0) = f(1) = 0 $ and $ f(a) > 0 $ for $ 0 < a < 1 $. Therefore, from formula (1), it follows that the cone $ S $ has the maximum volume at $ a = \sin \alpha = \frac{\sqrt{3}}{3} $.
Similarly, we observe that the function $ g(a) = a(1 - a) $ in the interval $ (0, 1) $ attains its maximum value at $ a = \frac{1}{2} $. Therefore, from formula (2), it follows that the sphere $ K $ has the maximum volume at $ a = \sin \alpha = \frac{1}{2} $, i.e., at $ \alpha = \frac{\pi}{6} $.
Next, we calculate the values of the function $ V(a) $ defined by formula (3) at the points $ a = \frac{\sqrt{3}}{3} $ and $ a = \frac{1}{2} $. We obtain:
It is easy to verify that $ \frac{4}{9} (2\sqrt{3} - 3) > \frac{13}{64} $, and thus $ V \left( \frac{\sqrt{3}}{3} \right) > V \left( \frac{1}{2} \right) $. Therefore, the sum of the volumes of $ S_2 $ and $ K_2 $ is greater than the sum of the volumes of $ S_1 $ and $ K_1 $. | V(\frac{\sqrt{3}}{3})>V(\frac{1}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXI - I - Task 1
Determine for which values of the parameter $ a $ a rhombus with side length $ a $ is a cross-section of a cube with edge length 2 by a plane passing through the center of the cube. | Suppose the rhombus $KLMN$ is a cross-section of the cube $ABCDA_1B_1C_1D_1$ by a plane passing through the center of the cube $O$. Since $O$ is also the center of symmetry of the cross-section, the opposite vertices of the rhombus lie on opposite edges of the cube. Let, for example, as in the figure, $K \in \overline{AB}$, $L \in \overline{A_1B_1}$, $M \in \overline{C_1D_1}$, $N \in \overline{CD}$. Since $O$ is the center of symmetry of the cube, $AK = MC_1$, $A_1L = NC$. Suppose that $K$ is not the midpoint of the edge $\overline{AB}$. Map the adjacent faces $ABB_1A_1$ and $A_1B_1C_1D_1$ onto the net of the cube.
om31_1r_img_6.jpgom31_1r_img_7.jpg
The right trapezoids $AKLA_1$ and $LB_1C_1M$ are congruent, as they are the sum of congruent rectangles and congruent right triangles. It follows that $A_1L = B_1L$, i.e., $L$ is the midpoint of the segment $\overline{A_1B_1}$, and therefore $N$ is the midpoint of $\overline{CD}$. In this way, we have shown that certain two opposite vertices of the rhombus are the midpoints of the edges of the cube on which they lie. The side of the rhombus is thus a segment connecting the midpoint of an edge with some point on the opposite side of the face, which is a square with side length $2$. The length of such a segment is no less than $2$ and no greater than $\sqrt{5}$, and any value in the interval $[2, \sqrt{5}]$ can be achieved by choosing the point $K$ appropriately on the side $\overline{AB}$. | [2,\sqrt{5}] | Geometry | math-word-problem | Yes | Yes | olympiads | false |
VIII OM - I - Zadanie 5
Jaki warunek powinna spełniać liczbą $ q $, aby istniał trójkąt, którego boki tworzą postęp geometryczny o ilorazie $ q $?
|
Weźmy pod uwagę trzy odcinki, których długości tworzą postęp geometryczny o ilorazie $ q $. Ponieważ stosunek dwóch odcinków jest liczbą niezależną od jednostki, jaką te odcinki mierzymy, więc jeśli za jednostkę obierzemy pierwszy z rozważanych trzech odcinków, to długości tych odcinków wyrażą się liczbami $ 1 $, $ q $, $ q^2 $. Aby istniał trójkąt, którego boki są odpowiednio równe tym odcinkom, potrzeba i wystarcza żeby suma każdych dwóch spośród tych odcinków była większa od trzeciego odcinka, tj. żeby spełnione były warunki: 1) $ 1 + q^2 > q $, 2) $ q + q^2 > 1 $, 3) $ 1 + q > q^2 $, którym możemy nadać postać:
Warunek 1) jest spełniony dla każdej wartości $ q $, gdyż wyróżnik trójmianu kwadratowego $ q^2 - q + 1 $ jest ujemny, $ \Delta = - 3 $, więc trójmian ma stale wartość dodatnią. Warunek 2) jest spełniony dla wartości $ q $ leżących po za pierwiastkami trójmianu $ q^2 + q - 1 $, tzn. gdy
Warunek 3) jest spełniony dla wartości $ q $ leżących między pierwiastkami trójmianu $ q^2 - q - 1 $, tzn. gdy
Ponieważ $ \frac{-1 - \sqrt{5}}{2} < \frac{1 - \sqrt{5}}{2} < \frac{-1 + \sqrt{5}}{2} < \frac{1 + \sqrt{5}}{2} $, więc ostatecznie otrzymujemy dla $ q $ warunek następujący:
Uwaga. Liczby $ \frac{- 1 + \sqrt{5}}{2} $ i $ \frac{1 + \sqrt{5}}{2} $ są odwrotnościami, gdyż ich iloczyn $ \frac{- 1 + \sqrt{5}}{2} \cdot \frac{1 + \sqrt{5}}{2} = 1 $. Stąd wynika, że jeśli pewna liczba $ q $ spełnia warunek (6), to jej odwrotność $ \frac{1}{q} $ też spełnia warunek (6) (rys. 3). Fakt ten był do przewidzenia; jeśli bowiem boki trójkąta wzięte w pewnej kolejności tworzą postęp geometryczny o ilorazie $ q $, to w odwrotnej kolejności dadzą one postęp geometryczny o ilorazie $ \frac{1}{q} $. Długość otrzymanego przedziału dla $ q $ wynosi $ \frac{1 + \sqrt{5}}{2} - \frac{- 1 + \sqrt{5}}{2} =1 $. Łatwo sprawdzić, ze liczby $ \frac{1 + \sqrt{5}}{2} $ i $ \frac{-1 + \sqrt{5}}{2} $ są jedyną parą liczb, które są odwrotnościami i których różnica równa się $ 1 $.
| \frac{-1+\sqrt{5}}{2}<\frac{1+\sqrt{5}}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LV OM - I - Task 7
Find all solutions to the equation $ a^2+b^2=c^2 $ in positive integers such that the numbers $ a $ and $ c $ are prime, and the number $ b $ is the product of at most four prime numbers. | There are three solutions $(a,b,c): (3,4,5), (5,12,13), (11,60,61)$. Let $a$, $b$, $c$ be numbers satisfying the conditions of the problem. Then
from the assumption that the number $a$ is prime, we obtain $c = b + 1$. Therefore, $a^2 = 2b + 1$, which implies that the number $a$ is odd.
Let $a = 2n + 1$. Then
if the number $n$ gives a remainder of 1 when divided by 3, then the number $a$ is divisible by 3. Therefore, it is composite, except for the case $n = 1$, which gives the solution $(a,b,c) = (3,4,5)$. If, on the other hand, the number $n$ gives a remainder of 2 when divided by 5, then the number $a$ is divisible by 5. Therefore, it is composite, except for the case $n = 2$. This case gives the second solution $(a,b,c) = (5,12,13)$.
If, however, the number $n$ gives a remainder of 1 or 3 when divided by 5, then the number $c$ is divisible by 5. Therefore, it is composite except for the case $n = 1$, which we have already considered.
There remains the case where the number $n$ gives a remainder of 0 or 2 when divided by 3, and a remainder of 0 or 4 when divided by 5. In this case, the number $b$ is divisible by $2 \cdot 6 \cdot 5 = 60$. On the other hand, the number $b$ is the product of at most four prime numbers, which implies that $b = 60$. This case leads to the third solution: $(a,b,c) = (11,60,61)$. | (3,4,5),(5,12,13),(11,60,61) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LX OM - III - Zadanie 2
Let $ S $ be the set of all points in the plane with both coordinates being integers. Find
the smallest positive integer $ k $ for which there exists a 60-element subset of the set $ S $
with the following property: For any two distinct elements $ A $ and $ B $ of this subset, there exists a point
$ C \in S $ such that the area of triangle $ ABC $ is equal to $ k $. | Let $ K $ be a subset of the set $ S $ having for a given number $ k $ the property given in the problem statement.
Let us fix any two different points $ (a, b), (c, d) \in K $. Then for some integers
$ x, y $ the area of the triangle with vertices $ (a, b) $, $ (c, d) $, $ (x, y) $ is $ k $, i.e., the equality
$ \frac{1}{2}|(a - c)(y - d) - (b - d)(x - c)| = k $ holds. From this, we obtain the condition that the equation
has for any fixed and different $ (a, b), (c, d) \in K $ a solution in integers $ x, y $.
We will prove that if the number $ m $ does not divide $ 2k $, then the set $ K $ has no more than $ m^2 $ elements.
To this end, consider pairs $ (a \mod m, b \mod m) $ of residues of the coordinates of the points of the set $ K $ modulo $ m $.
There are $ m^2 $ of them, so if $ |K| > m^2 $, then by the pigeonhole principle, we will find two different points
$ (a, b) \in K $ and $ (c, d) \in K $ such that $ a \equiv c \mod m $ and $ b \equiv d \mod m $. For such points,
the equation (1) has no solution, since the left side of the equation for any $ x $ and $ y $ is divisible by $ m $,
while the right side is not. Therefore, $ |K| \leqslant m^2 $ for any $ m $ that is not a divisor of $ 2k $.
From the above considerations, it follows that if $ |K| = 60 $, then $ 2k $ must be divisible by all numbers
$ m \leqslant 7 $, since $ 60 > 7^2 $. It is easy to check that the smallest natural number divisible by
2, 3, 4, 5, 6, 7 is $ 2^2 \cdot 3 \cdot 5 \cdot 7 = 420 $, so $ 210|k $.
We will show that for every $ k $ such that $ 210|k $, a 60-element set $ K $ having the property given in the problem statement can be constructed. Let $ K $ be the set of all elements of the set $ S $,
whose both coordinates are in the set $ \{0, 1,..., 7\} $. Fix any two different points
$ A =(a, b) $ and $ B=(c, d) $ from the set $ K $. Then $ a - c, b - d \in \{-7, -6,..., 6, 7\} $, so if
$ a \neq c $, then $ a-c|420 $ and if $ b = d $, then $ b-d|420 $. Without loss of generality, we can assume that $ b = d $.
Then $ b - d|2k $, since $ 420|2k $. The point $ C =(c + \frac{2k}{d-b} ,d) $ is therefore an element of the set $ S $,
and the area of the triangle $ ABC $ is
Moreover, $ |K| = 60 $. As the set $ K $, we can take any 60-element subset of the set $ K $.
Thus, we have shown that a 60-element set having the desired property exists only for positive integers $ k $
that are multiples of 210. The smallest such number is 210. | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
I OM - B - Task 1
Determine the coefficients $ a $ and $ b $ of the equation $ x^2 + ax + b=0 $ such that the values of $ a $ and $ b $ themselves are roots of this equation. | According to the theorems about the sum and product of the roots of a quadratic equation, the numbers $ a $ and $ b $ are the roots of the equation $ x^2 + ax + b = 0 $ if and only if they satisfy the system of equations
that is,
The second equation of the system (1) is satisfied when $ b = 0 $ or when $ a-1= 0 $, i.e., when $ a = 1 $.
System (1) therefore has two solutions:
The quadratic equations with the desired property are | 1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
I OM - B - Zadanie 1
Wyznaczyć współczynniki $ a $ i $ b $, równania $ x^2 + ax +b=0 $ tak, aby same wartości $ a $ i $ b $ były pierwiastkami tego równania.
|
Liczby $ a $ i $ b $ mają spełniać równanie $ x^2 + ax + b = 0 $. Zatem
czyli
Drugie z równań układu (2) daje $ b = 0 $ lub $ a + b + 1 = 0 $.
Biorąc $ b = 0 $ otrzymujemy rozwiązanie
Biorąc $ a + b + 1 = 0 $ mamy $ b = - a-l $; po podstawieniu do pierwszego z równań układu (2) otrzymujemy równanie
którego pierwiastkami są liczby $ a = 1 $ i $ a = -1 $, przy czym odpowiednimi wartościami $ b $ są liczby $ b = - 2 $ i $ b=-\frac{1}{2} $.
Stąd dalsze rozwiązania układu równań (2):
Ze znalezionych rozwiązań tylko liczby $ a = 0 $, $ b = 0 $ oraz
liczby $ a = 1 $, $ b = - 2 $ spełniają warunki zadania. Liczby $ a = -\frac{1}{2} $,
$ b = -\frac{1}{2} $ warunków tych nie spełniają, gdyż pierwiastkami równania
$ x^2 - \frac{1}{2}x -\frac{1}{2} = 0 $ są liczby $ -\frac{1}{2} $ i 1, a nie liczby $ -\frac{1}{2} $ i $ -\frac{1}{2} $.
Aby wyjaśnić ten paradoks, zauważmy, że układ równań (2)
daje tylko warunki konieczne dla poszukiwanych liczb $ a $ i $ b $,
Nie są to jednak warunki dostateczne: jeśli bowiem liczby
$ a $ i $ b $ spełniają układ (2), to każda z nich jest wprawdzie pierwiastkiem
równania $ x^2 + ax + b = 0 $, ale z tego nie wynika, że $ a $ i $ b $
dają oba pierwiastki tego równania. W przypadku, gdy $ a = b $,
jest to jeden i ten sam pierwiastek, jak to właśnie zachodzi w równaniu
$ x^2 -\frac{1}{2}x -\frac{1}{2} = 0 $.
| 0,0或1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXI OM - II - Problem 6
If $ A $ is a subset of set $ X $, then we define $ A^1 = A $, $ A^{-1} = X - A $. Subsets $ A_1, A_2, \ldots, A_k $ are called mutually independent if the product $ A_1^{\varepsilon_1} \cap A_2^{\varepsilon_2} \ldots A_k^{\varepsilon_k} $ is non-empty for every system of numbers $ \varepsilon_1, \varepsilon_2, \ldots, \varepsilon_k $, such that $ |\varepsilon_i| = 1 $ for $ i = 1, 2, \ldots, k $.
What is the maximum number of mutually independent subsets of a $ 2^n $-element set? | Since the number of all $n$-term sequences with values $-1$ and $1$ is $2^n$ (see preparatory problem D of series III), we can assume that the elements of the $2^n$-element set $X$ under consideration are such sequences.
Let $A_i$ for $i = 1, 2, \ldots, n$ denote the set of all sequences in $X$ whose $i$-th term is $1$. We will show that the subsets $A_1, A_2, \ldots, A_n$ are mutually independent.
A sequence $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)$, where $|\varepsilon_j| = 1$, belongs to each of the sets $A_i^{\varepsilon_i}$ for $i = 1, 2, \ldots, n$. If $\varepsilon_i = 1$, then the considered sequence has its $i$-th term equal to $1$, and thus belongs to $A_i^{\varepsilon_i} = A_i^1$; if, on the other hand, $\varepsilon_i = -1$, then the considered sequence has its $i$-th term equal to $-1$, and thus belongs to $X - A_i = A_i^{-1}$. Therefore, the set $A_1^{\varepsilon_1} \cap A_2^{\varepsilon_2} \cap \ldots \cap A_n^{\varepsilon_n}$ contains the sequence $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)$, and thus is non-empty.
We will now show that if $r > n$ and $B_1, B_2, \ldots, B_r$ are some subsets of the set $X$, then they are not mutually independent. If the sequences $(\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_r)$ and $(\varepsilon$ with values $-1$ and $1$ are different, then the sets
are disjoint. Specifically, if, for example, $\varepsilon_i = 1$, then we have $B \subset B_i^{\varepsilon_i} = B_i$ and $B \subset X - B_i$, and the sets $B_i$ and $X - B_i$ are obviously disjoint.
Thus, if the sets $B_1, B_2, \ldots, B_r$ were mutually independent, then all sets of the form $B_1^{\varepsilon_1} \cap B_2^{\varepsilon_2} \cap \ldots \cap B_r^{\varepsilon_r}$ would be non-empty and pairwise disjoint. There are $2^r > 2^n$ such sets. This is impossible, since the set $X$ contains only $2^n$ elements.
We have thus proved that $n$ is the maximum number of mutually independent subsets of a $2^n$-element set. | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
X OM - III - Task 3
Given is a pyramid with a square base $ABCD$ and apex $S$. Find the shortest path that starts and ends at point $S$ and passes through all the vertices of the base. | Let $ a $ denote the length of the side of square $ ABCD $; the lengths of the lateral edges of the pyramid will be denoted by the letters $ k $, $ l $, $ m $, $ n $ in such a way that $ k \leq l \leq m \leq n $.
Each path starting and ending at point $ S $ and passing through all the vertices of the square can be symbolically written as
where $ W_1, W_2, W_3, W_4 $ represent the vertices $ A $, $ B $, $ C $, $ D $ taken in some order. The part of the path from $ S $ to $ W_1 $ cannot be shorter than the edge $ SW_1 $, and the part of the path from $ W_4 $ to $ S $ cannot be shorter than the edge $ SW_4 $. These two parts of the path together cannot be shorter than the sum of the two shortest lateral edges, i.e., than $ k + l $. The part of the path from $ W_1 $ to $ W_2 $ cannot be shorter than the side of the square, i.e., than $ a $; the same applies to the path from $ W_2 $ to $ W_3 $ and from $ W_3 $ to $ W_4 $. The entire path (1) cannot, therefore, be shorter than
If we can find a path whose length is exactly $ k + l + 3a $, it will be the shortest path. To this end, we will determine the two shortest lateral edges of the pyramid.
