[quote="Dr Sonnhard Graubner"]hello, write the equation in the following form:
$ln\left(x^{2}+2x\right)=ln\left(x+1\right)^{2}$. From this we get:
$x^{2}+5x=x^{2}+2x+1$ with $x=\frac{1}{3}$.
Sonnhard.[/quote]

how did you cancel out the logs? what law did you use?

thanks for the help

[quote="Dr Sonnhard Graubner"]hello, i did use that $exp\left(ln(x)\right)=x$ for $x>0$.
Sonnhard.[/quote]

DO you mean?

$e^\left(ln(x)\right)$=$x$

[hide=" Step by step"]
First $2\ln(x+1)$ can be written $\ln(x+1)^{2}$ by the properties of logs.  Then $e^{\ln(x^{2}+5x)}= e^{\ln(x+1)^{2}}$.
Then we have
$x^{2}+5x = (x+1)^{2}$
$\implies x^{2}+5x = x^{2}+2x+1$
$\implies 3x = 1 \implies x = \frac{1}{3}$[/hide]

[quote="bos1234"]$ln(x^{2}+5x)=2ln(x+1)$

solve for x[/quote]

[hide="Full explanation"]Recall that $\ln(a)+\ln(b) = \ln(a\cdot b)$
[hide="If you want, here is a proof"]Let $x = \ln(a)$, $y = \ln(b)$, and $z = \ln(ab)$. We want to show that $x+y = z$.

From the definition of logs, $e^{x}= a$, $e^{y}= b$, $e^{z}= a b$. Now it is relatively obvious that $e^{x}\cdot e^{y}= e^{z}$. From a property of exponentials, $e^{x+y}= e^{z}$, so $x+y = z$, as desired.[/hide]

Let $a=b$ in that. Then:

$\ln(a)+\ln(a) = \ln(a^{2}$
$2\ln(a) = \ln(a^{2})$

Replace $a$ with $x+1$:
$2\ln(x+1) = \ln((x+1)^{2}) = \ln(x^{2}+2x+1)$

Now, we may substitute this equation into the given:
$\ln(x^{2}+5 x) = \ln(x^{2}+2x+1)$

Thus, $x^{2}+5x = x^{2}+2x+1$. So: $3x = 1 \Rightarrow \boxed{x = \frac{1}{3}}$.[/hide]

Then $e^{\ln(x^{2}+5x)}= e^{\ln(x+1)^{2}}$

I know that the law exists, but how can you make the logs indices? What permits us to do that?


algebra
polynomial
calculus
calculus computations

Let E be ther error when the Taylor Polynomial T(x) = x - [(x^3)/(3!)] is used to approximate f(x) = sinx at x = 5. Which of the following is true?

|E| < 0.0001 ******statement 1
0.0001 < |E| < 0.0003 *****statement 2
0.0003 < |E| < 0.005 *******statement 3
0.005 < |E| < 0.007 *******statement 4
0.07 < |E| ********statement 5

Shouldn't you be asking your calculator this question?

Anyway, the polynomial evaluates to -95/6, while sin is always between -1 and 1, so it's pretty clear that the error is enormous.

which would be satisfied by statement 5. Thank you.

Remark: There's of course no need for calculators:). Just look at the remainder (in Cauchy's or Lagrange's form) and estimate it.


[hide="answer"]one, two, three, four...so "S" for seven[/hide]

Try this one.

Fill in the missing term:

...G, M, k, h, D, 1, d, c, m, _, n...

[hide="Hint"]
If you didn't know, I could give you 52 tries and you probably couldn't guess it.
[/hide]

umm...[hide]Does it have to do with the metric system?
because it looks like the first letter of each measurement but I am not sure what comes between a millimeter and a nanometer.[/hide]

[hide="answer"]$\mu$
The letters are the prefixes for the metric system (giga, mega, kilo, hecto, deca, no prefix, deci, centi, milli, micro, nano)
[/hide]

[quote="math92"]What are the next five terms in the sequence:

O, T, T, F, F, S, ...

pretty easy[/quote]
[hide]
One, Two, Three, Four

The next letter would be S in correspondence to 7.[/hide]


calculus
derivative
function
complex analysis
complex analysis unsolved

Show that the successive derivatives of an analytic function at a point can never satisfy $ |f^{(n)}(z)| > n!n^n$.
Formulate a sharper theorem of same kind.

Let $ z_0$ be the point. By Taylor's Theorem, $ f$ can be represented as

$ f(z) \equal{} \sum_{j \equal{} 0}^{\infty} \frac {f^{n}(z_0)}{n!}(z \minus{} z_0)^n$

in a small neighborhood of $ z_0$. However, $ \left|\frac {f^{n}(z_0)}{n!}\right| > n^n$, so by the Cauchy–Hadamard theorem, the radius of convergence is $ 0$, a contradiction.


Solve the equation $ |y\minus{}6| \plus{} 2y \equal{} 9$.

If $ (y\minus{}6) \leq 0$, $ y \leq 6$ and the initial equation is $ 6 \minus{} y \plus{} 2y \equal{} 9$ or $ y \equal{} 3$.
If $ 0 \leq (y\minus{}6)$, $ 6 \leq y$ and the initial equation is $ y \minus{} 6 \plus{} 2y \equal{} 9$ or $ y \equal{} 5$, which isn't agreed upon the initial condition.
Therefore, the answer is $ y\equal{}3$


Find all functions $ f: R\minus{}>R$ such that $ f(x)f(x\plus{}y)\plus{}2f(x\plus{}2y)\plus{}f(2x\plus{}y)f(y)\equal{}x^{4}\plus{}y^{4}\plus{}x^{2}\plus{}y^{2}$.

There is no such function.

Assume, for a contradiction, that $ f$ does satisfy the functional equation.
Then, for $ x\equal{}y\equal{}0$, it leads to $ f(0)\equal{}0$ or $ f(0)\equal{}\minus{}1$.

For $ x\equal{}a,y\equal{}0$, it gives $ [f(a)]^2\plus{}2f(a)\plus{}f(2a)f(0)\equal{}a^4\plus{}a^2$. (1)
For $ x\equal{}0,y\equal{}a$, it gives $ [f(a)]^2\plus{}2f(2a)\plus{}f(a)f(0)\equal{}a^4\plus{}a^2$.
Thus, for all real $ a$, we must have $ 2f(a)\plus{}f(2a)f(0)\equal{}2f(2a)\plus{}f(a)f(0)$.
Since $ f(0) \neq 2$, it follows that $ f(a)\equal{}f(2a)$ for all $ a$. (2)

Back to (1), we deduce that $ [f(a)]^2\plus{}[2\plus{}f(0)]f(a)\equal{}a^4\plus{}a^2$.

Using (2), it follows that for each integer $ n$, we have
$ [f(a)]^2\plus{}[2\plus{}f(0)]f(a) \equal{} [f(2^na)]^2\plus{}[2\plus{}f(0)]f(2^na)\equal{}(2^na)^4\plus{}(2^na)^2$.
For $ a\equal{}1$, the LHS is constant and the RHS is not. A contradiction.

Pierre.


the orthocenter H is the symmetric of O wrt N. The circle $ \Gamma$ with center O and radius OA is the circumcircle. $ P: AH \cap \Gamma$. As well-know the side BC is the perpendicolar axes of HP that meet $ \Gamma$ on B and C.


geometry
AMC
USA(J)MO
USAMO
AIME
area of a triangle
Heron\u0027s formula
\/closed

For someone who completed Honors/Pre-AP Geometry, Introduction to Geometry may be too easy for you. Although you are using it for enrichment, it would be no good to waste 50 dollars on something you know everything of already, though, granted, you may not know everything completely and may pick up on one or two things the whole book although that would still be a waste of money.

My recommendation would be that you incorporate other subjects into your study of mathematics in addition to geometry. Think of bodybuilding. If you train nothing but your arms, you would look awkward, not get any awards, and overall be phsyically limited to your weakest link. The same principle applies to math. For you, perhaps an appropriate level book would be ACoPS (the Art and Craft of Problem Solving).

[quote="chopinftw"]Hi, I took Honors/Pre-AP Geometry several years ago and performed reasonably well. However, I could use a refresher. I also want to study geometry much more in depth than I did in high school. Would you recommend the "Introduction to Geometry" textbook by AoPS?  I'm not particularly prepping for any competitions; it's mostly for enrichment. Thanks in advance.[/quote]

I think Introduction to Geometry might still have some concepts an ordinary school course couldn't give you, so it might be worth to take a look at it if you've money to spare. However, it might be better to look at "Geometry Revisited" (there are a bunch of free copies floating around the Internet), because that still has a problem solving aspect like Intro to Geo, but covers advanced topics that would probably be better suited for you.

Um... I don't think you should do Vol.2  until you do the subjects in detail... Vol. 2 is more contest prep oriented

[quote="AIME15USAMO"]I am kind of in the same place as chpinftw, but I have also completed Intro to Geo. I am also taking precalculus (through AoPS) right now and my goal is to make USAMO this year. What are some recommendations? I was thinking of these three options. I would also be doing AIME and maybe a few Olympiad problems before the AIME for all 3 of these options

[u]Option A[/u]
Finish AoPS volume 1 and do Volume 2

[u]Option B[/u]
 Do ACoPS and maybe parts of Geometry Revisited

[u]Option C[/u]
 Geometry Revisited and practice AIME's for the rest of the time

If you have any other suggestions, feel free to give me your input. Thank You.[/quote]He doesn't want to prepare for any contests. He just wants to learn geometry in general.

As for a good way to learn, I hear this book: http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20
is good.

Also, if you ever wanna try a different type of problem, go here:
http://www.artofproblemsolving.com/Wiki/index.php/Category:Intermediate_Geometry_Problems

[quote]He doesn't want to prepare for any contests. He just wants to learn geometry in general. [/quote]

If by he, you mean me, then I am preparing for a contest, like I mentioned in my post. Also, I have finished the Intro Series (except for Intro Geo, which I'm almost done with), and I am doing precalc and Interm Alg right now.

[quote="Ihatepie"]As for a good way to learn, I hear this book: http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20
is good.[/quote]
I have just gotten that book (Challenging Problems in Geometry). Basically, it is just a problem book (as suggested by its name). But I think you actually have to know some of concepts there in order to solve those problems. The book has you do the problems, some of which proves a theorem. For instance, in one of the problems, it asks you to prove Heron's Formula, etc. And there are other problems that deal with Heron's Formula. It is a great problem book though! (The book also has solutions, so you can always learn the theorems that way too.)

[quote="chopinftw"]I wonder what Mr. Rusczyk thinks.  :maybe:[/quote]

Sorry I didn't answer sooner; I was on vacation.  I suspect the book would mostly be review, but you can take a look at the diagnostic tests here:

http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204

If you can't do most of the problems on the Post-test, then the book would likely be helpful.


Find all $ x\in Q$ satisfy
$ \{x^{2}\}\plus{}\{x\}\equal{}1$

[hide="Does the usual trick work?"]$ x\equal{} z\plus{}\frac{a}{b}$ with $ \{x\}\equal{}\frac{a}{b}$ and $ z\in\mathbb Z$ then $ x^{2}\equal{}z^{2}\plus{}z q\plus{}q^{2}$
$ \{\frac{z a b\plus{}a^{2}}{b^{2}}\}\equal{}\frac{b\minus{}a}{b}$[hide][/hide][/hide]

Er...
[hide]
$ x \equal{}\frac{p}{q}$, where $ (p, q) \equal{} 1$

$ x^{2}\plus{}x\in\mathbb{Z}$
$ \frac{p(p\plus{}q)}{q^{2}}$
Clearly this is irreducible.  Hence $ q \equal{} 1$.  And that yields $ \{x^{2}\}\plus{}\{x\}\equal{}0$, so there are no solutions.[/hide]

Something wrong here?

[quote="quangpbc"][quote="K81o7"]
$ \{x^{2}\}\plus{}\{x\}\equal{} 0$, so there are no solutions.
[/quote]

Try $ x \equal{}\frac{1}{2}$ so $ \{x^{2}\}\plus{}\{x\}\equal{}\frac{3}{4}$  :([/quote]

What do you mean?   :huh: Isn't what you're saying only confirming Krishanu's result?


Hi, i am having alot of trouble on these problems, could someone please please help me, ill be grateful!

