Given real numbers a, b the sequence $(x_{n})^{inf}_{n=0}$ is defined by $x_{0}= a$ and $x_{n+1}= x_{n}+b sin x_{n}$ for every $n \geq 0$

(a) If b=1, prove that for every a, the sequence $(x_{n})$ has a finite limit when n->$inf$, and find this limit.

(b) Prove that for every b>2 there exists a real number a for which the sequence $(x_{n})$ does not have a finite limit when n->{inf}

Consider $f(x)=x+bsin x$.
1. If $a=k\pi$, or b=0, then $x_{n}=a, \lim_{n\to \infty }x_{n}=a$.
2. If $0<b\le 2$, then $-1\le f'(x)$ and $f'=-1$ if and only if $x=(2k+1)\pi$. It prove, that exist limit and $\lim_{n\to \infty }=\pi (2[\frac{a}{2\pi }]+1)$,  if  $a\not =2k\pi$.
3. If $b>2$ and $a\not =k\pi$, then don't exist $\lim_{n\to \infty }x_{n}$.


Prove that, given 9 points on the plane, no three colinear, we can always get a convex pentagon.

This is a particular case of the famous Erds-Szekeres problem (the Happy ending problem). The solution of this problem is not really difficult but quite long to write.
An excellent survey of the problem, including the answer of this problem (if I remember correctly  ;)  ), and related results can be found here :
http://www.ams.org/bull/2000-37-04/

(Hope it works  ;)  )

Pierre.

[quote="pbornsztein"]An excellent survey of the problem, including the answer of this problem (if I remember correctly  ;)  ), and related results can be found here :
http://www.ams.org/bull/2000-37-04/[/quote]
Cool! You found links! I told about this article in another thread.


AMC
AIME
geometry
analytic geometry
geometric transformation
rotation
rhombus

the 21st AIME2 2003

problems 7, 11, and 14 I am having trouble with

7. ABCD is a rhombus. The circumradii of ABD, ACD are 12.5, 25. Find the area of the rhombus.

11. ABC has AC = 7, BC = 24, angle C = 90 degrees. M is the midpoint of AB, D lies on the same side of AB as C and had DA = DB = 15. Find area CDM.

13. ABCDEF is a convex hexagon with all sides equal and opposite sides parallel. Angle FAB = 120o. The y-coordinates of A, B are 0, 2 respectively, and the y-coordinates of the other vertices are 4, 6, 8, 10 in some order. Find its area.

7 and 11 I really can't see what to do. For 13, I got 2 equations:

a(radical 3)sin theta = 2
2a(cos theta) = 10

where a is length of hexagon and theta is the angle of rotation

when i solve this i get area of hexagon = 39.5 radical 3, but Kalva has a different answer

any help would be greatly appreciated

[hide="7"]Draw the rhombus $ABCD$. Rhombi have all sides equal to each other, and opposite angles equal to each other. $ABD$ will be an acute triangle, and $ACD$ will be an obtuse triangle. The circumradius of a triangle is the radius of the circumscribed circle, so we know that the long diagonal $AC=25\cdot 2=50$ and the short diagonal $BD=12.5\cdot 2=25$. Also, in a rhombus, the two diagonals are perpendicular and always bisect each other, so we can actually just take where they intersect, then form two triangles, each with base $25$ and height $25$. Therefore, we have $(\dfrac{1}{2}\cdot 25\cdot 25)\cdot 2=\boxed{625}$.[/hide]

[quote="JesusFreak197"]The circumradius of a triangle is the radius of the circumscribed circle, so we know that the long diagonal $AC=25\cdot 2=50$ and the short diagonal $BD=12.5\cdot 2=25$. [/quote]
I don't think this is right.  You have interchanged circumradius and altitude.  Your answer is accordingly incorrect.

This is how I did #7:
[hide]
Let the side length of the rhombus be denoted $x$.  By the Extended Law of Sines
$\dfrac{x}{\sin \angle ABD}=2R=25$
$\sin \angle ABD=\dfrac{x}{25}$
Similarly $\sin \angle ACD = \dfrac{x}{50}$
Since $\angle ABC +\angle BCD = 2 \angle ABD + 2 \angle ACD=\pi$, $\sin \angle ACD = \cos \angle ABD$
We can now use the pythagorean identity
$\dfrac{x^2}{25^2}+\dfrac{x^2}{50^2}=1$
$x=10\sqrt{5}$

We can also use the Extended Law Of Sines on $\angle BAD$ and $\angle ADC$.  We find that $2(BD) = CA$.

The diagonals of a rhombus are orthogonal.  Let $a=\left( \dfrac{1}{2} \right) BD$.  Thus, by the Pythagorean Theorem $5a^2=x^2=500$ which implies a=10.  The area of the rhombus equals twice the area of $\triangle ABC$, which is 200.  Thus the answer is $\boxed{400}$, which agrees with kalva.

[/hide]

Oh, right, I was assuming that for the circumscribed circle, one of the sides of the triangle would be the diameter of the circle. I don't really know why, since that's only true for right triangles. :huh:

Thanks for the explanations. I had never seen the Extended Law of  Sines - all I ever saw was the a/sin A  =  b/sin B = c/sin C... Shows how much I have to study. Curious, is there some other way to solve problem 7 without using the Extended Law of Sines but instead some other method that, while a bit more involved uses only basic knowledge of geometry? Because so far, I've been trying to solve these problems using only more basic tools/concepts, and while I'll certainly be trying to pick up more advanced laws/theorems, I'm not sure if I will be able to learn enough of them come time for the tests, in which case all I will have at my disposal are the simpler (though I imagine that for many of you on this forum the Extended Law of Sines is already basic knowledge) things taught in middle school. Also, could anyone explain to me as to how to solve the other problems?

[quote="Morgoth"]Curious, is there some other way to solve problem 7 without using the Extended Law of Sines but instead some other method that, while a bit more involved uses only basic knowledge of geometry?[/quote] There's always anothe way!
[hide]What we really need to do is find the circumradius of a general isosceles triangle. Let's look at triangle ADC. Let X be the midpoint of AC, and let AX=XC=a and DX=b.
Because the circumcenter of a triangle is equidistant from its vertices, the circumcenter O is the intersection of the perpendicular bisectors of the three sides. (Make sure you know this.) So if M is the midpoint of CD and O is the circumcenter, then O is on DX and DMO is a right triangle. In fact, DMO and DXC are similar!
We'd like to know the length of DO, so we use the similar triangles:
$\dfrac{DO}{DM}=\dfrac{DC}{DX}$. And we know that $DX=b$, $DC=\sqrt{a^2+b^2}$, and $DM=\tfrac{1}{2}\sqrt{a^2+b^2}$, so we immediately get $R=DO=\dfrac{a^2+b^2}{2b}$. Likewise, the circumradius of ABD is $\dfrac{a^2+b^2}{2a}$. These give the equations to find $a$ and $b$ and thus the area $2ab$.[/hide]

[quote="darktreb"]Yes, there's always another way!

Zabel I owned the WOOT test :).

Looking forward to grading yours  ;)[/quote]

Haha! Sounds good.  :)

Thanks for the explanation zabelman, it really helped. Can anyone help me with the other problems? I'm still stuck on those.

At first glance, for 11, [hide]I would coordinatize: place AB on the x-axis with M at the origin, find the coordinates of C and D, and get the area from there. (If you don't know how to get area from coordinates, just ask.[/hide]

How can you find the coordinates of D though?


superior algebra
superior algebra unsolved

prove that if $ x$ and $ y$ are in the same conjugacy class of a group $ G$ , then $ |C(x)|\equal{}|C(y)|$( the centralizer of   an element)

Ok, where are you stuck¿
Try to show that $ C(x)$ is conjugated to $ C(y)$ (or converse).

if $ G$ is finite , then it's trivially easy to prove using $ |Conj(a)|\equal{}[G: C(a)]$

Is that true ??

Well, I think that's harder to prove than the original statement  :D
But yes, it's true.

:D

What if $ G$ is infinite ? Does that statement still works or not?

Do u have another prove ?? :wink:

you [i]may[/i] read the first post of ZetaX.

I read it , but I couldn't do it .

I meant when I said "Do u have another proof ?" , Write it Plz.

 :)

Ti describe the first way in more exterior terms:
Consider the inner automorphisms $ z \mapsto szs^{\minus{}1}$ and see what happens.


I'm going to make a chess tournament since the last one pretty much failed. I will accept 8 or 16 competitors with either single, double or pool play elimination. I will join if necessary and I will create bracket. Please don't join if you don't think you can be on for a long time. Time control is: 20 minutes each player and no delay.

Aww delay is so much fun.  Oh well, I will join this CHESE tournament.

Since I don't know any of your skill levels I will pair you up according to conviencence. This should work, if you don't open a prod in 72 hours you lose by default.

I don't know what your talking about  :P  :wink:  :D


Antoine paid a tax of $ 10\%$ on his yearly income. How many dollars must his income have been if he had $ \$6,300$ remaining after taxes?

10% = 1/10, and he has 9/10 left after taxes. Thus we must multiply by 10/9 to 6300 = 7000 dollars before taxes.


Now i wish i could recollect (BLANK)
"Eureka" cried the great inventor.

who was the inventor? and what comes in blank

he was archimedes. what about the blank??

found the answer, its pi in the blank
and yeh it was archimedes so thnx


trigonometry
function
calculus
integration
number theory
real analysis
real analysis unsolved

Let $n\in\mathbb{N}^{*}$ and $a_{k},b_{k}\in\mathbb{R},k=1,2,\ldots, n$. Prove that the following equation
\[\sum_{k=1}^{n}(a_{k}\cos (kx)+b_{k}\sin (kx))=0, \]
has at least one solution in $(0,2\pi)$.

Oki, Gabi! I will try not post such easy and useless problems, But maybe there are here on the forum some users who like this kind of problems.
Don't worry for you I have prepared something special from analityc number theory( for example Bonferoni lemma  :wink: or Heat-Brown's lemma, etc). Dont' worry I will post them soon. :)

Don't scare me with such lemmas...  :wink:    Ok, I admit that from time to time we can post easy problems, but really they should say something...

In case you're looking at this and still don't see it:

[hide="Hints"]The function is continuous (why?), and the Mean Value Theorem for integrals (what's that? look it up) applies.[/hide]


function
superior algebra
superior algebra unsolved

Hello,
How to prove that we can't express alternative with only implication?
--
Regards.

I mean finding such a formula which consist of only '=>' and variables that is equivalent to alternative.

Let me state the problem in a more formal way.
We have two logical variables, $x$ and $y$, and build new formulas by combining two already constructed formulas with an implication sign, we can build $x \rightarrow y$ and then $y \rightarrow (x \rightarrow y)$. We want to show that we never get a formula that is equivalent to the equivalence of x and y, i.e. formula $x \leftrightarrow y$.

The brute force method is the following.
Since equivalence of formulas is equivalent to equality of the induced logical functions, you find all logical functions on two arguments that can be constructed the way I described. You get only 6 of the 16 functions. You prove that these are all you can get by combining any two of then and showing that you get one of the functions. Finally you point out that the function which maps $(x,y)$ to $x \leftrightarrow y$ is not one of these functions.

EDIT: I misunderstood the question. apc seems to be looking for "either x or y", $x \oplus y$, instead of "x iff y", $x \leftrightarrow y$. But that is also not one of the 6 functions above. :-)

LiMa


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Suppose that $A,B$ are matrices over a field such that any matrix commuting with $A$ also commutes with $B$. Does it always follow that $B$ is a polynomial in $A$?

I think Eugene gave a proof here http://www.mathlinks.ro/Forum/viewtopic.php?t=82634


How many integers can be written as the sum of three different members of the set:

$ {2, 4, 6, 8, 10, 12, 13}$

Any way other than brute force?

Ignore the $ 13$ for now.  Using $ 2, 4, 6, 8, 10, 12$ we can write any even integer from $ 2 \plus{} 4 \plus{} 6 \equal{} 12$ to $ 8 \plus{} 10 \plus{} 12 \equal{} 30$.  This is because at least one of the even numbers in a given triplet can be increased by $ 2$ to get the next even number.  This gives $ 10$ different sums.

Add the $ 13$ back in.  We can now also write any odd integer from $ 2 \plus{} 4 \plus{} 13 \equal{} 19$ to $ 10 \plus{} 12 \plus{} 13 \equal{} 35$, and for the same reasons.  This gives $ 9$ different sums.

In total, we have $ \boxed{19}$ different sums.


Four congruent rectangles are placed as shown to form a larger rectangle with area 1728 cm$ ^2$. How many centimeters are in the perimeter of one of the four congruent rectangles?

[asy]draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)); draw((1,0)--(1,3));
draw((1,1)--(4,1)); draw((1,2)--(4,2));[/asy]

The area of one rectangle is $ 432$.  And by the drawing, you can deduce that the $ 3l \equal{} w$.

$ lw \equal{} 432$

$ l \cdot 3l \equal{} 432$

$ 3l^2 \equal{} 432$

$ l^2 \equal{} 144$

$ l \equal{} 12$

Then, $ w$ must be $ 36$.