Draw 4 axes of symmetry of the square $ ABCD $ (Fig. 31). They divide the plane of the square into 8 angular regions $ \textrm{I}-\textrm{VIII} $. The projection $ O $ of the vertex $ S $ of the pyramid onto the plane of the square lies inside or on the boundary of one of these regions, say region $ \textrm{I} $. Based on the properties of the perpendicular bisector of a segment, the following inequalities then hold
Thus, $ OA $ and $ OB $ are the two shortest of the segments $ OA $, $ OB $, $ OC $, $ OD $. In this case, $ SA $ and $ SB $ are the two shortest lateral edges of the pyramid (the shorter projection corresponds to the shorter slant), namely $ SA = k $, $ SB = l $, from which it follows that the path $ SADCBS $, whose length $ SA + AD + DC + CB + BS = k + l + 3a $ is the sought shortest path.
When the point $ O $ lies on the ray $ MA $ but does not coincide with the center $ M $ of the square, there are two shortest paths $ SADCBS $ and $ SABCDS $. When the point $ O $ coincides with the point $ M $, there are four such paths. Each path can, of course, be traversed in two directions. | k++3a | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XIX OM - I - Problem 4
In the plane, given are parallel lines $ a $ and $ b $ and a point $ M $ lying outside the strip bounded by these lines. Determine points $ A $ and $ B $ on lines $ a $ and $ b $, respectively, such that segment $ AB $ is perpendicular to $ a $ and $ b $, and the angle $ AMB $ is the largest. | Let $ AB $ be a segment with endpoints $ A \in a $ and $ B \in b $, perpendicular to the lines $ a $ and $ b $, and $ N $ - the point symmetric to point $ M $ with respect to a line parallel to lines $ a $ and $ b $ and equidistant from them (Fig. 4).
The circle with diameter $ AB $ lies within the strip bounded by the lines $ a $ and $ b $, so point $ M $ lies outside this circle, which means $ \measuredangle AMB = \alpha $ is less than $ 90^\circ $. Therefore, the maximum of angle $ \alpha $ coincides with the maximum of $ \sin \alpha $, since for angles from $ 0^\circ $ to $ 90^\circ $, the sine of the angle increases with the angle.
When segment $ AB $ lies on line $ MN $, then $ \sin \alpha = 0 $.
When points $ A $, $ B $, and $ M $ are vertices of a triangle, and $ r $ denotes the radius of the circumcircle of this triangle, then $ \sin \alpha = \frac{AB}{2r} $. Segment $ MN $ is a chord of this circle, so $ MN \leq 2r $, hence
From this, it follows that the maximum of $ \sin \alpha $, and thus the maximum of $ \alpha $, occurs when $ \sin \alpha = \frac{AB}{MN} $, i.e., when
which means that points $ A $ and $ B $ lie on the circle with diameter $ MN $. | ABlieonthecirclewithdiameterMN | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - III - Task 4
In how many ways can a set of $ n $ items be divided into two sets? | The answer to the posed question will depend on whether - to the considered sets, we include the empty set, i.e., a set that does not have any element (object), or not; in the first case, the sought number is $ 1 $ greater than in the second. We agree to consider only non-empty sets.
Let $ p $ denote the number of all possible partitions of the set $ Z $ having $ n $ elements, into two sets, i.e., into two parts.
Then $ 2p $ denotes the number of all possible parts, i.e., subsets of the set $ Z $, with the condition that we only consider proper subsets, i.e., different from the set $ Z $ itself.
The number $ 2p $ of all subsets of a set with $ n $ elements can be found by adding the numbers of subsets with $ 1, 2, \ldots, (n - 1) $ elements.
The number of subsets with $ k $ elements is as many as there are combinations of $ n $ elements taken $ k $ at a time, i.e., $ \binom{n}{k} $. Therefore,
According to the binomial theorem of Newton,
Subtracting equations (1) and (2) and considering that $ \binom{n}{0} = \binom{n}{n} = 1 $, we obtain
and from this,
| 2^{n-1}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LV OM - III - Task 5
Determine the maximum number of lines in space passing through a fixed point and such that any two intersect at the same angle. | Let $ \ell_1,\ldots,\ell_n $ be lines passing through a common point $ O $. A pair of intersecting lines determines four angles on the plane containing them: two vertical angles with measure $ \alpha \leq 90^\circ $ and the other two angles with measure $ 180^\circ - \alpha $. According to the assumption, the value of $ \alpha $ is the same for every pair $ \ell_i, \ell_j $.
Consider a sphere $ S $ centered at $ O $ with an arbitrary radius, and denote by $ A_i $, $ B_i $ the points of intersection of the line $ \ell_i $ with this sphere. Each of the segments $ A_iA_j $, $ A_iB_j $, $ B_iB_j $ (for $ i \neq j $) is a chord of the sphere $ S $, determined by a central angle of measure $ \alpha $ or $ 180^\circ - \alpha $. Therefore, these segments have at most two different lengths $ a $ and $ b $.
Fix the notation so that $ \measuredangle A_iOA_n = \alpha $ for $ i = 1,\ldots,n-1 $. Then the points $ A_1, \ldots, A_{n-1} $ lie on a single circle (lying in a plane perpendicular to the line $ \ell_n $). Let $ A_1C $ be the diameter of this circle. Each of the points $ A_2,\ldots,A_{n-1} $ is at a distance $ a $ or $ b $ from the point $ A_1 $; hence, on each of the two semicircles with endpoints $ A_1 $ and $ C $, there are at most two points from the set $ \{A_2,\ldots ,A_{n-1}\} $. Therefore, this set has at most four elements; which means that $ n \leq 6 $.
On the other hand, if $ n = 6 $, we can place the points $ A_1,A_2,\ldots,A_5 $ at the vertices of a regular pentagon, and the plane of this pentagon at such a distance from the point $ A_6 $ that these six points, together with their antipodal points $ B_1,\ldots,B_6 $, are the vertices of a regular icosahedron. The segments $ A_iB_i $ (diameters of the sphere $ S $) connect opposite vertices of this icosahedron and any two of them form the same angle. Therefore, $ n = 6 $ is the largest possible number of lines $ \ell_i $. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - III - Task 5
Points $ A_1 $, $ B_1 $, $ C_1 $ divide the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $ in the ratios $ k_1 $, $ k_2 $, $ k_3 $. Calculate the ratio of the areas of triangles $ A_1B_1C_1 $ and $ ABC $. | According to the assumption
Hence
Let the symbol $ p (ABC) $ denote the area of triangle $ ABC $ (Fig. 21). Since
then
By applying the theorem that the ratio of the areas of two triangles having a common angle equals the ratio of the products of the sides of each triangle forming that angle, we obtain
Substituting into equation (2) the values of the ratios (1) gives the equality
after transforming the right side, we obtain | \frac{k_1k_2k_3}{(1+k_1)(1+k_2)(1+k_3)} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LVII OM - I - Problem 1
Determine all non-negative integers $ n $ for which the number
$ 2^n +105 $ is a perfect square of an integer. | If $ n $ is an odd number, then the number $ 2^n +105 $ gives a remainder of 2 when divided by 3. Since a number that is the square of an integer cannot give a remainder of 2 when divided by 3, the number $ n $ satisfying the conditions of the problem must be even.
Let's assume that $ n =2k $ for some non-negative integer $ k $. We want to solve the equation $ 2^{2k} +105 = m^2 $ in non-negative integers $ k $ and $ m $. We rewrite this equation as $ (m-2^k)(m+2^k) = 105 $.
From the obtained equation, it follows that $ m-2^k >0 $. Of course, the inequality $ m+2^k > m-2^k $ is also satisfied. Since $ 105=3\cdot 5\cdot 7 $, finding non-negative integers $ m $ and $ k $ that satisfy the relation $ (m-2^k)(m+2^k)=105 $ reduces to solving four systems of equations:
By subtracting the first equation from the second, we conclude that the number $ 2 \cdot 2^k =2^{k+1} $ equals 8, 16, 32, or 104. From this, we obtain three possible values for $ k $; they are: 2, 3, or 4. Thus, the possible values of the number $ n $ are 4, 6, or 8.
We directly check that for the three obtained values of $ n $, the number $ 2^n +105 $ is the square of an integer:
直接检查结果如下：
- 当 $ n = 4 $ 时, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- 当 $ n = 6 $ 时, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- 当 $ n = 8 $ 时, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $.
Directly checking the results:
- For $ n = 4 $, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- For $ n = 6 $, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- For $ n = 8 $, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $. | 4,6,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - II - Task 4
Find all prime numbers $ p $ such that $ 4p^2 +1 $ and $ 6p^2 + 1 $ are also prime numbers. | To solve the problem, we will investigate the divisibility of the numbers \( u = 4p^2 + 1 \) and \( v = 6p^2 + 1 \) by \( 5 \). It is known that the remainder of the division of the product of two integers by a natural number is equal to the remainder of the division of the product of their remainders by that number. Based on this, we can easily find the following:
\begin{center}
When \( p \equiv 0 \pmod{5} \), then \( u \equiv 1 \pmod{5} \), \( v \equiv 1 \pmod{5} \);\\
When \( p \equiv 1 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \);\\
When \( p \equiv 2 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 3 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 4 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \).\\
\end{center}
From the above, it follows that the numbers \( u \) and \( v \) can both be prime numbers only when \( p \equiv 0 \pmod{5} \), i.e., when \( p = 5 \), since \( p \) must be a prime number. In this case, \( u = 4 \cdot 5^2 + 1 = 101 \), \( v = 6 \cdot 5^2 + 1 = 151 \), so they are indeed prime numbers.
Therefore, the only solution to the problem is \( p = 5 \). | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - II - Problem 5
Determine all prime numbers $ p $ and natural numbers $ x, y $, for which $ p^x - y^3 = 1 $. | Suppose that the numbers $ p $, $ x $, $ y $ satisfy the given conditions. Then
Let's consider two cases:
1. $ y = 1 $. Then (1) reduces to the equation $ p^x = 2 $, from which $ p = 2 $, $ x = 1 $.
2. $ y > 1 $. Then both factors of the product on the right side of (1) are numbers greater than $ 1 $ and as divisors of the power of the prime number $ p $, they must themselves be positive powers of this prime number:
Now notice that
Thus, $ 3 $ is divisible by $ p $, which means that $ p = 3 $. Equation (3) takes the form
which is
The right side of the last equation is not divisible by $ 3 $. Hence $ l - 1 = 0 $ and $ 2k-1=k $, so $ k = l = 1 $ and by the relation (2) $ x = 2 $, $ y = 2 $.
In summary: the solutions $ (p, x, y) $ of the given equation can only be the triples $ (2, 1, 1) $ and $ (3, 2, 2) $. Verifying that these are indeed solutions is immediate. | (2,1,1)(3,2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXVI - I - Problem 5
Determine all integers $ m $ for which the polynomial $ x^3-mx^2+mx-(m^2+1) $ has an integer root. | Suppose that the integers $ x $ and $ m $ satisfy the equation $ x^3 - mx^2+mx-(m^2+1) = 0 $, which is $ (x^2+m)(x-m) = 1 $. Then $ x^2+m = 1 $ and $ x- m = 1 $ or $ x^2+m = -1 $ and $ x-m = -1 $. Adding the equations in each of these systems, we get $ x^2+x = 2 $ or $ x^2+x = - 2 $. The roots of the first equation are $ x = 1 $ and $ x = -2 $. They correspond to the numbers $ m = 0 $ and $ m = - 3 $. The second equation has no solutions in the set of integers. | =0=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XL OM - I - Task 6
Calculate the sum of the series $ \sum_{n\in A}\frac{1}{2^n} $, where the summation runs over the set $ A $ of all natural numbers not divisible by 2, 3, 5. | A natural number $ n $ is not divisible by $ 2 $, $ 3 $, and $ 5 $ if and only if, when divided by 30, it gives one of the following eight remainders:
Therefore, the sum we need to calculate is equal to
where $ s_i $ denotes the sum of an analogous series formed from terms corresponding to those values of $ n $ that have the form
Thus,
(The correctness of the applied transformation and the validity of equation (1) follow from the fact that the terms of the considered series are positive; they can therefore be grouped and rearranged in any way without affecting the convergence and value of the series sum.)
The right-hand side of equation (2) represents a geometric series with a common ratio of $ 2^{-30} $. Its sum is equal to
Substituting these numbers into equation (1), we get
This is the desired value. | \frac{1}{1-2^{-30}}(\frac{1}{2^1}+\frac{1}{2^7}+\frac{1}{2^{11}}+\frac{1}{2^{13}}+\frac{1}{2^{17}}+\frac{1}{2^{19}}+\frac{1}{2^{23}}+\frac{1}{2^{} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXIX OM - II - Problem 3
Inside an acute triangle $ ABC $, consider a point $ P $ and its projections $ L, M, N $ onto the sides $ BC, CA, AB $, respectively. Determine the point $ P $ for which the sum $ |BL|^2 + |CM|^2 + |AN|^2 $ is minimized. | Let $ P $ be any point inside the triangle $ ABC $. Considering the right triangles $ ANP $, $ BLP $, $ CMP $ (Figure 7), we observe that
Similarly, from the triangles $ AMP $, $ BNP $, $ CLP $ we get
Thus, by equating the right sides of the obtained equalities:
The left side of equation (1) is precisely the quantity we are considering in the problem; let's denote it by $ f(P) $. We need to find the point $ P $ for which $ f(P) $ reaches its minimum.
The sum of the left and right sides of equation (1) is $ 2f(P) $. Therefore,
Now notice that
The value of this expression is the smallest when the difference $ |BL| - |CL| $ is zero (since $ |BC| $ is constant) - that is, when $ L $ is the midpoint of segment $ BC $. Similarly, the other two sums in parentheses on the right side of (2) have the smallest value when points $ M $ and $ N $ are the midpoints of sides $ CA $ and $ AB $.
These conditions can be satisfied simultaneously: the projections of point $ P $ onto the sides of the triangle coincide with the midpoints of these sides if and only if $ P $ lies at the intersection of the perpendicular bisectors of the sides. Hence the answer: the quantity $ f(P) $ reaches its minimum when $ P $ is the circumcenter of triangle $ ABC $ (it lies in the interior of triangle $ ABC $, since the triangle is acute). | PisthecircumcenteroftriangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXXVI OM - I - Zadanie 9
W urnie jest 1985 kartek z napisanymi liczbami 1,2,3,..., 1985, każda lczba na innej kartce. Losujemy bez zwracania 100 kartek. Znaleźć wartość oczekiwaną sumy liczb napisanych na wylosowanych kartkach.
|
Losowanie $ 100 $ kartek z urny zawierającej $ 1985 $ kartek można interpretować jako wybieranie $ 100 $-elementowego podzbioru zbioru $ 1985 $-elementowego. Zamiast danych liczb $ 1985 $ i $ 100 $ weźmy dowolne liczby naturalne $ n $ i $ k $, $ n \geq k $. Dla dowolnego $ k $-elementowego zbioru $ X $ będącego podzbiorem zbioru $ Z = \{1,2,\ldots,n\} $ oznaczmy przez $ s(X) $ sumę liczb w zbiorze $ X $. Ponumerujmy wszystkie $ k $-elementowe podzbiory $ Z $ liczbami od $ 1 $ do $ ???????????????? = \binom{n}{k}: X_1, \ldots, X_N $. Wybieramy losowo jeden z tych zbiorów. Prawdopodobieństwo każdego wyboru jest takie samo, a więc równa się $ p = 1 /N $. Wartość oczekiwana sumy liczb w tym zbiorze równa się
Policzymy, w ilu zbiorach $ X_i $ występuje dowolnie ustalona liczba $ x \in Z $. Liczbie $ x $ towarzyszy w każdym z tych zbiorów $ k-1 $ liczb dowolnie wybranych ze zbioru $ Z-\{x\} $. Możliwości jest $ M = \binom{n-1}{k-1} $. Wobec tego każda
liczba $ x \in Z $ występuje w $ M $ zbiorach $ X_i $; tak więc w sumie każdy składnik $ x \in Z $ pojawia się $ M $ razy. Stąd $ s = M(1 + 2+ \ldots+n) = Mn(n+1)/2 $ i szukana wartość oczekiwana wynosi
W naszym zadaniu $ n = 1985 $, $ k = 100 $, zatem $ E = 99 300 $.
| 99300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LII OM - I - Task 4
Determine whether 65 balls with a diameter of 1 can fit into a cubic box with an edge of 4. | Answer: It is possible.
The way to place the balls is as follows.
At the bottom of the box, we place a layer consisting of 16 balls. Then we place a layer consisting of 9 balls, each of which is tangent to four balls of the first layer (Fig. 1 and 2). The third layer consists of 16 balls that are tangent to the balls of the second layer (Fig. 4 and 5). Similarly, we place two more layers (Fig. 6).
om52_1r_img_2.jpg
om52_1r_img_3.jpg
om52_1r_img_4.jpg
In total, we have placed $ 16 + 9 + 16 + 9 + 16 = 66 $ balls. It remains to calculate how high the fifth layer reaches.
om52_1r_img_5.jpg
om52_1r_img_6.jpg
om52_1r_img_7.jpg
Let's choose any ball from the second layer; this ball is tangent to four balls of the first layer. The centers of these five balls are the vertices of a regular square pyramid, each edge of which has a length of 1 (Fig. 3). By the Pythagorean theorem, the height of this pyramid is $ \frac{\sqrt{2}}{2} $. Therefore, the highest point that the fifth layer reaches is at a distance of $ \frac{1}{2} + 4 \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} = 1 + 2\sqrt{2} < 4 $ from the base plane. The 66 balls placed in this way fit into a cubic box with an edge of 4. | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
II OM - III - Task 2
What digits should be placed instead of zeros in the third and fifth positions in the number $ 3000003 $ to obtain a number divisible by $ 13 $? | Let the sought digits be denoted by $ x $ and $ y $, and write the number $ 30x0y03 $ in the form
We need to find such integers $ x $ and $ y $ that the number $ N $ is divisible by $ 13 $ and that the inequalities
are satisfied.