[b]1.[/b]

In triangle [i]ABC[/i], [i]D[/i] and [i]F[/i] are points on [i]BC[/i], and [i]E[/i] is a point on [i]AB[/i] such that [i]AD[/i] is parallel to [i]EF[/i], $\angle CEF \cong \angle ACB$, [i]AD[/i]=$15$ m, [i]EF[/i] = $8$ m and [i]BF[/i] = $8$ m. [b]Find the length of BC.[/b]

[b]2.[/b]

The vertices [i]A[/i] and [i]C[/i] of a triangle [i]ABC[/i] lie on the circumference of a circle. The circle cuts the sides [i]AB[/i] and [i]CB[/i] at points [i]X[/i] and [i]Y[/i] respectively. If [i]AC = XC[/i] and $\angle YXC = 40^{\circ}$, [b]Find the size of $\angle ABC$[/b]

[b]3. [/b]

In triangle [i]ABC[/i], $\angle ABC = 112^{\circ}$ and [i]AB = BC[/i]. [i]D[/i] is a point on the side [i]AB [/i]and [i]E[/i] is a point on the side [i]AC[/i] such that [i]AE = ED = DB[/i]. [b]Find the size of $\angle BCD$[/b]

I look forward to the dreaded solutions! Thanks alot  :)  and good luck with your maths  :D

keshin

[i][Edit: Re - texed to make clearer.][/i]

number one looks like... triangle ECF is isoceles.  if EF is 8, then FC is 8 too.
and... FC and BF are colinear, so...
At least, thats how it appears from how i drew it

1.
$\triangle ADB \sim \triangle EFB$
$\frac{15}{DB} = \frac{8}{8}$
$DB = 15$, $DF = 7$
$\angle CDA = \angle CFE$ (by congruent corresponding angles)
and the given angle proves by AA that $\triangle CDF \sim \triangle EFC$
using corresponding sides, $\frac{8}{CD} = \frac{7+CD}{15}$
$(CD)^{2} + 7(CD) - 8*15 = 0$
$(CD - 8)(CD + 15) = 0$
$CD = 8$
$8+7+8 = \boxed{23}$

2.
since an inscribed angle is half the arc it intercepts,
$arc YC = 2*40 = 80$
since AC = XC, \angle CAX = \angle CXA
since those angles are congruent, the arcs they intercept are congruent, so $arc AC = arc XYC$
by the secant-secant angle formula,
$\angle ABC = \frac{arc AC - arc XY}{2} = \frac{arc AC - (arc AC - 80)}{2} = \boxed{40}$

Hm for #3 woudln't AE be longer than AB?

Or are the points allowed to be on extensions?

Edit: whoops misread, thought E was on BC

It should be 17, here's my solution:

AED is similar to ABC, so its angles are 34-34-112 also.

Since the external angle of BDE is 34, and BCE is also 34, quadrilateral DECB is concylic. Since ADE is 34, angle EDB is 146, and since EDB is isosceles, DEB is 17. Since DECB is concylic, BCD=DEB=17.

3.
since $\triangle ABC$ is isosoles, $\angle BAC$ = $\angle BCA = 34$
since $\triangle AED$ is isosoles, $\angle DAE = \angle ADE = 34$, then $\angle DEA = 112$
since all of the angles are congruent, $\triangle ABC \sim \triangle AED$
by proportional sides, $\frac{AE}{AD} = \frac{AB}{AC}$ since $AB = BC$, and $AE = BD$

$\frac{BD}{AD} = \frac{BC}{AC}$

$\frac{BD}{BC} = \frac{AD}{AC}$
that shows the angle biscector theorem, so $\angle BCD = \angle ACD$
so $\angle BCD = \frac{34}{2} = \boxed{17}$


probability
geometry
3D geometry
tetrahedron
algebra
function
domain

Here are some moderate to hard ciphering questions
Comments are appreciated

1. The expression (1+i)^k is a real number when k is a multiple of what number? (i is defined as the square root of –1)

2. x, y, and z are distinct real numbers that sum to 0. What is the maximum possible value of
(xy + xz + yz)/(x^.5 + y^.5 + z^.5)

3. You are given a 4ft board. You go to the shop and receive 2 boards, each of which has a maximum possible length of 6ft. What is the probability that you can construct a triangle from the three boards?

4. xyz=1 , x+y+z=0 , and xy+yz+xz=0. Exactly one of x, y, and z is a real number. Find this value.

5. Two men are betting money on a game of tic-tac-toe. A man pays 3. On average, how much can the player expect to win or lose in dollars?

6. Find sqrt(3 + sqrt(3 + sqrt(3 + ….))).

7. In triangle ABC, D lies on side BC and AB=6, BD=3, DC=7 and AC=5. The length of the segment AD can be written in the form sqrt(a/b) where a/b is a fraction in lowest terms. What is a+b?

8. The sum of the coefficients of the expansion (a+b+c+d)^9 can be written in the form 4^k where k is an integer. Find k.

9. Find the ordered 5-tuple (a,b,c,d,e) that satisfies the following system of equations:
4a + b + c + d + e = 8
a + 4b + c + d + e = 14
a + b + 4c + d + e = 24
a + b + c + 4d + e = 34
a + b + c + d + 4e = 40

10. Find the volume of a regular tetrahedron with side length 1

11. The rational numbers a,b,c, and d have the property that (ax+b)/(cx+d)=1 has no solution in x. What is the value of (a^2)/(a^2 +c^2)?

12. The number X20191817…321 is divisible by 11 if and only if X equals what positive integer? (note that X is a digit of the number; this is not multiplication)

13. x/(y-6)=y/(z-8)=z/(x-10)=3. What is the value of x+y+z?

14. Find the sum

(sqrt(1+1) – sqrt(1))/(sqrt(1^2 +1) + (sqrt(2+1) – sqrt(2))/(sqrt(2^2 +2) +
(sqrt(3+1) – sqrt(3))/(sqrt(3^2 +3) + … + (sqrt(15+1) – sqrt(15))/(sqrt(15^2 +15)

15. Find the largest n such that (2004!)! is divisible by ((n!)!)!

16. Compute

(1*1! + 2*2! +3*3! + … +7*7! +1)/(6!)

17. Simplify

1(nC1) + 2(nC2) + 3(nC3) + … + n(nCn)

18. A teacher wants to divide un-graded papers into a collection of several stacks, each consisting of the same number of papers. He knows that there is somewhere between 1000 and 2000 papers. He tries dividing the stack into groups of 2, 3, 4, 5, 6,7, and 8, but repeatedly gets 1 paper left over. How many papers are there?

19. Find the sum of the series

1+ 2(1/5) + 3(1/5)^2 + 4(1/5)^3 + …

20. What is the smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 2^2^2 and 2^3^3.(note: a^b^c=(a^b)^c )

21. What is the remainder when 2^100 is divide by 15?

22. Find one value of a for which the triangle with vertices (20,30), (30,60), and (a,50) has area 20.

23. In triangle ADC, angle bisector DB is drawn. If AB=3, AD=6, and CD=8, find DB.

24. If f(x/(x-1)) for all x not equal to 0 or 1, and 0<y<(pi/2), then find f((secy)^2) in terms of y.

25. If f(x)=x^2 + x –1 for x greater than or equal to –2, and g(x)=x^2 – x for x<5, then what is the domain of g(f(x))?

26. Find a function that satisfies the functional equation
f(x+t) – f(x-t) = 4xt

27. Given f(x) such that f(1-x) + (1-x)f(x)=5, find f(5).

28. 20.41 Suppose you are given a triangle ABC with AB=5, BC=7, and AC=8. Find the length of median AD.

29. There are n people in a room. Suppose that every person shakes hands with exactly two other people. What is the minimum number of handshakes possible?

30. I am in line to get into a club. The bouncers always want the club to look crowded, so if there are n people in line, they admit rounddown(n/2) people every 5 minutes. Right after these people are admitted, another 5 people join the line. There are 99 people in line when I join the line (this is right after letting some people in, and no one else joins the line at this moment). How many minutes will I have to wait in order to get in?

31. The expressions 2*6*10*…*(4n-2) can be written in the form (s!)/(t!), where s and t are positive integers. Find s+t in terms of n.

32. There are 5 teams entered for a double elimination tournament. What is the minimum number of games that can be played in order to determine a winner?

33. Define a hypergon as a polygon drawn as follows: start at a single point, then draw a segment that is either one unit up or down. From there draw a second segment that is either two units to the left or right, then from there third segment that is either three units to the north or south, etc. Continue until the original point is reached again. What is the perimeter of the hypergon with the smallest possible perimeter? (no crossing or backtracking is allowed)

34. Find (x +1)(x^2 + 1)(x^4 +1)….., in terms of x where abs(x)<1. (abs(x) denotes the absolute value of x)

With Calculators:

35. Define the power tower function recursively by
f1(x) = x
fn(x) = (fn-1(x))^x
for n an integer greater than or equal to 2. Calculate lim(n approaches infinity) fn(1/2) to three decimal places.

36. What are the first five leading digits of 5^200

Didn't you post the exact same thing before? :?
If so, please avoid double-posting.

Also, I believe some of the problems are certainly beyond the GS level...

Sorry about the difficulty. Wasn't sure.
I posted it in this forum because I thought it may have been to easy for intermediate since no one in that forum was responding.


MATHCOUNTS
analytic geometry
percent
function
geometry
geometric transformation
reflection

1. a merchant has determined that if the selling price of peaches is 15 cents each, he will sell 400 of them each day. he has also determined that for each cent the price is increased, he will sell 16 fewer peachers each day. at waht price in cents should he sell each peach in order to maximize his income from peaches each day?

2. the points A (9,14) and R (a,b) are symetric with respect to the point P (5,3). what are the coordinates of point R?

3. if the graph of the equation y = 3(x-1)^2 - 1 is translated such that the equation of the resulting graph is y + 3(x+1) - 1, how many units and in what direction was the graph translated?

4. if the graph of the equation y = x^2 + 3 is reflected with respect to the line y = -2, what is the max value of the reflected graph?

5. a survey showed that 35% of the ppl surveyed read magazine X, 29% read Y, and 28% read Z. if 10% read X and Y, 8% read X and Z, 13% read Y and Z, and 4% read all three, what percent read X or Y, but not Z?

6. the function f(x) is called a nested function and is defined as:
f(x) = x-2 for x>5 and x-f(f(x+3)) for x :le: 5.
find f(1)

7. tennyson is building a circular corral for his pet dinosaur. to form the corral he pounds 20 posts equally spaced around the circumference. each post contributes 6 inches to the circumference and the fencing between adjacent posts contributes 8 ft to the circumference. to the nearest whol number, what is the number of square feet in the area of the corral? express your answer to the nearest integer.

8. find the value of m, express as an improper fraction, so that this relation is not a function. {(-2m+1, -4), (-6m+8, 0)}

9. a + 1/(b+1/(c+(1/d))) = 181/42. find a+b+c+d.

10. tom took five math tests and got an integer score on each of them. he never scored higher than 90 and his lowest score was on the fourth test. if his avg score, rounded to the nearest integer was 82, what is the lowest possible score he could have earned on the fourth test?

11. a science test has 10 questions worth 5 points each, 7 questions worth 6 points each, and 4 questions worth 2 points each. none of these questions will be given partial credit. how many scores between 0 and 100 are impossible to score?

12. a blank videotape has space for exactly two hours in SP mode, four hours in LP mode, and six hours in EP mode. vino has recorded three one hour episodes of party of five on the same tape: the first one in SP the second in LP and the third in EP. what is the number of min in SP mode that remain on the tape?

1. [hide]The number that he sells (15+x)(400-16x)=-16x^2+160x+6000=-16(x^2-10x+25)+6400=-16(x-5)^2+6400. So it is maximum when x=5, or when he sells his peaches for 20 cents each.[/hide]

2. [hide]Since A and R are symmetric around the point P, that means that P is the midpoint of points A and R. So (9+a)/2=5 and (14+b)/2=3. So a=1, b=-8 and the point is (1, -8)[/hide]

BTW check #3 there may be something wrong with it

4. [hide]The lowest point on the graph is 3. When you reflect it across the line y=-2, the highest point on the graph would be -7.[/hide]

Could you rephrase #5??

#2  (9,14) is right 4 and up 11 of (5,3), so the new point is the opposite: it is [b](1,-8)[/b]

#3 Old vertex: (1,-1); New vertex: (-1,-1).  So the graph moves[b] 2 units left[/b]

#4 Assuming you mean "greatest value of y":

This will occur at the vertex of the new graph.  Translating (0,3) across the line, we get (0,-7), so the answer is [b]-7[/b]

#5 From a Venn diagram, we see that the value of the sums of the non-Z zones is [b]37%[/b].

#6 If you play with the function enough, you see that it [b]never resolves itself[/b].  Is that a good answer?

#10  Let us say that the 4 other tests were all 90's.  Then, we have:

x+360/5=82

x+360=410
x=[b]50[/b]

7. [hide]There are twenty posts and twenty spaces between the posts which makes for 20*(8.5)=170 feet in the circumference. So there are 85/pi feet in the radius, which makes for 7225/(pi^2) or about 732 square feet.[/hide]

8. [hide]Since it can't be a function, -2m+1=-6m+8 (in a function, a value of x cannot have two different values for y) That gies 7/4=1 3/4[/hide]

9. [hide]The largest integer smaller than 181/42 is 4, so a=4. Then 1/(b+1/(c+(1/d)))=13/42, and b+1/(c+(1/d))=42/13. The greatest integer smaller than that is 3, so b=3. Then 1/(c+(1/d))=3/13 and c+(1/d)=13/3. So c=4 and d=3. a+b+c+d=4+3+4+3=14[/hide]

could someone explain how to draw the graphs for 3 and 4? and how to do 6 and 11? thanks!
heres two more:

1. What is the probability that a randomly chosen factor of 15^90 is a multiple of 15^65? Express your answer as a common fraction.