Thus, the perimeter of one rectangle is $ 2 (12 \plus{} 36)$, or $ \boxed{96}$.


[quote="amirhtlusa"]i just remembered this problem, it's pretty easy, it's from a very old number theory book...
what is the smallest integer that can be written in a form of $13x+73y$ for three diffrent $(x,y)$? $(x,y\in\mathbb{N})$[/quote]

Here's my best attempt at it...
[hide]
If there is a solution (x,y), then two other solutions are (x-73, y+13) and (x+73, y-13)

Now we just need to find a value of (x, y) such that x-73 is positive and y-13 is positive. Obviously, (74,14) works. So the smallest number that can be expressed is 74*13+14*73= 1984.

[/hide]

Never mind didn't read the for three different x y bit.


let $x.y.z$ be reel numbers satisfaying the tree conditions
$|x+y+z|\leq1$
$|x-y+z|\leq1$
$|4x+2y+z|\leq8$
$|4x-2y+z|\leq8$
prove that $|x|+3|y|+|z|\leq7$.good luck :wink:


Let ABC be a triangle. We denote by  $A_1$,  $B_1$ and $C_1$ the points of
intersection of the bissectors of the angles of ABC with the respective
opposite sides.
We assume that the points $B, A_1, B_1$ and $C_1$ are cocyclic.
Prove that :

$\frac{BC}{AB+AC}=\frac{AC}{AB+BC}-\frac{AB}{BC+AC}$

Denoting a = BC, b = CA, c = AB, we have to show $\frac{b}{c + a} - \frac{c}{a + b} = \frac{a}{b + c}.$ Let the circumcircle of the triangle $\triangle A_1B_1C_1$ meet the triangle sides at $M \in a, N \in b, P \in c$ different from $A_1, B_1, C_1.$  According to the more general problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=82460]Circle (a'b'c')[/url],

$x = MA' = \frac 1 2 \left(\frac{b(b + c)}{c + a} - \frac{c(b + c)}{a + b} - \frac{a(b - c)}{b + c}\right)$

According to the condition of this problem, the points $M \equiv B$ are identical, hence x > 0 and

$\frac{ca}{b + c} = BA_1 \equiv MA_1 = x$

$(b + c) \left(\frac{b}{c + a} - \frac{c}{a + b}\right) = \frac{a(b - c)}{b + c} + \frac{2ca}{b + c} = \frac{a(b + c)}{b + c} = a$

$\frac{b}{c + a} - \frac{c}{a + b} = \frac{a}{b + c}$


as 102^35^17,is divisible by 102, 3a+1 is congruent to 0 mod 102.but 3 divides 102, so 3a+1 should be 3k for some k,a contradiction if a is a natural number.


A sum of a list of positive integers is 20. What is the maximum possible product of the integers in this list?

[hide]So it appears we'll probably have to use numbers more than once. Our choices are $ 2^{10}$ or $ 2(3^6)$ since otherwise there are too few numbers. Since $ 2^3<3^2$, the first choice is a smaller one. 1458.[/hide]


trigonometry
analytic geometry
graphing lines
slope

If $ x^2\plus{}y^2\equal{}12x\plus{}8y\minus{}27$ then what is the max. value of $ 3x\plus{}4y$?

Beaten again. I had, after completing the squares, a circle with centre at $ (6, 4)$ and radius $ 5$ and then the tangent $ 3x \plus{} 4y \equal{} c$ with greater y-intercept.

Thanks, kunny. I had the following.

The normal has slope $ \frac {4}{3}$ and passes through $ (6, 4)$ and so has y-intercept at $ (0, \minus{} 4)$ and intercepts the circle again at $ (3, 0)$ so point of tangency must be $ (6 \plus{} 3, 4 \plus{} 4) \equal{} (9, 8)$ and so $ \max(3x \plus{} 4y) \equal{} 27 \plus{} 32 \equal{} 59$. I have a diagram with three congruent $ 3 \minus{} 4 \minus{} 5$ triangles with vertices at $ (0, \minus{} 4)$, $ (0, 0)$ and $ (3, 0)$; $ (3, 0)$, $ (6, 0)$ and $ (6, 4)$; and $ (6, 4)$, $ (9, 4)$ and $ (9, 8)$ but don't know how to post a diagram (yet!).

[quote="vinskman"][quote="kunny"]$ x^2 \plus{} y^2 \equal{} 12x \plus{} 8y \minus{} 27\Longleftrightarrow (x \minus{} 6)^2 \plus{} (y \minus{} 4)^2 \equal{} 25$. Let $ x \minus{} 6 \equal{} X,\ y \minus{} 4 \equal{} Y$, we have $ 3x \plus{} 4y \equal{} 3X \plus{} 4Y \plus{} 34$ with $ X^2 \plus{} Y^2 \equal{} 25^2.$ Since $ (3X \plus{} 4Y)^2 \plus{} (4X \minus{} 3Y)^2 \equal{} 25\Longleftrightarrow \minus{} 25\leq 3X \plus{} 4Y\leq 25$, yielding $ 9\leq 3x \plus{} 4y\leq 59$.

Ans. $ Max(3x \plus{} 4y) \equal{} 59,\ (X,\ Y) \equal{} (3,\ 4)\Longleftrightarrow (x,\ y) \equal{} (9,\ 8)$.[/quote]

$ (3X \plus{} 4Y)^2 \plus{} (4X \minus{} 3Y)^2 \equal{} 25\Longleftrightarrow \boxed{\minus{} 25\leq 3X \plus{} 4Y\leq 25}$, yielding $ 9\leq 3x \plus{} 4y\leq 59$.

How comes?[/quote]
Use the substitution $X=x-6$ and $Y=y-4$.

I have a problem that from given problem we could simplify as,
$x^2$ $+$ $(y+4)^2$ $+$ $11$/$4$ $=$ $3x$ $+$ $4y$

So, then how could we find the maximum value??


How many moles of phosphorus atoms are in 100g of P4S10

1)0.225

2)0.900

3)4.00

3)400

4)0.487

Molar mass of $P\sub{4}S\sub{10}$ is $123.88+320.70=444.58 g/mol$

$\frac{100}{444.58}\approx0.225$

Since there are 4 atoms in Phosphorus pentasulfide

$0.225\times4=0.900=\boxed{2)0.900}$


$ a\plus{}b\equal{}\boxed{117^\circ}$.
By the exterior angle theorem.

[hide="alternately"]
Since $ a$ and $ 138^{\circ}$ are supplementary, they sum to $ 180^{\circ},$ so $ a\equal{}42^{\circ}.$ Similarly, $ b\plus{}105^{\circ}\equal{}180^{\circ}\Rightarrow b\equal{}75^{\circ}.$ Therefore, $ a\plus{}b\equal{}42^{\circ}\plus{}75^{\circ}\equal{}\boxed{117^{\circ}}.$[/hide]


induction
number theory proposed
number theory

I hope it's not again one of my "masterpieces", but I think the following is a very nice and difficult problem:
   Let $ a,b,c$ some positive integers and $ x,y,z$ some integer numbers such that we have
   a) $ax^2+by^2+cz^2= abc+2xyz-1$;
   b) $ab+bc+ca\geq x^2+y^2+z^2$.
   Prove that $ a,b,c$ are all sums of three squares of integer numbers.

I guess they can be represented even as sum of two squares.
Excuse me if it's wrong:

Firstly a well-known lemma:
If $xy=z^2+1$ then we can find $u,v,w,t \in \mathbb{N}$ with $x=u^2+v^2,y=w^2+t^2, z=uw+vt$ and hence $ut-vw=1$. This lemma can be proven by induction on $x+y+z$ by noticing that if $x<y$ then $x<z$ (unless trivially) and that $x(x+y-2z)=(z-x)^2+1$ is a smaller triple satisfying the condition. Moreover, the lemma can be extended even to $z<0$ by an easy argument, just in this case we have $u,v,w,t \in \mathbb{Z}$

Next, we transform the condition into $(ax-yz)^2+1=(ac-y^2)(ab-z^2)$ which leads us by using the lemma to the existence of desired $u,v,w,t$ and $t=\frac {vw+1}u$ from the condition $ut-vw=1$.
This leads us to the system
$ax=uw+yz+\frac {v(vw+1)}u$ (1)
$ac=u^2+v^2+y^2$ (2)
$ab=w^2+z^2+\frac {(vw+1)^2}{u^2}$ (3)

Now let $p|a$. Multiplyiing the first relation with $u$ and the third with $u^2$ we get in $\mathbb{Z}_p$
$u^2w+uyz+v^2w+v=0$ (1')
$u^2+v^2+y^2=0$ (2)
$u^2w^2+u^2z^2+w^2v^2+2vw+1=0$ (3')

Now substract $w$ times (2) from (1') to get
$v=y^2w-uyz$
Substract $w^2$ times (2) from (3') to get
$2vw=y^2w^2-1-u^2z^2$

Comparing the last two identities gives us
$2w(y^2w-uyz)=y^2w^2-1-u^2z^2$ hence $y^2w^2-2ywuz+u^2z^2+1=0$ thus $(yw-uz)^2+1=0$ which means $p$ is of form $4k+1$ and a sum of two squares.
Since $p$ is any prime dividing $a$, so does $a$ and analogously $b$, $c$.

Here's another nice problem: find all solutions of the above equation. I think I've found a solution.

[quote = iura]Next, we transform the condition into $(ax-yz)^2+1=(ac-y^2)(ab-z^2)$ which leads us by using the lemma to the existence of desired $u,v,w,t$ and $t=\frac {vw+1}u$ from the condition $ut-vw=1$.[/quote]

Should the LHS not be $(ax-yz)^2 + a$? That would spoil the rest of the argument, which is really nice :(


Let $\{x_n\}$ be a geometric progression satisfying $x_1=1$ and
\[ \sum_{k=0}^{\infty}a_{3k+1}=\frac89 \]
Find $\sum_{n=1}^{\infty}a_n$.

Could you please show us your work, tµtµ?

[hide]$\{a_{3k+1}\}$ is geometric sequence with ratio: $\frac{a_{3.1+1}}{a_{3.0+1}}=a_4$, if $S_n$ is the sum of the first $n$ terms of this sequence, we get:

$S_n=\frac{a_1(1-a_4^n)}{1-a_4}=\frac{a_1}{1-a_4}-\frac{a_1.a_4^n}{1-a_4}$ if $-1<a_4<1 \implies \sum_{k=0}^{\infty}a_{3k+1}= \lim_{n \rightarrow \infty}S_n=\frac{a_1}{1-a_4}=\frac{1}{1-a_4}$ consequently:

$\frac{8}{9}=\frac{1}{1-a_4} \implies a_4=-\frac{1}{8}$

$\{a_n\}$ is geometric sequence with ratio $q$ then:$a_4=1.q^3 \implies q=-\frac{1}{2}$ and:

$\sum_{n=1}^{\infty}a_{n}=\frac{1}{1-q}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$ [/hide]


Problem is paraphrased.

Pedro starts at the center of a circle with radius 120. He walks halfway to the circle, makes a 90 degree right turn, walks halfway to the circle, makes a 90 degree right turn, walks halfway to the circle, makes a 90 degree right turn, then walks halfway to the circle a final time. How far is he from the center of the circle? Round to the nearest whole number.

[hide="hint"]Build some right triangles by drawing in some radii, but you have to be creative in what you use as legs.[/hide]

[hide="bigger hint"]
Power of a point kills this.  Just not in a nice way.
[/hide]

[hide=":P"]PM me if you find this!
[/hide]

I remember this one.  It was in All-Time Greatest right?  Like Teki-Teki said, power of a point is really helpful.  The first time I did it with triangles and it took way more than 6 minutes...

Ugh, I hate this problem. The solution is awful w/o pics which i suck at making/doing stuff with so I'm not gonna post it.

I did it with  Power of a Point and right triangles (Use Power of a Point :ninja: ). I suck at LaTeX, so forget about a solution.

i think i did this and was way off, whats the answer and how to u find it?

I wrote this up in the coaches forum once upon a time, and it was discussed before with a couple other explanations here :

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=77717
[hide]

[quote="frost13"]My solution was painful, but no trig.  I think I used the pythagorean theorem 4 times, and had to solve a couple simple 2 step equations.

Here's what I did if you can follow.

First you travel 60 up

Turn right and head until you hit the circle.  Draw the radius to the point where you hut circle.  Find the horizontal leg(103.9) and cut it in half(51.9).

From that halfway point draw a line down until you hit the circle. Draw a radius to that new point on the circle.  Also draw from the center out to that new vertical line.  You know the distance to the right of center is 51.9, use the P.T. to find how far below the center(108.2) the line hits the circle, and add the 60 you had to travel to get back below the center(168.2)  Now cut that in half(84.1) to see you far you are supposed to travel down, but subtract 60 to see how far below center you are(24.1)

You now need to repeat that whole process to see how far left of center it is to the circle(117.5) plus the 51.9 to the right of center you were to see how far you travel total(169.4).  Cut that in half(84.7) and subtract the 51.9 to get how far left of center you are(32.8).  Once you have that you use the P.T. one more time with the 24.1 and the 32.8 to get how far from center you are (40.7) and then round to 41.