The numbers $ 10^6 $, $ 10^4 $, $ 10^2 $ give remainders $ 1 $, $ 3 $, $ 9 $ respectively when divided by $ 13 $, so
\[ 3 \cdot 10^6 = 13k_1 + 3, \]
\[ x \cdot 10^4 = 13k_2 + 3x, \]
\[ y \cdot 10^2 = 13k_3 + 9y, \]
where $ k_1 $, $ k_2 $, $ k_3 $ are natural numbers. Therefore,
\[ 30x0y03 = 3 \cdot 10^6 + x \cdot 10^4 + y \cdot 10^2 + 3 = 13k_1 + 3 + 13k_2 + 3x + 13k_3 + 9y + 3 \]
\[ = 13(k_1 + k_2 + k_3) + 3 + 3x + 9y + 3 \]
\[ = 13(k_1 + k_2 + k_3) + 3x + 9y + 6 \]
where $ k $ is a natural number.
For the number $ N $ to be divisible by $ 13 $, it is necessary and sufficient that the number $ x + 3y + 2 $ is divisible by $ 13 $, i.e., that $ x $ and $ y $ satisfy the equation
\[ x + 3y + 2 = 13m \]
where $ m $ is a natural number.
From the inequalities $ x \leq 9 $, $ y \leq 9 $, it follows that
\[ 0 \leq x + 3y + 2 \leq 38 \]
so the number $ m $ must satisfy the inequality $ 13m \leq 38 $, i.e., $ m $ can only be $ 1 $ or $ 2 $.
1° Taking $ m = 1 $, we get the equation for $ x $ and $ y $
\[ x + 3y + 2 = 13 \]
Under the constraints $ 0 \leq x \leq 9 $, $ 0 \leq y \leq 9 $, this equation has three integer solutions:
\[ (x, y) = (8, 1), (4, 2), (2, 3) \]
2° Taking $ m = 2 $, we get the equation
\[ x + 3y + 2 = 26 \]
and we obtain the solutions:
\[ (x, y) = (9, 5), (6, 6), (3, 7), (0, 8) \]
Thus, the problem has $ 7 $ solutions, corresponding to the numbers: $ 3080103 $, $ 3040203 $, $ 3020303 $, $ 3090503 $, $ 3060603 $, $ 3030703 $, $ 3000803 $.
Note. In problem 35, it was about changing the digits of a certain specified number. Let's consider two more general examples of such problems:
I. Given a natural number $ N $ not divisible by a prime number $ p $ and having a digit $ 0 $ in the $ k $-th decimal place. Can the zero be replaced by a digit $ x $ so that the new number is divisible by $ p $?
First, note that if $ p $ is $ 2 $ or $ 5 $, the problem has no solution. Therefore, we assume in the following that $ p \ne 2 $ and $ p \ne 5 $.
When we place the digit $ x $ in the $ k $-th decimal place of the number $ N $, the resulting number $ N(x) $ is given by the formula
\[ N(x) = N + x \cdot 10^{k-1} \]
Consider the values of the function $ N(x) $ when $ x $ takes the values $ 0, 1, 2, \ldots, (p - 1) $:
\[ N(0), N(1), N(2), \ldots, N(p-1) \]
Each of the $ p $ numbers in the sequence (2) gives a different remainder when divided by $ p $. Indeed, according to formula (1), the difference between two numbers in the sequence (2) is given by
\[ N(i) - N(j) = (i - j) \cdot 10^{k-1} \]
When $ i \ne j $, this number is not divisible by $ p $, because the factor $ (i - j) $ is non-zero and less in absolute value than $ p $, and the factor $ 10^{k-1} $ has only the prime divisors $ 2 $ and $ 5 $. Therefore, the numbers $ N(i) $ and $ N(j) $ give different remainders when divided by $ p $.
Since the remainders of the numbers in the sequence (2) are different and there are $ p $ of them, one and only one of these remainders is zero, i.e., one and only one of the numbers in the sequence (2) is divisible by $ p $; let this number be $ N(x_0) $.
If $ x_0 \leq 9 $, then $ x_0 $ gives a solution to the problem.
If $ x_0 > 9 $, the problem has no solution.
Hence the conclusion:
a) When $ p = 3 $, the problem always has three solutions; the sought digits are either $ 1 $, $ 4 $, $ 7 $ or $ 2 $, $ 5 $, $ 8 $.
b) When $ p = 7 $, the problem has either two solutions - the digits $ 1 $, $ 8 $ or $ 2 $, $ 9 $, or one solution - one of the digits $ 3 $, $ 4 $, $ 5 $, $ 6 $.
c) When $ p \geq 11 $, the problem may have one solution or no solution. If $ p = 11 $, the first of these cases occurs, for example, for the number $ 10 $, and the second case is given by the number $ 109 $.
II. Given a natural number $ N $ not divisible by a prime number $ p \geq 11 $ and having a digit $ 0 $ in the $ k $-th and $ l $-th decimal places. Can the zeros be replaced by digits $ x $ and $ y $ so that the new number is divisible by $ p $?
When we place the digit $ x $ in the $ k $-th decimal place and the digit $ y $ in the $ l $-th decimal place of the number $ N $, we obtain the number $ N(x, y) $ given by the formula
\[ N(x, y) = N + x \cdot 10^{k-1} + y \cdot 10^{l-1} \]
Consider the values of the function $ N(x, y) $ obtained by substituting $ 0, 1, 2, \ldots, (p - 1) $ for $ x $ and $ y $. We get $ p^2 $ numbers:
\[ N(0, 0), N(0, 1), \ldots, N(0, p-1), N(1, 0), N(1, 1), \ldots, N(1, p-1), \ldots, N(p-1, 0), N(p-1, 1), \ldots, N(p-1, p-1) \]
We state, as in problem I, that in each row and each column of this table there is one and only one number divisible by $ p $; let the number in the $ i $-th row be denoted by $ N(i, y_i) $.
The table thus contains $ p $ numbers divisible by $ p $; they are the numbers
\[ N(0, y_0), N(1, y_1), N(2, y_2), \ldots, N(p-1, y_{p-1}) \]
where the sequence of numbers $ y_0, y_1, y_2, \ldots, y_{p-1} $ contains the same numbers as the sequence $ 0, 1, 2, \ldots, (p - 1) $, but in a different order. Note also that $ y_0 $ is certainly not equal to $ 0 $, since the number $ N(0, 0) = N $, by assumption, is not divisible by $ p $.
The number $ N(i, y_i) $ in the sequence (4) gives a solution to the problem only if $ i \leq 9 $ and $ y_i \leq 9 $. It is easy to count how many such numbers the sequence (4) can contain. They should be sought among the numbers
\[ N(0, y_0), N(1, y_1), N(2, y_2), \ldots, N(9, y_9) \]
In the case where none of the numbers $ y_0, y_1, \ldots, y_9 $ is greater than $ 9 $, the problem has $ 10 $ solutions given by the numbers in the sequence (5). This is the maximum possible number of solutions.
If among the numbers $ y_0, y_1, \ldots, y_9 $ there are numbers greater than $ 9 $ (there can be at most $ p - 10 $ such numbers), the corresponding terms should be crossed out from the sequence (5). The problem thus has at least $ 10 - (p - 10) $, i.e., $ 20 - p $ solutions.
Hence the conclusion:
a) When $ p = 11 $, the problem has at least $ 9 $ solutions. An example of the case where there are the maximum number of $ 10 $ solutions is the number $ 101015 $.
b) When $ p = 13 $, the problem has at least $ 7 $ solutions; the case where there are only $ 7 $ solutions was in problem 35.
c) When $ p = 17 $, we have at least three solutions.
d) When $ p = 19 $, the problem has at least one solution.
e) When $ p \geq 23 $, the problem may have no solutions. An appropriate example for the case $ p = 23 $ can be found easily if one considers that the number $ 10^{22} $ gives a remainder of $ 1 $ when divided by $ 23 $. We propose this as an exercise. | 3080103,3040203,3020303,3090503,3060603,3030703,3000803 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Task 5
Find all pairs of positive integers $ x $, $ y $ satisfying the equation $ y^x = x^{50} $. | We write the given equation in the form $ y = x^{50/x} $. Since for every $ x $ being a divisor of $ 50 $, the number on the right side is an integer, we obtain solutions of the equation for $ x \in \{1,2,5,10,25,50\} $. Other solutions of this equation will only be obtained when $ x \geq 2 $ and for some $ k \geq 2 $, the number $ x $ is simultaneously the $ k $-th power of some natural number and a divisor of the number $ 50k $. If $ p $ is a prime divisor of such a number $ x $, then $ p^k|50^k $. Since $ p^k > k $, it cannot be that $ p^k|k $, so $ p \in \{2,5\} $. If $ p = 2 $, then $ 2^k|2k $, from which $ k = 2 $. If, however, $ p = 5 $, then $ 5^k|25k $, from which again $ k = 2 $. Therefore, $ x $ must simultaneously be the square of some natural number and a divisor of the number $ 100 $.
We obtain two new values of $ x $ in this case: $ x = 4 $ and $ x = 100 $. Thus, the given equation has 8 solutions: | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
VIII OM - I - Problem 8
In the rectangular prism $ ABCDA_1B_1C_1D_1 $, the lengths of the edges are given as $ AA_1 = a $, $ AB = b $, $ AD = c $. On the face $ A_1B_1C_1D_1 $, a point $ M $ is chosen at a distance $ p $ from the side $ A_1B_1 $, and at a distance $ q $ from the side $ A_1D_1 $, and a parallelepiped is constructed with the base $ ABCD $ and the lateral edge $ AM $. Calculate the area of the lateral faces of this parallelepiped. | Accepting the notation indicated in figure $5$, we calculate the area of the parallelogram $AMQD$ which is a face of the parallelepiped $ABCDMNPQ$.
The parallelogram $AMQD$ and the rectangle $AKLD$ have a common base $AD$ and height $AK$, thus
But $AK = \sqrt{AA_1^2 + A_1K^2} = \sqrt{a^2 + q^2},\ KL = AD = c$; hence
Similarly, we calculate that the area of $AMNB = b \sqrt{a^2 + p^2}$. | \sqrt{^2+} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - I - Problem 2
How many digits do all natural numbers with at most $ m $ digits have in total? | The number of one-digit natural numbers is $ c_1 = 9 $; the number of two-digit numbers is $ c_2= 10^2 -10= 10 \cdot 9 $, they have $ 2 \cdot 10 \cdot 9 $ digits; the number of three-digit numbers is $ c_3=10^3-10^2= 10^2 \cdot 9 $, the number of their digits is $ 3 \cdot 10^2 \cdot 9 $, etc. Finally, the number of $ m $-digit numbers is $ c_m = 10^m -10^{m-1} = 10^{m-1} \cdot 9 $; they have $ c_m =m \cdot 10^{m-1} \cdot 9 $ digits.
All numbers with at most $ m $ digits have in total
digits. To calculate the value of the expression in parentheses in formula (1), we will first solve a more general problem, namely find the formula for the sum
where $ a \ne 1 $, depending on $ a $. From equality (2) we get
Subtracting equality (3) from (2) we have
from which
After arranging the terms, we get the formula
Taking $ a = 10 $ in (2) and (4) and substituting the obtained value of the sum (2) into equality (1), we get the desired number of
Note. We derived formula (4) under the assumption that $ a \ne 1 $. When $ a = 1 $, formula (4) makes no sense; in this case, the value of $ S $ can be obtained by substituting $ a = 1 $ in formula (2):
thus, according to the known formula for an arithmetic progression,
We can, however, arrive at formula (5) based on formula (4) by the following reasoning.
Let $ a -1= x $; substituting $ a = 1 + x $ in formula (4) we get
Expanding $ (1 + x)^{m+1} $ and $ (1 + x)^m $ according to the binomial theorem of Newton. We will only calculate the terms containing $ x^0 $, $ x $, and $ x^2 $; the remaining terms form a polynomial in $ x $, in which all terms are of at least the third degree and we can write
where $ f(x) $ denotes some polynomial in $ x $.
Hence,
or
Substituting $ a = 1 $ in formula (6) we get formula (5). It turns out, therefore, that formula (6) gives the value of the sum (2) for any value of $ a $. However, it is useful for calculating this sum only when $ a = 1 $; when $ a \ne 1 $, it must be restored to the form (4). | 9\cdot\frac{10^-1}{9}\cdot\frac{(+1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 8
On three pairwise skew edges of a cube, choose one point on each in such a way that the sum of the squares of the sides of the triangle formed by them is minimized. | Let $ABCD$ be the base of a cube and $AA_1$, $BB_1$, $CC_1$, $DD_1$ its edges perpendicular to $ABCD$. Let $X$, $Y$, $Z$ be points chosen on three pairwise skew edges of the cube. It is known that by rotating the cube around certain axes passing through its center, the cube can be superimposed onto itself in such a way that any chosen edge can cover any other edge. Therefore, without loss of generality, we can assume that point $X$ lies on edge $A_1$. The skew edges to $AA_1$ are $BC$, $CD$, $B_1C_1$, $C_1D_1$. (Fig. 8); among them, the mutually skew ones are $BC$ and $C_1D_1$ as well as $B_1C_1$ and $CD$. Thus, we have two sets of three pairwise skew edges: $AA_1$, $BC$, $C_1D_1$ and $AA_1$, $B_1C_1$, $CD$, which transform into each other when the cube is reflected through the plane $AA_1C_1C$, which is a plane of symmetry of the cube. Therefore, we can assume that the considered set of three pairwise skew edges is $AA_1$, $BC$, $C_1D_1$ and that, for example, point $Y$ lies on $BC$ and point $Z$ on $C_1D_1$. Let $AX = x$, $BY = y$, $C_1Z = z$, $AA_1 = a$. Then
from the obtained formula, it is clear that the sum of the squares of the sides of triangle $XYZ$ reaches a minimum when $x = y = z = \frac{a}{2}$, i.e., when points $X$, $Y$, $Z$ are chosen at the midpoints of the respective edges. The minimum value is $\frac{9}{2} a^2$. | \frac{9}{2}^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLVII OM - I - Problem 1
Determine all integers $ n $ for which the equation $ 2 \sin nx = \tan x + \cot x $ has solutions in real numbers $ x $. | Assume that the given equation has solutions in real numbers: let $ \alpha $ be one of these solutions. Of course, $ \alpha $ cannot be an integer multiple of $ \pi/2 $, because for such numbers the right side loses its meaning. According to the equation,
hence $ 2\sin n\alpha \cdot \sin 2\alpha=2 $, which means
(we use the formula: $ \cos u - \cos v = 2\sin \frac{1}{2}(u+v)\sin\frac{1}{2}(v-u)) $. The difference of two cosines is equal to $ 2 $ if and only if the first one has the value $ 1 $ and the second one $ -1 $. We thus obtain the system of equalities
from which it follows that
($ p, q $ are integers).
In the second of the equalities (1), we have a non-zero number on the right side; hence the left side also has a non-zero value. We can therefore divide the first equality by the second. We get the relation $ \frac{n-2}{n+2}=\frac{2q}{2p+1} $.
By introducing the notation $ n-2 = m $, we rewrite this relation as $ (2p+1)m = 2q(m+4) $, or
We see that the number $ m $ multiplied by an odd number gives a result divisible by $ 8 $, which is possible only if the number $ m $ itself is divisible by $ 8 $. Since $ n = m + 2 $, this means that the number $ n $ gives a remainder of $ 2 $ when divided by $ 8 $.
Taking into account the condition of the problem that the number $ n $ must be positive, we arrive at the following conclusion:
Conversely, if the number $ n $ has such a form, then the considered equation has solutions in real numbers --- one of them is, for example, the number $ x = \frac{1}{4} \pi $; indeed: for $ x=\frac{1}{4} \pi $ and $ n = 8k + 2 $ we have $ nx = 2k\pi + \frac{1}{2} \pi $, from which $ \sin nx = 1 $; both the left and the right side of the equation have the value $ 2 $.
Conclusion: the numbers whose determination was the goal of the problem are characterized by the statement (2).
Note: We will also mention a variant of the solution based on the observation (right at the beginning) that for any number $ x $ the right side of the equation, as a number of the form $ t + (1/t) $, has an absolute value not less than $ 2 $, while the absolute value of the left side is not greater than $ 2 $. The equation can therefore be satisfied only when both sides have the value $ 2 $ or $ -2 $. Each of these possibilities gives a system of two simple trigonometric equations. Completing the reasoning is an easy task and leads to conclusion (2), which characterizes the sought numbers $ n $. We leave the elaboration of the details to the reader. | 8k+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 5
Given a segment $ AD $. Find points $ B $ and $ C $ on it, such that the product of the lengths of the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $ is maximized. | We consider such positions of points $ B $ and $ C $, for which
this means, we assume that points $ A $, $ B $, $ C $, $ D $ lie on a line in this exact order (we do not lose generality, as the product being studied in the problem does not change value when the roles of $ B $ and $ C $ are swapped); the weak inequalities in (1) indicate that we allow for boundary cases where some of the points $ A $, $ B $, $ C $, $ D $ coincide.