2. If a motorist increases her speed by 8 mph she will cover a distance of 120 miles in 30 fewer min. What is her original rate of speed in mph?

[quote="topper"]#6 If you play with the function enough, you see that it [b]never resolves itself[/b].  Is that a good answer?[/quote]

I'm pretty sure it does resolve itself.. how about you post some working and we can see what happens..

I'll get to it next week, the musical I'm in is this week.  Thanks.

[hide=11]Because everything is literally just a reflection of another number, the only impossible numbers are 1, 3, 97, and 99. Use symmetry.[/hide]

10 is wrong

[hide]its 48[/hide]

[hide]Average is 81.5[/hide]


[quote="Simba"]I've done question 1, trying to do question 2 at the moment, but I don't think there is enough information.

It says:

"In a certain figure, triangle ABC is inscribed in a circle O with AB = 4, BC = 8 and AC = 9. If segment BY bisects AC, find DC."

Thanks,

~~Simba

Edit: Also question 6 looks to be a bit dodgy.

"In the diagram below, points B and C are centers of two tangent circles, points E and F are points of tangency, BF + 2, and BD = 10. Lines BD and BF intersect at A. Find AE."[/quote]I just went through both of the problems and they both have enough info. I'll edit problem 2 to make it clearer though, without changing the4 problem.  :)

Thanks for pointing that out.

What about question 7?

Does the last part mean log to base 16 of (a + b), or log to base 16 of a, then plus b? Is the 'b' including in the logarithm?

Thanks,

~~Simba

Reply: It's log 16 of (a+b). I'll fix that one too.  :roll:  :)

[quote="shobber"]I think that in problem 6, $BF=2$.[/quote]

Ok, thanks :) ...

(Problem Fixed.  :roll: )

for question 3 is simplest form mean factored, expanded or something else

For simplest form, I just want it to have the least number of terms.  :)

For question 9, are you referring to 'a' and 'b' as the roots of the equation?

For instance, if the roots were x = 7 and x = 9, you would want 2(7 + 9) = 2 * 16 = 32?

Thanks,
Simba

Reply: A and B are the two unknown coefficients of the problem. You need to find those given the conditions.  :)

[quote="Simba"]Reply: A and B are the two unknown coefficients of the problem. You need to find those given the conditions.  :)[/quote]

So in the question, 'A' and 'B' are the same as 'a' and 'b'?

Thanks,

~~Simba


The circle $ C_1$ meets a circle  $ C_2$ at $ A$ and $ B$ .The tangent of $ C_2$ at $ A$ meets $ C_1$ at $ C$ and the tangent of $ C_1$ at $ A$ meets $ C_2$ at $ D$. Let $ (d)$ be a line pass through $ A$ and $ C$ the circum-circle of $ \Delta ACD$. $ (d)$ meets $ C_1,C_2$ and $ C$ at $ M,N$ and $ P$ respectively. Prove that $ AM\equal{}NP$


quadratics
number theory unsolved
number theory

let $n = 4m^2 + 3$, where m is an integer not divisible by 3. show that there exists a prime p | n with p = 7 mod 12

all primes dividing it are bigger than three

using quadratic reciprocity you can prove that $4 m^2+3\equiv 0\bmod{p}$ implies $p\equiv 1\bmod{3}$

now as $4 m^2+3\equiv 3 \bmod{4}$ there has to be a prime p that is $p\equiv 3\bmod{4}$

this prime will then satisfy $p\equiv 7\bmod{12}$


induction
inequalities proposed
inequalities

Let $x_1, x_2,...,x_n$ be the positive real numbers and $x_i=|=x_j$.Prove that:
           $x_1^3+x_2^3+...+x_n^3\geq(x_1+x_2+...+x_n)^2$

I think the inequlity is right.
and induction works

He has forgotten to tell that $x_1$, $x_2$, ..., $x_n$ should be [b]integers[/b]. The problem was solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=54151 and probably elsewhere, too (I remember seeing it a number of times).

  darij


function
integration
algorithm
real analysis
real analysis unsolved

Let $C$ be a real number. Let $f_{1},f_{2},f_{3}$ be continuous functions defined on $[0,1]$ with $\int_{0}^{1}f_{i}= C$ for each $i$. Prove that there exist mutually disjoint subsets $P_{1},P_{2},P_{3}$ of $[0,1]$ such that $P_{1}\cup P_{2}\cup P_{3}= [0,1]$ and $\int_{P_{i}}f_{i}\ge C/3$ for each $i$.


I can't seem to figure out how to solve this one:

If $abc=1$, then prove

$\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \ge \frac{3}{2}$.

Hints or solutions are welcome!

By http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 , we have

$ \left( 1+abc\right) \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) \geq 3$.

But since abc = 1, we have 1 + abc = 2, and thus

$ 2 \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) \geq 3$,

so that

$ \frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\geq\frac32$.

This is, of course, exactly your inequality.

  Darij

Thanks! That is indeed a very nice inequality you proved in the other thread.

See

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=inequalities&t=19469

also

Alternatively, take $a = \frac{x}{y}$, $b = \frac{y}{z}$, and $c = \frac{z}{x}$. We get

$\sum_{cyc} \frac{1}{a+ab} = \sum_{cyc} \frac{1}{\frac{x}{y} + \frac{x}{z}} = \sum_{cyc} \frac{yz}{xy + xz}$

But adding three to both sides, we find the equivalent

$\left(xy+yz+zx\right)\left(\frac{1}{xy+yz} + \frac{1}{yz+zx} + \frac{1}{zx+xy}\right) \ge \frac{9}{2}$

which follows from Cauchy-Schwartz or AM-HM.

[quote="Xixas"]Can someone please give a link, where Po-Shen Loh's Notes on Inequalities could be found. Thank you very much.[/quote]

[url=http://www.cco.caltech.edu/~po/docs/math/2-inequalities-solns.pdf]http://www.cco.caltech.edu/~po/docs/math/2-inequalities-solns.pdf[/url] -- it's mentioned in the Inequalities Theorems and Formulas forum.


What is the exact problem here? To find the relation between $j,k,m,n$? Or to find $x,y$ in terms of $j,k,m,n$?

Express $x$ and $y$ in terms of the other four. That [i]is[/i] messy. :ninja:

Use the fact that the form of all of the equations is a circle.
Or...
If you expand for example, the first term in the first equation, you get:

x^2 - (j^2 - k^2)/8 - (j^2 - k^2)/16x^2

Then, do this to all of them and put all of the variable free terms, such as (j^2 - k^2)/8, on one side and it is pretty striaghtforward to solve through subbing in, adding equations, and seperating variables and constant.

[quote="Farenhajt"]That particular feat can't be accomplished, as there's actually only one independent equation.

On the other hand, the relation between $j,k,m,n$ can be found easily: $k^{2}+m^{2}=j^{2}+n^{2}$[/quote]

And you can use what Farenhajt said as one of the "subs".

[quote="Altheman"]i believe this is similar to AIME II 2006 #15, try and find a geometric interpretation[/quote]

Thanks for repeating:
"Use the fact that the form of all of the equations is a circle.
Or... " -neelnal

 :roll:  :wink:

Umm, actually I don't think they are equations of circles. Look at them more carefully; there are $x$ and $y$ terms in the denominators.


(The ghost of) r15s11z55y89w21 presents |-|!dden wol2cl (if you can't figure this out, maybe you should consider something else...no offense intended). I give a picture, you figure out which word (or sometimes number) is hidden in the picture! Sometimes the word would be something you wouldn't find in a dictionary, like "macromedia". [u]All words used are at least four letters long, and all numbers are at least three-digit numbers. [/u]PM me (me=Hopkinsmathclub) your answer, [color=red][size=134][u][b]do not post your answers or anything that might give a hint towards the answer on the thread[/b][/u][/size][/color]! Those who answer correctly receive 5 points. If there are more incorrect submissions than correct submissions, the first two to submit incorrectly will receive 1 point each, and if there are no correct submissions, the first to submit will receive 3 points. If your submission contains inappropriate language, you will receive -15 points. None of the answers are inappropriate words. 420, 666, and 1488 will not be considered inappropriate numbers. On submission PMs, title your submission [color=green]"Hidden #[insert number]: [insert word you're submitting]"[/color] for example [color=blue]"Hidden #11: bourgeoisie"[/color]. [i]Please[/i] do not share answers with others. [i]Please[/i] don't sabotage this game by deleting PMs before r15s11z55y89w21 reads them if you know the password to this account; I am, after all, working in your favor because I'm working towards Hopkinsmathclub-being-able-to-have-a-blog-ness. [i]Please[/i] play this game fairly. All right, enough begging for playing it fair, enjoy the game!

[size=200]Number One[/size]

Difficulty: Moderately Easy

Find the hidden word.

[u]Results[/u]

Submissions: 4
Correct Answers: 4 ([color=green][b][i]100.00%[/i][/b][/color])
Score changes: [b]Dojo [color=green]+5[/color], dragon96 [color=green]+5[/color], jjfun1 [color=green]+5[/color], bbgun34 [color=green]+5[/color][/b]
[i]Extra point for jjfun1 for noticing an error![/i]
Current Top Scores:
jjfun1 6
Dojo 5
dragon96 5
bbgun34 5

Answer: [b][color=brown]Hidden[/color][/b]

[size=200]Number Two[/size]

Difficulty: Moderate

[u]Results[/u]

Submissions: 2
Correct Answers: 0 ([color=red][b][i]0.00%[/i][/b][/color])
Score changes: [b]rd5493 [color=green]+3[/color], dragon96 [color=green]+1[/color][/b]
Current Top Scores:
dragon96 6
jjfun1 6
Dojo 5
bbgun34 5
rd5493 3

Answer: [b][color=blue]Green[/color][/b]

[size=200]Number Three[/size]

Difficulty: Moderate

[u]Results[/u]

Submissions: 7 (2: Careful, 3: Cape, 1: Chapel, 1: Carpool)
Correct Answers: 2 ([color=orange][b][i]28.57%[/i][/b][/color])
Score changes: [b]Yoshi [color=green]+5[/color], bbgun34 [color=green]+5[/color], dragon96 [color=orange]+2[/color], vahalla [color=orange]+2[/color], Dojo [color=orange]+2[/color], PowerOfPi [color=orange]+2[/color], westiepaw [color=orange]+2[/color][/b]
Current Top Scores:
bbgun34 10
dragon96 8
Dojo 7
jjfun1 6
Yoshi 5
rd5493 3
vahalla 2
PowerOfPi 2
westiepaw 2

Answer: [b][color=orange]Careful[/color][/b]

[size=200]Number Four[/size]

Difficulty: Easy

[u]Results[/u]

Submissions: 5 (3: Easy, 1: Ask, 1: Fog)
Correct Answers: 3 ([color=indigo][b][i]60.00%[/i][/b][/color])
Score changes: [b]PowerOfPi [color=green]+5[/color], vahalla [color=green]+5[/color], dragon96 [color=green]+5[/color][/b]
Notes: I think I'll start taking off points for not following the submission format.
Current Top Scores:
dragon96 13
bbgun34 10
Dojo 7
vahalla 7
PowerOfPi 7
jjfun1 6
Yoshi 5
rd5493 3
westiepaw 2

Answer: [b][color=olive]Easy[/color][/b]

[size=200]Number Five[/size]

Difficulty: Moderately Hard

[size=200]Number Six[/size]

Just giving it ahead of time. Number five submissions are still accepted.

Difficulty: Moderate

[u]Results[/u]

Submissions: 10 (10: Truth)
Correct Answers: 10 ([color=green][b][i]100.00%[/i][/b][/color])
Score changes: [b]james4l [color=green]+5[/color], vahalla [color=green]+5[/color], PowerOfPi [color=green]+5[/color], dragon96 [color=green]+5[/color], Dojo [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+5[/color], Yoshi [color=green]+5[/color], meewhee009 [color=green]+4[/color], bbgun34 [color=green]+4[/color][/b]
Notes: As I have warned, I took off a point for not following the submission format. From now on, I'll take off two points.
Current Scores:
dragon96 18
bbgun34 14
Dojo 12
vahalla 12
Yoshi 10
PowerOfPi 7
jjfun1 6
james4l 5
math_explorer 5
bluecarneal 5
meewhee009 4
rd5493 3
westiepaw 2

Answer: [b][color=Blue]Truth[/color][/b]

[color=red][u][b]Number Six is still open. [/b][/u][/color]

Sorry, but how is the ansewr truth? Doesn't make much sense.

[quote="andersonw"]Lower-right hand corner  :wink:[/quote]

A green square?

[quote="AIME15"]You're looking at the wrong puzzle.  Puzzle 5 has the answer "truth".[/quote]

Oh, I see. Thanks!

[size=200]Number Seven[/size]

Difficulty: Hard

Number Six Results coming soon.