Like I said painful, but no trig.  They made th first Target in that pair very simple so you could spend most of the 6 minutes on this one.

Sorry if that didn't help any, or wasn't clear enough.  Pretty hard without drawing the picture as you go.[/quote]
[/hide]


algebra
polynomial
function
algebra proposed

Basically the same idea: the graph of the function $f(x)=x^3+mx^2+nx+p$ is symmetric across the point $A(-m/3,f(-m/3))$. In our case, $A(2,18)$. Therefore, the points $B(a,16),C(b,20)$ are also symmetric across $A$. It follows that the midpoint of the interval $(a,b)$ is $2$, that is, $a+b=4.$
P.S. I used the fact that $f$ is increasing, therefore the equations $f(x)=16,f(x)=20$ have unique real solutions.

I cant see how Rust say that " if f(a)+f(b)=36 then a+b=4 ? "

wat i do is

Letting $g(x)=x^3-6x^2+17x-18=(x-2)(x^2-4x+9)$ and $g(a)=-2$ , $g(b)=2$

so $g(a)=(a-2)(a^2-4a+9)=-2$ $\implies a-2=\frac{-2}{a^2-4a+9}$

similarly $b-2=\frac{2}{b^2-4b+9}$ . Add up to get

$a+b-4=\frac{2(a-b)(a+b-4)}{(a^2-4a+9)(b^2-4b+9)}$

so $a+b=4$

but I dont know if $a+b\not =4$ , would there be a solution or not for $a,b$

are you sure that a,b are reals?
if they can be complex,there are a pile of solutions!


As usual everyone will vote for Euler... (he's very popular nowadays)  :P

Vote for Archimedes even  he is old, he is still great

[quote="miyomiyo"][quote="_funcia"]everyone vote for erdos!!! you have to!

[hide="10,9,8,7,6,5,4,3,2,1...."]
Welcome. You have just registred on the Art of Problem Solving Math Forums.
In your first 1 hour(s) on the board you can only post a message once every 10 minutes
You will be able to submit this message in 1 minute(s)[/hide][/quote]
Who is this new _funcia person?[/quote]

[b]She's[/b] [url=http://www.artofproblemsolving.com/Forum/usercp.php?u=19209&mode=viewprofile]funcia[/url], but [b]she's[/b] not using that account anymore, because [b]she[/b] wants to keep it at 1337 posts.  :rotfl:

Posting in GFF does not increase post number. :lol:

[quote="Temperal"][quote="miyomiyo"][quote="_funcia"]everyone vote for erdos!!! you have to!

[hide="10,9,8,7,6,5,4,3,2,1...."]
Welcome. You have just registred on the Art of Problem Solving Math Forums.
In your first 1 hour(s) on the board you can only post a message once every 10 minutes
You will be able to submit this message in 1 minute(s)[/hide][/quote]
Who is this new _funcia person?[/quote]

He's [url=http://www.artofproblemsolving.com/Forum/usercp.php?u=19209&mode=viewprofile]funcia[/url], but he's not using that acocunt anymore, because he wants to keep it at 1337 posts.  :rotfl:[/quote]
funcia is a girl

[quote="pianoforte"][/quote][quote="Temperal"][quote="miyomiyo"][quote="_funcia"]everyone vote for erdos!!! you have to!

[hide="10,9,8,7,6,5,4,3,2,1...."]
Welcome. You have just registred on the Art of Problem Solving Math Forums.
In your first 1 hour(s) on the board you can only post a message once every 10 minutes
You will be able to submit this message in 1 minute(s)[/hide][/quote]
Who is this new _funcia person?[/quote]

He's [url=http://www.artofproblemsolving.com/Forum/usercp.php?u=19209&mode=viewprofile]funcia[/url], but he's not using that acocunt anymore, because he wants to keep it at 1337 posts.  :rotfl:[/quote][quote="pianoforte"]
funcia is a girl[/quote]
funcia is a girl, how right!
I am funcia and here I come to tell you that I am also _funcia coz i have 1337 posts and this is g&ff after all

Well, now you have 1338 posts, so you might as well go back to funcia.

you don't understand _funcia was stopped  :(

[quote="Temperal"]How are you posting that, then?[/quote]
funcia wasn't suspended-only her reincarnation :(

[quote="funcia"][quote="Temperal"]How are you posting that, then?[/quote]
funcia wasn't suspended[/quote]Yet!  :!:

Yikes. :maybe:

EDIT: Isn't this getting off topic?

Ramanujan!

[hide="  :wink: "]
[hide=" :maybe: "]
[hide=" :o "]
Ramanujan!!!
[/hide][/hide][/hide]

[color=white]LOL that was random[/color]

[quote="Temperal"]Why does everyone like Euler?[/quote]
dude don't instantiate and then generalise

THAT IS AGAINST THE RULES!!!! WARNING WARNING!!!!

I like Erdos.

Riemann owns. Why can't anybody see that?

[quote="vishalarul"]Gauss pwns!

Consider that if he were still alive, mathematics would be 50 years advanced from now.

Go Gauss![/quote]
Exactly how do you know that?

Time Machine. Which reminds me, Einstein should be up there as an option (like spillovers from physics ... maybe Bohr etc.)

9 more votes and we will be with EULER!!! w00t go my little ERDOS friends...


[b]SOLVE[/b]
[img]http://i305.photobucket.com/albums/nn231/mrlonely09xxx/cats-1.jpg[/img] [size=150]=0[/size]

thanks  :!:  :?:

hello, can you  use $ \text{\LaTeX}$ please i don't know what you mean.
Sonnhard.

Latex-fied version, though I'm no quite sure what he means by the second part.... Because the $ \plus{}......\plus{}(n\plus{}\sqrt{x})(n\plus{}1)$ part doesn't make sense to me....

$ \frac{\sqrt{x(1\plus{}\sqrt{x})}}{2\plus{}\sqrt{(1\minus{}x)}}\plus{}\frac{(1\plus{}\sqrt{x})(2\plus{}\sqrt{x})}{2.3\plus{}\sqrt{1\minus{}x}}\plus{}......\plus{}(n\plus{}\sqrt{x})(n\plus{}1)\equal{}0$.

I think the first radical is only for the x, so
$ \frac{\sqrt{x}(1\plus{}\sqrt{x})}{2\plus{}\sqrt{(1\minus{}x)}}\plus{}\frac{(1\plus{}\sqrt{x})(2\plus{}\sqrt{x})}{2.3\plus{}\sqrt{(1\minus{}x)}}\plus{}......\plus{}(n\plus{}\sqrt{x})(n\plus{}1)\equal{}0$

I can see the pattern in the numberator, but the denominator makes no sense to me.

my teacher gave it to me :(
not spam  :chief:
i hate spam x-(


[quote="Ignite168"][quote="kohjhsd"][quote="Ignite168"]A right triangle ABC with hypotenuse AB has side AC=15. Altitude CH divides AB into segments AH and HB with HB = 16. Find the area of triangle ABC.[/quote]

[hide]In triangle $ABC$, let $AH=x$.

Then, $CH=\sqrt{AH \cdot HB}=4\sqrt{x}$.

$AH \cdot AB=AC^{2}$, therefore $AH \cdot 16=x(x+16)=225$

Solving, knowing that $x>0$, $x=25$.

Then the area is $\frac{1}{2}\cdot CH \cdot AB=\frac{1}{2}\cdot (20) \cdot (41)= \boxed{410}$.[/hide][/quote]

Correct up until you solved $x(x+16)=225$ ;)[/quote]

[hide="revision :oops: "]
Actually, $x=9$, then the area is $\frac12 \cdot CH \cdot AB=\frac12 \cdot 12 \cdot 25=6 \cdot 25=\boxed{150}$.[/hide]


number theory
greatest common divisor
prime numbers
algebra unsolved
algebra

I had this one at an exam and I could not find an answer for it.
I'ts probably quite easy, but I would really like to know. Please help me. :?

The problem:
Let $a,b,x,y \in N$ With gcd (a,b)=1 and $x^a=y^b$
Prove that a $z$ exist for which $x=z^b$ and $y=z^a$

Consider prime decompositions of $x$ and $y$. They contain the same prime numbers, so $x=p_1^{a_1}...p_n^{a_n}$ and $y=p_1^{b_1}...p_n^{b_n}$. Due to $x^a=y^b$ we have $aa_i=bb_i$, but $\gcd(a,b)=1$. It follows $b\mid a_i$ and $a\mid a_i$, i.e. $a_i=ba_i'$ and $b_i=ab_i'$. Therefore $x=(p_1^{a'_1}...p_n^{a'_n})^b$ and $y=(p_1^{b'_1}...p_n^{b'_n})^a$.


1. let a,b,c be positive real numbers with a^2+b^2+c^2=1 prove that
  1<=a/(1+bc) + b/(1+ac) + c/(1+ab) <=2^1/2

 thank you :)

It follows from proving
\[ a^2\leq \frac{a}{1\plus{}bc}\leq \frac{a\sqrt{2}}{a\plus{}b\plus{}c}\]
For the left inequality, start with AM-GM on $ b^2,c^2$. For the right inequality, start with RMS-AM on $ a,b\plus{}c$.

[quote="scorpius119"]It follows from proving
\[ a^2\leq \frac {a}{1 \plus{} bc}\leq \frac {a\sqrt {2}}{a \plus{} b \plus{} c}
\]
For the left inequality, start with AM-GM on $ b^2,c^2$. For the right inequality, start with RMS-AM on $ a,b \plus{} c$.[/quote]
i am grateful but please explain more :wink:

The left inequality is equivalent to $ a(1\plus{}bc)\leq 1$. First, $ bc\leq \frac{b^2\plus{}c^2}{2}\equal{}\frac{1\minus{}a^2}{2}$ so it reduces to $ a(3\minus{}a^2)\leq 2$ for $ a\in [0,1]$ which can just be factored.

The right inequality is equivalent to $ a\plus{}b\plus{}c\leq (1\plus{}bc)\sqrt{2}$. This follows from $ a\plus{}b\plus{}c\leq \sqrt{2(a^2\plus{}(b\plus{}c)^2)}\equal{}\sqrt{2(1\plus{}2bc)}$ and another simple one-variable inequality $ \sqrt{1\plus{}2bc}\leq 1\plus{}bc$.

[quote="scorpius119"]The left inequality is equivalent to $ a(1 \plus{} bc)\leq 1$. First, $ bc\leq \frac {b^2 \plus{} c^2}{2} \equal{} \frac {1 \minus{} a^2}{2}$ so it reduces to $ a(3 \minus{} a^2)\leq 2$ for $ a\in [0,1]$ which can just be factored.

The right inequality is equivalent to $ a \plus{} b \plus{} c\leq (1 \plus{} bc)\sqrt {2}$. This follows from $ a \plus{} b \plus{} c\leq \sqrt {2(a^2 \plus{} (b \plus{} c)^2)} \equal{} \sqrt {2(1 \plus{} 2bc)}$ and another simple one-variable inequality $ \sqrt {1 \plus{} 2bc}\leq 1 \plus{} bc$.[/quote]
 thank you  scorpius119 :blush:


function
calculus
derivative
inequalities
limit
real analysis
real analysis solved

Let $g: R\to R$ be a differentiable function whose derivative $g'$ satisfies the inequality $|g'(x)|\leq 2 \forall x \in R$.
Show that if $\delta$ is small enough, then $f: R\to R$ defined by $f(x)=x+\delta g(x)$ is bijective.

$\delta<\frac12$ suffices. That would give us $f'(x)=1+g'(x)\ge 1-2\delta$ for all $x.$ By the Mean Value Theorem, that shows that $f$ is strictly increasing (hence one-to-one) and that for $x>0,\,f(x)\ge f(0)+(1-2\delta)x$ and for $x<0,\,f(x)\le f(0)+(1-2\delta)x.$

Thus $\lim_{x\to\infty}f(x)=\infty$ and $\lim_{x\to-\infty}f(x)=-\infty.$ Since $f$ is continuous (it's differentiable, after all) the Intermediate Value Theorem now shows that $f$ maps $\mathbb{R}$ onto $\mathbb{R}.$

It is not necessary that $\delta$ be a constant, only that it be small (e.g. less than 1/2 in absolute value).

[quote="Kent Merryfield"]for $x>0,\,f(x)\ge f(0)+(1-2\delta)x$ and for $x<0,\,f(x)\le f(0)+(1-2\delta)x.$[/quote]
How you do get this?

[quote="derekhohk"]How you do get this?[/quote]
This is the Mean Value Theorem in operation. Suppose that $f$ is differentiable and $f'(x)\ge a$ everywhere. Then for $x> x_0,\,f(x)-f(x_0)=f'(\xi)(x-x_0)$ for some $x_0<\xi<x.$ Thus, $f(x)-f(x_0)\ge a(x-x_0)$ or $f(x)\ge f(x_0)+a(x-x_0).$ A similar result applies for $x<x_0;$ we merely have to be careful with the direction of the intequality signs.

fleeting_guest was trying to milk a little more out of the exact same argument. One does have to be a little bit careful. If $\sup |\delta(x)|<\frac12,$ we're fine. But merely $|\delta(x)|<\frac12.$ for all $x$ is insuffient.