Let us take the length of segment $ AD $ as the unit: $ |AD| = 1 $. Furthermore, let us denote:
then $ |AB| + |CD| = 1 - x $, from which
The product of the lengths of the segments in question is
this inequality becomes an equality if and only if $ y = 0 $ - that is, when $ |AB| = |CD| $.
The maximum value of the polynomial
on the interval $ \langle 0;\ 1\rangle $ can be found by differentiation:
we see that $ f $ achieves its maximum at $ x = \frac{1}{5} \sqrt{5} $.
[Instead of differentiating, one can apply a more elementary (and thus more elegant) method: using the inequality between the geometric mean and the arithmetic mean of five numbers. These five numbers will be: $ 4x^2 $ and the number $ 1-x^2 $ repeated four times. We obtain the inequality
or
with equality achieved when the averaged numbers are equal; that is, when
$ 4x^2 = 1 - x^2 $, or for $ x = \frac{1}{5} \sqrt{5} $.]
In accordance with the estimate (2) and the remark following it, the expression $ I $ achieves its maximum for $ x = \frac{1}{5} \sqrt{5} $, $ y = 0 $. This corresponds to placing segment $ BC $ of length $ x = \frac{1}{5} \sqrt{5} $ symmetrically relative to the midpoint of segment $ AD $ - while maintaining the order (1), or by swapping the roles of points $ B $ and $ C $. | \frac{1}{5}\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LX OM - I - Task 12
Given a prime number $ p $. On the left side of the board, the numbers $ 1, 2, 3, \cdots, p - 1 $ are written,
while on the right side, the number $ 0 $ is written. We perform a sequence of $ p - 1 $ moves, each of which proceeds
as follows: We select one of the numbers written on the left side of the board, add it to all the
remaining numbers on the board, and then erase the selected number.
Determine for which values
of $ p $ it is possible to choose the numbers in such a way that the number remaining on the board after
all moves have been performed is divisible by $ p $. | Answer: For all prime numbers $p$ except 2 and 3.
Let $ t_i $ for $ i=1,2, \cdots,p-1 $ denote the initial value of the number that
was erased from the board in the $ i $-th move. Thus, the sequence $ (t_1,t_2,t_3, \cdots,t_{p-1}) $ is a permutation
of the sequence $ (1,2,3, \cdots,p-1) $ and uniquely determines the way the erased numbers are chosen in subsequent moves.
Let us introduce the notation
We will prove by induction that after the $ i $-th move, all numbers on the board are greater
by $ s_i $ than their initial values.
Indeed: this statement is true for $ i = 1 $, since in the first move the number
$ t_1 $ was erased, increasing all other numbers by $ s_1 = t_1 $. If, however, after
the $ i $-th move all numbers on the board were greater by $ s_i $ than their initial values,
then in particular the number initially equal to $ t_{i+1} $ had the value
$ t_{i+1} + s_i $ after the $ i $-th move. Therefore, as a result of the $ (i+1) $-th move, the amount
$ t_{i+1} + s_i $ was added to all remaining numbers. Since the numbers were greater by $ s_i $ than their initial values after the $ i $-th move,
they became greater by $ t_{i+1}+s_i+s_i $ than their initial values after the $ (i+1) $-th move.
It remains to observe that
which completes the inductive proof. We have thus shown that after all $ p $ moves, the number
remaining on the right side of the board is
We have thus reduced the problem to finding all prime numbers $ p $ for which there exists a permutation
$ (t_1,t_2,t_3, \cdots,t_{p-1}) $ of the sequence $ (1,2,3, \cdots,p-1) $ such that the number (1) is divisible by $ p $.
We directly check that for $ p=2 $ and $ p=3 $ it is not possible to obtain a number divisible by $ p $:
for $ p = 2 $ the number (1) is $ 1 $, and for $ p = 3 $ it can be $ 1+2\cdot 2=5 $ or $ 2+2\cdot 1 = 4 $.
We will show that if $ p $ is a prime number greater than $ 3 $, then for the permutation
formed from the sequence $ (1,2,3,4, \cdots,p - 2,p - 1) $ by swapping the positions in pairs $ (2j-1,2j) $ for
$ j \geqslant 2 $ and then reversing the order, the number (1) is divisible by $ p $.
Indeed, for the permutation (2), the number (1) has the value
For $ j =2,3, \cdots, \frac{1}{2}(p-1) $, the relation
holds, from which it follows that $ L = M - N $, where $ M $ is the sum of numbers of the form $ 2^{i-1}i $ for
$ i =1,2, \cdots,p-1 $, and $ N =2^2 +2^4 +2^6 + \cdots +2^{p-3} $. Using the formula for the sum of consecutive
terms of a geometric sequence, we calculate that
and
Thus,
By Fermat's Little Theorem (see LI Mathematical Olympiad, Main Committee Report,
Warsaw 2001, Appendix A, p. 102), the number $ 2^{p-1} -1 $ is divisible by $ p $. Since $ p $
is a prime number greater than $ 3 $, the fraction appearing on the right side of relation (3) is divisible by $ p $. Therefore, the right side of this relation is divisible by $ p $, which completes
the solution of the problem. | Forallpexcept23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXIX OM - I - Problem 1
For each positive number $ a $, determine the number of roots of the polynomial $ x^3+(a+2)x^2-x-3a $. | Let's denote the considered polynomial by $ F(x) $. A polynomial of the third degree has at most three real roots. We will show that the polynomial $ F $ has at least three real roots - and thus has exactly three real roots (for any value of the parameter $ a > 0 $).
It is enough to notice that
If a continuous function with real values takes values of different signs at two points, then at some point in the interval bounded by these points, it takes the value zero (Darboux property).
Hence, in each of the intervals $ (-a-3; -2) $, $ (-2; 0) $, $ (0; 2) $, there is a root of the polynomial $ F $. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLVIII OM - II - Problem 4
Determine all triples of positive integers having the following property: the product of any two of them gives a remainder of $ 1 $ when divided by the third number. | Let $ a $, $ b $, $ c $ be the numbers we are looking for. From the given conditions, these numbers are greater than $ 1 $ and are pairwise coprime. The number $ bc + ca + ab - 1 $ is divisible by $ a $, by $ b $, and by $ c $, so it is also divisible by the product $ abc $:
After dividing both sides by $ abc $:
The sum on the left side of equation (1) is greater than $ 1 $, so $ k > 0 $. The sum on the left side of (2) is less than $ 2 $, so $ k < 2 $. Therefore, $ k = 1 $. The right side of (2) is greater than $ 1 $, so the largest of the numbers $ 1/a $, $ 1/b $, $ 1/c $ exceeds $ 1/3 $. Assuming $ a > b > c $, we conclude that $ c = 2 $. Equation (1) then takes the form $ 2a + 2b + ab = 2ab + 1 $, or $ (a - 2)(b - 2) = 3 $. Hence, $ a = 5 $, $ b = 3 $, $ c = 2 $. These numbers satisfy the conditions of the problem and are (up to permutation) its only solution. | (5,3,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LVIII OM - I - Problem 2
Determine all pairs of positive integers $ k $, $ m $, for which each of the numbers $ {k^2+4m} $, $ {m^2+5k} $ is a perfect square. | Suppose the pair $ (k,m) $ satisfies the conditions of the problem.
If the inequality $ m\geq k $ holds, then
$$(m+3)^2=m^2+6m+9>m^2+5m\geq m^2+5k>m^2,$$
and since $ m^2+5k $ is a square of an integer, it follows that one of the equalities $ m^2+5k=(m+1)^2 $ or $ m^2+5k=(m+2)^2 $ must hold.
If $ m^2+5k=(m+1)^2=m^2+2m+1 $, then $ 2m=5k-1 $ and by the conditions of the problem, the number $ k^2+4m=k^2+2(5k-1)=k^2+10k-2 $ must be a square of an integer. We have $ k^2+10k-2< k^2+10k+25=(k+5)^2 $, so the following relationship holds
from which we obtain $ 2k\leq 18 $ and $ k\le 9 $. Considering the equality $ 2m=5k-1 $, we see that $ k $ must be an odd number. By direct calculation, we check that the values of the expression $ k^2+10k-2 $ for $ k=1 $, $ 3 $, $ 5 $, $ 7 $, $ 9 $ are respectively 9, 37, 73, 117, 169. We obtain squares of integers only for $ k=1 $ and $ k=9 $; the corresponding values of $ m={1\over 2}(5k-1) $ are 2 and 22.
If $ m^2+5k=(m+2)^2=m^2+4m+4 $, then $ 4m=5k-4 $, so the number $ k^2+4m=k^2+5k-4 $ is a square of an integer. The inequality
proves that $ k^2+5k-4\leq (k+2)^2=k^2+4k+4 $, which gives $ k\le 8 $. Moreover, the number $ m={5\over 4}k-1 $ is an integer, so the number $ k $ is divisible by $ 4 $. The expression $ k^2+5k-4 $ for the values $ k=4 $, $ 8 $ takes the values 32, 100 respectively, and only for $ k=8 $ do we obtain a square of an integer; the corresponding value of $ m={5\over 4}k-1 $ is 9.
The case $ m<k $ remains to be considered. In this case, we can write
and since $ k^2+4m $ is a square of an integer, we obtain the relationship $ k^2+4m=(k+1)^2=k^2+2k+1 $, i.e., $ 2k=4m-1 $. However, this equality cannot be true, because its left side is an even number, and the right side is odd. Therefore, there are no solutions in this case.
It is sufficient to verify directly that the three pairs obtained indeed satisfy the conditions of the problem. | (1,2),(9,22),(8,9) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XX OM - II - Task 2
Find all four-digit numbers in which the thousands digit is equal to the hundreds digit, and the tens digit is equal to the units digit, and which are squares of integers. | Suppose the number $ x $ satisfies the conditions of the problem and denote its consecutive digits by the letters $ a, a, b, b $. Then
The number $ x $ is divisible by $ 11 $, so as a square of an integer, it is divisible by $ 11^2 $, i.e., $ x = 11^2 \cdot k^2 $ ($ k $ - an integer), hence
Therefore,
The number $ a+b $ is thus divisible by $ 11 $. Since $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $, then $ 0 < a+b \leq 18 $, hence
Therefore, we conclude that $ b \ne 0 $, $ b \ne 1 $; since $ b $ is the last digit of the square of an integer, it cannot be any of the digits $ 2, 3, 7, 8 $. Thus, $ b $ is one of the digits $ 4, 5, 6, 9 $. The corresponding values of $ a $ are $ 7, 6, 5, 2 $, so the possible values of $ x $ are only the numbers $ 7744 $, $ 6655 $, $ 5566 $, $ 2299 $. Only the first one is a square of an integer.
The problem has one solution, which is the number $ 7744 $. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - III - Problem 6
A plane is covered with a grid of regular hexagons with a side length of 1. A path on the grid is defined as a sequence of sides of the hexagons in the grid, such that any two consecutive sides have a common endpoint. A path on the grid is called the shortest if its endpoints cannot be connected by a shorter path. Find the number of shortest paths on the grid with a fixed starting point and a length of 60. | Attention. The phrase "consecutive sides have a common end" used in the text should be understood to mean that directed segments are considered, and the end of the previous one is the start of the next (thus, it does not form a path, for example, a star of segments with one common end). Despite the lack of precision, this formulation was understood by all solvers in this way, in accordance with the intuitive sense of the word "path."
Let us establish one of the directions of the sides of the considered hexagonal grid ($ \Sigma_6 $), let's call it horizontal and distinguish the directions on it: left, right. From each node of the grid $ \Sigma_6 $, exactly one horizontal side extends. Let $ W $ denote the set of nodes from which the horizontal side extends to the right, and $ W $ the set of the remaining nodes, that is, those from which the horizontal side runs to the left. Each side of the grid connects a point of the set $ W $ with a point of the set $ W $.
Let $ O $ be a selected fixed node, the starting point of the considered paths. We can assume that $ O \in W $. Starting from point $ O $, after an even number of steps, we will find ourselves at a point of the set $ W $, and after an odd number - at a point of the set $ W $. Therefore, two paths with a common start and end have the same parity of length; if a path can be shortened - then by at least $ 2 $.
Two different points of the set $ W $ will be called adjacent if they are connected by a path of length $ 2 $; this path is then unique. Let us connect each pair of adjacent points of the set $ W $ with a segment. In this way, a grid $ \Sigma_3 $ of equilateral triangles with side $ \sqrt{3} $ will arise; the nodes of the grid $ \Sigma_3 $ are the points of the set $ W $. We define the concept of a path on the grid $ \Sigma_3 $, the length of the path, and the concept of the shortest path analogously to the corresponding concepts for the grid $ \Sigma_6 $, with the exception that we take the side of the grid $ \Sigma_3 $ as the unit of length.
In the following, we will consider only paths of even length on the grid $ \Sigma_6 $, starting at point $ O $ and satisfying the following condition: if $ OP_1P_2 \ldots P_{2k-1}P_{2k} $ is the considered path, then $ P_{2i-2} \neq P_{2i} $ for $ i = 1, \ldots, k $ (we assume $ P_0 = O $); in other words, no side of the grid is traversed "there and back" in two consecutive steps, the first of which has an odd number and the second an even number (of course, every shortest path satisfies this condition). By following such a path, in each subsequent pair of steps, one moves from a point of the set $ W $ to an adjacent point of the set $ W $. Therefore, each path on the grid $ \Sigma_6 $ that satisfies this condition and has length $ 2k $ determines a path of length $ k $ on the grid $ \Sigma_3 $. This is a one-to-one correspondence, because the ends of each side of the grid $ \Sigma_3 $ are connected by exactly one path of length $ 2 $ on the grid $ \Sigma_6 $. Shortest paths on one grid correspond to shortest paths on the other grid.
We will try to count the shortest paths of length $ k $ on the grid $ \Sigma_3 $ (starting at $ O $).
Consider six paths (on the grid $ \Sigma_3 $) of length $ k $ starting from $ O $ and running along rays. Their ends are the vertices of a regular hexagon, which we will denote by $ H_k $. By starting from point $ O $ and making $ k $ moves (on the grid $ \Sigma_3 $), we will not leave the strip bounded by the lines containing any two parallel sides of the hexagon $ H_k $ - and thus we will not leave the area of this hexagon. We can, however, reach any point lying on its boundary. Moreover, any node of the grid $ \Sigma_3 $ lying inside $ H_k $ belongs to some hexagon $ H_j $, where $ j < k $, and therefore is reachable by a path of length $ j $. Therefore, only the points on the boundary of $ H_k $ are reached by the shortest paths of length $ k $ (on the grid $ \Sigma_3 $).
How many such paths are there? Any side $ AB $ of the hexagon $ H_k $ contains $ k+1 $ points of the set $ W $, and if we number them $ Q_0, \ldots, Q_k $ ($ Q_0 = A $, $ Q_k = B $), we will notice that $ \binom{k}{r} $ shortest paths lead from point $ O $ to point $ Q_r $; this follows by easy induction from the observation that the shortest paths to any of the points $ Q_0, \ldots, Q_{k-1} $ are extensions of the shortest paths to two adjacent nodes on the boundary of the hexagon $ H_{k-1} $, and to points $ A $ and $ B $ there is one path of length $ k $ each. (In other words: by writing in each node of the grid $ \Sigma_3 $ the number of shortest paths from point $ O $ to the given node, we obtain six copies of Pascal's triangle, one in each of the six sectors of the plane with vertex $ O $, determined by the directions of the sides of the grid $ \Sigma_3 $). There are thus $ 2^k $ shortest paths connecting $ O $ with the points of side $ AB $.
The hexagon $ H_k $ has six sides. We therefore have $ 6 \cdot 2^k - 6 $ shortest paths to points on the boundary of $ H_k $; the term $ -6 $ prevents the double counting of paths leading to the vertices of $ H_k $.
The task is to determine the number of shortest paths (starting at $ O $) of length $ 60 $ on the grid $ \Sigma_6 $, i.e., the number of shortest paths (starting at $ O $) of length $ 30 $ on the grid $ \Sigma_3 $. According to the above considerations, this number is $ 6 \cdot 2^{30} - 6 $. | 6\cdot2^{30}-6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXV OM - II - Problem 1
For a given natural number $ n $, find the number of solutions to the equation $ \sqrt{x} + \sqrt{y} = n $ in natural numbers $ x, y $. | For $ n = 1 $, the given equation has no solutions in natural numbers. For $ n > 1 $, each pair of numbers $ x_1 = 1^2, y_1 = (n-1)^2, x_2 = 2^2, y_2 = (n-2)^2, \ldots, x_{n-1} = (n-1)^2, y_{n-1} = 1^2 $ is obviously a solution to the given equation, since $ \sqrt{k^2} + \sqrt{(n - k)^2} = k + (n - k) = n $ for $ 0 < k < n $.