[u]Results[/u]

Submissions: 16 (16: Sight)
Correct Answers: 16 ([color=green][b][i]100.00%[/i][/b][/color])
Score changes: [b]BOGTRO, vahalla, Yoshi, PowerOfPi, jjfun1, james4l, meewhee009, cf249, dragon96, zapi2007, math_explorer, westiepaw, bluecarneal, gaussintraining [color=green]ALL +5[/color], bbgun34 [color=green]+3[/color], rd5493 [color=green]+3[/color][/b]
Notes: Please follow the format! You only get 60% of the score if you don't.
Current Scores:
dragon96 23
bbgun34 17
vahalla 17
PowerOfPi 17
Yoshi 15
Dojo 12
jjfun1 11
james4l 10
math_explorer 10
bluecarneal 10
meewhee009 9
rd5493 7
westiepaw 7
BOGTRO 5
cf249 5
zapi2007 5
gaussintraining 5

Answer: [b][color=Violet]Sight[/color][/b]

[color=red][u][b]Number Seven is still open. [/b][/u][/color]

[hide="Results for Puzzle 4"]
[quote="Hopkinsmathclub"][u]Results[/u]

Submissions: 5 (3: Easy, 1: Ask, 1: Fog)
Correct Answers: 3 ([color=indigo][b][i]60.00%[/i][/b][/color])
Score changes: [b]PowerOfPi [color=green]+5[/color], vahalla [color=green]+5[/color], dragon96 [color=green]+5[/color][/b]
Notes: I think I'll start taking off points for not following the submission format.
Current Top Scores:
dragon96 13
bbgun34 10
Dojo 7
vahalla 7
[b]PowerOfPi 7[/b]
jjfun1 6
Yoshi 5
rd5493 3
westiepaw 2[/quote][/hide]
[hide="Results for Puzzle 5"]
[quote="Hopkinsmathclub"][u]Results[/u]

Submissions: 10 (10: Truth)
Correct Answers: 10 ([color=green][b][i]100.00%[/i][/b][/color])
Score changes: [b]james4l [color=green]+5[/color], vahalla [color=green]+5[/color], PowerOfPi [color=green]+5[/color], dragon96 [color=green]+5[/color], Dojo [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+5[/color], Yoshi [color=green]+5[/color], meewhee009 [color=green]+4[/color], bbgun34 [color=green]+4[/color][/b]
Notes: As I have warned, I took off a point for not following the submission format. From now on, I'll take off two points.
Current Scores:
dragon96 18
bbgun34 14
Dojo 12
vahalla 12
Yoshi 10
[b]PowerOfPi 7[/b]
jjfun1 6
james4l 5
math_explorer 5
bluecarneal 5
meewhee009 4
rd5493 3
westiepaw 2[/quote][/hide]
You forgot to add 5 to my score (see my bolded score).

[quote="Hopkinsmathclub"]Correct Answers: 9 ([color=green][b][i]100.00%[/i][/b][/color])[/quote]
You forgot to change that.

Oh, whoops. LOL.

[size=200]Number Sixteen[/size]

Made by [b][color=green]fortenforge[/color][/b]

Difficulty: Moderate

[color=red][u][b]Number Fifteen still open. [/b][/u][/color]

[u]Results[/u]

Submissions: 6 (3: Exit, 1: Edit, 1: Fist, 1: Fact)
Correct Answers: 3 ([color=orange][b][i]50.00%[/i][/b][/color])
Score changes: [b]fortenforge [color=green]+5[/color], math_explorer [color=green]+5[/color], PowerOfPi [color=green]+5[/color], zapi2007 [color=green]+3[/color][/b]
Current scoreboard:
[b]james4l 45[/b]
[b]vahalla 42[/b]
[i]math_explorer 39[/i]
[i]dragon96 37[/i]
[i]PowerOfPi 37[/i]
[i]zapi2007 28[/i]
BOGTRO 22
bbgun34 17
Dojo 15
Yoshi 15
westiepaw 12
jjfun1 11
fortenforge 10
cf249 10
bluecarneal 10
meewhee009 9
rd5493 7
gaussintraining 5

Answer: [b][color=blue]Exit[/color][/b]

[color=red][u][b]Number Sixteen is still open. [/b][/u][/color]

[size=200]Number Seventeen[/size]

Difficulty: Moderately Easy

[u][b][color=red]Number Sixteen is still open. [/color][/b][/u]

[u]Results[/u]

Submissions: 3 (2: Symbol, 1: Uyombs) <--Lol?
Correct Answers: 2 ([color=indigo][b][i]66.67%[/i][/b][/color])
Score changes: [b]james4l [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+1[/color][/b]
Current scoreboard:
[b]james4l 50[/b]
[b]math_explorer 44[/b]
[b]vahalla 42[/b]
[i]dragon96 37[/i]
[i]PowerOfPi 37[/i]
[i]zapi2007 28[/i]
BOGTRO 22
bbgun34 17
Dojo 15
Yoshi 15
fortenforge 14
westiepaw 12
bluecarneal 11
jjfun1 11
cf249 10
meewhee009 9
rd5493 7
gaussintraining 5

Answer: [b][color=black]Symbol[/color][/b]

[color=red][u][b]Number Seventeen is still open. [/b][/u][/color]

[size=200]Number Eighteen[/size]

Difficulty: Hard

The answer is a number. Bonus point if you can point out the significance of the number.

[u]Results[/u]

Submissions: 2 (1: Oversize, 1: Pursue) <--Lol?
Correct Answers: 1 ([color=orange][b][i]50.00%[/i][/b][/color])
Score changes: [b]math_explorer [color=green]+5[/color], james4l [color=green]+1[/color][/b]
Current scoreboard:
[b]james4l 51[/b]
[b]math_explorer 49[/b]
[i]vahalla 42[/i]
[i]dragon96 37[/i]
[i]PowerOfPi 37[/i]
zapi2007 28
BOGTRO 22
bbgun34 17
Dojo 15
Yoshi 15
fortenforge 14
westiepaw 12
bluecarneal 11
jjfun1 11
cf249 10
meewhee009 9
rd5493 7
gaussintraining 5

Answer: [b][color=red]Oversize[/color][/b]

[color=red][u][b]Number Eighteen is still open. [/b][/u][/color][/quote]

[u]Results[/u]

Submissions: 4 (3: 5040, 1: 504) (I just realized I still forgot to delete the <--Lol label from ages ago)
Correct Answers: 3 ([color=green][b][i]75.00%[/i][/b][/color])
Score changes: [b]PowerOfPi [color=green]+5,+1[/color], westiepaw [color=green]+5[/color], math_explorer [color=green]+5,+1[/color], james4l [color=green]+3[/color][/b]
Current scoreboard:
[color=orange][b]math_explorer 55[/b]
[b]james4l 54[/b][/color]
[i]PowerOfPi 43[/i]
[i]vahalla 42[/i]
dragon96 37
zapi2007 28
BOGTRO 22
westiepaw 17
bbgun34 17
Dojo 15
Yoshi 15
fortenforge 14
bluecarneal 11
jjfun1 11
cf249 10
meewhee009 9
rd5493 7
gaussintraining 5

Answer: [b]5040[/b]

Significance of the number: 7!, ALSO ACCEPTABLE: according to Plato, "the ideal population of a city," due to easy divisibility when dividing into groups

[size=200]Number Nineteen[/size]

Difficulty: Moderately Easy

The answer is a number. Bonus point if you can point out the significance of the number. Some of you will immediately recognize this number, while others might have a really tough time. It's one of those numbers that are life-crucial in some occupations and totally unimportant in others.

[u]Results[/u]

Submissions: 6 (6: 69314718)
Correct Answers: 6 ([color=green][b][i]100.00%[/i][/b][/color])

Current scoreboard:
[color=orange][b]math_explorer 60[/b]
[b]james4l 59[/b][/color]
PowerOfPi 48
dragon96 42
vahalla 42
zapi2007 33
BOGTRO 22
westiepaw 17
bbgun34 17
cf249 15
Dojo 15
Yoshi 15
fortenforge 14
bluecarneal 11
jjfun1 11
meewhee009 9
rd5493 7
gaussintraining 5

Answer: [b]69314718[/b]

Significance of the number: The first few decimal places in the decimal representation of ln 2 and ln (1/2). This is used in banking and accounting as a dividend for an expression to calculate doubling time. (Credit only given for both parts.)

[size=200]Number Twenty[/size]

Now for something fun to test your guessing skills! Match the person with their body text message for their submission to Number Six!

[color=white]Hint: Twenty is an even number. [/color]

BOGTRO
vahalla
Yoshi
PowerOfPi
jjfun1
james4l
meewhee009
cf249
dragon96
zapi2007
math_explorer
westiepaw
bluecarneal
gaussintraining
bbgun34
rd5493
[hide=" "]-2[/hide]
Sight
Sight
Hidden #6: sight
loldarn
Hidden #6: Sight
Hidden #6: sight
Hidden #6: Sight
*insert filler text here*
Not sure if you received it on FTW.
hey.
sight
No way that was moderate. I got it in 10 seconds.
Hello. Am I supposed to put anything of value in the body of this message?
 :lol:
Hidden #6: SIGHT
Unless, of course, I missed a huge very important part of the picture.

You should make people PM PhireKalk6781 instead since I accidentally found out who you are and each time I log in to HMC, I see this PM pop up thing.

EDIT: Also, what do we put as our title?

[21:27] PhireKaLk6781: Just put Hidden #20

[u]Results[/u]

Submissions: 2 (james4l [color=green][b]+2[/b][/color], dragon96 [color=green][b]+4[/b][/color])

Final scoreboard:
[color=blue]
[b]james4l 61[/b] :winner_first:
[b]math_explorer 60[/b] :winner_second:
[b]PowerOfPi 48[/b] :winner_third:[/color]
[i]dragon96 46
vahalla 42[/i]
zapi2007 33
BOGTRO 22
westiepaw 17
bbgun34 17
cf249 15
Dojo 15
Yoshi 15
fortenforge 14
bluecarneal 11
jjfun1 11
meewhee009 9
rd5493 7
gaussintraining 5

[u]Answers[/u]
BOGTRO is Hidden #6: SIGHT
vahalla is Unless, of course, I missed....of the picture.
For everyone else, just move two up on the list. This solution could have been derived by quoting.

Future hidden puzzles, if any, will be posted on PhireKaLk6781's blog.

Very nice waste of your ti... er, I mean, very amusing thing to do when bored.


[quote="nonie"]1. How many postitve integers less than 36 are equal to 4 times an odd integer?

2. A long distance phone call costs 1.80 for the first 3 minutes, and 40 cents for each additional minute. If an $x$ minute phone call costs 4.20, then $x$=?

3.After $1/8$ of a ribbon is thrown away, the rest is cut into 2 pieces that are of the ratio 4:5. If 9 inches were thrown away, then what is the length of the shorter piece?

4. What's the least number of squares of side length 2 inches required to cover, without overlapping, a larger square of a side length of 8 inches?

5.In the games that the Bengals played the Lions, the Bengals won $2/3$ of their games,and lost $3/$  of the othergames. If the teams tied in 2 games, how many games did the Bengals tie the Lions?[/quote]

[hide="number 1"]1, 3, 5, and seven, each multiplied by four give us 4, 12, 20, and 28. Four integers. [/hide]

[hide="number 2"]After three minutes we have 4.20 - 1.80 0r 2.40. That gives us 2.40/.40 = 6 minutes left. 3+6=9=x. [/hide]

[hide="number 3"]The original ribbon must've been 72 inches. There was 63 inches left after 1/8 was taken away. The smaller piece was 4/9 or this, or 28 inches. [/hide]

[hide="number 4"]the area of the smaller square is four sq.in. The area of the big square is 64 sq.in. 64/4=16. [/hide]

Number 5: The bengals lost 3/what of the other games?

Those are all right, and as for number 5 sorry it was 3/4 i edited a few minutes ago.

[quote="236factorial"][hide="number 5"]2 games - it's right in the problem :rotfl:  I think the question is wrong... [/hide][/quote]

OMG i was copying the question from my notebook and my wrinting was really messy  so i must have not realized what i was writing... It is supposed to be how many games did they PLAY.. So i guess I'll change that now.

[hide="new answer"]They tied in 1/12 of the games. So they played 24 games. [/hide]

[quote="nonie"]1. How many postive integers less than 36 are equal to 4 times an odd integer?

2. A long distance phone call costs 1 dollar and 80 cents for the first 3 minutes, and 40 cents for each additional minute. If and $x$ minute phone call costs 4 dollars and 20 cents , then $x$=?

3.After $1/8$ of a ribbon is thrown away, the rest is cut into 2 pieces that are of the ratio 4:5. If 9 inches were thrown away, then what is the length of the shorter piece?

4. What's the least number of squares of side length 2 inches required to cover, without overlapping, a larger square of a side length of 8 inches?

5.In the games that the Bengals played the Lions, the Bengals won $2/3$ of their games,and lost $3/4$  of the other games. If the teams tied in 2 games, how many games did the Bengals PLAY the Lions?[/quote]
[hide]
1)1, 3, 5, 7 are odd integers that can be multiplied by 4 but are stilll less than 36.
1x4=4, 3x4=12, 5x4=20, 7x4=28
The numbers are 4, 12, 20, 28.