Let $g(x)=2x$ and $\delta(x)=\frac{-x}{2x+1}.$ Then $f(x)=\frac{x}{2x+1}.$ We do have that $f$ is one-to-one, but it fails to map $\mathbb{R}$ onto $\mathbb{R}.$

[quote="Kent Merryfield"]By the Mean Value Theorem, that shows that $f$ is strictly increasing[/quote]
How do you get this?

Theorem (Mean Value Theorem):

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b),$ then there exists $\xi\in(a,b)$ such that $f(b)-f(a)=f'(\xi)(b-a).$

Corollary:

If $f$ is differentiable on an interval and $f'(x)>0$ for each $x$ in that interval, then for any $a,b$ in that interval such that $a<b,$ we have $f(a)<f(b).$ In other words, $f$ is strictly increasing.

-

Try putting these two pieces together yourself.


I write these letters on a row of cards in the following order, one letter per card:
A,B,C,D,E,F,G,H
I rearrange the cards' order and rearrange them again using the same algorithm as the first.
After the 2 rearrangements, I get:
F,A,E,G,D,H,C,B
What do the order of cards look like after my first rearrangement?

[hide="Hint"]Hint: Write the initial order of the cards on one line, skip a line and write the final order of the cards on the following line. Then realize that if you move a card to one of the eight possible spots, it must move to where that card had moved previously... so you will always end up with a pattern like this:
*******A****
***A***B****
***B********
Notice how in both cases B is below A.[/hide]


[quote=DreamTeam]Consider a point $ M$ found in the same plane with the triangle $ ABC$, but not found on any of the lines $ AB,BC$ and $ CA$. Denote by $ S_1,S_2$ and $ S_3$ the areas of the triangles $ AMB,BMC$ and $ CMA$, respectively. Find the locus of $ M$ satisfying the relation:
$ (MA^2\plus{}MB^2\plus{}MC^2)^2\equal{}16(S_1^2\plus{}S_2^2\plus{}S_3^2)$[/quote]

Anyone?


Let $ a,\ b,\ c$ be positive real numbers. Prove that $ \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac{1}{\sqrt {ca}} \plus{} 2\sqrt {3}\sqrt {a \plus{} b \plus{} c}\geq 9$.

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Prove that $ \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac {1}{\sqrt {ca}} \plus{} 2\sqrt {3}\sqrt {a \plus{} b \plus{} c}\geq 9$.[/quote]

$ \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac {1}{\sqrt {ca}} \plus{} 2\sqrt {3(a \plus{} b \plus{} c)}$
$ \ge \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac {1}{\sqrt {ca}} \plus{} 2 (\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})$
$ \equal{}(\frac {1}{\sqrt {ab}}\plus{}\sqrt{a}\plus{}\sqrt{b})\plus{}(\frac {1}{\sqrt {bc}}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}(\frac {1}{\sqrt {ca}}\plus{}\sqrt{c}\plus{}\sqrt{a})$
$ \ge 3\plus{}3\plus{}3\equal{}9$ by AM-GM


Could anyone tell me what is the famous and very useful mixing variable? thanks

Why noone? It is a secret ? Please one post this theorem.

Yeah, it is a closely guarded secret, you must go through the murderous ritual of searching for the secret magical words "mixing variables" (impossible to guess!) and with lots and lots of luck you will get the answer...

Well, let me make this unmanageable work a bit easier for you:
http://www.mathlinks.ro/Forum/viewtopic.php?t=23892 post #4

  Darij


$P,Q$ is on the circumcircle of $ABC$ and its center is $O$. $\angle POQ$ is fixed angle.
$X$ is a intersection of $P$, $Q$'s Simson line. Find the locus of $X$ as $P$ varies.

when $\angle POQ=180$, then locus is nine-point circle (well-known)

I conjectured it by GSP and proved. I spented about 2 hours.

Anyone have a solution?
Try to find a circle containing X.


linear algebra
matrix
IMC
college contests

For $ n\geq 0$ define the matrices $ A_n$ and $ B_n$ as follows: $ A_0 \equal{} B_0 \equal{} (1)$, and for every $ n>0$ let
\[ A_n \equal{} \left( \begin{array}{cc} A_{n \minus{} 1} & A_{n \minus{} 1} \\
A_{n \minus{} 1} & B_{n \minus{} 1} \\
\end{array} \right) \ \textrm{and} \ B_n \equal{} \left( \begin{array}{cc} A_{n \minus{} 1} & A_{n \minus{} 1} \\
A_{n \minus{} 1} & 0 \\
\end{array} \right).
\]
Denote by $ S(M)$ the sum of all the elements of a matrix $ M$. Prove that $ S(A_n^{k \minus{} 1}) \equal{} S(A_k^{n \minus{} 1})$, for all $ n,k\geq 2$.

well i wouldn't post anything but the case is that i got the author's solution. it's a combinatorical problem s ((a_n)^k) counts the number of  n row x m column tables with 0 or 1 entries which do not contain  submatrix of four ones 1,1 and below them 1,1.  this nuber clearly corresponds to the nuber of n tuples of natural nubers whose decimal expansion contais of no more than m digits  . s ((a_n)^k) is in fact  sum(sum (sum (sum( .....(sum  a_i_j*a_j_k*a_k_l*....*a_s_t*a_t_u) where the a_i_j is 0 or 1 so the summand can be 1 only if all the a_i_j 's are 1 so the indices i,j,k,l,......,s,t,u form an admissible -k tuple. (the recurrence relations force it to be so) now clearly the number of such m x n tables is equal to the number of n x m tables


1) show that the ring R=M2(F) of two by two matrices over a field F has no proper 2-sided ideals.

2) Suppose that F is a field with |F*|=p for p prime. Prove that p=2^n-1 for some n. Describe how to construct all such F's.

For point 2) : Since $F$ is a field $p+1$ must be the power of a prime , and if $p$ is a prime different from $2$ the $p+1$ is even so $p=2^{n}-1$ . what if $p=2$,and $F=\mathbb{F}_{3}$?!

[quote="math beginner"]
2) Suppose that F is a field with |F*|=p for p prime. Prove that p=2^n-1 for some n. Describe how to construct all such F's.[/quote]
$p+1 = q^{n}$
If $p$ is odd then $q=2$. So $p=2^{n}-1$
Since there is only one field upto isomorphism all fields having this property are $\mbox{GF}(zm_{p})$
Where $M_{p}$ are Mersenne primes.

If $p$ is even, i.e. $p=2$ then $q=3,n=1$.
So $\mathbb{Z}_{3}$


geometry
geometric transformation
perimeter
inequalities
ratio
linear algebra
matrix

Here is the IMO shortlist problems 1984, I got it from internet.

Here are the translations of the Russian problems from the "English Parts" file:

[u][b]Mongolia[/b][/u]

[b]Problem 1.[/b] Let $d$ be the sum of all diagonals of a convex $n$-gon, and let $p$ be the perimeter of this $n$-gon. Prove that

$\displaystyle \frac{n-3}{2}<\frac{d}{p}<\frac{\left[ \frac{n}{2}\right] \left[ \frac{n+1}{2}\right] -2}{2}$,

and show that these bounds cannot be improved.

[i]Note.[/i] This was actually problem 5 of the IMO 1984.

[b]Problem 2.[/b] A country has $n$ cities, and every pair of cities is connected by a railway line. A railway tester has to cruise every of these railway lines once and only once. If the tester cannot go from one city to another one by railway anymore since he has already cruised the railway line joining these two cities, then he can use the airplane. Find the minimal number of airplane tickets the tester needs in order to have all railway lines cruised.

[i]Note.[/i] Cruising a railway line means taking this line in order to pass from one of the cities this line connects to the other one. The tester can cruise the railway line only once, thus he can use it only in one of the two directions.

[b]Problem 3.[/b] Prove the existence of different natural numbers $m_{1}$, $m_{2}$, ..., $m_{k}$ satisfying the inequality

$\displaystyle \pi ^{-1984}<25-\left( \frac{1}{m_{1}}+\frac{1}{m_{2}}+...+\frac{1}{m_{k}}\right) <\pi ^{-1960}$,

where $\pi $ is the ratio between the circumference of a circle and its diameter.

[b]Problem 4.[/b] The set $\left\{ 1;\;2;\;...;\;49\right\} $ is subdivided into three subsets. Show that at least one of these three subsets contains three different numbers $a$, $b$, $c$ such that $a+b=c$.

[u][b]Soviet Union[/b][/u]

[b]Problem 1.[/b] Given an $m\times n$ matrix $\left\| p_{ij}\right\| $ whose elements are real numbers. For every $i\in \left\{1;\;2;\;...;\;m\right\}$, let

$\displaystyle \textbf{(1)}\;\;\;\;\;\;\;\;a_{i}=\sum_{j=1}^{n}p_{ij}$,

and for every $j\in \left\{ 1;\;2;\;...;\;n\right\}$, let

$\displaystyle \textbf{(2)}\;\;\;\;\;\;\;\;b_{j}=\sum_{i=1}^{n}p_{ij}$.

[i]Rounding[/i] a non-integer number will mean replacing it by one of the two next integers (i. e. replacing it either by its floor or by its ceiling).

Prove that one can round the numbers $a_{i}$, $b_{j}$, $p_{ij}$ in such a way that the equations (1) and (2) remain valid.

[b]Problem 2.[/b] A (not necessarily regular) tetrahedron is inscribed into a sphere with radius $1$ such that the center of the sphere lies inside the tetrahedron.

Prove that the sum of the edgelengths of the tetrahedron is greater than $6$.

[b]Problem 3.[/b]

Let $x_{1}$, $x_{2}$, ..., $x_{k}$ be the nontrivial divisors of a natural number $n$, arranged such that $x_{1}<x_{2}<...<x_{k}$. Find all numbers $n$ such that $x_{5}^{2}+x_{6}^{2}-1=n$.

[i]Note.[/i] The ''nontrivial'' divisors of the number $n$ are all natural divisors of $n$ except for $1$.

[b]Problem 4.[/b]

From the medians of an acute-angled triangle, we form another triangle. Let $R$ and $R_{m}$ be the circumradii of the first and the second triangles, respectively.

Prove that $R_{m}>\frac{5}{6}R$.

[b]Problem 5.[/b]

In the Martian language, each finite chain of Latin letters is considered to be a word.

The Martian dictionary consists of infinitely many tomes: The first tome contains all one-letter words; the second tome contains all two-letter words; etc.. All words are numbered in lexicographical order, and the numeration in each tome is continued in the next tome.

Find the word whose number equals the sum of the numbers of the words [i]Prague[/i], [i]Olympiad[/i], [i]Mathematics[/i].

For Orl, here is the TeX source:

[code]\underline{\textbf{Mongolia}}

\bigskip

\textbf{Problem 1.} Let $d$ be the sum of all diagonals
of a convex $n$-gon, and let $p$ be the perimeter of
this $n$-gon. Prove that
\begin{equation*}
\frac{n-3}{2}<\frac{d}{p}<\frac{\left[ \frac{n}{2}\right]
\left[ \frac{n+1}{2}\right] -2}{2},
\end{equation*}
and show that these bounds cannot be improved.

\textit{Note.} This was actually problem 5 of the IMO 1984.

\bigskip

\textbf{Problem 2.} A country has $n$ cities, and every
pair of cities is connected by a railway line. A railway
tester has to cruise every of these railway lines once
and only once. If the tester cannot go from one city to
another one by railway anymore since he has already
cruised the railway line joining these two cities, then
he can use the airplane. Find the minimal number of
airplane tickets the tester needs in order to have all
railway lines cruised.

\textit{Note.} Cruising a railway line means taking
this line in order to pass from one of the cities this
line connects to the other one. The tester can cruise
the railway line only once, thus he can use it only in
one of the two directions.

\bigskip

\textbf{Problem 3.} Prove the existence of different
natural numbers $m_{1},$ $m_{2},$ $...,$ $m_{k}$
satisfying the inequality
\begin{equation*}
\pi ^{-1984}<25-\left( \frac{1}{m_{1}}
+\frac{1}{m_{2}}+...+\frac{1}{m_{k}}\right) <
\pi ^{-1960},
\end{equation*}
where $\pi $ is the ratio between the circumference
of a circle and its diameter.

\bigskip

\textbf{Problem 4.} The set
$\left\{ 1;\;2;\;...;\;49\right\} $ is subdivided
into three subsets. Show that at least one of these
three subsets contains three different numbers $a,$
$b,$ $c$ such that $a+b=c.$

\bigskip

\underline{\textbf{Soviet Union}}

\bigskip

\textbf{Problem 1.} Given an $m\times n$ matrix
$\left\| p_{ij}\right\| $ whose elements are real
numbers. For every
$i\in \left\{1;\;2;\;...;\;m\right\} ,$ let
\begin{equation}
a_{i}=\dsum\limits_{j=1}^{n}p_{ij},  \tag{(1)}
\end{equation}
and for every
$j\in \left\{ 1;\;2;\;...;\;n\right\} ,$ let
\begin{equation}
b_{j}=\dsum\limits_{i=1}^{n}p_{ij}.  \tag{(2)}
\end{equation}

\textit{Rounding} a non-integer number will mean
replacing it by one of the two next integers (i. e.
replacing it either by its floor or by its ceiling).