We will show that there are no other solutions. If a pair of natural numbers $ x $, $ y $ were a solution to the given equation, and, for example, the number $ x $ were not a square of a natural number, then the equality $ y = (n - \sqrt{x})^2 $ would hold, from which it would follow that $ y - n^2 - x = 2n\sqrt{x} $. On the left side of this equation, there is an integer, so $ 2n \sqrt{x} $ is also an integer. Therefore, $ \sqrt{x} $ is a rational number, $ \sqrt{x} = \frac{m}{n} $. Hence, $ x = \left( \frac{m}{n} \right)^2 $, $ xn^2 = m^2 $. In the prime factorization of the numbers $ n^2 $, $ m^2 $, each prime factor appears with an even exponent. Therefore, due to the uniqueness of the prime factorization of a natural number, it follows that in the prime factorization of the number $ x $, each prime factor also appears with an even exponent,
Thus, $ x $ is a square of a natural number, contrary to the assumption. Similarly, we conclude that no pair of natural numbers $ (x, y) $, in which $ y $ is not a square of a natural number, is a solution to the given equation.
Therefore, the given equation has $ n - 1 $ solutions in natural numbers. | n-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Task 2
Investigate when the sum of the cubes of three consecutive natural numbers is divisible by $18$. | Let $ a - 1 $, $ a $, $ a + 1 $, be three consecutive natural numbers; the sum of their cubes
can be transformed in the following way:
Since one of the numbers $ a - 1 $, $ a $, $ a + 1 $ is divisible by $ 3 $, then one of the numbers $ a $ and $ (a + 1) (a - 1) + 3 $ is also divisible by $ 3 $. Therefore, the sum $ S $ of the cubes of three consecutive natural numbers is always divisible by $ 9 $.
The sum $ S $ is thus divisible by $ 18 = 9 \cdot 2 $ if and only if it is divisible by $ 2 $. Since $ S = 3a^3 + 6a $, $ S $ is even if and only if $ a $ is even, i.e., when $ a - 1 $ is odd. Hence, the conclusion:
The sum of the cubes of three consecutive natural numbers is divisible by $ 18 $ if and only if the first of these numbers is odd. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - III - Problem 5
Let $ n $ and $ k $ be natural numbers. From an urn containing 11 slips numbered from 1 to $ n $, we draw slips one by one, without replacement. When a slip with a number divisible by $ k $ appears, we stop drawing. For a fixed $ n $, determine those numbers $ k \leq n $ for which the expected value of the number of slips drawn is exactly $ k $. | Let's establish a natural number $ k \in \{1,2,\ldots,n\} $. Consider the following events:
Let's denote the probabilities of these events: $ p_j = P(\mathcal{A}_j) $, $ q_j = P(\mathcal{B}_j) $. Of course, $ \mathcal{B}_0 $ is a certain event. Therefore, $ q_0 = 1 $.
The occurrence of event $ \mathcal{A}_j $ is equivalent to drawing $ j $ tickets, after which the drawing is stopped - that is, the random variable $ X $ (the number of tickets drawn) takes the value $ j $. Thus,
The occurrence of event $ \mathcal{A}_j $ is also equivalent to the fact that a number divisible by $ k $ did not appear in the first $ j-1 $ draws, but did appear in the first $ j $ draws - that is, event $ \mathcal{B}_{j-1} $ occurred, but event $ \mathcal{B}_j $ did not. In other words, event $ \mathcal{A}_j $ is equal to the difference between events $ \mathcal{B}_{j-1} $ and $ \mathcal{B}_j $ (considered as subsets of the appropriate probability space). Therefore,
(we use the fact that $ \mathcal{B}_j \subset \mathcal{B}_{j-1} $).
The expected value of the random variable $ X $ is equal to
In the set $ \{1,2,\ldots,n\} $, there are $ m $ numbers divisible by $ k $ and $ l $ numbers not divisible by $ k $, where
(symbol $ [t] $ denotes, as usual, the greatest integer not exceeding $ t $). If $ j > l $, then in one of the first $ j $ draws, a number divisible by $ k $ must have appeared, which means that $ \mathcal{B}_j $ is an impossible event. Therefore, $ q_j = 0 $ for $ j > l $, and thus,
From the conditions of the problem, it follows that for $ j \leq l $, the probabilities $ q_j $ are given by the formula
Indeed: in the first draw, we draw a number not divisible by $ k $ with probability $ l/n $; there remain $ n-1 $ tickets in the urn, of which $ l-1 $ are tickets with numbers not divisible by $ k $; in the second draw, we draw a number not divisible by $ k $ with probability $ (l-1)/(n-1) $; and so on. Drawing numbers not divisible by $ k $ in $ j $ consecutive draws occurs with the probability equal to the product written on the right side of formula (2).
We transform this product as follows:
(because $ l + m = n $). After substituting these expressions, equality (1) takes the form
(we applied a change of the summation index: $ s = n-j $). The last sum can be simplified in a known way:
(see the solution to problem 11 from the first round of this olympiad: formula (2) and the Remark on page 51). Hence,
In the problem, we need to determine the conditions under which the value $ E(X) $ is exactly equal to $ k $ - that is, under which the equality
holds.
If relation (3) holds, then obviously $ k $ is a divisor of the number $ n + 1 $. Conversely, if $ n+1 $ is divisible by $ k $, then from the inequality
(whose extreme members are integers) we infer that
which means that equality (3) holds.
Therefore, the answer is: the sought values of $ k $ are all divisors of the number $ n + 1 $ belonging to the set $ \{1,2,\ldots,n\} $. | k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LIV OM - I - Task 7
At Aunt Reni's, $ n \geq 4 $ people met (including Aunt Reni). Each of the attendees gave at least one gift to at least one of the others. It turned out that everyone gave three times as many gifts as they received, with one exception: Aunt Reni gave only $ \frac{1}{6} $ of the number of gifts she received. Determine, depending on $ n $, the smallest possible number of gifts Aunt Reni could have received. | Let $ x_i $ ($ i = 1,2, \ldots ,n-1 $) denote the number of gifts received by the $ i $-th guest of Aunt Renia, and let $ x_R $ denote the number of gifts Aunt Renia received. Then the total number of gifts is
Therefore, the number $ x_R $ is divisible by $ 12 $, say $ x_R = 12k $, where $ k $ is some non-negative integer.
Since each person gave at least one gift, they must have also received at least one gift. Hence, we infer that $ x_i \geq 1 $ for $ i = 1,2,\ldots ,n-1 $, which, by the second equality (1), gives $ 5k \geq n-1 $. Since $ k $ is an integer, the last inequality is equivalent to the inequality $ k \geq [(n-1+4)/5] $, where $ [a] $ denotes the greatest integer not greater than $ a $. Therefore,
It remains to show that under the conditions of the problem, it is possible for Aunt Renia to receive exactly $ 12 [(n+3)/5] $ gifts.
Assume that $ n - 1 = 5k - r $, where $ r = 0,1,2,3,4 $, and $ k $ is some positive integer. Divide the $ n-1 $ guests of Aunt Renia into (at most) three disjoint groups:
(Since $ n \geq 4 $, then $ 2k-r \geq 0 $.) Next, seat all the people $ A_i $ and $ C_i $ around a round table and assume that:
Each person $ A_i $ and $ C_i $ gives 1 gift to the person sitting to their right;
Each person $ A_i $ gives 2 gifts to Aunt Renia;
Each person $ B_i $ gives 3 gifts to Aunt Renia;
Each person $ C_i $ gives 5 gifts to Aunt Renia;
Aunt Renia gives 1 gift to each person $ B_i $ and $ C_i $.
In this way, each person $ A_i $ and $ B_i $ gave 3 gifts and received one, while each person $ C_i $ gave 6 gifts and received two. Thus, each guest of Aunt Renia gave three times as many gifts as they received. On the other hand, Aunt Renia received
gifts, while giving $ 2k $ gifts, which is six times less.
Answer: Aunt Renia could receive at least $ 12[(n+3)/5] $ gifts. | 12[(n+3)/5] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXXV OM - III - Problem 1
Determine the number of functions $ f $ mapping an $ n $-element set to itself such that $ f^{n-1} $ is a constant function, while $ f^{n-2} $ is not a constant function, where $ f^k = f\circ f \circ \ldots \circ f $, and $ n $ is a fixed natural number greater than 2. | Suppose that $ A = \{a_1, a_2, \ldots, a_n \} $ is a given $ n $-element set, and $ f $ is a function satisfying the conditions of the problem. Consider the sets $ A, f(A), f^2(A), \ldots, f^{n-1}(A) $. From the assumption, it follows that the set $ f^{n-1}(A) $ is a singleton, while the set $ f^{n-2}(A) $ has more than one element. Furthermore, $ f^{k-1}(A) \supset f^k(A) $ for $ k = 1,2, \ldots, n - 1 $, since every element of the form $ f^k(x) $ is of the form $ f^{k-1}(y) $, where $ y = f(x) $, and $ f^{k-1}(A) \ne f^k(A) $. If $ f^{k-1}(A) = f^k(A) = \{a_{i_1}, a_{i_2}, \ldots, a_{i_r}\} $, then the function $ f $ would map the subset $ \{a_{i_1}, a_{i_2}, \ldots, a_{i_r}\} $ onto itself and would be injective on this subset. Each iteration of the function $ f $ would thus be injective on this subset, contradicting the fact that $ f^{n-1} $ is a constant function.
Therefore, the inclusions
hold, none of which is an equality. The set $ A $ has $ n $ elements, and the set $ f^{n-1}(A) $ has one element. It follows that each set in the sequence (*) has one element fewer than the previous set.
We have shown that each function $ f $ satisfying the conditions of the problem determines a sequence of subsets of $ A $, in which each subsequent subset is obtained from the previous one by discarding one element, and thus uniquely determines a permutation of the set $ A $.
On the other hand, each such sequence of subsets of $ A $ uniquely determines the function $ f $ because if
then
Functions satisfying the conditions of the problem are thus uniquely determined by permutations of the set $ A $. Their number is therefore $ n! $. | n! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXI OM - I - Problem 12
Find the smallest positive number $ r $ with the property that in a regular tetrahedron with edge length 1, there exist four points such that the distance from any point of the tetrahedron to one of them is $ \leq r $. | Let $ O $ be the center of the sphere circumscribed around the regular tetrahedron $ ABCD $ with edge $ 1 $ (Fig. 9). The radius of the sphere circumscribed around $ ABCD $ is $ R = \frac{1}{4} \sqrt{6} $. The tetrahedron $ ABCD $ is divided into tetrahedra with vertices: $ O $, one of the vertices of the tetrahedron, the midpoint of one of the edges emanating from that vertex of the tetrahedron, and the foot of the perpendicular from one of the faces containing that edge. Since from each of the $ 4 $ vertices of the tetrahedron, $ 3 $ edges emanate, and each of them belongs to two faces of the tetrahedron, the number of considered tetrahedra is $ 4 \cdot 3 \cdot 2 = 24 $.
We will prove that each of these $ 24 $ tetrahedra is inscribed in one of the four spheres with radius $ \frac{R}{2} $ and centers $ P_A $, $ P_B $, $ P_C $, $ P_D $, which are the midpoints of the segments $ \overline{OA} $, $ \overline{OB} $, $ \overline{OC} $, $ \overline{OD} $, respectively. Let $ E $ be the midpoint of the edge $ \overline{AB} $, and $ F $ the foot of the perpendicular dropped from point $ D $ to the plane of triangle $ ABC $. Consider the tetrahedron $ AEFO $. Since angle $ AFO $ is a right angle, $ P_AF = AP_A = OP_A = \frac{R}{2} $.
Similarly, to prove that $ P_AE = \frac{1}{2} R $, it suffices to state that angle $ AEO $ is a right angle. This follows from the fact that the plane defined by points $ D $, $ E $, $ F $ is perpendicular to the edge $ \overline{AB} $, because $ DE \bot AB $, $ CE \bot AB $, and $ S \in CE $.
In this way, we prove that each of the $ 6 $ tetrahedra, one of whose vertices is $ A $, is contained in a sphere with center $ P_A $ and radius $ \frac{R}{2} $. By symmetry, it follows that the remaining tetrahedra are also contained in the corresponding spheres with radius $ \frac{R}{2} $.
Thus, every point of the tetrahedron $ ABCD $ is at a distance from one of the points $ P_A $, $ P_B $, $ P_C $, $ P_D $ not greater than $ \frac{R}{2} $.
Suppose that certain points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ and a number $ S $ have the property that every point of the tetrahedron $ ABCD $ is at a distance from one of the points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ not greater than $ S $. In particular, each of the five points $ O $, $ A $, $ B $, $ C $, $ D $ is at a distance from one of the points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ not greater than $ S $. Therefore, two of the points $ O $, $ A $, $ B $, $ C $, $ D $ are at a distance from some point $ Q_i $ not greater than $ S $. If these points are vertices of the tetrahedron, for example, $ A $ and $ B $, then we obtain $ 1 = AB \leq AQ_i + BQ_i \leq 2S $, i.e., $ S \geq \frac{1}{2} $. If, however, these points are $ O $ and one of the vertices of the tetrahedron, for example, $ A $, then we obtain $ R = OA \leq OQ_i + AQ_i \leq 2S $, i.e., $ S \geq \frac{R}{2} $.
Since $ \frac{R}{2} < \frac{1}{2} $, in each case we obtain that $ S \geq \frac{R}{2} $. Therefore, the smallest positive number $ r $ with the property given in the problem is $ r = \frac{R}{2} = \frac{1}{8} \sqrt{6} $.
Note. It can be proved that the points $ P_A $, $ P_B $, $ P_C $, $ P_D $ are the only points such that every point of the tetrahedron is at a distance from one of them not greater than $ \frac{\sqrt{6}}{8} $. | \frac{\sqrt{6}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LVI OM - I - Task 2
Determine all natural numbers $ n>1 $ for which the value of the sum $ 2^{2}+3^{2}+\ldots+n^{2} $ is a power of a prime number with a natural exponent. | From the formula $ 1+2^{2}+3^{2}+ \ldots+n^{2}=\frac{1}{6}n(n+1)(2n+1) $, true for any natural number $ n $, we obtain
The task thus reduces to finding all natural numbers $ n>1 $, prime numbers $ p $, and positive integers $ k $ for which the equality
is satisfied.
Since the factor $ 2n^2 +5n+6 $ is greater than $ 6 $, it follows from equation (1) that it is divisible by $ p $. We will show that the factor $ n-1 $ is not divisible by $ p $.
Suppose $ p|n-1 $. Since the number
is also divisible by $ p $, $ p $ is a prime divisor of the number $ 13 $. Therefore, $ p = 13 $. Hence $ n - 1 \geq 13 $, which implies the inequality $ M \geq 36\cdot 13 $. From equation (1), we infer that $ M $ is a divisor of the number $ 6\cdot13^k $, and since $ M > 6\cdot13 $, it must be that $ 13^2 |M $. From relation (2), we then obtain that the number $ 9(n-1)+13 $ is divisible by $ 13^2 $, and thus
The number $ (n-1) $ is — by equation (1) — a divisor of the number $ 6\cdot13^k-1 $ and is simultaneously — by divisibility (3) — not divisible by 13. It follows that the number $ \frac{1}{13}(n-1) $ is equal to 1, 2, 3, or 6. However, in none of these four cases is divisibility (3) satisfied. We have obtained a contradiction, which proves that the factor $ n-1 $ is not divisible by $ p $.
From equation (1), we then infer that the factor $ n-1 $ must be one of the numbers: 1, 2, 3, or 6. Hence the number $ n $ is equal to 2, 3, 4, or 7. Direct verification shows that indeed each of these numbers satisfies the conditions of the problem:
Answer: The number $ n $ is equal to 2, 3, 4, or 7. | 2,3,4,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XL OM - II - Task 3
Given is a trihedral angle $ OABC $ with vertex $ O $ and a point $ P $ inside it. Let $ V $ be the volume of the parallelepiped with two vertices at points $ O $ and $ P $, whose three edges are contained in the rays $ OA^{\rightarrow} $, $ OB^{\rightarrow} $, $ OC^{\rightarrow} $. Calculate the minimum volume of the tetrahedron, whose three faces are contained in the faces of the trihedral angle $ OABC $, and the fourth face contains the point $ P $. | Let $ \mathbf{Q} $ denote the considered parallelepiped with volume $ V $. Let $ OA $, $ OB $, $ OC $ be its three edges. We draw a plane through point $ P $ intersecting the rays $ OA^\rightarrow $, $ OB^\rightarrow $, $ OC^\rightarrow $ at points $ K $, $ L $, $ M $ respectively (thus forming the tetrahedron $ OKLM $; its volume is the quantity we need to minimize).
Consider the following three numbers (ratios of triangle areas):
Of course, $ x, y, z > 0 $, $ x + y + z = 1 $.
Let the distances from points $ P $ and $ M $ to the plane $ ABC $ be denoted by $ d $ and $ h $, and the distances from these points to the line $ KL $ - by $ d $ and $ h $. Due to the parallelism of the line $ CP $ to the plane $ OAB $, the proportion holds:
Similarly, we can interpret the numbers $ x $ and $ y $. Thus,
From the formula for the volume of a pyramid and the relationships (1) and (2), we obtain the equality:
Twice the area of triangle $ OAB $ is the area of the face of the parallelepiped $ \mathbf{Q} $ contained in the plane $ \mathbf{OAB} $; the number $ d $ is the length of the corresponding height of this parallelepiped. Therefore, $ d \cdot 2\mathrm{area}(OAB) = \mathrm{volume}(\mathbf{Q}) = V $. From this and the arithmetic mean-geometric mean inequality for the numbers $ x $, $ y $, $ z $, it follows that:
The obtained inequality becomes an equality when $ x = y = z = 1 $, i.e., when the segments $ OK $, $ OL $, $ OM $ are three times longer than $ OA $, $ OB $, $ OC $, respectively. This happens when the plane $ KLM $ is drawn through point $ P $ parallel to the plane $ ABC $.