2)(4.20-1.80)/.40=each minute over 3 minutes.
=2.4/.40=6 minutes over 3, (3+6) so the call was 9 minutes long

3)if 9=1/8 of the total, the total was 72 inches. After being cut, the remainder was 72-9=63 inches. 4/9 of 63=28 inches

4)the small square has area 4 sq in. The large square is 64 sq in. 64/4=16. It requires 16 small squares.

5)2/3x + 3/12 x +2 = x
11/12x+2=x
1/12x=2
x=24[/hide]

[quote="nonie"][hide]
[quote]
3)if 9=1/8 of the total, the total was 72 inches. After being cut, the remainder was 72-9=63 inches. 4/9 of 63=36 inches
[/quote]

b4k4ka,
Your answersfor 1,2,4,+5 are correct
but for number 3 you made a simple computation error 4/9ths of 63 is 28 not 36[/hide][/quote]
oh oops, sorry i'll change that right away


Putnam
algebra
polynomial
logarithms
quadratics
college contests

A [i]repunit[/i] is a positive integer whose digits in base $ 10$ are all ones. Find all polynomials $ f$ with real coefficients such that if $ n$ is a repunit, then so is $ f(n).$

There are a two-dimensional family of polynomials with this property. For nonnegative integers $ m$ and integers $ r\ge 1\minus{}m$, the polynomial $ f_{m,r}(x)\equal{}\frac{10^r(9x\plus{}1)^m\minus{}1}{9}$ takes the $ k$-digit repunit $ \frac{10^k\minus{}1}{9}$ to the $ (mk\plus{}r)$-digit repunit $ \frac{10^{mk\plus{}r}\minus{}1}{9}$. As it happens, these polynomials have integer coefficients when $ r$ is nonnegative.

Verifying that these polynomials work is simple arithmetic. We must also show that these are the only such polynomials.
First, note that all constant polynomials $ f(x)\equal{}\frac{10^r\minus{}1}{9}$ with $ r\ge 1$ work. We assume that $ f$ has degree at least $ 1$ in the following.
For a general polynomial $ f(x)\equal{}a_mx^m\plus{}a_{m\minus{}1}x^{m\minus{}1}\plus{}\cdots$, we have $ \log(f(x))\equal{}m\log x\plus{}\log a_m\plus{}o(1)$ as $ m\to\infty$. Applying $ f$ to the $ k$-digit repunit $ x_k\equal{}\frac{10^k\minus{}1}{9}$, $ \log(f(x_k))\equal{}m(\log(10^k)\minus{}\log(9))\plus{}\log(a_m)\plus{}o(1)$
$ \equal{}mk\minus{}m\log(9)\plus{}\log(a_m)\plus{}o(1)\equal{}mk\plus{}\log(9^{\minus{}m}a_m)\plus{}o(1)$ as $ k\to\infty$ (We specialize to the base 10 logarithm here for convenience).

We want this to be the logarithm of some $ g(k)$-digit repunit, which is $ g(k)\minus{}\log(9)\plus{}o(1)$. For large $ k$, we therefore have $ mk\plus{}\log(9^{\minus{}m}a_m)\minus{}g(k)\plus{}\log(9)\plus{}o(1)\equal{}mk\plus{}\log(9^{1\minus{}m}a_m)\minus{}g(k)\plus{}o(1)\equal{}0$. Since the possible values of $ mk\plus{}\log(9^{1\minus{}m}a_m)\minus{}g(k)$ differ by integers, they must eventually be equal to zero, and thus $ g_k\equal{}mk\plus{}r$ for $ r\equal{}\log(9^{1\minus{}m}a_m)$. Two polynomials which are equal for infinitely many values (large enough repunits) are equal, and $ f(x)\equal{}\frac{10^r(9x\plus{}1)^m\minus{}1}{9}$ for some positive $ m$ and some $ r$. For the values to actually be repunits all the way down to $ f(1)$, we must have $ mk\plus{}r\ge 1$ for all $ k\ge 1$, and thus $ r\ge 1\minus{}m$.

Hm...this problem was given at MOP this year...what was it doing on the Putnam? Am I right in thinking that the Putnam occurs after MOP? (i.e. it occurred sometime in the Fall of 2007, as opposed to the Spring of 2007?)

The Putnam happens at the beginning of December. I doubt the question-writers for the Putnam know anything about the questions used at MOP.

Oops, looks like I was mistaken.

The MOP problem was as follows:

Find all quadratic polynomials with integer coefficients that transform repunits into repunits.

My apologies.

[quote=duongbgbg32][hide=Solution]. Let $f$ be such a polynomial and observe that from the hypothesis it follows that there exists a sequence $\left(a_{n}\right)_{n \geq 1}$ of positive integers such that $f\left(\frac{10^{n}-1}{9}\right)=\frac{10^{a n}-1}{9} .$ But this sequence $\left(a_{n}\right)_{n \geq 1}$ cannot be really arbitrary:
actually we can find precious information from an asymptotic study. Indeed, suppose that $\operatorname{deg}(f)=d \geq 1$. Then there exists a nonzero number $A$ such that $f(x) \approx A x^{d}$ for large values of $x .$ Therefore $f\left(\frac{10^{n}-1}{9}\right) \approx \frac{A}{9^{d}} \cdot 10^{n d} .$ Thus $10^{a_{n}} \approx \frac{A}{9^{d-1}} \cdot 10^{n d} .$ This shows that the sequence $\left(a_{n}-n d\right)_{n \geq 1}$ converges to a limit $l$ such that $A=9^{d-1} \cdot 10^{l}$. Because this sequence consists of integers, itbecomes eventually equal to the constant sequence $l$. Thus from a certain point we have $f\left(\frac{10^{n}-1}{9}\right)=\frac{10^{n d+l}-1}{9}$. If $x_{n}=\frac{10^{n}-1}{9},$ we deduce that the equation $f(x)=\frac{(9 x+1)^{d} \cdot 10^{6}-1}{9}$ has infinitely many solutions, so $f(X)=\frac{(9 X+1)^{d} \cdot 10^{t}-1}{9}$[/quote]bruh why 13 year bump

It's encouraged to post complete solutions to past problems. There is nothing wrong with bumping this thread.


Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack? (blackjack means one of the card is an ace, and the other is either ten, jack, queen, or king)

[hide]The chances that one of you will get it is 1/13*4/13*2=8/169, so the probability that you won't get it is 161/169, so that squared is [b]25921/28561[/b][/hide]

but after I get 2 cards, the dealer will only have 50 cards from the deck to choose from?

Oh yeah.

Hint
[hide]Use choose numbers[/hide]

but when calculating the probability of the 2nd person getting a black jack somewhat depends on what 2 cards the frist person got. Like the first person can get an Ace and something else and not form a black jack, or he can get two Aces, or he can get two Tens,jacks,queens,kings etc....I am confused

I am not sure if this is right:
[hide] There are 52 choose 2 ways to randomly pick 2 cards out of 52 (I will just refer to a choose b as aCb, sorry I don't know how to Latex "choose").  Now, there are 4 aces in a deck and 16 other royals (including 10).  Thus, there are 64 ways for the dealer to give the first person a blackjack.  Therefore, the chance is 64/52C2 that the first person will recieve a blackjack.  Assume that the first person was not given a blackjack.  This chance is 1-64/52C2.
Since there are 50 cards left and you want to give the second person 2 of them, there are 50C2 ways to do this.  Now, there are still 4 aces and 16 other royals (including 10), thus there are 64 ways to give the second person a blackjack.  Thus, the chance that the second person recieves a blackjack is 64/50C2, so the chance that he won't is 1-64/50C2.  Therefore, the chance that none of them will recieve a blackjack is (1-64/50C2)(1-64/52C2).  [/hide]

Another hint

[hide]Find the probability that the [b]four[/b] cards dealt are the "right " cards[/hide]

by "right " cards I suppose you mean all four cards are not either ace, ten,jack,queen,king. But there is always a possibility that you get an Ace and a two, or a ten and a ten?

Call the 10 Jack, Queen King face cards. You need 2 face cards and 2 aces. Find the probability of that happening

ok, that takes care of probability of bothing getting a black jack..but the question is neither of them getting a black jack

subtract each of the individual probabilities from 1, and then multiply them together!

what if only one of them gets a black jack?


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Prove or disprove that : If $ A$ are matrices with rational numbers, and $ A^5\equal{}I$, and $ 1$ is not eigenvalue of $ A$, then  the order of $ A$ must be divisible by $ 4$. Thanks in advance.

The characteristic polynomial $ p_A(x)$ has rational coefficients. Since we have ruled out $ 1$ as an eigenvalue, that means that all roots of $ p_A$ must also be roots of $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1.$ But that polynomial is irreducible over the rationals. Hence, we must have that $ p_A(x) \equal{} (x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)^k$ for some $ k\in\mathbb{N},$ from which we see that $ A$ is a $ 4k\times 4k$ matrix.

Second question: does any such $ A$ actually exist? Can we find a $ 4\times 4$ rational matrix with $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ as its characteristic polynomial?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28984]here[/url] for an almost identical problem. The solution there also works here.

Ah, yes. Should have thought of that.

$ A\equal{}\begin{bmatrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \minus{}1&\minus{}1&\minus{}1&\minus{}1\end{bmatrix}$ works as an example.

$ A$ is the (transpose of the) [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url] of $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$.  (A companion matrix is its own rational canonical form.)


calculus
integration
function
derivative
inequalities
real analysis
real analysis unsolved

Let $ f: [0,\infty]\rightarrow \mathbb{R}$ a derivable function, convex, with $ f(0)\equal{}0$.
Prove that $ \int_{0}^{x}f(t)dt \leq \frac{x^2}{2}f'(x)$, $ \forall x \in [0, \infty)$.

I think it will be sufficient to show $ \int\limits_0^x {f(t)dt} \le \int\limits_0^x {tf'(t)dt}$, which is a consequence of the hypotheses. It's not that difficult; the main thing being tested is whether you know about convex functions.

[quote="§outh§tar"] $ \int\limits_0^x {f(t)dt} \le \int\limits_0^x {tf'(t)dt}$[/quote]

Ok. I proved that, but it isn't very clear for me  why that is a consequence of the hypotheses  :P  something to enlight me would be great. Thanks  :)

[quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac{x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt\minus{}\int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P

[quote="Aalexandru"][quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac {x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt \minus{} \int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]
There is a proof already but I wanna say your argument here is incorrect

@bogdanno: Yes, you are right  :oops:
@ith_power: Your solution is quite simple and natural, but you haven't used the fact that $ f(0)\equal{}0$. The $ f$ you are refering to in your proof is not the same with the $ f$ from the hypotheses. Is there a problem or that condition is not necessary?

[quote="ith_power"][quote="Aalexandru"][quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac {x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt \minus{} \int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]

To prove that $ \int\limits_0^x {tf'(t)dt} \le \frac {x^2}{2}f'(x) \forall x \in [0, \infty)$.

Let $ f(x)$denote $ rhs \minus{} lhs$. then
$ f'(x) \equal{} \frac {x^2}{2}f''(x) \plus{} f'(x)x \minus{} xf'(x) \ge 0 \forall x \in [0, \infty)$ since $ f$ is convex.
thus $ f(x) \ge f(0) \equal{} 0$.[/quote]

Remind me again why f is twice differentiable  :?:

[quote="§outh§tar"][/quote][quote="ith_power"][quote="Aalexandru"][quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac {x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt - \int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]

To prove that $ \int\limits_0^x {tf'(t)dt} \le \frac {x^2}{2}f'(x) \forall x \in [0, \infty)$.

Let $ f(x)$denote $ rhs - lhs$. then
$ f'(x) = \frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \ge 0 \forall x \in [0, \infty)$ since $ f$ is convex.
thus $ f(x) \ge f(0) = 0$.[/quote][quote="§outh§tar"]

Remind me again why f is twice differentiable  :?:[/quote]

I sensed that something was wrong with ith_power's proof  :)
Ok. Other ideas?:P ( I've noticed that the function $ f(x)=x$ holds the equality. Does that help in some way?)

I showed you what the first step is. Next you note that the derivative is nondecreasing and use corresponding properties of the integral to obtain your upper bound.

[quote="ith_power"][quote="Aalexandru"][quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac {x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt \minus{} \int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]

To prove that $ \int\limits_0^x {tf'(t)dt} \le \frac {x^2}{2}f'(x) \forall x \in [0, \infty)$.

[b]Let $ f(x)$denote $ rhs \minus{} lhs$. then [/b]
$ f'(x) \equal{} \frac {x^2}{2}f''(x) \plus{} f'(x)x \minus{} xf'(x) \ge 0 \forall x \in [0, \infty)$ since $ f$ is convex.
thus $ f(x) \ge f(0) \equal{} 0$.[/quote]

why can [b]you let f=lhs-rhs[/b] ?

[quote="1234567a"][/quote][quote="ith_power"][quote="Aalexandru"][quote="BlueVelvet"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]

Like this: $ \int_{0}^{x}f(t)dt \leq \frac {x^2}{2}f'(x) \Leftrightarrow \int_{0}^{x}tf'(x)dt - \int_{0}^{x}f(t)dt \geq 0$. (1)
But $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$.
And, from (1), we have that $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]

To prove that $ \int\limits_0^x {tf'(t)dt} \le \frac {x^2}{2}f'(x) \forall x \in [0, \infty)$.