Prove that one can round the numbers $a_{i},$
$b_{j},$ $p_{ij}$ in such a way that the equations
(1) and (2) remain valid.

\bigskip

\textbf{Problem 2.} A (not necessarily regular)
tetrahedron is inscribed into a sphere with radius
$1$ such that the center of the sphere lies inside
the tetrahedron.

Prove that the sum of the edgelengths of the
tetrahedron is greater than $6.$

\bigskip

\textbf{Problem 3.}

Let $x_{1},$ $x_{2},$ $...,$ $x_{k}$ be the
nontrivial divisors of a natural number $n,$
arranged such that $x_{1}<x_{2}<...<x_{k}.$ Find
all numbers $n$ such that $x_{5}^{2}+x_{6}^{2}-1=n.$

\textit{Note.} The ''nontrivial'' divisors of the
number $n$ are all natural divisors of $n$ except
for $1.$

\bigskip

\textbf{Problem 4.}

From the medians of an acute-angled triangle, we
form another triangle. Let $R$ and $R_{m}$ be the
circumradii of the first and the second triangles,
respectively.

Prove that $R_{m}>\frac{5}{6}R.$

\bigskip

\textbf{Problem 5.}

In the Martian language, each finite chain of Latin
letters is considered to be a word.

The Martian dictionary consists of infinitely many
tomes: The first tome contains all one-letter words;
the second tome contains all two-letter words; etc..
All words are numbered in lexicographical order, and
the numeration in each tome is continued in the next
tome.

Find the word whose number equals the sum of the
numbers of the words \textit{Prague},
\textit{Olympiad}, \textit{Mathematics}.[/code]

Thanks for translating the problems, Darij.  :)

[b][size=150]IMO 1984 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012303#p2012303]1. (FRA 1)[/url] Find all solutions of the following system of $n$ equations in $n$ variables:
\[\begin{array}{c}\  x_1|x_1| - (x_1 - a)|x_1 - a| = x_2|x_2|,x_2|x_2| - (x_2 - a)|x_2 - a| = x_3|x_3|,\ \vdots \ x_n|x_n| - (x_n - a)|x_n - a| = x_1|x_1|\end{array}\]
where $a$ is a given number.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012306#p2012306]2. (CAN 2)[/url] Prove:

(a) There are infinitely many triples of positive integers $m, n, p$ such that $4mn - m- n = p^2 - 1.$

(b) There are no positive integers $m, n, p$ such that $4mn - m- n = p^2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012308#p2012308]3. (USS 3)[/url] Find all positive integers $n$ such that
\[n=d_6^2+d_7^2-1,\]
where $1 = d_1 < d_2 < \cdots < d_k = n$ are all positive divisors of the number $n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366611&sid=c1de64b9288e42ff40d79056f6961512#p366611]4. (MON 1)[/url] Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:
\[ n-3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n+1\over 2}\Bigr]-2,\]
where $ [x]$ denotes the greatest integer not exceeding $ x$.[IMO Problem 5]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366604&sid=c1de64b9288e42ff40d79056f6961512#p366604]5. (FRG 1)[/url] Prove that $0\le yz+zx+xy-2xyz\le{7\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.[IMO Problem 1]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012311#p2012311]6. (CAN 3)[/url] Let $c$ be a positive integer. The sequence $\{f_n\}$ is defined as follows:
\[f_1 = 1, f_2 = c, f_{n+1} = 2f_n - f_{n-1} + 2 \quad  (n \geq 2).\]
Show that for each $k \in \mathbb N$ there exists $r \in \mathbb N$  such that $f_kf_{k+1}= f_r.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012317#p2012317]7. (FRG 5)[/url] (a) Decide whether the fields of the $8 \times 8$ chessboard can be numbered by the numbers $1, 2, \dots , 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms

[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28446[/img][/list]

is divisible by four.

(b) Solve the analogous problem for

[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28447[/img][/list]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366609&sid=c1de64b9288e42ff40d79056f6961512#p366609]8. (ROM 2)[/url] Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$, where $\angle AOX$ is measured in radians in the range $[0,2\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$. [IMO Problem 3]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012318#p2012318]9. (POL 2)[/url] Let $a, b, c$ be positive numbers with $\sqrt a +\sqrt b +\sqrt c = \frac{\sqrt 3}{2}$. Prove that the system of equations
\[\sqrt{y-a}+\sqrt{z-a}=1,\] \[\sqrt{z-b}+\sqrt{x-b}=1,\] \[\sqrt{x-c}+\sqrt{y-c}=1\]
has exactly one solution $(x, y, z)$ in real numbers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012322#p2012322]10. (GBR 1)[/url] Prove that the product of five consecutive positive integers cannot be the square of an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012324#p2012324]11. (CAN 1)[/url] Let $n$ be a positive integer and $a_1, a_2, \dots , a_{2n}$ mutually distinct integers. Find all integers $x$ satisfying
\[(x - a_1) \cdot (x - a_2) \cdots (x - a_{2n}) = (-1)^n(n!)^2.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366605&sid=c1de64b9288e42ff40d79056f6961512#p366605]12. (NET 1)[/url] Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.[IMO Problem 2]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012327#p2012327]13. (BUL 5)[/url] Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume $1$ does not exceed $\frac{2}{3 \pi}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366610&sid=c1de64b9288e42ff40d79056f6961512#p366610]14. (ROM 5)[/url] Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel. [IMO Problem 4]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012329#p2012329]15. (LUX 2)[/url] Angles of a given triangle $ABC$ are all smaller than $120^\circ$. Equilateral triangles $AFB, BDC$ and $CEA$ are constructed in the exterior of $ABC$.

(a) Prove that the lines $AD, BE$, and $CF$ pass through one point $S.$

(b) Prove that $SD + SE + SF = 2(SA + SB + SC).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=165811&sid=c1de64b9288e42ff40d79056f6961512#p165811]16. (POL 1)[/url]Let $a,b,c,d$ be odd integers such that $0<a<b<c<d$ and $ad=bc$. Prove that if $a+d=2^k$ and $b+c=2^m$ for some integers $k$ and $m$, then $a=1$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012330#p2012330]17. (FRG 3)[/url]In a permutation $(x_1, x_2, \dots , x_n)$ of the set $1, 2, \dots , n$ we call a pair $(x_i, x_j )$ discordant if $i < j$ and $x_i > x_j$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012331#p2012331]18. (USA 5)[/url] Inside triangle $ABC$ there are three circles $k_1, k_2, k_3$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_1, k_2, k_3$ are $1, 4$, and $9$. Determine the radius of $k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012334#p2012334]19. (CAN 5)[/url] The triangular array $(a_{n,k})$ of numbers is given by $a_{n,1} = \frac 1n$, for $n = 1, 2, \dots , a_{n,k+1} = a_{n-1,k} - a_{n,k}$, for $1 \leq k \leq n - 1$. Find the harmonic mean of the $1985^{th}$ row.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012335#p2012335]20. (USA 2)[/url] Determine all pairs $(a, b)$ of positive real numbers with $a \neq 1$ such that
\[\log_a b < \log_{a+1} (b + 1).\]

I have compiled the IMO longlist problems of the year $1984$. I request any moderator to update the contests page of this

[b][size=150]IMO 1984 Longlisted problems (Posted in Forums) :[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043827#p2043827]Problem 1:[/url] $(AUS 1)$ The fraction $\frac{3}{10}$ can be written as the sum of two positive fractions with numerator $1$ as follows: $\frac{3}{10} =\frac{1}{5}+\frac{1}{10}$ and also $\frac{3}{10}=\frac{1}{4}+\frac{1}{20}$. There are the only two ways in which this can be done. In how many ways can $\frac{3}{1984}$ be written as the sum of two positive fractions with numerator $1$?
Is there a positive integer $n,$ not divisible by $3$, such that $\frac{3}{n}$ can be written as the sum of two positive fractions with numerator $1$ in exactly $1984$ ways?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045422#p2045422]Problem 2:[/url] $(AUS 2)$ Given a regular convex $2m$- sided polygon $P$, show that there is a $2m$-sided polygon $\pi$ with the same vertices as $P$ (but in different order) such that $\pi$ has exactly one pair of parallel sides.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045442#p2045442]Problem 3:[/url] $(AUS 3)$ The opposite sides of the reentrant hexagon $AFBDCE$ intersect at the points $K,L,M$ (as shown in the figure). It is given that $AL = AM = a, BM = BK = b$, $CK = CL = c, LD = DM = d, ME = EK = e, FK = FL = f$.
[img]http://imgur.com/LUFUh.png[/img]
$(a)$ Given length $a$ and the three angles $\alpha, \beta$ and $\gamma$ at the vertices $A, B,$ and $C,$ respectively, satisfying the condition $\alpha+\beta+\gamma<180^{\circ}$, show that all the angles and sides of the hexagon are thereby uniquely determined.
$(b)$ Prove that
\[\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d}\]
Easier version of $(b)$. Prove that
\[(a + f)(b + d)(c + e)= (a + e)(b + f)(c + d)\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047656#p2047656]Problem 4:[/url] $(BEL 1)$ Given a triangle $ABC$, three equilateral triangles $AEB, BFC$, and $CGA$ are constructed in the exterior of $ABC$. Prove that:
$(a) CE = AF = BG$;
$(b) CE, AF$, and $BG$ have a common point.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047661#p2047661]Problem 5:[/url] $(BEL 2)$ For a real number $x$, let $[x]$ denote the greatest integer not exceeding $x$. If $m \ge 3$, prove that
\[\left[\frac{m(m+1)}{2(2m-1)}\right]=\left[\frac{m+1}{4}\right]\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2028585#p2028585]Problem 6:[/url] $(BEL 3)$ Let $P,Q,R$ be the polynomials with real or complex coefficients such that at least one of them is not constant. If $P^n+Q^n+R^n = 0$, prove that $n < 3.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047664#p2047664]Problem 7:[/url] $(BUL 1)$ Prove that for any natural number $n$, the number $\dbinom{2n}{n}$ divides the least common multiple of the numbers $1, 2,\cdots, 2n -1, 2n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047667#p2047667]Problem 8:[/url] $(BUL 2)$ In the plane of a given triangle $A_1A_2A_3$ determine (with proof) a straight line $l$ such that the sum of the distances from $A_1, A_2$, and $A_3$ to $l$ is the least possible.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2045507#p2045507]Problem 9:[/url] $(BUL 3)$ The circle inscribed in the triangle $A_1A_2A_3$ is tangent to its sides $A_1A_2, A_2A_3, A_3A_1$ at points $T_1, T_2, T_3$, respectively. Denote by $M_1, M_2, M_3$ the midpoints of the segments $A_2A_3, A_3A_1, A_1A_2$, respectively. Prove that the perpendiculars through the points $M_1, M_2, M_3$ to the lines $T_2T_3, T_3T_1, T_1T_2$ meet at one point.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047671#p2047671]Problem 10:[/url] $(BUL 4)$ Assume that the bisecting plane of the dihedral angle at edge $AB$ of the tetrahedron $ABCD$ meets the edge $CD$ at point $E$. Denote by $S_1, S_2, S_3$, respectively the areas of the triangles $ABC, ABE$, and $ABD$. Prove that no tetrahedron exists for which $S_1, S_2, S_3$ (in this order) form an arithmetic or geometric progression.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012327&#p2012327]Problem 11:[/url] $(BUL 5)$ Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume $1$ does not exceed $\frac{2}{3 \pi}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012324&#p2012324]Problem 12:[/url] $(CAN 1)$ Let $n$ be a positive integer and $a_1, a_2, \dots , a_{2n}$ mutually distinct integers. Find all integers $x$ satisfying
\[(x - a_1) \cdot (x - a_2) \cdots (x - a_{2n}) = (-1)^n(n!)^2.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012306&#p2012306]Problem 13:[/url] $(CAN 2)$ Prove:

(a) There are infinitely many triples of positive integers $m, n, p$ such that $4mn - m- n = p^2 - 1.$

(b) There are no positive integers $m, n, p$ such that $4mn - m- n = p^2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012311&#p2012311]Problem 14:[/url] $(CAN 3)$ Let $c$ be a positive integer. The sequence $\{f_n\}$ is defined as follows:
\[f_1 = 1, f_2 = c, f_{n+1} = 2f_n - f_{n-1} + 2 \quad  (n \geq 2).\]
Show that for each $k \in \mathbb N$ there exists $r \in \mathbb N$  such that $f_kf_{k+1}= f_r.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044164#p2044164]Problem 15:[/url] $(CAN 4)$ Consider all the sums of the form
\[\displaystyle\sum_{k=1}^{1985} e_kk^5=\pm 1^5\pm 2^5\pm\cdots\pm1985^5\]
where $e_k=\pm 1$. What is the smallest nonnegative value attained by a sum of this type?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012334&#p2012334]Problem 16:[/url] $(CAN 5)$ The harmonic table is a triangular array:

$1$

$\frac 12 \qquad \frac 12$

$\frac 13 \qquad \frac 16 \qquad \frac 13$

$\frac 14 \qquad \frac 1{12} \qquad \frac 1{12} \qquad \frac 14$

Where $a_{n,1} = \frac 1n$ and $a_{n,k+1} = a_{n-1,k} - a_{n,k}$ for $1 \leq k \leq n-1.$ Find the harmonic mean of the $1985^{th}$ row.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012303&#p2012303]Problem 17:[/url]$(FRA 1)$ Find all solutions of the following system of $n$ equations in $n$ variables:
\[\begin{array}{c}\  x_1|x_1| - (x_1 - a)|x_1 - a| = x_2|x_2|,x_2|x_2| - (x_2 - a)|x_2 - a| = x_3|x_3|,\ \vdots \ x_n|x_n| - (x_n - a)|x_n - a| = x_1|x_1|\end{array}\]
where $a$ is a given number.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048045#p2048045]Problem 18:[/url] $(FRA 2)$ Let $c$ be the inscribed circle of the triangle $ABC$, $d$ a line tangent to $c$ which does not pass through the vertices of triangle $ABC$. Prove the existence of points $A_1,B_1, C_1$, respectively, on the lines $BC,CA,AB$ satisfying the following two properties:
$(i)$ Lines $AA_1,BB_1$, and $CC_1$ are parallel.
$(ii)$ Lines $AA_1,BB_1$, and $CC_1$ meet $d$ respectively at points $A' ,B'$, and $C'$ such that
\[\frac{A'A_1}{A' A}=\frac{B'B_1}{B 'B}=\frac{C'C_1}{C'C}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044173#p2044173]Problem 19:[/url] $(FRA 3)$ Let $ABC$ be an isosceles triangle with right angle at point $A$. Find the minimum of the function $F$ given by $F(M) = BM +CM-\sqrt{3}AM$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366604&#p366604]Problem 20:[/url] $(FRG 1)$Prove that $0\le yz+zx+xy-2xyz\le{7\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043844#p2043844]Problem 21:[/url] $(FRG 2)$
$(1)$ Start with $a$ white balls and $b$ black balls.
$(2)$ Draw one ball at random.
$(3)$ If the ball is white, then stop. Otherwise, add two black balls and go to step $2$.
Let $S$ be the number of draws before the process terminates. For the cases $a = b = 1$ and $a = b = 2$ only, find $a_n = P(S = n), b_n = P(S \le n), \lim_{n\to\infty} b_n$, and the expectation value of the number of balls drawn: $E(S) =\displaystyle\sum_{n\ge1} na_n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012330&#p2012330]Problem 22:[/url] $(FRG 3)$ In a permutation $(x_1, x_2, \dots , x_n)$ of the set $1, 2, \dots , n$ we call a pair $(x_i, x_j )$ discordant if $i < j$ and $x_i > x_j$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043846#p2043846]Problem 23:[/url] $(FRG 4)$ A $2\times 2\times 12$ box fixed in space is to be filled with twenty-four $1 \times 1 \times 2$ bricks. In how many ways can this be done?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012317&#p2012317]Problem 24:[/url] $(FRG 5)$ (a) Decide whether the fields of the $8 \times 8$ chessboard can be numbered by the numbers $1, 2, \dots , 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms

[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28446[/img][/list]

is divisible by four.

(b) Solve the analogous problem for

[list][img]http://www.artofproblemsolving.com/Forum/download/file.php?id=28447[/img][/list]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012322&#p2012322]Problem 25:[/url] $(GBR 1)$ Prove that the product of five consecutive positive integers cannot be the square of an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048049#p2048049]Problem 26:[/url] $(GBR 2)$ A cylindrical container has height $6 cm$ and radius $4 cm$. It rests on a circular hoop, also of radius $4 cm$, fixed in a horizontal plane with its axis vertical and with each circular rim of the cylinder touching the hoop at two points.
The cylinder is now moved so that each of its circular rims still touches the hoop in two points. Find with proof the locus of one of the cylinder’s vertical ends.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047673#p2047673]Problem 27:[/url] $(GBR 3)$ The function $f(n)$ is defined on the nonnegative integers $n$ by: $f(0) = 0, f(1) = 1$, and
\[f(n) = f\left(n -\frac{1}{2}m(m - 1)\right)-f\left(\frac{1}{2}m(m+ 1)-n\right)\]
for $\frac{1}{2}m(m - 1) < n \le \frac{1}{2}m(m+ 1), m \ge 2$. Find the smallest integer $n$ for which $f(n) = 5$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048055#p2048055]Problem 28:[/url] $(GBR 4)$ A “number triangle” $(t_{n, k}) (0 \le k \le n)$ is defined by $t_{n,0} = t_{n,n} = 1 (n \ge 0),$
\[t_{n+1,m} =(2 -\sqrt{3})^mt_{n,m} +(2 +\sqrt{3})^{n-m+1}t_{n,m-1} \quad (1 \le m \le n)\]
Prove that all $t_{n,m}$ are integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044215#p2044215]Problem 29:[/url] $(GDR 1)$ Let $S_n = \{1, \cdots, n\}$ and let $f$ be a function that maps every subset of $S_n$ into a positive real number and satisfies the following condition: For all $A \subseteq S_n$ and $x, y \in S_n, x \neq y, f(A \cup \{x\})f(A \cup \{y\}) \le f(A \cup \{x, y\})f(A)$. Prove that for all $A,B \subseteq S_n$ the following inequality holds:
\[f(A) \cdot f(B) \le f(A \cup B) \cdot f(A \cap B)\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046274#p2046274]Problem 30:[/url] $(GDR 2)$ Decide whether it is possible to color the $1984$ natural numbers $1, 2, 3, \cdots, 1984$ using $15$ colors so that no geometric sequence of length $3$ of the same color exists.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048065#p2048065]Problem 31:[/url] $(LUX 1)$ Let $f_1(x) = x^3+a_1x^2+b_1x+c_1 = 0$ be an equation with three positive roots $\alpha>\beta>\gamma > 0$. From the equation $f_1(x) = 0$, one constructs the equation $f_2(x) = x^3 +a_2x^2 +b_2x+c_2 = x(x+b_1)^2 -(a_1x+c_1)^2 = 0$. Continuing this process, we get equations $f_3,\cdots, f_n$. Prove that
\[\lim_{n\to\infty}\sqrt[2^{n-1}]{-a_n} = \alpha\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012329&#p2012329]Problem 32:[/url] $(LUX 2)$ Angles of a given triangle $ABC$ are all smaller than $120^\circ$. Equilateral triangles $AFB, BDC$ and $CEA$ are constructed in the exterior of $ABC$.

(a) Prove that the lines $AD, BE$, and $CF$ pass through one point $S.$

(b) Prove that $SD + SE + SF = 2(SA + SB + SC).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366611&#p366611]Problem 33:[/url] $(MON 1)$ Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:
\[ n-3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n+1\over 2}\Bigr]-2,\]
where $ [x]$ denotes the greatest integer not exceeding $ x$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043850#p2043850]Problem 34:[/url] $(MON 2)$ One country has $n$ cities and every two of them are linked by a railroad. A railway worker should travel by train exactly once through the entire railroad system (reaching each city exactly once). If it is impossible for worker to travel by train between two cities, he can travel by plane. What is the minimal number of flights that the worker will have to use?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044224#p2044224]Problem 35:[/url] $(MON 3)$ Prove that there exist distinct natural numbers $m_1,m_2, \cdots , m_k$ satisfying the conditions
\[\pi^{-1984}<25-\left(\frac{1}{m_1}+\frac{1}{m_2}+\cdots+\frac{1}{m_k}\right)<\pi^{-1960}\]
where $\pi$ is the ratio between a circle and its diameter.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046271#p2046271]Problem 36:[/url] $(MON 4)$ The set $\{1, 2, \cdots, 49\}$ is divided into three subsets. Prove that at least one of these subsets contains three different numbers $a, b, c$ such that $a + b = c$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048074#p2048074]Problem 37:[/url] $(MOR 1)$ $(MOR 1)$ Denote by $[x]$ the greatest integer not exceeding $x$. For all real $k > 1$, define two sequences:
\[a_n(k) = [nk]\text{ and  } b_n(k) =\left[\frac{nk}{k - 1}\right]\]
If $A(k) = \{a_n(k) : n \in\mathbb{N}\}$ and $B(k) = \{b_n(k) : n \in \mathbb{N}\}$, prove that $A(k)$ and $B(k)$ form a partition of $\mathbb{N}$ if and only if $k$ is irrational.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046270#p2046270]Problem 38:[/url] $(MOR 2)$ Determine all continuous functions $f$ such that
\[f(x + y)f(x - y) = (f(x)f(y))^2\:\:\:\: \forall(x, y) \in\mathbb{R}^2\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048083#p2048083]Problem 39:[/url] $(MOR 3)$ Let $ABC$ be an isosceles triangle, $AB = AC, \angle A = 20^{\circ}$. Let $D$ be a point on $AB$, and $E$ a point on $AC$ such that $\angle ACD = 20^{\circ}$ and $\angle ABE = 30^{\circ}$. What is the measure of the angle $\angle CDE$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366605&#p366605]Problem 40:[/url] $(NET 1)$ Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049190#p2049190]Problem 41:[/url] $(NET 2)$ Determine positive integers $p, q$, and $r$ such that the diagonal of a block consisting of $p\times q\times r$ unit cubes passes through exactly $1984$ of the unit cubes, while its length is minimal. (The diagonal is said to pass through a unit cube if it has more than one point in common with the unit cube.)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049194#p2049194]Problem 42:[/url] $(NET 3)$ Triangle $ABC$ is given for which $BC = AC + \frac{1}{2}AB$. The point $P$ divides $AB$ such that $BP : PA = 1 : 3$. Prove that $\angle CAP = 2\angle CPA$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=165811&#p165811]Problem 43:[/url] $(POL 1)$ Let $a,b,c,d$ be odd integers such that $0<a<b<c<d$ and $ad=bc$. Prove that if $a+d=2^k$ and $b+c=2^m$ for some integers $k$ and $m$, then $a=1$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=665420#p665420]Problem 44:[/url] $(POL 2)$ Let $a,b,c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}= \frac{\sqrt{3}}{2}$
Prove that the system of equations
\[\sqrt{y-a}+\sqrt{z-a}=1\]
\[\sqrt{z-b}+\sqrt{x-b}=1\]
\[\sqrt{x-c}+\sqrt{y-c}=1\]
has exactly one solution $(x,y,z)$ in real numbers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049147#p2049147]Problem 45:[/url] $(POL 3)$ Let $X$ be an arbitrary nonempty set contained in the plane and let sets $A_1, A_2,\cdots,A_m$ and $B_1, B_2,\cdots, B_n$ be its images under parallel translations. Let us suppose that $A_1\cup A_2 \cup \cdots\cup A_m \subset B_1 \cup B_2 \cup\cdots\cup B_n$ and that the sets $A_1, A_2,\cdots,A_m$ are disjoint. Prove that $m \le n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048093#p2048093]Problem 46:[/url] $(ROM 1)$ Let $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ be two sequences of natural numbers such that $a_{n+1} = na_n + 1, b_{n+1} = nb_n - 1$ for every $n\ge 1$. Show that these two sequences can have only a finite number of terms in common.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366609&#p366609]Problem 47:[/url] $(ROM 2)$ Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$, where $\angle AOX$ is measured in radians in the range $[0,2\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2047724#p2047724]Problem 48:[/url] $(ROM 3)$ Let $ABC$ be a triangle with interior angle bisectors $AA_1, BB_1, CC_1$ and incenter $I$. If $\sigma[IA_1B] + \sigma[IB_1C] + \sigma[IC_1A] = \frac{1}{2}\sigma[ABC]$, where $\sigma[ABC]$ denotes the area of $ABC$, show that $ABC$ is isosceles.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048107#p2048107]Problem 49:[/url] $(ROM 4)$ Let $n > 1$ and $x_i \in \mathbb{R}$ for $i = 1,\cdots, n$. Set
\[S_k = x_1^k+ x^k_2+\cdots+ x^k_n\]
for $k \ge 1$. If $S_1 = S_2 =\cdots= S_{n+1}$, show that $x_i \in \{0, 1\}$ for every $i = 1, 2,\cdots, n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366610&#p366610]Problem 50:[/url] $(ROM 5)$ Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049199#p2049199]Problem 51:[/url] $(SPA 1)$ Two cyclists leave simultaneously a point $P$ in a circular runway with constant velocities $v_1, v_2 (v_1 > v_2)$ and in the same sense. A pedestrian leaves $P$ at the same time, moving with velocity $v_3 = \frac{v_1+v_2}{12}$ . If the pedestrian and the cyclists move in opposite directions, the pedestrian meets the second cyclist $91$ seconds after he meets the first. If the pedestrian moves in the same direction as the cyclists, the first cyclist overtakes him $187$ seconds before the second does. Find the point where the first cyclist overtakes the second cyclist the first time.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046268#p2046268]Problem 52:[/url] $(SPA 2)$ Construct a scalene triangle such that
\[a(\tan B - \tan C) = b(\tan A - \tan C)\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2048109#p2048109]Problem 53:[/url] $(SPA 3)$ Find a sequence of natural numbers $a_i$ such that $a_i = \displaystyle\sum_{r=1}^{i+4} d_r$, where $d_r \neq d_s$ for $r \neq s$ and $d_r$ divides $a_i$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049154#p2049154]Problem 54:[/url] $(SPA 4)$ Let $P$ be a convex planar polygon with equal angles. Let $l_1,\cdots, l_n$ be its sides. Show that a necessary and sufficient condition for $P$ to be regular is that the sum of the ratios $\frac{li}{l_{i+1}} (i = 1,\cdots, n; l_{n+1}= l_1)$ equals the number of sides.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049156#p2049156]Problem 55:[/url] $(SPA 5)$ Let $a, b, c$ be natural numbers such that $a+b+c = 2pq(p^{30}+q^{30}), p > q$ being two given positive integers.
$(a)$ Prove that $k = a^3 + b^3 + c^3$ is not a prime number.
$(b)$ Prove that if $a\cdot b\cdot c$ is maximum, then $1984$ divides $k$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049160#p2049160]Problem 56:[/url] $(SWE 1)$ Let $a, b, c$ be nonnegative integers such that $a \le b \le c, 2b \neq a + c$ and $\frac{a+b+c}{3}$ is an integer. Is it possible to find three nonnegative integers $d, e$, and $f$ such that $d \le e \le f, f \neq c$, and such that $a^2+b^2+c^2 = d^2 + e^2 + f^2$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043852#p2043852]Problem 57:[/url] $(SWE 2)$ Let $a, b, c, d$ be a permutation of the numbers $1, 9, 8,4$ and let $n = (10a + b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046264#p2046264]Problem 58:[/url] $(SWE 3)$ Let $(a_n)_1^{\infty}$ be a sequence such that $a_n \le a_{n+m} \le a_n + a_m$ for all positive integers $n$ and $m$. Prove that $\frac{a_n}{n}$ has a limit as $n$ approaches infinity.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046262#p2046262]Problem 59:[/url] $(USA 1)$ Determine the smallest positive integer m such that $529^n+m\cdot 132^n$ is divisible by $262417$ for all odd positive integers $n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012335#p2012335]Problem 60:[/url] $(USA 2)$ Determine all pairs $(a, b)$ of positive real numbers with $a \neq 1$ such that
\[\log_a b < \log_{a+1} (b + 1).\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043853#p2043853]Problem 61:[/url] $(USA 3)$ A fair coin is tossed repeatedly until there is a run of an odd number of heads followed by a tail. Determine the expected number of tosses.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049209#p2049209]Problem 62:[/url] $(USA 4)$ From a point $P$ exterior to a circle $K$, two rays are drawn intersecting $K$ in the respective pairs of points $A, A'$ and $B,B' $. For any other pair of points $C, C'$ on $K$, let $D$ be the point of intersection of the circumcircles of triangles $PAC$ and $PB'C'$ other than point $P$. Similarly, let $D'$ be the point of intersection of the circumcircles of triangles $PA'C'$ and $PBC$ other than point $P$. Prove that the points $P, D$, and $D'$ are collinear.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012331&#p2012331]Problem 63:[/url] $(USA 5)$ Inside triangle $ABC$ there are three circles $k_1, k_2, k_3$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_1, k_2, k_3$ are $1, 4$, and $9$. Determine the radius of $k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2049210#p2049210]Problem 64:[/url] $(USS 1)$ For a matrix $(p_{ij})$ of the format $m\times n$ with real entries, set
\[a_i =\displaystyle\sum_{j=1}^n p_{ij}\text{  for  }i = 1,\cdots,m\text{  and  }b_j =\displaystyle\sum_{i=1}^m p_{ij}\text{  for  }j = 1, . . . , n\longrightarrow(1)\]
By integering a real number, we mean replacing the number with the integer closest to it. Prove that integering the numbers $a_i, b_j, p_{ij}$ can be done in such a way that $(1)$ still holds.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2046259#p2046259]Problem 65:[/url] $(USS 2)$ A tetrahedron is inscribed in a sphere of radius $1$ such that the center of the sphere is inside the tetrahedron. Prove that the sum of lengths of all edges of the tetrahedron is greater than 6.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=141613#p141613]Problem 66:[/url] $(USS 3)$ Let $1=d_1<d_2<....<d_k=n$ be all different divisors of positive integer n written in ascending order. Determine all n such that:
\[d_6^{2} +d_7^{2} - 1=n\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2044228#p2044228]Problem 67:[/url] $(USS 4)$ With the medians of an acute-angled triangle another triangle is constructed. If $R$ and $R_m$ are the radii of the circles circumscribed about the first and the second triangle, respectively, prove that
\[R_m>\frac{5}{6}R\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2043854#p2043854]Problem 68:[/url] $(USS 5)$ In the Martian language every finite sequence of letters of the Latin alphabet letters is a word. The publisher “Martian Words” makes a collection of all words in many volumes. In the first volume there are only one-letter words, in the second, two-letter words, etc., and the numeration of the words in each of the volumes continues the numeration of the previous volume. Find the word whose numeration is equal to the sum of numerations of the words Prague, Olympiad, Mathematics.