The number $ \frac{9}{2} V $ is thus the sought minimum.
Note. When the segments $ OA $, $ OB $, $ OC $ are pairwise perpendicular and have equal length, they can be taken as the unit vectors of a Cartesian coordinate system. The entire reasoning can then be elegantly expressed in the language of analytic geometry. For: $ A = (1,0,0) $, $ B = (0,1,0) $, $ C = (0,0,1) $, $ P = (1,1,1) $; any plane passing through point $ P $ is given by an equation of the form $ ax + by + cz = 1 $, with positive coefficients $ a $, $ b $, $ c $ chosen so that $ a + b + c = 1 $. (It is not hard to notice that the numbers $ a $, $ b $, $ c $ are nothing else but the numbers $ x $, $ y $, $ z $ from the previous solution method; this time we need the letters $ x $, $ y $, $ z $ to denote the variables of the coordinate system; hence the change in notation.) The considered plane intersects the coordinate axes at points $ K = (1/a, 0,0) $, $ L = (0,1/b,0) $, $ M = (0,0, 1/c) $.
The parallelepiped $ \mathbf{Q} $ is now a unit cube; its volume $ V $ equals $ 1 $. The tetrahedron $ OKLM $ has a volume of $ (1/6)(1/a)(1/b)(1/c) $. As before, applying the inequality between means, we observe that (with the constant sum $ a + b + c = 1 $) the product $ abc $ reaches its maximum when $ a = b = c = \frac{1}{3} $; the volume of the tetrahedron $ OKLM $ then reaches its minimum, equal to $ \frac{9}{2} $.
Thus, the thesis of the problem is proven in a very special case, where $ |OA| = |OB| = |OC| = 1 $, $ OA \perp OB \perp OC \perp OA $. This "very special" case is, however, in a certain sense, quite general. It should be obvious to those familiar with the concept of affine transformations and their simplest properties. Any parallelepiped can be affinely transformed into a unit cube; and the ratio of volumes is an invariant of affine transformations of three-dimensional space. And it is precisely these two observations that suffice to deduce the thesis of the problem in full generality from the special case considered above. | \frac{9}{2}V | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLVIII OM - I - Problem 7
Calculate the upper limit of the volume of tetrahedra contained within a sphere of a given radius $ R $, one of whose edges is the diameter of the sphere. | Let $ABCD$ be one of the considered tetrahedra. Assume that the edge $AB$ is the diameter of the given sphere. Denote the distance from point $C$ to the line $AB$ by $w$, and the distance from point $D$ to the plane $ABC$ by $h$. The volume of the tetrahedron $ABCD$ is given by
Of course, $w \leq R$, $h \leq R$, and therefore $V \leq \frac{1}{3} R^3$.
The tetrahedron $ABCD$ was chosen arbitrarily; hence, the upper bound of the volumes of the considered tetrahedra does not exceed the number $\frac{1}{3} R^3$.
When $ABC$ is an isosceles right triangle with hypotenuse $AB$, then $w = R$. When, on the other hand, $ABD$ is an isosceles right triangle with hypotenuse $AB$, lying in a plane perpendicular to the plane $ABC$, then also $h = R$. When both these conditions are met, the equality $V = \frac{1}{3} R^3$ holds.
Conclusion: the upper bound of the considered volumes is equal to $\frac{1}{3} R^3$. | \frac{1}{3}R^3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXI OM - I - Problem 10
Find the largest natural number $ k $ with the following property: there exist $ k $ different subsets of an $ n $-element set, such that any two of them have a non-empty intersection. | Let the sought number be denoted by $ k(n) $. We will prove that $ k(n) = 2^{n-1} $. To each $ n $-term sequence with values $ 0 $ or $ 1 $, we assign the set of indices such that the corresponding terms of the sequence are equal to $ 1 $. This assignment establishes a one-to-one correspondence between the set of all $ n $-term sequences with values $ 0 $ or $ 1 $ and the set of all subsets of an $ n $-element set. It follows that the number of all subsets of an $ n $-element set is the same as the number of sequences with $ n $ terms, i.e., $ 2^n $ (see preparatory problem D, series III).
Consider in an $ n $-element set $ X $ the family of its subsets containing a fixed element $ a $. Any two subsets in this family have a non-empty intersection, as the element $ a $ belongs to it. The number of subsets in this family is equal to the number of all subsets of the $ (n-1) $-element set $ X - \{a\} $, which is $ 2^{n-1} $. Therefore, $ k(n) \geq 2^{n-1} $.
On the other hand, if a set $ A $ belongs to some family $ R $ of subsets of the set $ X $, in which any two subsets have a non-empty intersection, then the set $ X - A $ does not belong to $ R $. Therefore, the number of subsets belonging to the family $ R $ does not exceed half the number of all subsets of the set $ X $, i.e., $ k(n) \leq \frac{1}{2} \cdot 2^n = 2^{n-1} $.
From both proven inequalities, it follows that $ k(n) = 2^{n-1} $. | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - II - Task 1
Determine the maximum possible length of a sequence of consecutive integers, each of which can be expressed in the form $ x^3 + 2y^2 $ for some integers $ x, y $. | A sequence of five consecutive integers -1, 0, 1, 2, 3 satisfies the conditions of the problem: indeed, we have
On the other hand, among any six consecutive integers, there exists a number, say
$ m $, which gives a remainder of 4 or 6 when divided by 8. The number $ m $ is even; if there were a representation
in the form $ m = x^3 +2y^2 $ for some integers $ x, y $, then the number $ x $ would be even. In this case, however,
we would obtain the divisibility $ 8|x^3 $ and as a result, the numbers $ m $ and $ 2y^2 $ would give the same remainder (4 or 6) when divided by 8.
Therefore, the number $ y^2 $ would give a remainder of 2 or 3 when divided by 4. This is impossible: the equalities
prove that the square of an integer can give a remainder of only 0 or 1 when divided by 4.
We have thus shown that among any six consecutive integers, there exists a number that cannot be represented
in the form $ x^3 +2y^2 $.
Answer: The maximum possible length of such a sequence is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 11
In triangle $ ABC $, angle $ A $ is $ 20^\circ $, $ AB = AC $. On sides $ AB $ and $ AC $, points $ D $ and $ E $ are chosen such that $ \measuredangle DCB = 60^\circ $ and $ \measuredangle EBC = 50^\circ $. Calculate the angle $ EDC $. | Let $ \measuredangle EDC = x $ (Fig. 9). Notice that $ \measuredangle ACB = \measuredangle $ABC$ = 80^\circ $, $ \measuredangle CDB = 180^\circ-80^\circ-60^\circ = 40^\circ $, $ \measuredangle CEB = 180^\circ - 80^\circ-50^\circ = \measuredangle EBC $, hence $ EC = CB $. The ratio $ \frac{DC}{CE} $ of the sides of triangle $ EDC $ equals the ratio of the sides $ \frac{DC}{CB} $ of triangle $ BDC $, so the ratios of the sines of the angles opposite the corresponding sides in these triangles are equal:
The right side of the obtained equation can be transformed:
We need to find the convex angle $ x $ that satisfies the equation
or the equation
By transforming the products of sines into differences of cosines, we obtain an equivalent equation
Considering the condition $ 0 < x < 180^\circ $, we get $ x = 30^\circ $.
Note. The last part of the solution can be slightly shortened. Specifically, from the form of equation (1), it is immediately clear that it has a root $ x = 30^\circ $. No other convex angle satisfies this equation; if
then
so
thus, if $ 0 < x < 180^\circ $ and $ 0 < y < 180^\circ $, then $ x = y $. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - II - Problem 6
A square with side length $ n $ is divided into $ n^2 $ unit squares. Determine all natural numbers $ n $ for which such a square can be cut along the lines of this division into squares, each of which has a side length of 2 or 3. | If $ n $ is an even number, then a square of dimensions $ n \times n $ can be cut into $ 2 \times 2 $ squares. Let's assume, then, that $ n $ is an odd number. The square $ n \times n $ consists of $ n $ rows, with $ n $ cells (unit squares) in each row. We paint these rows alternately black and white (in a "zebra" pattern). Due to the oddness of $ n $, both the top and bottom rows will be the same color—let's assume black. The difference between the number of black cells and the number of white cells in the entire checkerboard is $ n $.
Suppose that a division into $ 2 \times 2 $ and $ 3 \times 3 $ squares is possible. Each $ 2 \times 2 $ square has two white and two black cells. Each $ 3 \times 3 $ square has three white and six black cells, or vice versa. The difference between the number of black cells and the number of white cells in a $ 2 \times 2 $ square is $ 0 $, and in a $ 3 \times 3 $ square, it is $ 3 $ or $ -3 $. By adding all these differences (corresponding to the individual squares in the division), we obtain a number divisible by $ 3 $. However, this sum is also equal to the difference between the number of black cells and the number of white cells in the entire checkerboard, which is $ n $.
Conclusion: If the considered division is possible, then the number $ n $ is divisible by $ 2 $ or by $ 3 $. Conversely, if $ n $ is divisible by $ 2 $ or by $ 3 $, then the $ n \times n $ square can be cut in the desired way (into only $ 2 \times 2 $ squares or $ 3 \times 3 $ squares).
Note 1. When $ n $ is divisible by $ 6 $, the checkerboard can, of course, be cut into both $ 2 \times 2 $ and $ 3 \times 3 $ squares. Moreover, mixed divisions are possible; the reader can verify (this is an easy exercise) that if $ n $ is a number greater than $ 6 $ and divisible by $ 2 $ or by $ 3 $, then the $ n \times n $ square can be cut into $ 2 \times 2 $ and $ 3 \times 3 $ squares, using both types. | nisdivisible2or3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXVI - III - Task 2
On the surface of a regular tetrahedron with edge length 1, a finite set of segments is chosen in such a way that any two vertices of the tetrahedron can be connected by a broken line composed of some of these segments. Can this set of segments be chosen so that their total length is less than $1 + \sqrt{3}$? | On the surface of a regular tetrahedron with edge length $1$, one can choose a finite set of segments such that any two vertices of the tetrahedron can be connected by a broken line composed of some of these segments, and the total length of the segments belonging to this set is less than $1 + \sqrt{3}$.
Consider, in particular, the rhombus $ABDC$ formed by unfolding two faces of the given tetrahedron onto a plane (Fig. 10). Then $AB = 1$ and $\measuredangle BAC = 60^\circ$. Let $P$ be the midpoint of segment $\overline{BC}$, and $Q$ be a point in triangle $ABP$ from which the sides $\overline{AP}$ and $\overline{BP}$ are seen at an angle of $120^\circ$. Then $\measuredangle AQB = 120^\circ$ as well. Let $x = AQ$, $y = BQ$, $z = PQ$. Since $AP = \frac{1}{2}\sqrt{3}$, $BP = \frac{1}{2}$, and $\measuredangle APB = 90^\circ$, comparing the area of triangle $ABP$ to the sum of the areas of triangles $AQB$, $BQP$, and $AQP$ gives $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} (xy + yz + zx) \cdot \sin 120^\circ$, which simplifies to
Since $\cos 120^\circ = -\frac{1}{2}$, applying the law of cosines to triangles $AQB$, $BQP$, and $AQP$ yields
Adding the equations (2), (3), (4), and equation (1) multiplied by $3$ gives $2(x + y + z)^2 = \frac{7}{2}$, and thus $x + y + z = \frac{\sqrt{7}}{2}$.
If $Q'$ is the point symmetric to $Q$ with respect to point $P$, then the set of segments $\overline{AQ}$, $\overline{BQ}$, $\overline{QP}$, $\overline{PQ'}$, $\overline{GQ}$, $\overline{DQ}$ has the property mentioned in the problem, and the sum of the lengths of these segments is $\sqrt{7}$. This number is less than $1 + \sqrt{3}$. | \sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVI OM - I - Problem 7
On the side $ AB $ of triangle $ ABC $, a point $ K $ is chosen, and points $ M $ and $ N $ are determined on the lines $ AC $ and $ BC $ such that $ KM \perp AC $, $ KN \perp BC $. For what position of point $ K $ is the area of triangle $ MKN $ the largest? | Let $ BC = a $, $ AC = b $, $ \measuredangle ACB = \gamma $, $ KM = p $, $ KN = q $ (Fig. 2).
Notice that $ \measuredangle MKN = 180^\circ - \gamma $.
Indeed, if from a point $ K_1 $ chosen on the bisector of angle $ ACB $ we drop perpendiculars $ K_1M_1 $ and $ K_1N_1 $ to the sides of this angle, then the rays $ KM $ and $ K_1M_1 $ are parallel and similarly directed since points $ K $ and $ K_1 $ lie on the same side of the line $ AC $.
The same applies to the rays $ KN $ and $ K_1N_1 $. Hence, angles $ MKN $ and $ M_1K_1N_1 $ are equal, as one can be obtained from the other by a parallel shift. Angles $ M_1K_1N_1 $ and $ ACB $ are opposite angles of a convex quadrilateral, whose two other angles are right angles. Therefore, $ \measuredangle M_1K_1N_1 = 180^\circ - \gamma $ and similarly $ \measuredangle MKN = 180^\circ - \gamma $, thus
From this equality, we conclude that the area of $ MKN $ is maximized when the product $ pq $ reaches its maximum value; simultaneously, the product $ abpq $ also reaches its maximum. However, the sum of the numbers $ bp $ and $ aq $ is constant, since
Therefore, the maximum of the product $ bp \cdot aq $ occurs when $ bp = aq $, i.e., when $ \mathrm{area}\ AKC = \mathrm{area}\ BKC $, which happens when point $ K $ is the midpoint of side $ AB $.
The reasoning above can be slightly modified by calculating $ p $ and $ q $ from triangles $ AKM $ and $ BKN $:
The maximum of $ pq $ occurs when the product $ AK \cdot KB $ is the largest. Since the sum $ AK + KB = AB $ is constant, this happens when $ AK = KB $. | KisthemidpointofsideAB | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XVII OM - I - Problem 1
Present the polynomial $ x^5 + x + 1 $ as a product of two polynomials of lower degree with integer coefficients. | Suppose the desired factorization exists. None of the factors of this factorization can be of the first degree. For if for every $x$ the equality held
where the numbers $m, n, a, b, \ldots$ are integers, then it would follow that $ma = 1$, so $m = a = \pm 1$, hence the polynomial $x^5 + x + 1$ would have an integer root $-\frac{n}{m}$. This is impossible, however, because for every integer $x$ the number $x^5 + x + 1$ is odd.
Hence, the desired factorization must have the form
where the numbers $m, n, p, a, b, c, d$ are integers.
In this case, $ma = 1$, $pd = 1$, so $m = a = \pm 1$, $p = d = \pm 1$.
Thus, there are four possible cases, but it suffices to consider two of them; the remaining cases reduce to these by changing the sign of both factors in the factorization.
a) When $m = a = 1$, $p = d = 1$, then by comparing the coefficients of like powers of $x$ on both sides of the equality
we obtain
The system of equations (1) - (4) has one solution $n = 1$, $b = -1$, $c = 0$, which gives the factorization
b) When $m = a = 1$, $p = d = -1$, we similarly obtain the system of equations
It is easy to check that this system has no solutions. The solution to the problem is therefore the formula
where $\varepsilon = \pm 1$.
Note. The factorization sought in the problem can be found quickly in the following way:
Such a solution to the problem still requires a proof that there is no other factorization besides the one obtained by changing the signs of both found factors. The easiest way to show this is by using complex numbers. Let $x_1$, $x_2$ denote the roots of the polynomial $x^2 + x + 1$, and $x_3$, $x_4$, $x_5$ the roots of the polynomial $x^3 - x^2 + 1$.
Then
Any quadratic divisor of the polynomial $x^5 + x + 1$ with integer coefficients is of the form $m(x - x_i)(x - x_k)$, where $m$ is an integer and $x_i$, $x_k$ are two roots of the polynomial. It is easy to see that such a divisor can only be $m(x - x_1)(x - x_2)$. The detailed proof of this is left as an exercise.
Hint. One of the numbers $x_3$, $x_4$, $x_5$ is an irrational real number, and the other two are complex conjugates, whose sum and product are irrational. | x^5+x+1=(x^2+x+1)(x^3-x^2+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - III - Zadanie 3
Dana jest liczba pierwsza $ p \geq 3 $. Określamy ciąg $ (a_n) $ wzorami
Wyznaczyć resztę z dzielenia liczby $ a_{p^3} $ przez $ p $.
|
Podany wzór rekurencyjny
możemy zastosować do każdego ze składników jego prawej strony (pod warunkiem, że numer $ n - p $ jest nie mniejszy od $ p $, czyli że $ n \geq 2p $); dostajemy zależność
Stosując ponownie wzór (1) do każdego ze składników ostatniej sumy otrzymujemy równość
(jeśli tylko $ n \geq 3p $). Po $ k $-krotnym zastosowaniu analogicznego przekształcenia uzyskujemy związek
sensowny i słuszny pod warunkiem że $ n \geq kp $.
(Wyprowadzenie wzoru (2) ma charakter indukcyjny; proponujemy Czytelnikowi, jako proste ćwiczenie, przeprowadzenie szczególowego uzasadnienia, przez indukcję względem $ k = 1,2,\ldots,[n/p] $, z wielokrotnym wykorzystaniem wzoru (1).)