[b]Let $ f(x)$denote $ rhs - lhs$. then [/b]
$ f'(x) = \frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \ge 0 \forall x \in [0, \infty)$ since $ f$ is convex.
thus $ f(x) \ge f(0) = 0$.[/quote][quote="1234567a"]

why can [b]you let f=lhs-rhs[/b] ?[/quote]

I am very sorry , actually it was a typo.  :oops: assume $ g(x)$ denote rhs-lhs.. then
$ g'(x) = \frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \ge 0 \forall x \in [0, \infty)$ since $ f$ is convex.
thus $ g(x)\ge g(0) = 0$

[quote="§outh§tar"]
Remind me again why f is twice differentiable  :?:[/quote]
since $ f$ is convex, i used the fact that $ f''(x) \ge 0$. :maybe:

[quote="Aalexandru"]
@ith_power: Your solution is quite simple and natural, but you haven't used the fact that $ f(0) = 0$. The $ f$ you are refering to in your proof is not the same with the $ f$ from the hypotheses. Is there a problem or that condition is not necessary?[/quote]
is it really necessary to have $ f(0)=0$? :maybe:

Question: is it true for a convex function $ f$ the inequality: $ f(t) \leq t\frac{f(x)}{x}$, $ \forall t \in [0,x]$ and $ x \in  [0, \infty]$? ( I saw this result in another problem, but I don't know how to prove it).
If it's true, I think I have a solution for the problem, finally :)
Ok, so I want to prove that    $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$.
$ f$ convex $ \Rightarrow f(t) \leq t\frac{f(x)}{x} \Rightarrow \int_{0}^{x}f(t)dt \leq \frac{f(x)}{x}\int_{0}^{x}tdt\equal{}\frac{xf(x)}{2}$ [b] (1) [/b] $ \Rightarrow \minus{}\int_{0}^{x}f(t) \geq \minus{}\frac{xf(x)}{2}$
On the other hand $ \int_{0}^{x}tf'(t)dt\equal{}tf(t)|_{0}^{x} \minus{}\int_{0}^{x}f(t)dt \geq xf(x)\minus{}\frac{xf(x)}{2}\equal{}\frac{xf(x)}{2}$ [b] (2) [/b]
From (1) and (2) we have that     $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$
Because $ f$ is convex $ f'$ is non decreasing so $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$ and from here  think it's obvious.

[quote]since $ f$ is convex, i used the fact that $ f''(x) \ge 0$. :maybe: [/quote]

how do you know f'' exists?

[quote="Aalexandru"]@southstar [quote="Aalexandru"]Let $ f: [0,\infty]\rightarrow \mathbb{R}$ a [b]derivable[/b] function, convex .[/quote][/quote]
Yes but its not aparent that $ f$ is twice differentiable. :(

[quote="Aalexandru"]Question: is it true for a convex function $ f$ the inequality: $ f(t) \leq t\frac {f(x)}{x}$, $ \forall t \in [0,x]$ and $ x \in [0, \infty]$? ( I saw this result in another problem, but I don't know how to prove it).
If it's true, I think I have a solution for the problem, finally :)
Ok, so I want to prove that    $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$.
$ f$ convex $ \Rightarrow f(t) \leq t\frac {f(x)}{x} \Rightarrow \int_{0}^{x}f(t)dt \leq \frac {f(x)}{x}\int_{0}^{x}tdt \equal{} \frac {xf(x)}{2}$ [b] (1) [/b] $ \Rightarrow \minus{} \int_{0}^{x}f(t) \geq \minus{} \frac {xf(x)}{2}$
On the other hand $ \int_{0}^{x}tf'(t)dt \equal{} tf(t)|_{0}^{x} \minus{} \int_{0}^{x}f(t)dt \geq xf(x) \minus{} \frac {xf(x)}{2} \equal{} \frac {xf(x)}{2}$ [b] (2) [/b]
From (1) and (2) we have that     $ \int_{0}^{x}tf'(t)dt \geq \int_{0}^{x}f(t)dt$
Because $ f$ is convex $ f'$ is non decreasing so $ \int_{0}^{x}tf'(x)dt \geq \int_{0}^{x}tf'(t)dt$ and from here  think it's obvious.[/quote]

I ask again if this solution is correct ( my proof doesn't rely on $ f$ being twice differentiable) :)

Obviously, given a point $ x > 0$, there exists $ \xi\in[0,x]$ such that
\[ \int_0^x tf'(t) dt \equal{} f'(\xi) \int_0^x t dt \equal{} \frac {x^2}2 f'(\xi).
\]
Since $ f'$ is increasing, $ f'(\xi) \le f'(x)$, yielding
\[ \int_0^x tf'(t) dt \le \frac {x^2}2 f'(x).
\]
That's all.


Hi, I noticed some of these people's on the site are very good at making their own avatars. My avatar is also given as a present by one of users of this site and I wonder if there's any site that lets you to draw these comic figures.

I'm asking this because I have a biology exam and in this project, I need to draw a comic strip on terms of DNA. But problem is, my drawing is -10^9999 worse than my $\LaTeX$.

Thanks.

if you have this thing called paint, i think you  can draw you own thing

Paint is kind of hard to use.  My photoshop trial version expired, but at least some computers at our school have it.

You coud try [url=http://www.gimp.org/]The GIMP[/url] though it's kinda hard to use and takes a bit of practice (probably not so good if you need it for something immediate), but it seems incredibly powerful.  It's almost like a GNU clone of Photoshop but a little less user friendly and more powerful.

I draw my frames in MSPAINT and compile them with a horrible device :P . See my signature for shiny animations.

I didn't make this, but look! [img]http://img199.exs.cx/img199/2918/newbie0tr.gif[/img]


In the diagram, $ OA$ is a radius of the large circle and a diameter of the smaller circle. What is the number of square centimeters in the area inside the larger circle but outside the smaller circle? Express your answer in terms of $ \pi$.
[asy]pair O,A; A = (2,0); draw(O--A); draw(Circle(O,2)); draw(Circle((1,0),1));
label("O",O,W); label("A",A,E); label("4 cm",(1,0),N);[/asy]

The area of the larger circle is $ 16\pi$, and the area of the smaller circle is $ 4\pi$. Thus we have $ 16\pi\minus{}4\pi\equal{}\boxed{12\pi}$.


Prove that it will always take even number of moves for a knight to move back to it's original position).

EDIT-oops! I meant to post this in the Intermediate Forum

This is really easy, since each move changes the color of the square the knight is on, if the knight starts on square of color "a", next move will be to a square of color "b", and then back to "a", so the knight will be on a space of the same color only after an even number of moves. Therefore, it can only get back to the original square after an even number of moves.

probability1.01 wrote:Extension:

Prove that if lattice points on the coordinate plane are made into squares, and the knight starts at (0, 0), then the sum of the coordinates of a square and the number of moves it takes to get there will be of the same parity.

Here's my solution:

[hide]A knight's move can be represented as the vector (+-1, +-2) or the vector (+-2, +-1). Either way you're either adding +-1 or +-3 to the sum of the knight's coordinates each move. Thus, each move, the parity of the sum of the coordiantes changes. Both the number of moves and the sum of the coordinates start off at the same parity (even). Each move, the parity of the number of moves changes and so does the parity of the sum of the coordinates.[/hide]

Yasha's solution is correct (and surprisingly, the first one I thought of, despite having been a chess player for 4-5 years), and Demon's answer is the one that chess coaches would give (and it also proves the extension).  Good job!

The first is [hide]just 64!/32! then divide by permutations of equivalent placements (e.g., bishop and bishop have two ways of being placed).[/hide]

The second is incredibly hard, even rather challenging to program for a computer to do.

The first one is very big: [hide]64!/32!/(8!*2!*2!*2)^2 (8 for the pawns, 2 for the each of the rooks, knights, and bishops, this is for each color).[/hide]

This is equal to [hide]46,347,266,955,878,096,411,920,459,823,285,670,400,000.[/hide]

Go TI-89!

I think you would have to calculate the entire game tree in order to calculate the second one. I don't think it's possible to be able to quickly recognize every impossible situation. Some impossible situations are obvious, like two white bishops both on white spaces, but there are probably others for which it is harder to tell.

[quote="Hexaditidom"]Interesting!
Heres another one:

How many different ways are there to arrange all the chess pieces on the 8x8 board:

1. Not following the rules (like a lone bishop, two kings adjacent, all pawns on a rank, etc.)

2. (very hard) Yes following the rules.

(I thought i had the first one already figured out, but i realized i was slightly off)[/quote]

Is it all the chess pieces for a single side (16 white pieces) or ALL the pieces, black and white (32 pieces)?

Whoops! I made a mistake in the previous post. I divided by the equivalent placements of only one color of pieces. I'll go edit it.

Yeah.  You would have discount everything with the pawns in the way, like the b and d pawns must be moved before white queens bishop can move...etc, situations with stalemate or checkmate, situations where both kings are in check, situations with pawns on the back rows...there are just so many.  It would be quite hard.


Find last 2 digits of $1234^23$ i think its 04 but i'm not sure  :blush:
PS: the power is 23  ...

:?:  :?:  :?:  :(
Just tell me if it's correct or not... plz


The infinite sequence of 2's and 3's \[\begin{array}{l}2,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3, \\ 3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,\cdots \end{array}\] has the property that, if one forms a second sequence that records the number of 3's between successive 2's, the result is identical to the given sequence. Show that there exists a real number $r$ such that, for any $n$, the $n$th term of the sequence is 2 if and only if $n = 1+\lfloor rm \rfloor$ for some nonnegative integer $m$.

We claim that $ m$ th $ 2$ appears as $ [(2 \plus{} \sqrt {3})(m \minus{} 1)] \plus{} 1$th term is this sequence

Prove by induction,the foundation is trivial,and for every positive integer $ \le m$ is true.

Let's take a look at thr case of $ m \plus{} 1$

$ \exists$ $ r,r\in N$ s.t. $ [(2 \plus{} \sqrt {3})(r \minus{} 1)] \plus{} 1\le m < [(2 \plus{} \sqrt {3})r] \plus{} 1$

After some simple calculation.We obtain $ r \equal{} [\frac {m}{2 \plus{} \sqrt {3}}] \plus{} 1$

This yields the number of $ 2$ $ 3$ in the first $ m$ terms are $ r$ and $ m \minus{} r$ respectively

So the position of $ m \plus{} 1$th $ 2$ is $ 2r \plus{} 3(m \minus{} r) \plus{} m \plus{} 1 \equal{} 4m \minus{} r \plus{} 1 \equal{} [(2 \plus{} \sqrt {3})m] \plus{} 1$ which complete yhe

induction.

QED

This is sequence [url=http://www.research.att.com/~njas/sequences/A007538]A007538[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


There are no rational numbers so that:

$x^{2005}-3x-1=0$

Here's a start maybe?:
[hide]
Through Vieta's formulae:

$(r_{1}+r_{2}+r_{3}+...+r_{2005}) = 1$

$(r_{1}r_{2}r_{3}r_{4}...r_{2005}) = 1$

Therefore
$r_{1}+r_{2}+r_{3}+...+r_{2005}= r_{1}r_{2}r_{3}r_{4}...r_{2005}$

and then somehow that fact should prove that none of $r_{i}$ are rational?
[/hide]

EDIT: The fact that I jumped to this conclusion before thinking of the rational root theorem.... oh man that is not good.

If there exists a rational root $x=\frac{p}{q},\ \gcd (p,q)=1$, then $p|(-1)\wedge q|1$.  These conditions do not lead to roots.  And the conclusion follows.

[quote="max_tm"]Therefore
$r_{1}+r_{2}+r_{3}+...+r_{2005}= r_{1}r_{2}r_{3}r_{4}...r_{2005}$

and then somehow that fact should prove that none of $r_{i}$ are rational?[/quote]

Actually, a variant of that was a USAMO problem last year.  This fact alone is insufficient; consider

$r_{1}= r_{2}= ... = r_{2003}= 1, r_{2004}= a, r_{2005}= b$

Then we merely require that $2003+a+b = ab \implies (a-1)(b-1) = 2004$ which has infinite rational (not to mention several integer!) solutions, although none of them satisfy the original two conditions.

[hide]From the Rational Roots Theorem, the only possible rational roots are $\pm1$. Neither work, so there are no rational number solutions.[/hide]


calculus
integration
trigonometry
calculus computations

Help me to calculate the following integral: $ \int_{\minus{}\frac \pi 2 }^{\frac \pi 2}\frac{\sin{\theta}\sin{2 \theta}}{\sqrt{R^2\plus{}|x|^2\minus{}2R|x|\cos{\theta}}}d \theta$

Maybe I'm thinking about this in the complete wrong way but this reminds me of Poisson's integral formula.

hello, here can you read somthing about the Poisson's integral formula:
http://msvlab.hre.ntou.edu.tw/paper/tmes/ijmest2006.pdf
Sonnhard.