I know the most obvious answers - MIT, Harvard, and all of the other ivy league schools in the area. Unfortunately, I'm sure that those are, for the most part, beyond me financially, not to mention I would have a slim chance of being accepted. So I have two questions:

1) I know that these are the "dream" schools. But really, is it that big of a deal if you can't get into them? (in terms of jobs after college, expected salaries, etc). And if the answer to this is "yes", then how hard is it to transfer into them for graduate school?  :lol:

2) Going back to my original subject, what are some other schools in the New England area that are good for math / possibly computer science related majors?

I really am passionate about math, but seeing as how it was somewhat of a late discovery (spring of my junior year), I haven't had much of a chance to build an impressive resume to qualify for the more competitive schools.

What are your thoughts on liberal arts schools?  They're programs in math and computer science don't match up to the likes of Harvard and MIT, but since the schools are so small, you'll get a lot of close attention in those departments.  I was thinking Williams and Amherst.

Make sure that these schools really are beyond you financially. Harvard is providing quite generous financial aid which might make it cheaper to go to Harvard than other schools.

http://www.hno.harvard.edu/gazette/2007/12.13/99-finaid.html

It's a misconception to think that any school is beyond you financially.  Just apply and wait for the aid package.  If you get into all those really good schools, presumably at least one will give you a nice deal.  And even if you have a pretty nice aid package, you can ask for more.  I was offered 32k/year for 4 years, but I said "I think I need more"  and they gave me 35k/year for 4 years instead.  $ \$$12000 is quite a bit of money.

1. No, it doesn't matter as much as your abilities matter.

2. Rensselaer Polytechnic Institute.  Many people here are attending because they ot rejected by MIT and CalTech.  I'd argue it's the third best tech school.


geometry
trigonometry
inequalities
triangle inequality
trig identities
Law of Sines
geometry unsolved

The incircle of $ ABC$ is tangent to the sides $ BC$, $ CA$ and $ AB$ respectively at the points: $ A_1$, $ B_1$ and $ C_1$. It is also known that the line $ A_1B_1$ pass through the middle point of the segment $ CC_1$. Find the angles of the triangle if their $ \text{sin}$'s form an arithmetic progression.

My solution is quite unelegant. I hope there is a better approach.
Write $ a\equal{}\left|BC\right|$, $ b\equal{}\left|CA\right|$, $ c\equal{}\left|AB\right|$ and $ \alpha\equal{}\angle CAB$, $ \beta\equal{}\angle ABC$, $ \gamma\equal{}\angle BCA$. W.l.o.g. $ \sin\alpha\leq \sin\beta$. Now there are two cases:
[b]1:[/b] $ \sin\gamma\leq \sin\alpha\leq \sin\beta$. By the law of sines and by the initial conditions $ \sin\beta\minus{}\sin\alpha\equal{}\sin\alpha\minus{}\sin\gamma\Leftrightarrow \minus{}2a\plus{}b\plus{}c\equal{}0$ holds. Since $ A_{1}B_{1}$ passes through the midpoint of $ CC_{1}$ $ C_{1}A_{1}B_{1}$ and $ A_{1}CB_{1}$ have the same altitudes with respect to $ A_{1}B_{1}$, hence the same area ($ A(\cdot)$ denotes the area):
$ A\left(ABC\right)\minus{}A\left(AC_{1}B_{1}\right)\minus{}A\left(C_{1}BA_{1}\right)\minus{}A\left(A_{1}CB_{1}\right)\equal{}A\left(A_{1}CB_{1}\right)$
$ \Leftrightarrow bc\sin\alpha\minus{}\left(\frac{\minus{}a\plus{}b\plus{}c}{2}\right)^{2}\sin\alpha\minus{}\left(\frac{a\minus{}b\plus{}c}{2}\right)^{2}\sin\beta\minus{}2\left(\frac{a\plus{}b\minus{}c}{2}\right)^{2}\sin\gamma\equal{}0$
$ \Leftrightarrow abc\minus{}a\left(\frac{\minus{}a\plus{}b\plus{}c}{2}\right)^{2}\minus{}b\left(\frac{a\minus{}b\plus{}c}{2}\right)^{2}\minus{}2c\left(\frac{a\plus{}b\minus{}c}{2}\right)^{2}\equal{}0$
$ \Leftrightarrow 0\equal{}ab\left(2a\minus{}b\right)\minus{}\frac{a^{3}}{4}\minus{}b\left(\frac{3a}{2}\minus{}b\right)^{2}\minus{}2\left(2a\minus{}b\right)\left(b\minus{}\frac{a}{2}\right)^{2}\equal{}\left(b\minus{}a\right)\left(2b\minus{}5a\right)\left(2b\minus{}a\right)/4$.
Because of $ a\leq b$ either $ a\equal{}b$ or $ 2b\minus{}5a\equal{}0$. In the first case the triangle is equilateral, in the second case the triangle inequality is not fulfilled.
[b]2:[/b]$ \sin\alpha\leq \sin\gamma\leq \sin\beta$. Analogously to case 1 we get:
$ 0\equal{}bc\left(2c\minus{}b\right)\minus{}\left(2c\minus{}b\right)\left(b\minus{}\frac{c}{2}\right)^{2}\minus{}b\left(\frac{3c}{2}\minus{}b\right)^{2}\minus{}\frac{c^{3}}{2}\equal{}\minus{}c\left(c\minus{}b\right)^{2}$.
Again $ ABC$ has to be equilateral.


A gold shipment consists of $ N > 1$ bars containing a total of $ 2^{100}$ gold atoms. The number of atoms in each bar is also a power of $ 2$ (which can vary from bar to bar). Show the bars can be loaded onto two ships so that each carries the same amount of gold. (What is the value of the shipment?)

We will prove this by induction after $ n \geq 1$, for a total of $ 2^n$ gold atoms. For $ n\equal{}1$ it is clear, since the only possibility is $ N\equal{}2$, two bars of one atom each. Assume the hypothesis proven up to $ n > 1$.
Let the $ N$ bars contain $ 2^{a_i}$ atoms, with $ \sum_{i\equal{}1}^N 2^{a_i} \equal{} 2^{n\plus{}1}$, and $ 0 \leq a_N \leq \cdots \leq a_1$.

If $ a_N > 0$, take $ b_i \equal{} a_i \minus{} 1$, so $ \sum_{i\equal{}1}^N 2^{b_i} \equal{} 2^{n}$. We can apply the induction hypothesis to obtain a partition $ I \cap J \equal{} \emptyset$, $ I \cup J \equal{} \{1,2,\ldots,N\}$ and $ \sum_{i \in I} 2^{b_i} \equal{} \sum_{j \in J} 2^{b_j} \equal{} 2^n$. Multiplying by $ 2$ proves that this is a valid partition for $ n\plus{}1$ too.

If $ a_N \equal{} 0$, let $ k$ be the least index such that $ a_k \equal{} 0$. Then $ N \minus{} k \plus{} 1 \equal{} 2M$ needs be even. Take $ a'_k \equal{} a'_{k\plus{}1} \equal{} \cdots \equal{} a'_{N\minus{}M} \equal{} 1$ ("pack" the one-atom bars in pairs) and $ a'_i \equal{} a_i$ for $ i \equal{} 1, 2, \ldots, k\minus{}1$, so $ \sum_{i\equal{}1}^{N\minus{}M} 2^{a'_i} \equal{} 2^{n\plus{}1}$. We find ourselves now in the previous case (with $ a'_{N\minus{}M} \equal{} 1 > 0$ and $ N\minus{}M > 1$), so we can find a valid partition.


quadratics
number theory solved
number theory

[color=green]Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.[/color]

Thanks very much guys

For large even $n$, $2^{n}$ is a square and it will not be "close" to another square.

For odd $n$, the LHS is $\equiv 2 \bmod 3$, not a quadratic residue.


How many rectangles (of any size and in any position) are there on a standard 8 x 8 checkerboard?

[hide]you can check ur solution here: http://www.projecteureka.org/problem/question/377[/hide]

There are [b]1296[/b] rectangles:
To choose the locations of the vertical sides, there are 9C2=36 ways to pick them.  The same goes for horizontal sides, so the total ways to make a rectangle is 36*36=1296.


[hide="da anwer"]
$-10+15-20+25=10$
    She made 10 dollars[/hide]

[quote="MCrawford"]A girl bought a dog for $\$10, sold it for $\$15, bought it back for $\$20, and finally sold it for $\$25.  Did the girl make or lose money, and how much did she make or lose?[/quote]

[hide]$-10+15-20+25=10$[/hide]


16 hours and 160 bottles of beer after the New Year's movie, Bobert decides to walk up the Leaning Tower of Treething with Caddlebot and drop him off. Every time Caddlebot hits the ground, he bounces to 1/4 of his previous height. In total, he moves 350 feet. How tall is the Leaning Tower of Treething?

[hide="Hmmm"]The LToT is x feet tall. The amount that Caddlebot moves after his original fall can be represented by $2\sum_{i=1}^{\infty} \left(\frac 14\right)^i\cdot x$, or at least I hope it can because otherwise my solution is wrong. So
$x+2\sum_{i=1}^{\infty} \left(\frac 14\right)^i\cdot x=350$
The first term of the infinite geometric series is x/4, and the common ratio is 1/4. So plugging into the formula for such a series,
$x+2\cdot\frac{x/4}{1-1/4}=350$
$x+2\cdot\frac x3=350$
$5x/3=350$
$x=210$
Is 210 feet the right answer? Probably not.
[/hide]
P.S. This is more Intermediate than Getting Started I think...

Yeah, I think summing an infinite geometric series is just across the line into (easy) Intermediate Forum level, so I'll move this.

No offense but does the problem has to involve FBI or CIA actions? (a.k.a. crime :? ) It's fine with me since it's fictional and funny but I'm just wondering. ;)

I must resort to chatspeak to express my reaction to what Silverfalcon said.

"wtf?"

Yes, it does involve the CIA. I paid the CIA agent Bobert five hundred dollars plus a cat poison bottle to push Caddlebot off of the Leaning Tower of Treething.

[hide]

Yep, it is 210 feet high.

[/hide]

He means, why do all of your problems involve people killing or trying to kill other people and animals for no good reason? You could just have easily had Bobert kick a basketball off the top of the LToT.  :P
I solved a problem involving an infinite geometric series correctly?! I'm going to go celebrate!  :lol:

Geez treething. I didn't mean for it to be violent to the point that it's reputation is gonna go down like this. I mean god your next problem gonna be like this:

"Frankinfueter shoots 15000000 AK - 47 rounds into Caddlebot, if the AK - 47's rate of fire is 300 rounds per sec how long does it take to shoot the rounds into Caddlebot?"

And that is EVIL. But I still hope that everyone catches on. So if you're posting a word problem please put in Albatross and friends. I'll soon have a name for all 26 letters of the alphabet. Also if we could get this in the AMC it would be so awesome.


Questions is : What would be the amount of compound interest on $8,000. invested for one year at 6%, compounded quarterly? Round your answer to the nearest dollar??????  :what?:

In order to do this, you need the compound interest formula.

This is: (if i remember it correctly)

P(1 + r/n) ^ nt

P = principal, or the initial amount invested
r = rate, in a decimal (e.g. 6% = .06)
n = number of years over which the investment occurrs
t = number of compoundings in one year. Note that quarterly = 4 compoundings/yr.

Knowing this, plugging the formula into your calculator should solve this.

[hide="Hint"]Let me give an example to illustrate the concept of compound interest. Say the initial amount is 1000 dollars, with 10% interest compounded monthly.

Since interest is compounded every month:
After 1 month, there would be $ 1000d + 1000d*10\% = 1100d$.
After 2 months, there would be $ 1100d + 1100d*10\% = 1210d$.

Basically, interest is compounded on the total amount of money at each point in time (including whatever interest has been earned so far), rather than in simple interest where only the initial amount counts.[/hide]

[hide="Solution"]There are 4 quarters in a year, so the interest of 6% would be compounded 4 times.

$ 1 + 6\% = 1.06$
\begin{align*}8000d*1.06^4 - 8000d & = 8000d*1.26247696 - 8000d \\
& = 0.26247696*8000d \\
& \approx\boxed{2100d}\end{align*}[/hide]

ZhangPeijin is correct, I misunderstood what the compound interest of 6% referred to. So since the 6% is yearly interest, if it's compounded quarterly then each quarter there would be 1.5% interest added, for 4 quarters. So, just evaluate $ 8000*1.015^4\minus{}8000$ to get the solution.


summer program
PROMYS
calculus
calculus computations

Can anyone check whether this proof has enough detail and is rigorous enough?

Prove that for any $a, b, c, d \in R$, $-(a-b) = b-a$. I know this sounds trivial, but the proof has to be detailed as possible.

Proof: $-(a-b) = (-1)(a+(-b)) = (-1)(a)+(-1)(-b) =-a+b = b-a$ by commutative field property.

Is this a reasonable proof?

2ndly, If $S_{1}, S_{2}$ are nonempty subsets of $R$ that are bounded from above, prove that $lub \{x+y: x \in S_{1}, y \in S_{2}\}= lub S_{1}+lub S_{2}$. lub stands for least upper bound.
Again here's my proof:

The conditions given imply that $x \le lub S_{1}$ and $y \le lub S_{2}$. Adding the two gives $x+y \le lub S_{1}+lub S_{2}$.

This almost seems correct, BUT, how can I show that $lub S_{1}+lub S_{2}$ is smaller than ANY OTHER upper bound (let's call it $n$)?

As always, help is appreciated.

[quote="bythecliff"]Proof: $-(a-b) = (-1)(a+(-b))$[/quote]

If you want a rigorous proof from axioms, this is unacceptable.  Prove this.

[quote="bythecliff"]$(-1)(a)+(-1)(-b) =-a+b$[/quote]

This also needs to be proven.  Namely, that $(-1)(-b) = b$.

[hide="A simpler rigorous proof"] We want to show that $a-b$ and $b-a$ are additive inverses.

$(a-b)+(b-a) = (a+(-b))+(b+(-a)) = \left( a+(-b+b) \right)+(-a) = \left( a+0 \right)+(-a) = a+(-a) = 0$

Before this is acceptable, we need to show that the additive identity is unique, and then that the additive inverse is unique.  This is left as an exercise, since it is very simple. [/hide]

OK, let me do the first proof again:

let's prove a simpler case of $-a = (-1)(a)$ for all $a \in R$. $(-1)(a)+a = a((-1)+1) = a(0) = 0$, so $(-1)(a)$ is a solution to the equation $x+a = 0$.

Also, $x+a = 0 \Rightarrow  x+a+(-a) = 0+(-a) \Rightarrow  x+0 = (-a) \Rightarrow  x =-a$.

So both $x = (-1)(a)$ and $x =-a$ are solutions to the equation $x+a = 0$, and because the solution is unique, $(-1)(a) = (-a)$. Hence, $-(a-b) = (-1)(a+(-b))$.

[quote="t0rajir0u"][quote="bythecliff"]and because the solution is unique[/quote]

Needs to be proven.  (You asked for rigor!  :P )[/quote]

Not so much that it needs to be proven as doesn't need to be there at all.  If $x = y$ and $x = z$, $y = z$ by transitivity of equality.

[quote="JBL"]Not so much that it needs to be proven as doesn't need to be there at all.  If $x = y$ and $x = z$, $y = z$ by transitivity of equality.[/quote]

No, no, his statement was that if $t+a = 0$ and $t+b = 0$ it follows that $a = b$.  To prove this statement rigorously you need to show that the additive identity is unique.   :P

This is super PROMYS-level picky, though.


at twelve oclock the second minut and hour hands all coincide. What is the next time, in hours, minutes and seconds(to the nearest second), the minute and hour hands will coincide

The hour hand coicides with the minute hand 11 times in 12 hours, therefore if we say the hour is $ h$, the time would be $ h$ hours $ h$ minutes and $ \frac{12}{11}h$ seconds. we want to find a time where $ \frac{12}{11}h$ is equal to $ h$. then we get $ \frac{1}{11}h=0$ so $ h=0$. that means the only time that the minute, second, and hour hand will coincide, will be 12 hours from 12:00.

[quote="dingzhou"]are you sure, it only asks when hour and minute not second.[/quote]
The question asks for the nearest second.

Dude next time you post a question use a search button (or you can just scroll down on this page)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157457

this post was on [b]this[/b] page (if you scroll down you will find it) and ive seen this question atleast 10 times here.

Some people might not have seen the question before and wouldn't know to search it... it's not like he's spamming or anything.

Anyways, the wording of the question should be improved.. but it asks for when just the minute and hour hands coincide next.
[hide]We can easily see that they will coincide after one o'clock. Let $ x$ equal the number of minutes that will pass before the minute and hour hands coincide. Since 12 minutes will pass before the hour hand moves one mark, we have
$ x=\frac{x}{12}+5$
$ x=\frac{60}{11}$ minutes $ \approx 5$ minutes $ 27$ seconds.
So the answer is 1 hour 5 minutes 27 seconds.[/hide]

isnt there a formula to this. like the $ n$th time past 12 that the hand will coincide will be $ \frac{12}{11}n$ because there are 12 hours and the hours hand makes 1 full circle when the minute hand makes 12 making it 12 11ths hours the first time they coincide.


Given a circle $(O_{k})$ and a triangle $ABC$.
Let the circle $(O_{k})$ meet the line $BC$ at $A_{1}$ and $A_{2}$, the line $CA$ at $B_{1}$ and $B_{2}$, and the line $AB$ at $C_{1}$ and $C_{2}$.
We construct the three circles with diameters $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$.

Prove that the radical center of these three circles is the isogonal conjugate of the center of $(O_{k})$ with respect to triangle $ABC$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16404 ; please tell me if you want the solution to be somewhat more detailed.

  darij


function
calculus
derivative
inequalities
algebra unsolved
algebra

$ f: \mathbb{R^{ \plus{} }}\rightarrow\mathbb{R}$

$ h(x) \equal{} f(x) \minus{} x^3$ and $ g(x) \equal{} f(x) \minus{} 3x$ are increasing in $ \mathbb{R}$

show that $ T(x) \equal{} f(x) \minus{} x^2 \minus{} x$ is incrising too

NB: $ f$ can be note derivative

What does 'greated' means?
Do you mean 'increasing'?

Pierre.

[hide="@Pierre"]Bsr mr.Pierre et desolé pour mon anglais;
je vx dire par greatest croissante j ai pas su la traduction exacte.
ps: f peux etre non derivable.
merci ;)[/hide]

Let $ a,b$ be two arbitrary reals with $ a > b$. We are given that

$ f(a) \minus{} f(b) > a^3 \minus{} b^3$
$ f(a) \minus{} f(b) > 3a \minus{} 3b$
and we want to show that
$ f(a) \minus{} f(b) > a^2 \plus{} a \minus{} b^2 \minus{} b$

Let $ C \equal{} \frac{f(a) \minus{} f(b)}{a \minus{} b}$ (note $ a \minus{} b > 0$). The given inequalities become
$ C > a^2 \plus{} ab \plus{} b^2$
$ C > 3$
and what we want is
$ C > a \plus{} b \plus{} 1$

If we show this holds true for $ a > b \geq 1$ and $ 1 \geq a > b$, then the result for the remaining cases follows by transitivity ($ f(a) > f(1) > f(b)$).

$ a > b \geq 1$: $ C > a^2 \plus{} ab \plus{} b^2 \equal{} (a\plus{}b)a \plus{} b^2 > (a\plus{}b)(1) \plus{} 1^2 \equal{} a \plus{} b \plus{} 1$
$ 1 \geq a > b$: $ C > 3 > a \plus{} b \plus{} 1$


I found this problem in The Art and Craft of Problem Solving:

Prove that for any positive integer n, there exists a positive multiple of n that only uses the digits 7 and 0.

Consider the remainders when $ 7, 77, 777, ....$ are divided by $ n$.

Thank you! That helped.

Full proof:

Consider the first $ n$ numbers of the sequence 7, 77, 777, 7777...

We have two cases:

1. One of them is a multiple of $ n$.
In this case we are done.

2. None of them are multiples of $ n$.
Then there are $ n\minus{}1$ possible remainders upon division by $ n$. By the pigeonhole principle, two of the numbers must have the same remainder. So their difference is a multiple of n. But their difference only contains the digits 7 and 0, so we are done.


im finishing my freshman year of high school at north ridge high school. already i have been asked about colleges. i know that i want to go to a christian college but i also want to leave home. the only christian colleges i know of are here in indiana or in the surrounding states. does anyone know of good christian colleges away from home?

This depends a lot on your definition of Christian in what kind of colleges you are looking for. I assume, because you are posting here on AoPS, that you might be interested in a college with a strong math or science program besides, but I'm not sure how Christian is Christian enough for your requirements. (My point of reference is going to a secular state university for my undergraduate degree, where I met plenty of very devout Christians from all over the world.) What colleges do you have in mind so far? That might give me an idea of colleges in other regions of the country with similar characteristics. Or just tell me in more detail what characteristics you are looking for.

Samford University. Its in Birmingham, Alabama. Where I live.