Jeśli teraz $ n \geq p^2 $, to możemy w równości (2) podstawić $ k = p $, otrzymując
Dla $ i = 1,2,\ldots,p-1 $ współczynniki dwumianowe $ {{p}\choose{i}} $ są liczbami podzielnymi przez $ p $ (patrz: Uwaga). Jeśli więc z sumy napisanej po prawej stronie równości (3) odrzucimy wszystkie składniki z wyjątkiem dwóch skrajnych (odpowiadających wartościom wskaźnika $ i = 0 $ oraz $ i = p $), to otrzymamy liczbę przystającą do $ a_n $ modulo $ p $:
(dla $ n \geq p^2 $). Przyrównując (modulo $ p $) prawe strony związków (1) i (4) stwierdzamy, że $ a_{n-1}+a_{n-p} \equiv a_{n-p} + a_{n-p^2} $, czyli
Oznaczmy przez $ r_j $ resztę z dzielenia liczby $ a_j $ przez $ p $ (dla $ j = 0,1,2,\ldots $). Uzyskaną zależność przepisujemy w postaci
To znaczy, że ciąg ($ r_j $) jest okresowy, z okresem $ p^2-1 $; zatem
Podstawiamy $ n = p^3 $, $ k = p $, i otrzymujemy wynik:
Pozostaje zauważyć, że (w myśl wzoru (1))
i wobec tego $ r_{p^3} = r_p = p-1 $. Tak więc reszta z dzielenia liczby $ a_{p^3} $ przez $ p $ równa się $ p-1 $.
Uwaga. Fakt, na który powołaliśmy się w rozwiązaniu:
liczba pierwsza $ p $ jest dzielnikiem liczb $ {{p}\choose{1}} $, $ {{p}\choose{2}} $, $ \ldots $, $ {{p}\choose{p-1}} $
jest dobrze znany (oraz banalnie prosty: w równości $ {{p}\choose{i}} \cdot (i!) \cdot ((p-i)!) = p! $ czynniki $ i! $ oraz $ (p-i)! $ są dla $ i \in \{1,2,\ldots,p-1\} $ niepodzielne przez $ p $, więc czynnik $ {{p}\choose{i}} $ musi dzielić się przez $ p $).
| p-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXVIII - II - Task 3
In a hat, there are 7 slips of paper. On the $ n $-th slip, the number $ 2^n-1 $ is written ($ n = 1, 2, \ldots, 7 $). We draw slips randomly until the sum exceeds 124. What is the most likely value of this sum? | The sum of the numbers $2^0, 2^1, \ldots, 2^6$ is $127$. The sum of any five of these numbers does not exceed $2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 124$. Therefore, we must draw at least six slips from the hat.
Each of the events where we draw six slips from the hat, and the seventh slip with the number $2^{n-1}$ ($n = 1, 2, \ldots, 7$) remains in the hat, is equally likely. The probability of such an event is thus $\displaystyle \frac{1}{7}$.
The sum of the numbers on the drawn slips is equal to $127 - 2^{n-1}$. If $n = 1$, this sum is $126$; if $n = 2$, it is $125$; if $n = 3, 4, 5, 6$ or $7$, the sum is less than $124$ and we must draw a seventh slip. In this last case, the sum of the numbers on all the drawn slips will be $127$. Therefore, the probability that the sum of the numbers on all the slips drawn according to the conditions of the problem is $125$, $126$, or $127$, is $\displaystyle \frac{1}{7}$, $\displaystyle \frac{1}{7}$, $\displaystyle \frac{5}{7}$, respectively.
Thus, the most probable value of the sum is $127$. | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
I OM - B - Task 12
How many sides does a polygon have if each of its angles is $ k $ times larger than the adjacent angle? What values can $ k $ take? | If each angle of a polygon is $ k $ times larger than the adjacent angle,
then the sum of all $ n $ angles of the polygon, which - as
we know - equals $ (n - 2) \cdot 180^{\circ} $, is also $ k $ times larger
than the sum of the corresponding adjacent angles (i.e., the exterior angles),
which means it is $ k $ times larger than $ 360^{\circ} $. Therefore
from which
Since $ n - 2 $ must be a natural number, then $ 2k $ should
be a natural number, i.e., $ k $ must be half
of a natural number, so $ k = \frac{1}{2}, 1, \frac{3}{2}, 2, \dots $ | \frac{1}{2},1,\frac{3}{2},2,\dots | Geometry | math-word-problem | Yes | Yes | olympiads | false |
IV OM - III - Task 3
Through each vertex of a tetrahedron of a given volume $ V $, a plane parallel to the opposite face of the tetrahedron has been drawn. Calculate the volume of the tetrahedron formed by these planes. | To make it easier to find a solution, let's first consider an analogous problem on the plane:
Let's draw through each vertex of the triangle $ABC$ a line parallel to the opposite side of the triangle. These lines form a triangle $A_1B_1C_1$ similar to triangle $ABC$ (Fig. 46). The center of similarity is the point $S$, where the lines $AA_1$, $BB_1$, $CC_1$ intersect; point $S$ is the centroid of triangle $ABC$, as well as of triangle $A_1B_1C_1$. The similarity is inverse, because corresponding points, e.g., $A$ and $A_1$, lie on the rays $SA$ and $SA_1$ in opposite directions. The ratio of similarity is
Since the ratio of the areas of similar figures equals the square of the ratio of similarity, the area of triangle $A_1B_1C_1$ is $4$ times larger than the area of triangle $ABC$, which is also directly visible.
Returning to the main topic, let's briefly recall the properties of similar figures in space. Let a figure $F$ be given in space, whose points are $A, B, C, \ldots$ Choose any point $S$ and a positive number $k$. When we measure segments $SA, SB, SC, \ldots$ on the rays $SA, SB, SC, \ldots$, the points $A$ form a figure $F$ directly similar to figure $F$ relative to point $S$ in the ratio $k$. If, however, we measure the same segments $SA_1 = k \cdot SA, SB_1 = k \cdot SB, SC_1 = k \cdot SC, \ldots$ on the extensions of the rays $SA, SB, SC, \ldots$, we obtain a figure $F_1$ inversely similar to figure $F$ relative to point $S$ in the ratio $k$ (Fig. 47). From this definition, it easily follows that the figure similar to the segment $AB$ is the segment $A_1B_1$, with the lines $AB$ and $A_1B_1$ being parallel (or coinciding, which happens when the line $AB$ passes through point $S$). The figure similar to the triangle $ABC$ is, therefore, the triangle $A_1B_1C_1$, with the planes of the triangles being parallel (or coinciding in the case where the plane $ABC$ contains point $S$). The figure similar to the tetrahedron $ABCD$ is the tetrahedron $A_1B_1C_1D_1$. The ratio of similar segments equals the ratio of similarity, the ratio of the areas of similar triangles (or generally: planar figures) equals the square of the ratio of similarity, and the ratio of the volumes of similar tetrahedra (or generally: solids) equals the cube of the ratio of similarity.
To solve the problem, we will also need the following theorem:
The segments connecting the vertices of a tetrahedron with the centroids of the opposite faces intersect at one point, which divides each of these segments in the ratio $3 \colon 1$.
The proof of this theorem is simple; it suffices to prove it for one pair of such segments. Let $M$ be the centroid of the face $BCD$, and $N$ be the centroid of the face $ACD$ of the tetrahedron $ABCD$ (Fig. 47). The lines $BM$ and $AN$ intersect at the midpoint $P$ of the edge $CD$.
The segments $AM$ and $BN$ connecting the vertices $A$ and $B$ of the triangle $ABP$ with the points $M$ and $N$ of the sides $BP$ and $AP$ intersect at a point $S$ inside this triangle. Since $AP = 3 \cdot NP$ and $BP = 3 \cdot MP$, it follows that $MN \parallel AB$ and $AB = 3 MN$; hence, $AS = 3 \cdot SM$ and $BS = 3 \cdot SN$, which is what we had to prove. The point $S$ is the centroid of the tetrahedron.
Applying the above knowledge, we can briefly state the solution to the problem: We construct the figure inversely similar in the ratio $3$ to the tetrahedron $ABCD$ relative to its centroid $S$. This figure is the tetrahedron $A_1B_1C_1D_1$, whose vertices lie on the extensions of the rays $SA$, $SB$, $SC$, $SD$, with $SA_1 = 3 \cdot SA$, $SB_1 = 3 \cdot SB$, $SC_1 = 3 \cdot SC$, $SD_1 = 3 \cdot SD$.
The plane $B_1C_1D_1$ is parallel to the corresponding plane $BCD$, and it passes through the vertex $A$, since from the equality $AS = 3 \cdot SM$ it follows that point $A$ is the point similar to point $M$ of the face $BCD$. Therefore, the tetrahedron $A_1B_1C_1D_1$ is the tetrahedron mentioned in the problem. Since it is similar to the tetrahedron $ABCD$ in the ratio $3$, its volume is $3^3 \cdot V$, or $27 V$. | 27V | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 8
Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by a translation,
(3) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) has a point in common with $ W_0 $,
(4) the polyhedra $ W_0, W_1, \ldots, W_n $ have pairwise disjoint interiors. | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$ under the translation by $\overrightarrow{\mathbf{v}}$, is the center of symmetry of $W_1$. Figure 3 illustrates a planar variant of the considered problem (a representation of the spatial configuration would obscure this illustration); polyhedra $W_0$ and $W_1$ are depicted as centrally symmetric polygons.
om42_1r_img_3.jpg
By condition (3), polyhedra $W_0$ and $W_1$ have common points (possibly many). Let $K$ be a common point of $W_0$ and $W_1$ (arbitrarily chosen). Denote by $L$ the image of point $K$ under the central symmetry with respect to $O_1$; thus, $L \in W_1$. Let $N$ be a point such that $\overrightarrow{NL} = \overrightarrow{\mathbf{v}}$ and let $M$ be the midpoint of segment $NK$ (Figure 4). Therefore, $N \in W_0$. According to condition (1), the set $W_0$ is convex; this means that with any two points belonging to $W_0$, the entire segment connecting these points is contained in $W_0$. Since $K \in W_0$ and $N \in W_0$, it follows that $M \in W_0$. Segment $MO_1$ connects the midpoints of segments $KN$ and $KL$, and thus $\overrightarrow{MO_1} = \frac{1}{2} \overrightarrow{NL} = \frac{1}{2} \overrightarrow{\mathbf{v}}$, which means $M$ is the midpoint of segment $O_0O_1$.
Let $U$ be the image of polyhedron $W_0$ under a homothety with center $O_0$ and scale factor 3. We will show that $W_1 \subset U$. Take any point $P \in W_1$: let $Q \in W_0$ be a point such that $\overrightarrow{QP} = \overrightarrow{\mathbf{v}}$ and let $S$ be the center of symmetry of parallelogram $O_0O_1PQ$. The medians $O_0S$ and $QM$ of triangle $O_0O_1Q$ intersect at a point $G$ such that $\overrightarrow{O_0G} = \frac{2}{3}\overrightarrow{O_0S} = \frac{1}{3}\overrightarrow{O_0P}$ (Figure 5). This means that $P$ is the image of point $G$ under the considered homothety. Since $G$ is a point on segment $QM$ with endpoints in the set (convex) $W_0$, it follows that $G \in W_0$. Therefore, $P \in U$ and from the arbitrariness of the choice of point $P \in W_1$ we conclude that $W_1 \subset U$.
om42_1r_img_4.jpg
om42_1r_img_5.jpg
In the same way, we prove that each of the sets $W_i (i=1, \ldots, n)$ is contained in $U$. Of course, also $W_0 \subset U$. Thus, the set $W_0 \cup W_1 \cup \ldots \cup W_n$ is a polyhedron contained in $U$. By conditions (2) and (4), its volume equals the volume of $W_0$ multiplied by $n+1$. On the other hand, the volume of $U$ equals the volume of $W_0$ multiplied by 27. Therefore, $n \leq 26$.
It remains to note that the value $n = 26$ can be achieved (example realization: 27 cubes $W_0, \ldots, W_{26}$ arranged like a Rubik's cube). Thus, the sought number is $26$.
Note. We obtain the same result assuming that $W_0, \ldots, W_n$ are any bounded, closed convex bodies (not necessarily polyhedra), with non-empty interiors, satisfying conditions (1)-(4); the reasoning carries over without any changes. Moreover, condition (4) turns out to be unnecessary. This was proven by Marcin Kasperski in the work 27 convex sets without a center of symmetry, awarded a gold medal at the Student Mathematical Paper Competition in 1991; a summary of the work is presented in Delta, issue 3 (1992). | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XI OM - I - Task 3
Each side of a triangle with a given area $ S $ is divided into three equal parts, and the points of division are connected by segments, skipping one point to form two triangles. Calculate the area of the hexagon that is the common part of these triangles. | The solution to the problem can be easily read from Fig. 1. The area of the hexagon $123456$ equals $\frac{2}{3}S$, as this hexagon is obtained from the triangle $ABC$ by cutting off three equal corner triangles with an area of $\frac{1}{9}S$. The sides of triangles $135$ and $246$ bisect each other into equal segments because, for example, in the trapezoid $1234$, the diagonals $13$ and $24$ divide into parts in the same ratio as the bases $23$ and $14$, i.e., in the ratio $1 \colon 2$, so, for instance, segment $II\ 2$ equals one-third of segment $42$. The hexagon $123456$ consists of $18$ triangles of equal area, and the shaded hexagon $I\ II\ III\ IV\ V\ VI$ consists of $6$ triangles of the same area. Therefore, the area of the shaded hexagon is $\frac{2}{9} S$. | \frac{2}{9}S | Geometry | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Task 11
Given are two intersecting lines $ a $ and $ b $. Find the geometric locus of point $ M $ having the property that the distance between the orthogonal projections of point $ M $ on lines $ a $ and $ b $ is constant, equal to a given segment $ d $. | Let $A$ and $B$ denote the projections of point $M$ onto lines $a$ and $b$ intersecting at point $O$, and let $\alpha$ denote the convex angle $AOB$ (Fig. 27). The circle with diameter $OM$ passes through points $A$ and $B$, since angles $OAM$ and $OBM$ are right angles; this circle is thus the circumcircle of triangle $AOB$ and $AB = OM \sin \alpha$. If $AB = d$, then $OM = \frac{d}{\sin \alpha}$. Therefore, point $M$ lies on the circle $k$ with center $O$ and radius $\frac{d}{\sin \alpha}$.
Conversely, if a point $N$ lies on this circle $k$ (Fig. 28), then its projections $C$ and $D$ onto lines $a$ and $b$ lie on the circle with diameter $ON = \frac{d}{\sin \alpha}$; this circle is the circumcircle of triangle $COD$, so $CD = \frac{d}{\sin \alpha} \cdot \sin \measuredangle COD$. But $\measuredangle COD$ is equal to $\alpha$ or $180^\circ - \alpha$, so $\sin \measuredangle COD = \sin \alpha$ and $CD = d$.
The sought geometric locus is thus the circle drawn with the intersection of the given lines as the center and a radius equal to $\frac{d}{\sin \alpha}$, where $\alpha$ denotes one of the angles between the given lines. | The\geometric\locus\is\the\circle\with\center\O\\radius\\frac{}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
X OM - I - Task 8
The sides of a triangle are the medians of another triangle. Calculate the ratio of the areas of both triangles. Can a triangle be constructed from the medians of any triangle? | Let $ S $ be the centroid of triangle $ ABC $. Extend the median $ AD $ by a segment $ DM = SD $ (Fig. 15). We obtain triangle $ BSM $, in which $ BS = \frac{2}{3} BE $, $ SM = \frac{2}{3} AD $, and $ BM = SC = \frac{2}{3} CF $ (since quadrilateral $ SBMC $ is a parallelogram). Triangle $ BSM $ is thus similar to triangle $ T $ formed by the medians of triangle $ ABC $, with a scale factor of $ \frac{2}{3} $; the area of triangle $ BSM $ is therefore $ \frac{4}{9} $ of the area of $ T $. However, the area of $ \triangle BSM = \text{area of }\triangle ASB = \frac{1}{3} \cdot \text{area of }\triangle ABC $. It follows that $ \text{area of }T = \frac{3}{4} \cdot \text{area of }\triangle ABC $. | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXV - I - Problem 11
Let $ X_n $ and $ Y_n $ be independent random variables with the same distribution $ \left{ \left(\frac{k}{2^n}, \frac{1}{2^n}\right) : k = 0, 1, \ldots, 2^n-1\right} $. Denote by $ p_n $ the probability of the event that there exists a real number $ t $ satisfying the equation $ t^2 + X_n \cdot t + Y_n = 0 $. Calculate $ \lim_{n\to\infty} p_n $. | A favorable event consists in the discriminant of the quadratic polynomial $ t^2 + X_n t + Y_n $ being a non-negative number. Let $ \displaystyle X_n = \frac{r}{2^n} $, $ \displaystyle Y_n = \frac{s}{2^n} $, where $ r, s = 0, 1, \ldots, 2^n - 1 $. The inequality $ X_n^2 - 4 Y_n \geq 0 $ is equivalent to the inequality $ r^2 - 4 \cdot 2^n s \geq 0 $, that is,
We will find the number of such pairs $ (r, s) $ for which (1) holds. For any $ a $, the number of solutions to the inequality $ 0 \leq x \leq a $ in integers $ x $ is $ [a] + 1 $, where $ [a] $ is the greatest integer not exceeding $ a $. Therefore, the number of pairs $ (r, s) $ satisfying
(1) is equal to
Consider the sum
By the formula
we have
Since for any real number $ x $ the inequality
holds, we have
The number of all elementary events, i.e., the considered pairs $ (r, s) $, is $ 2^{2n} $. Therefore, $ \displaystyle p^n = \frac{A^n}{2^{2n}} $ and from (3) it follows that
From (2) we obtain that $ \displaystyle \lim_{n \to \infty} \frac{B_n}{2^{2n}} = \frac{1}{12} $ and therefore from (4) we conclude that $ \displaystyle \lim_{n \to \infty} p_n = \frac{1}{12} $ | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XXXVIII OM - II - Task 1
From an urn containing one ball marked with the number 1, two balls marked with the number 2, ..., $ n $ balls marked with the number $ n $, we draw two balls without replacement. We assume that drawing each ball from the urn is equally probable. Calculate the probability that both drawn balls have the same number. | To make the content of the task meaningful, we need to assume that $ n \geq 2 $.