I've seen two derivations of it: one manipulating Cauchy's integral formula with complex conjugates and another one exploiting properties of Fourier series.


Let us call P(n) - the product of all digits of number n (in decimal notation).
For example, P(1243)=1*2*4*3=24; P(198501243)=0.
Let us call n to be a good number, if (p(n)<>0) and (n mod P(n)=0).
Let us call n to be a perfect number, if both n and n+1 are good numbers.

You are to write a program, which, given the number K(k<=100000), counts all such
numbers n that n is perfect and n contains exactly K digits in decimal notation. Time limit 0.2 sec

If K>1, then both P(n) and P(n+1) are multiples of (n with the last digit removed)..what does that tell you about all of ns digits except the last?
After that, its just checking all possibilities for the last digit.


whenever i use the forums now, all the text is magnified really big... how do i make it regular again
and sometimes when i click on a link the text will be ez to read for a sec

thx

erm.. internet explorer most of the time
im not too sure i know what u mean by skin?

Click [url=http://www.mathlinks.ro/Forum/usercp.php]here[/url]. The first option (the dropdown menu) is the Board Layout option. You can choose between MathLinks and AoPS. A picture would really help at this point to understand what's going on ...  :? (use the Prt Scr button on your keyboard to take a picture of your screen).

[quote="sheepwarrior"]erm.. the print screen didnt work...
maybe its an ie problem not an aops problem[/quote]What skin do you use? In IE 6.0 go to View -> Text Size and try to find the appropriate measure ;)

oh thx!! it worked! but still whenever i click a link the text shrinks...

edit: wait nvm

thx again! :D

2nd edit: wait.... it still shrinks when i click a link.. it goes back to normal once the page comes up but still...
allnd when i click reveal hidden content all teh text spreads apart or overlaps   ?/???


Hi I'm having a hard time finding the solution to the following problem. In triangle ABC, point D is on AC such that AB=AD.  If the measure of angle ABC - measure of angle ACB = 30, find the measure of CBD.
(Note that the triangle is NOT drawn to scale.)

[img]http://i13.photobucket.com/albums/a260/ContraForce/Triangle.jpg[/img]

Since AB=AD then by Isosceles Triangle Theorem mADB = mABD.
Then using Euclidean Parallel Postulate, I construct line EA such that EA is parallel to CB and EC is perpendicular to CB, and construct AG in the same fashion so that AG is perpendicular to CB.  Then a quadralateral is created with sides CG, AG,EA,EC, and diagonal AC.  Using the converse to the alternate interior angles theorem (CAIA) we see that mCEA = mACG, and since mECG = mGAE = 90  this implies that mACG=mACE by angle addition axiom.  Then by AIA , EC is parallel to AG.  WE can show trianlge EAC is congruent to triangle GCA using Angle Side Angle.  At this point the work I'm doing doesn't seem to be leading me to a solution for the question.  I've spent a lot of time on this problem and its driving me crazy so any help is greatly appreciated.  Please note that I am restricted to certain Theorems, axioms, etc.. ( i.e. I cannot use things we haven't learned in class as a justification.)

I'm sorry, I meant to post this in the unsolved section, I do know that there is a solution I just can't seem to get to it.  Can a mod please move this thread?  Thanks

No extra lines, no nothing...
Put $\angle CBA=\beta$ and $\angle ACB=\gamma$. $\angle BDA=\angle ABD=\delta$, and suppose that position of points on line $AC$ is $A-D-C$, it's the same for other position. Then $\delta = \gamma+\beta-\delta$ (from triangle $CBD$), and $\beta-\gamma = 30^{o}$. From this $\beta-\delta =\angle CBD =15^{o}$.
Bye


Find all natural numbers $x,y$ such that
\[6(x!+3)=y^2+5\]

[quote="Palytoxin"][quote="paladin8"][quote="s0mp"][hide] Rewrite the equation as 6x!+13=y^2. Starting with x=5, 6x!+3 ends in 3. However, the last digits of a square can only be 0, 1, 4, 5, 6, and 9. So, x must be 4 or less. We can simply check x=1, 2, 3, and 4, and we see that the solutions are (2, 5) and (3,7). [/hide][/quote]

Couple of typos, but I think that's it. $6x!+18$, not $6x!+13$ or $6x!+3$ and so it ends in $8$, which also works with your argument. Nice solution.[/quote]
s0mp is correct, paladin8 you got it wrong because $6(x!+3)=6x!+18=y^2+5$ which means that $6x!+13=y^2$. and not only that he is doing $\mod 10$ so $6x!+13\equiv 6x!+3$[/quote]

Oh, I see. Yeah I kept the $5$ on the other side when using the $6x!+18$. I thought he had a typo because if you use $6x!+13 = y^2+5$ you have no solutions for $x \ge 5$ either. My bad.

we can work (mod 8) , if x>3 we have y²=5(mod 8)  witch is impossible ,so  x is 0,1,2 or 3, we check that 2 and 3 works...


Is there an improvement award for the AMCs ?

Yes, but the award goes to schools based on team score, not to individuals.
Recall that the school team score is the sum of the top 3 scores from a school.

From the award letter that goes to schools:
"The American Mathematics Competitions program of the Mathematical Association of America annually sponsors an pair of challenging mathematics contests at the high school level, the AMC 10 and the AMC 12.  Your school and mathematics teachers have participated in these challenging contests for at least the last two years.  As a result of your school's improvement in performance from the 2006 AMC contests to the 2007 AMC contests, your school is the State/Provincial Winner of the 2007 Pedagoguery Award.  One award goes to to the school in each state or province with the best difference in team score on the AMC 10 and also to to the school in each state or province with the best difference in team score on the AMC 12 from 2006 to 2007.  The actual award is a set of die-cast aluminum mathematical models of the five platonic solids and three additional truncated platonic solids.  This award is provided courtesy of Pedagoguery Software Inc. of 4446 Lazelle Avenue, Terrace, British Columbia, Canada V8G 1R8.  Through the generosity of Pedagoguery Software, Inc,  the American Mathematics Competitions is making this award to stimulate continued year-to-year interest in the AMC contests and to reward teachers who coach their teams to greater levels of achievement on the contests and in mathematics generally. "

A brief description of the award is on page 8 of the AMC 10/AMC 12 Teachers' Manual in the section on awards.

No, we do not have an award for individual students.

Steve Dunbar
MAA Director, American Mathematics Competitions

Which is a shame, because next year I have a shot at winning the Most Unimproved award with a score of -150.


If $a,b,c$ are the sides of a triangle, prove that
\[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]

[i]Proposed by Hojoo Lee, Korea[/i]

[quote="mattilgale"]If $ a$, $ b$, $ c$ are the sides of a triangle, prove that
\[ \sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]
[/quote]
$ \sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3\Leftrightarrow\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\sqrt a+\sqrt b-\sqrt c}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)}\geq0.$
Let's $ a\geq b\geq c.$  Then
$ \frac{1}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)}\geq\frac{1}{(\sqrt{ac}+\sqrt{b(a+c-b)})(\sqrt a+\sqrt c-\sqrt b)}$
since, $ \sqrt{ac}+\sqrt{b(a+c-b)}\geq\sqrt{bc}+\sqrt{a(b+c-a)}$ and $ \sqrt a+\sqrt c-\sqrt b\geq\sqrt b+\sqrt c-\sqrt a.$
Thus, your inequality is proved. :)

My solution is the same with yours :wink:

[quote="zibi"]I don't understand this one [quote]$\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0 \Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\sqrt a+\sqrt b-\sqrt c}\geq 0$[/quote][/quote]
You are right:
$\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0 \Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\left(\sqrt a+\sqrt b-\sqrt c\right)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{a+b-c}\right)}\geq 0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}\right)}\geq0.$
But $\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}\geq\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}$ true too. Thank you. :)

What was the idea behind the first (mistaken) simplification?

[quote="mattilgale"]If a, b, c are sides of a triangle prove that
\[\sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]
[/quote]

$\sum \frac{\sqrt{-a+b+c}}{-\sqrt{a}+\sqrt{b}+\sqrt{c}}$
$\leq \sqrt{3\sum \frac{-a+b+c}{(-\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}}$ by Cauchy's
$=\sqrt{3\sum \frac{2x^{2}+2xy+2xz-2yz}{4x^{2}}}$
(where $x=\frac{-\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}$, and so on)
$=\sqrt{3\sum \frac{x^{2}+xy+xz-yz}{2x^{2}}}$
$=\sqrt{3 \frac{3x^{2}y^{2}z^{2}+\sum_{sym}x^{3}y^{2}z-\sum_{cyc}x^{3}y^{3}}{2x^{2}y^{2}z^{2}}}$
$\leq \sqrt{3 \frac{6x^{2}y^{2}z^{2}}{2x^{2}y^{2}z^{2}}}=3$ by Schur's for $xy,yz,zx$.

Let me post the solution I had for this problem (I was hoping that I would get to coordinate this problem if it were chosen in the competition last year :) )

Since $ a,b,c$ are the sides of a triangle, it is clear that $ x=\sqrt a, y=\sqrt b, z= \sqrt c$ are also the sides of a triangle (so the numbers $ -x+y+z$, $ x-y+z$ and $ x+y-z$ are positive). We first apply a Cauchy to get rid of the square roots in the numerators of the fractions
\[ \left(\sum \frac{ \sqrt{-x^{2}+y^{2}+z^{2}}}{-x+y+z}\right)^{2}\leq 3 \sum \frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}.\] Therefore what we want to prove is
\[ \sum \frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}\leq 3. \]
Moving all into one side (and distributing the 3 to each of the fractions equally) we get
\[ 0 \leq \sum \frac{ x^{2}+yz-xy-xz }{ (-x+y+z)^{2}}= \sum \frac 1{(-x+y+z)^{2}}(x-y)(x-z) , \]
which of course is nothing else than Vornicu-Schur, because if $ x \geq y\geq z$ then $ \frac 1{(-x+y+z)^{2}}\geq \frac 1{(x-y+z)^{2}}\geq \frac 1{(x+y-z)^{2}}$ and we are done.

[quote="silouan"]Who is the proposer of that problem ? It's from Korea according to IMO compedium so Hojoo Lee ?[/quote]

According to  http://ultrametric.googlepages.com/proposals.pdf
this problem indeed has been constructed by Hojoo Lee.

Is It True ?

[tex]$ \frac{\sqrt[n]{a\plus{}b\minus{}c}}{\sqrt[n]{a}\plus{}\sqrt[n]{b}\minus{}\sqrt[n]{c}}\plus{} \frac{\sqrt[n]{b\plus{}c\minus{}a}}{\sqrt[n]{b}\plus{}\sqrt[n]{c}\minus{}\sqrt[n]{a}}\plus{} \frac{\sqrt[n]{c\plus{}a\minus{}b}}{\sqrt[n]{c}\plus{}\sqrt[n]{a}\minus{}\sqrt[n]{b}}\leq 3$    ? :D

Put $ a\equal{}x\plus{}y, b\equal{}x\plus{}z, c\equal{}y\plus{}z$. Then we should prove that $ \sum\frac{\sqrt{2x}}{\sqrt{x\plus{}y}\plus{}\sqrt{x\plus{}z}\minus{}\sqrt{y\plus{}z}}\le 3$.

By Cauchy, we get that

$ LHS\le \frac{(\sqrt[4]{2x}\plus{}\sqrt[4]{2y}\plus{}\sqrt[4]{2z})^2} {\sqrt{x\plus{}y}\plus{}\sqrt{x\plus{}z}\plus{}\sqrt{y\plus{}z}}\le 3\frac{\sqrt{2x}\plus{}\sqrt{2y}\plus{}\sqrt{2z}} {\sqrt{x\plus{}y}\plus{}\sqrt{x\plus{}z}\plus{}\sqrt{y\plus{}z}}\equal{} 3\frac{2\sqrt{x}\plus{}2\sqrt{y}\plus{}2\sqrt{z}} {\sqrt{2x\plus{}2y}\plus{}\sqrt{2x\plus{}2z}\plus{}\sqrt{2y\plus{}2z}}$.

$ \sqrt{x}\plus{}\sqrt{y}\le \sqrt{2x\plus{}2y}$, because its squared is true: $ x\plus{}y\plus{}2\sqrt{xy}\le x\plus{}y\plus{}x\plus{}y$ (we have positive numbers here), so numerator$ \le$denominator. Thus

$ 3\frac{2\sqrt{x}\plus{}2\sqrt{y}\plus{}2\sqrt{z}} {\sqrt{2x\plus{}2y}\plus{}\sqrt{2x\plus{}2z}\plus{}\sqrt{2y\plus{}2z}}\le 3$.

We're done.

Uh.. sorry, my solution above is wrong..

we take $ a\equal{}x^{2}$ $ b\equal{}y^{2}$ $ c\equal{}z^{2}$
then $ x,y,z$ are also side of triangle
so there are exist uniqe triple such that $ x\equal{}a'\plus{}b' y\equal{}b'\plus{}c' z\equal{}c'\plus{}a'$
so only condition for triples $ (a',b',c')$ should be  $ a',b',c' \in R^\plus{}$
the we rewrite our inequality
$ \sum\frac{\sqrt{4a'^2\minus{}2(a'\minus{}b')(a'\minus{}c')}}{2a'}$
but we have $ 4a'^2\minus{}2(a'\minus{}b')(a'\minus{}c')$ is negatif for example triple $ (1,10,2)$
where is my fault? can anyone tell me

[quote="mestav"]I find my mistake.I use Ozandro's Lemma for this solution but it is not true[/quote]
what is Ozandro's Lemma?

other solution:
We have that
\[ \frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} =\sqrt{1-2\frac{(\sqrt{c}-\sqrt{a})(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2}} \le 1-\frac{(\sqrt{c}-\sqrt{a})(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2} .\]
So we prove that if $x=\sqrt{b}+\sqrt{c}-\sqrt{a}>0$,  $y=\sqrt{a}+\sqrt{c}-\sqrt{b}>0$,  $z=\sqrt{a}+\sqrt{b}-\sqrt{c}>0$ then \[ \sum x^{-2}(x-y)(x-z) \ge 0. \]
It is obviously true by Schur's inequality.

Hey mathuz Schur's Inequality tells $r>0$ but you have $r=-2$

[quote="TripteshBiswas"]Hey mathuz Schur's Ineuality tells r>0 but you have r=-2[/quote]
See here the  proof when $r$ is negative:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=415872
the last post

thanks mudok it was not known to me when r is negative

[b]Lemma[/b]:
$\sum_{cyc} \frac {(x-z)(x-y)}{x^2}\ge 0$. Since $2(x-y)(x-z)=(x-y)^2+(x-z)^2-(y-z)^2$ the inequality is reduced to the inequality $\sum_{cyc} (x-y)^2(\frac {1}{x^2}+\frac {1}{y^2}-\frac{1}{z^2})$ where we deonte $S_x=\frac {1}{x^2}+\frac {1}{y^2}-\frac{1}{z^2}$ this is indeed true by the SOS method just assume WLOG $x\ge y\ge z$ and put $x-y=a,y-z=b$ the inequality becomes $a^2(S_z+S_y)+b^2(S_x+S_y)+2abS_y$ but we can easily see that $S_z+S_y,S_x+S_y,S_y$ are all greater than $0$ so our inequality is proven.
[b]Back to the problem[/b]
Denote by $S_a,S_b,S_c$ those expressions in the inequality we need to prove. Now notice that the inequality $\sum_{cyc}(1-(S_a)^2)\ge 0$ reduces to the lemma for numbers $\sqrt{a}+\sqrt{b}-\sqrt{c},\sqrt{a}-\sqrt{b}+\sqrt{c},-\sqrt{a}+\sqrt{b}+\sqrt{c}$.
This gives $\sum_{cyc} (S_a)^2\le 3$ and from here we easily get the desiret result

My solution is easy

Since $a,b,c$ are the sides of a triangle, we can set
$a=(q+r)^2$, $b=(p+r)^2$, $c=(p+q)^2$

Thus the inequality can be rewritten as
$\sum_{cyclic} \frac{\sqrt{2p^2+2pr+2pq-2qr}}{p} \le 6$

Thus is it is sufficient to show that
$\sum_{cyclic} \frac{2p^2+2pr +2pq -2qr}{p^2} \le 12$
$\iff \sum_{cyclic} \frac{pr+pq-qr}{p^2} \le 3$
$\iff \sum_{cyclic} q^2r^2(pr+pq-qr) \le 3p^2q^2r^2$

This is third degree Schur's inequality for $pq,qr,rp$

Hence proved

Let [b]a,b,c>0[/b],and [b]a+b+c=3[/b] prove that:
${\frac {a\sqrt {\sqrt {b}+\sqrt {c}}}{\sqrt {a+\sqrt {bc}}}}+{\frac {b
\sqrt {\sqrt {c}+\sqrt {a}}}{\sqrt {b+\sqrt {ac}}}}+{\frac {c\sqrt {
\sqrt {a}+\sqrt {b}}}{\sqrt {c+\sqrt {ab}}}}\leq 3.$
http://www.artofproblemsolving.com/community/c6h1098581p4979124

[quote=mattilgale]If $a,b,c$ are the sides of a triangle, prove that
\[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \][i]Proposed by Hojoo Lee, Korea[/i][/quote]
[url=http://www.artofproblemsolving.com/community/c6h149013p842206]TST PERU - IMO 2007[/url]

my solution:
consider a function $ f(x)=\sqrt{x} $ and three point A(a,$ \sqrt{a} $),B(b,$ \sqrt{b} $),C(c,$ \sqrt{c} $)
choose D,E,F s.t. ABDC,ABCE,and AFBC are all parallelogram
choose $ G_{D} $,$ G_{E} $,and $ G_{F} $ on f(x) s.t. $ DG_{D} $,$ EG_{E} $,and $ FG_{F} $ are plumb lines
now,we only show that $ \frac{y_{G_{D}}}{y_{D}}+\frac{y_{G_{E}}}{y_{E}}\leq 2 $
that is $ \frac{DG_{D}}{y_{D}}\geq \frac{EG_{E}}{y_{E}} $
it is obious because $ y_{D}\geq y_{E} $且$ DG_{D}\leq EG_{E} $

Solution from [i]Twitch Solves ISL[/i]:

To deal with the long denominators, let $x = \sqrt b + \sqrt c - \sqrt a$, etc.\ for brevity. Note that $x > 0$, and similarly for the others. Then we have that \[ 0 < b+c-a = \left( \frac{x+y}{2} \right)^2 	+ \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 	= x^2+xy+xz-yz \] Thus the problem asks us to prove that \[ \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \le 3\sqrt 2 \] for positive real numbers $x$, $y$, $z$ satisfying the extra condition that all radicals are positive.
By Cauchy-Schwarz though, we have that \[ 	\left( \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 	\le 3 \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2} \] and expanding fully, we see it suffices to show \[ \sum_{\text{sym}} xy^2z^3 \le 	(xy)^3 + (yz)^3 + (zx)^3 + 3(xyz)^2 \] which is Schur's inequality applied to $xy$, $yz$, $zx$.

[quote=v_Enhance]Solution from Twitch stream:

Then we have that \[ 0 < b+c-a = \left( \frac{x+y}{2} \right)^2 	+ \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 	= x^2+xy+xz-yz \] [/quote]
Sir, it would be $b+c-a=\left(\frac{x+y}{2}\right)^2+\left(\frac{z+x}{2}\right)^2-\left(\frac{y+z}{2}\right)^2=\frac{2x^2+2xy+2zx-2yz}{4}=\frac{x^2+xy+xz-yz}{2}$

Solved with [b]eisirrational[/b], [b]goodbear[/b], [b]nukelauncher[/b], [b]Th3Numb3rThr33[/b].

Let \(a=(q+r)^2\), \(b=(r+p)^2\), \(c=(p+q)^2\), so that the desired inequality is \[\sum_\mathrm{cyc}\frac{\sqrt{p(p+q+r)-qr}}p\le3\sqrt2.\]

Then observe that \begin{align*} \left(\sum_\mathrm{cyc}\frac{\sqrt{p(p+q+r)-qr}}p\right)^2 &\le3\sum_\mathrm{cyc}\frac{p(p+q+r)-qr}{p^2}\\ &=3\left(3+\sum_\mathrm{sym}\frac pq-\sum_\mathrm{cyc}\frac{qr}{p^2}\right), \end{align*} so it suffices to show \[3+\sum_\mathrm{cyc}\frac{qr}{p^2}\ge\sum_\mathrm{sym}\frac pq.\] This is just Schur's inequality on \(p^{-2/3}q^{1/3}r^{1/3}\), \(p^{1/3}q^{-2/3}r^{1/3}\), \(p^{1/3}q^{1/3}r^{-2/3}\): \begin{align*}     3\prod\sqrt[3]{\frac{qr}{p^2}}+\sum_\mathrm{cyc}\frac{qr}{p^2}     &\ge\sum_\mathrm{cyc}\sqrt[3]{\frac{qr}{p^2}}\sqrt[3]{\frac{rp}{q^2}}\left(\sqrt[3]{\frac{qr}{p^2}}+\sqrt[3]{\frac{rp}{q^2}}\right)\\     &=\sum_\mathrm{cyc}\sqrt[3]{\frac{r^2}{pq}}\left(\frac{q\sqrt[3]r+r\sqrt[3]q}{\sqrt[3]{p^2q^2}}\right)\\     &=\sum_\mathrm{cyc}\frac{qr+rp}{pq}=\sum_\mathrm{sym}\frac pq. \end{align*}

This inequality is just rationalization and schur.

Let $a=(x+y)^2$, $b=(y+z)^2$, $c=(z+x)^2$, $a$, $b$, $c\ge0$. It suffices to show that
\[S=\sum_{\text{cyc}}{\frac{\sqrt{x^2+xy+xz-yz}}{x} }\le 3\sqrt{2}\]
but by Schur's, $\sum_{\text{cyc}}{(xy)^2(yz+zx-xy)}\le 3x^2y^2z^2$ so \[S^2 \le (1+1+1)\left(\sum_{cyc}{\frac{z^2}{z^2}+\frac{yz+zx-xy}{z^2}}\right)\le (1+1+1)(3+3)=18\] which gives the desired result.

Let $x = \sqrt b + \sqrt c - \sqrt a$ and cyclic permutations. The inequality boils down to $$\sum \frac{\sqrt{x^2+xy+xz-yz}}x \leq 3\sqrt 2.$$  But by Cauchy $$\left(\sum \frac{\sqrt{x^2+xy+xz-yz}}x\right)^2 \leq 3\sum \frac{x^2+xy+xz-yz}{x^2} \leq 18.$$ The last inequality $$\sum \frac{xy+xz-yz}{x^2} \leq 3$$ expands to Schur with $r=2$.

Substitute $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$ and so on. This then simplifies as \[ \frac13 \sum_{\text{cyc}} \frac{\sqrt{x^2 + xz + xy - yz}}{x} \le \sqrt{2} \]
By Power-Mean and Schur on $r = -2$ it follows that \[ \frac13 \sum_{\text{cyc}} \frac{\sqrt{x^2 + xz + xy - yz}}{x} \le \left(\frac13 \sum_{\text{cyc}} \frac{x^2 + xz + xy - yz}{x^2}\right)^{\frac 12} \le \left(\frac 13 (3 + 3) \right)^{\frac 12}. \]

We can first make the substitution $x = \sqrt b + \sqrt c - \sqrt a$, and cyclically for $y,z$, noting all of these variables are also positive. Then the left hand side can be expressed as
\[\sum_{\text{cyc}} \sqrt{\frac{x^2+xy+xz-yz}{2x^2}} \leq \sqrt{3 \cdot \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{2x^2}}\]

using Cauchy. Hence it suffices to show
\[\sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2} \leq 6\]
\[\iff (xy)^3 + (yz)^3 + (zx)^3 + 3(xy)(yz)(zx) \ge \sum_{\text{sym}} (xy)^2(yz),\]

which is indeed just Schur's for $t=1$ on $xy,yz,zx$. $\blacksquare$

Make the substitution $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$, with $y$ and $z$ defined cyclically. Now note,
\begin{align*}
\sqrt{b+c-a} = \sqrt{\frac{x^2+xy+xz-yz}{2}}
\end{align*}
Thus we wish to prove,
\begin{align*}
 \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \leq 3\sqrt{2}\\
\left( \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \leq 18
\end{align*}
However Cauchy gives,
\begin{align*}
\left( \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \leq 3\left(\sum_{cyc} \frac{x^2+xy+xz-yz}{x^2} \right)
\end{align*}
Substituting this into our original expression we find,
\begin{align*}
\sum_{cyc} \frac{x^2+xy+xz-yz}{x^2}  &\leq 6\\
\end{align*}
which is just Schurs upon expansion.

Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}>0$ (positive by the triangle inequality), etc., so $y+z=2\sqrt{a}$ and the LHS becomes
$$\sum_{\mathrm{cyc}} \frac{\sqrt{(\frac{x+y}{2})^2+(\frac{x+z}{2})^2-(\frac{y+z}{2})^2}}{x}=\sum_{\mathrm{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x\sqrt{2}}.$$
It thus suffices to prove that
$$\sum_{\mathrm{cyc}} \sqrt{1+\frac{y}{x}+\frac{z}{x}-\frac{yz}{x^2}} \leq 3\sqrt{2}.$$
By Jensen's, it suffices to show that
$$1+\frac{1}{3}\left(\sum_{\mathrm{cyc}} \frac{y}{x}+\frac{z}{x}-\frac{yz}{x^2}\right) \leq 2 \iff \sum_{\mathrm{cyc}} \frac{y}{x}+\frac{z}{x} \leq 3+\sum_{\mathrm{cyc}} \frac{yz}{x^2} \iff 3x^2y^2z^2+3x^3y^3+3y^3z^3+3z^3x^3 \geq x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3 \iff [3,3,0]+[2,2,2] \geq 2[3,2,1].$$
This is just first-degree Schur on $xy,yz,zx$, so we're done. $\blacksquare$