In total, there are $ N = 1+2+ ... +n = n(n+1)/2 $ balls in the urn. Two balls can be chosen from $ N $ balls in $ q $ ways, where
Let us fix $ k \in \{2, \ldots,n\} $. From the $ k $ balls marked with the number $ k $, we can choose a pair of balls in $ \binom{k}{2} $ ways. The total number of ways to choose a pair of balls with the same number (from all $ N $ balls) is
Here is an inductive proof of formula (2): the formula is correct for $ n = 2 $; assuming its validity for some $ n \geq 2 $, we have for $ n+1 $
Thus, formula (2) is proved.
The sought probability is the quotient of the found number (formula (2)) by $ q $ (formula (1)), i.e., it equals $ \dfrac{4}{3(n+2)} $. | \frac{4}{3(n+2)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
LX OM - I - Task 5
For each integer $ n \geqslant 1 $, determine the largest possible number of different subsets of the set $ \{1,2,3, \cdots,n\} $ with the following property: Any two of these subsets are either disjoint or one is contained in the other. | Answer: $ 2n $ subsets.
Let the number of sought subsets be denoted by $ a_n $.
It is not difficult to observe that for $ n =1,2,3,\cdots $ the following $ 2n $
subsets of the set $ \{1,2,3, \cdots,n\} $ have the property described in the problem statement:
Indeed, any of the first $ n+1 $ sets has at most one element, so it is contained in any set with which it is not disjoint. On the other hand, for any two of the remaining
$ n - 1 $ sets, the set that is listed earlier in the above sequence is contained in the other set.
In this way, we have proved that
On the other hand, let $ B_1, B_2, \cdots, B_t $ be different
subsets of the set $ \{1, 2, \cdots,m\} $ with the property described in the problem statement. If among the listed subsets the set
$ \{1,2, \cdots,m\} $ does not appear, we can add it, increasing the number of these
subsets by $ 1 $ without violating the validity of the postulated condition.
Let us assume, then, that $ B_t = \{1,2, \cdots,m\} $.
Among the sets $ B_1, B_2, \cdots, B_{t-1} $, let us choose the one that has the most elements. For the sake of argument, let it be $ B_1 $
and let $ d $ denote the number of its elements. Then $ d < n $.
Moreover, the equality $ d = 0 $ means that $ B_1 $ is the empty set, which, given the choice of the set $ B_1 $, is possible only if
$ t = 2 $. In the further part of the reasoning, we assume, then, that $ 0 < d < n $.
By the definition of the set $ B_1 $, each of the sets
$ B_i $ ($ i=1,2,\cdots,t - 1) $ is either a subset of the set $ B_1 $,
or is disjoint from it. By renumbering, we can assume that the sets $ B_1, B_2, \cdots, B_r $ are subsets of the set
$ B_1 $, while the sets $ B_{r+1}, B_{r+2}, \cdots, B_{t-1} $ are
disjoint from $ B_1 $.
The sets $ B_1, B_2, \cdots, B_r $ are different subsets of
the $ d $-element set $ B_1 $, any two of which are either
disjoint or one is contained in the other. It follows that
$ r \leqslant a_d $. Among the $ t-r-1 $ sets
$ B_{r+1}, B_{r+2}, \cdots, B_{t-1} $, any two are either disjoint,
or one is contained in the other, and they are subsets of the
$ (m-d) $-element set $ {1, 2, \cdots,m}\setminus B_1 $,
and moreover, the empty set does not appear among them, as it
is a subset of $ B_1 $. Hence we obtain
$ t-r -1 \leqslant a_{m-d} -1 $ and consequently
As a result, the number of subsets of the set $ \{1,2, \cdots,m\} $
with the desired property does not exceed the number $ a_d +a_{m-d} $.
We have thus shown that $ a_m $ does not exceed the maximum of the numbers
If, then, in the inequality (1) equality holds for
$ n =1, 2, \cdots,m - 1 $, then it also holds for $ n = m $.
Since for $ n = 1 $ the inequality (1) becomes an equality, using
induction, we finally conclude that $ a_n =2n $ for $ n =1,2,3, \cdots $. | 2n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
X OM - II - Task 2
Under what relationship between the sides of a triangle is it similar to the triangle formed by its medians? | Let's denote the medians drawn to the sides $a$, $b$, $c$ of triangle $ABC$ by $m_a$, $m_b$, $m_c$ respectively. We need to investigate in which triangles the medians are proportional to the sides of the triangle.
First, it is important to note that the larger of two unequal medians is drawn to the shorter side. To verify this, we can use the known formulas for the length of medians:
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}, \quad m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}, \quad m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
\]
From these formulas, it follows that, for example, if $m_a > m_b$, then $m_a^2 > m_b^2$, and according to the above formula, $b^2 > a^2$, hence $b > a$.
Suppose $a \leq b \leq c$, then $m_c \leq m_b \leq m_a$. The triangle with sides $m_a$, $m_b$, $m_c$ is similar to the triangle with sides $a$, $b$, $c$ if and only if
\[
\frac{m_a}{a} = \frac{m_b}{b} = \frac{m_c}{c}
\]
In problem 8, we proved that the area of the triangle equals $\frac{4}{3}$ of the area of the triangle formed by its medians. If these triangles are similar, the ratio of their areas equals the square of the ratio of corresponding sides, so from (2) we have $m_a^2 = \frac{3}{4} c^2$. Therefore, using the first formula from (1),
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\]
Thus,
\[
\frac{3}{4} c^2 = \frac{1}{4} (2b^2 + 2c^2 - a^2)
\]
Simplifying, we get
\[
3c^2 = 2b^2 + 2c^2 - a^2
\]
\[
c^2 = 2b^2 - a^2
\]
Conversely, if condition (3) is satisfied, then the triangle with sides $a$, $b$, $c$ is similar to the triangle formed by its medians. Indeed, considering the relationship (3) in the formulas (1), we obtain
\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} = \frac{1}{2} \sqrt{2b^2 + 2(2b^2 - a^2) - a^2} = \frac{1}{2} \sqrt{6b^2 - 3a^2} = \frac{\sqrt{3}}{2} \sqrt{2b^2 - a^2} = \frac{\sqrt{3}}{2} c
\]
\[
m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} = \frac{1}{2} \sqrt{2a^2 + 2(2b^2 - a^2) - b^2} = \frac{1}{2} \sqrt{4b^2 - b^2} = \frac{1}{2} \sqrt{3b^2} = \frac{\sqrt{3}}{2} b
\]
\[
m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - (2b^2 - a^2)} = \frac{1}{2} \sqrt{3a^2} = \frac{\sqrt{3}}{2} a
\]
From this, it follows that $m_a$, $m_b$, $m_c$ are proportional to $c$, $b$, $a$.
Thus, we have the following result:
A triangle is similar to the triangle formed by its medians if and only if the square of the middle side of the triangle is equal to the arithmetic mean of the squares of the shortest and longest sides.
Note 1. In the solution, we could have avoided referring to problem 8 and derived condition (3) from formulas (1) and (2). This straightforward calculation is proposed as an exercise. | ^2=2b^2-^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 12
For any natural number $ n $, consider the rectangular prism $ K_n $ with edge lengths $ 1,1,n $ and the set $ R_n $ consisting of $ 4n+1 $ points: the vertices of the rectangular prism $ K_n $ and the points dividing its longer edges into unit segments. We randomly select three different points from the set $ R_n $. Let $ p_n $ be the probability that the selected points are the vertices of an obtuse triangle. Calculate $ \lim_{n\to \infty} p_n $.
Note: Three collinear points do not form a triangle. | Let $ Q(n) $ denote the number of all three-element subsets of the set $ R_n $, $ T(n) $ - the number of non-degenerate triangles with vertices in the set $ R_n $, and $ P(n) $ - the number of obtuse triangles. The probability of interest is the fraction $ p_n = P(n)/Q(n) $. Notice that
(Justification of the second equality: from the number of all triples, we subtract the number of collinear triples of points - that is, triples of points lying on any of the four edges, each of which contains $ n +1 $ points of the set $ R_n $.)
After expanding the binomial coefficients, we obtain the formulas
Let us divide the set $ R_n $ into four-element "layers" $ H_0, H_1, \ldots, H_n $: the vertices of one of the square faces of the cube $ K_n $ form the layer $ H_0 $, the vertices of the other - the layer $ H_n $; the layer $ H_k $ consists of points of the set $ R_n $ at a distance of $ k $ units from the plane of the layer $ H_0 $.
Since we are interested in the limiting (as $ n \to \infty $) behavior of the quantities under consideration, we can assume that $ n \geq 4 $.
Fix a natural number $ r $, $ 3 \leq r \leq n-1 $, and choose any numbers $ k, m \in \{0,1,\ldots,n\} $ such that $ m - k = r + 1 $. There are thus $ r $ layers between layers $ H_k $ and $ H_m $. By choosing any points $ A \in H_k $, $ C \in H_m $, and $ B \in H_l $, where $ l \in \{k + 1,\ldots,m-1\} $, we obtain either an obtuse triangle or a collinear triple of points (since $ m-k \geq 4 $). We have $ 16 $ ways to position the pair of points $ A $ and $ C $ and $ 4r $ possible positions for point $ B $ - a total of $ 64r $ possibilities; $ 4r $ of them give collinear triples. The remaining $ 60r $ are obtuse triangles.
Since the numbers $ k $, $ m $ (with a given difference $ r+1 $) can be chosen in $ n-r $ ways, the number $ P(n) $ of all possible obtuse triangles is at least the sum
Thus, we have a two-sided estimate
The quantities $ Q(n) $, $ T(n) $, $ S(n) $ are third-degree polynomials (in $ n $). The coefficient of $ n^3 $ in $ Q(n) $ is $ 32/3 $, while in $ T(n) $ and $ S(n) $ the coefficient of $ n^3 $ is $ 10 $. Therefore,
and by the squeeze theorem,
| \frac{15}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XIII OM - II - Task 6
Find a three-digit number with the property that the number represented by these digits and in the same order, but in a different base of numeration than $ 10 $, is twice as large as the given number. | If $ x $, $ y $, $ z $ denote the consecutive digits of the sought number, and $ c $ - the new base of numeration, then
the equation can be replaced by the equation
The task is to solve equation (1) in integers $ x $, $ y $, $ z $, $ c $, satisfying the conditions $ 1 \leq x \leq 9 $, $ 0 \leq y \leq 9 $, $ 0 \leq z \leq 9 $, $ c > 0 $.
We will first prove that the only possible value of $ c $ is $ 15 $. Indeed, when $ x $, $ y $, $ z $ satisfy the above conditions, and $ 0 < c \leq 14 $, then
while if $ c > 16 $, then
Substituting $ c = 15 $ into equation (1), we obtain the equation
If integers $ x $, $ y $, $ z $ satisfy equation (2), then $ z $ is divisible by the common factor $ 5 $ of the first two terms, and since $ 0 \leq z \leq 9 $, there are only two possibilities:
a) $ z = 0 $; from equation (2) we get
from which it is clear that $ x $ and $ y $ can only have the values $ x = 1 $, $ y = 5 $, which gives the sought number the value $ 150 $. This number indeed satisfies the condition of the problem, as in base $ 15 $ numeration it represents the number $ 15^2 + 5 \cdot 15 = 300 = 2 \cdot 150 $.
b) $ z = 5 $; to determine $ x $ and $ y $ we get from (2) the equation:
from which it is clear that $ x < 3 $, since for $ x \geq 3 $ we have $ -5x + y + 1 \leq -15 + 9 + 1 = -5 $. When $ x = 1 $, then $ y = 4 $, and when $ x = 2 $, we have $ y = 9 $; the corresponding values of the sought number are $ 145 $ and $ 295 $. Both these numbers satisfy the condition of the problem, as
The problem thus has $ 3 $ solutions, which are the numbers $ 145 $, $ 150 $, $ 295 $. The appropriate base of numeration for each of them is $ 15 $. | 145,150,295 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXII - II - Task 4
Given are natural numbers $ k, n $. We define inductively two sequences of numbers $ (a_j) $ and $ (r_j) $ as follows:
First step: divide $ k $ by $ n $ and obtain the quotient $ a_1 $ and the remainder $ r_1 $,
j-th step: divide $ k + r_{j-1} $ by $ n $ and obtain the quotient $ a_j $ and the remainder $ r_j $. Calculate the sum $ a_1 + \ldots + a_n $. | According to the definition
Adding these equalities we get
thus
The number $ r_n $ as the remainder of the division of $ k+r_{n-1} $ by $ n $ satisfies the condition $ 0 \leq r_n < n $, but from the last equality, it follows that $ r_n $ is a number divisible by $ n $. Therefore, $ r_n = 0 $ and
thus | k | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XXXIX OM - I - Problem 8
For a given cube with edge length 1, find the set of segments with a total length not exceeding $1 + 3\sqrt{3}$, having the property that any two vertices of the cube are the endpoints of some broken line composed of segments from this set. | We will provide an example of a set of segments with the required property. This example can be considered the spatial counterpart of the "optimal road network connecting the vertices of a rectangle," constructed in preparatory task $C$. (A hint for the construction is the observation that it is purposeful to search for such an arrangement of segments so that in each "network node" exactly three segments meet, lying in the same plane and forming angles of $120^\circ$, because the existence of a node of any other form allows for a local modification of the network in the vicinity of such a node, leading to a shortening of the network.)
Let $A_1B_1C_1D_1$ and $A_2B_2C_2D_2$ be two opposite faces of a cube, labeled so that segments $A_1A_2$, $B_1B_2$, $C_1C_2$, $D_1D_2$ are four parallel edges of the cube. Denote the midpoints of edges $A_1B_1$, $C_1D_1$, $A_2B_2$, $C_2D_2$ by $K_1$, $L_1$, $K_2$, $L_2$, respectively, and the midpoints of segments $K_1L_1$ and $K_2L_2$ (i.e., the midpoints of the two mentioned faces) by $O_1$ and $O_2$ (see Figure 4).
om39_1r_img_4.jpg
On segment $O_1O_2$, we find points $P_1$ and $P_2$ such that $|O_1P_1| = |O_2P_2| = \sqrt{3}/6$. Then $|P_1P_2| = 1 - \sqrt{3}/3$. (Segments $K_1P_1$, $L_1P_1$, $P_1P_2$, $P_2K_2$, $P_2L_2$ form a network in the square $K_1L_1L_2K_2$ as described in task $C$; the angles at points $P_1$ and $P_2$ are each $120^\circ$.)
The length of segment $K_1P_1$ is $\sqrt{3}/3$. Let $M_1$ be the midpoint of this segment; thus, $|K_1M_1| = |M_1P_1| = \sqrt{3}/6$. (It is not difficult to verify that now segments $M_1A_1$ and $M_1B_1$ form angles of $120^\circ$ with segment $M_1P_1$). Segments $A_1M_1$ and $B_1M_1$ have lengths equal to $\sqrt{3}/3$. Similarly, let $N_1$, $M_2$, $N_2$ be the midpoints of segments $L_1P_1$, $K_2P_2$, $L_2P_2$, respectively. We consider the set of the following thirteen segments:
Sum of their lengths is $1 + 3\sqrt{3}$. In this way, we have found a set of segments satisfying the given conditions. | 1+3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - III - Problem 5
Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $. | The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in particular
By adding these inequalities side by side, we get that $a_1 + a_2 + \ldots + a_{100} = 1 + 2 + \ldots + 100 < 10B$, which means $505 < B$.
Thus, the number $A$ defined in the problem satisfies the inequality
On the other hand, consider the following permutation $a_1, a_2, \ldots, a_{100}$ of the set of natural numbers not greater than $100$
This permutation can be defined by the formulas:
We will prove that the sum of any $10$ consecutive terms of this permutation is not greater than $505$.
Indeed, if the first of the considered $10$ terms has an even number $2k$, then
If, however, the first of the considered terms has an odd number $2k + 1$, then
Thus, the sum of any $10$ consecutive terms of this permutation is not greater than $505$. Therefore, the number $A$ defined in the problem satisfies the inequality
From (1) and (2), it follows that $A = 505$.
Note 1. The problem can be generalized as follows: Find the largest integer $A$ such that for any permutation of the set of natural numbers not greater than an even number $n = 2t$, the sum of some $m = 2r$ (where $r$ is a divisor of $t$) consecutive terms is at least $A$.
By making minor changes in the solution provided above, consisting in replacing the number $100$ with $2t$ and the number $10$ with $2r$, it can be proved that $A = \frac{1}{2} m(n + 1)$.
Note 2. In the case where $m \leq n$ are any natural numbers, it is generally not true that every permutation of the set of natural numbers not greater than $n$ contains $m$ consecutive terms with a sum not less than $\frac{1}{2} m(n + 1)$. For example, for $n = 6$, $m = 4$, the permutation $6, 4, 1, 2, 3, 5$ does not contain four consecutive terms with a sum not less than $14$. | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits