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algebra

The set of n-varaible formula  is asubset of functions of n variables $x_1,x_2,...,x_n$ defined recursively
\[ (x_1,x_2,...,x_n) \mapsto max {f_1(x_1,..,x_n),...,f_k(x_1,x_2,...x_n)}\]
or
\[ (x_1,x_2,...,x_n) \mapsto min {f_1(x_1,..,x_n),...,f_k(x_1,x_2,...x_n)}\]
which $f_i$ are n-varaible formulas For example
\[ max(x_2,x_3,min(x_1,max(x_4,x_5)))\]
is a f-variable formula.
Suppose $P$ and $Q$ are two n-variable formulas that:
\[ P(x_1,x_2,...,x_n)=Q(x_1,x_2,...,x_n)\]
for all $x_1,...,x_n \in {0,1}$ Prove the above equation holds also for all $x_1,...,x_n \in \mathbb{R}$

Bilieve,me the problem is easy think on it

Assume there is an $n$-tuple $X=(x_1,x_2,\ldots,x_n)$ s.t. $P(X)\ne Q(X)$. This means that $P(X')\ne Q(X')$, where $X'=aX+b,\ a,b\in\mathbb R,a\ne 0$. We can clearly choose $a,b$ s.t. $X'\in[0,1]^n$, which gets us a contradiction.

I understand exactly nothing from this solution.

I can't see why, though. $aX+b$ is defined very naturally as $(ax_1+b,ax_2+b,\ldots,ax_n+b)$.

What I mean is that we take the system $n$ points on the real line represented by $x_1,x_2,\ldots,x_n$ and we apply a homothety of center $O(0,0)$ and a translation in such a way that the image of the system of points lies in $(0,1)$. The relations of $\max$ and $\min$ are kept between these points, because the new system of points is just a smaller version of the initial one.

Dear grobber, as far as I understood, the problem says that for all x_i whose values are exactly 0 or 1. But of course I might be wrong.

Oh my..  :blush:

I guess I saw $0\ 1$ and I automatically completed it to $(0,1)$. Moving the cursor over $0,1$, I see that Omid meant $\{0,1\}$, so you're right and I made a very stupid mistake.

I can't apologize enough.. Maybe it's better this way, though: it makes the problem a lot more challenging, doesn't it? :)

i think we have the following(i'm not totally sure, just thought of it for a very short moment):
if $P(0,...0,1,0,...,0)=1$, where the 1 is at i-th place, then $P(x_1,...,x_n)=x_i$ if $x_i$ is maximal(it may be equal to some other number). if $P(1,...,1,0,1,...,1)=0$, then $P(x_1,...,x_n)=x_i$ if $x_i$ is minimal(it may be equal to some other number, again). else the whole thing does not depend on $x_i$.
if this is true, the problem is easy. assume $Q(x_1,...,x_n)=x_i\neq x_j=P(x_1,...,x_n)$. by the above, $x_i$ has to be maximal and $x_j$ the minimal(or the other way round) element. but then $P(1,...,1,0,1,...,1)=0$ and $Q(1,...1,0,1,...1)=1$, where everything is 1 except for the j-th element, contradiction.
i hope you may understand what i want to say.

Peter

sorry for that nonsense, a counterexample for my claim is $\min\{x_1,\max\{x_2,x_3\}\}$. choose $x_1=2,x_2=1,x_3=0$ to get 1.

I think I have something, but it's really hard for me to explain it. That's why I'll only give an outline.

First of all, transform $P$ into its simplest form, meaning that there's no $\min(\min(),\ldots)$ (the same for $\max$, of course).

It's easier now to talk in terms of trees. We construct a tree corresponding to $P$ as follows: a node that's not a leaf is either $\min$ or $\max$, and the leafs have the labels $1,2,\ldots,n$ (representing $x_1,\ldots$). Each leaf (and each node) is connected to the node that's immediately above it. For example, if we have $\min(x_1,\ldots)$, then $x_1$ is connected to the node representing that $\min$.

I think that two functions which are equal on $\{0,1\}$ have the same tree, and we can show this by induction after $n$.

Let's assume leaf $x_1$  is connected to two different types of nodes in the two trees (a $\min$ node and a $\max$ node in the $P$ and $Q$ trees, respectiely). In this case we can choose the values of the other variables s.t. $P\ne Q$ (I'm not proving this, as you might have noticed; I'll explain it later if someone wants me to :)).

After we have established that $1$ is connected to the same type of node (say $\max$), we can take it to be $0$, and it doesn't influence the outcome anymore, so we're left with two trees with $n-1$ leafs, so, by using the induction hypothesis, we show that they are the same, except maybe for the position of leaf $1$.

Then it's not that hard to show that $1$ has the same position in the two trees. Here's where we'll use the fact that between two $\max$ nodes on different levels there must be at least a $\min$ node, and vice-versa (i.e. that's why it's important for the function to be represented in its "simplest" form).

Well of course you think very difficult.
First we know that for every permutation $\sigma$ and formula $f$ there exist k that for every :  $x_{\sigma(1)} \geq x_{\sigma(2)} \geq ... \geq x_{\sigma(n)}$ then:
$f(x_1,x_2,...,x_n)=x_k$
Now suppse for a permutation $\sigma$ and $x_{\sigma(1)} \geq x_{\sigma(2)} \geq ... \geq x_{\sigma(n)}$ then $P(x_1,...,x_n) \ne Q(x_1,...,x_n)$
then suppose $P(x_1,...,x_n)=x_i$ and $Q(x_1,...,x_n)=x_j$.Suppose $i=\sigma(m),j=\sigma(k),m>k$ now consider$y_{\sigma(1)}=1, y_{\sigma(2)}=1,y_{\sigma(k)}=1,y_{\sigma(k+1)}=0,...,y_{\sigma(n)}=0$
then $(y_1,...y_n) \in (0,1)^n$ and $P(y_1,...y_n) \ne Q(y_1,...,y_n)$ Contradiction.


geometry
number theory
prime factorization
modular arithmetic
prime numbers

this really shouldn't be in the Getting Started forum... more like in the Intermediate Topics forum... maybe the moderator(s) should move this.

[quote="MarkSmith888"]How many different (x,y) solve the following equation:
7X+13Y = 495. (x and y are both positive intergers).
Is the only way to do this is with Diopahtine equations or is there something sqecial about this equation?
[/quote]

It's not that hard w/ Diophantine's...
X......Y
2......37
15....30
28....23
41....16
54....9
67....2

for the first one, can you use a graphing calc??? :)

[quote="MarkSmith888"]Just some quick question I need help on.

Is there some quick way to factor 51,890,256  without going through tons of numbers.  My area tournament always has some big number like that with some wierd prime number factors in it and then wants us to find the number of factors.

When 2^500 is divided by 13, the remainder in the division is?  Usually the remainder repeats but I went up to 2^12 and still couldn't find anything.

How many different (x,y) solve the following equation:
7X+13Y = 495. (x and y are both positive intergers).
Is the only way to do this is with Diopahtine equations or is there something sqecial about this equation?[/quote]
1) It's really easy to do a prime factorization usually.
51890256=2*25945128=2^2*12972564=2^3*6486282= 2^4*3243141. Now add up the digits to get 18, so it's divisible by 3. 2^4*3*1001047. Digits again, 2^4*3^2*360349. That's not divisible by 5 or 7, but it is by 11. The rest of the factorization is 17,41,47, and I used a TI89 somewhere in there just because it got boring. But anyway, the stuff you factor on contests usually isn't like that. It would probably be more like 1058400, which can easily be factored by hand.

2) Learn and use modular arithmetic for that.

3) Factor 495.

For your 2^500 problem - what did you get when you hit 2^12 (what was your remainder)?

As for how to knock these off quickly in general, go grab your AoPS Volume 2 & read about Fermat's Theorem (and Euler's generalization while you're at it).

On the remainder one:
Hmmm...I do these a little different, but I could be wrong
I find the units digit of A^(n-1) and multiply it by the units digit of A^n, divide , and get the remainder. I've never seen this not work, but I've only tried it on a few problems.
So in this case : 8*6=48../13=3r9 so answer is 9?

I know I'm probably just trying to get something when there is nothing, but is there some easy way to factor 51,890,256 without using a graphing calculator.  This question is actually taken from a real test that doesn't allow calculators.  Usually, there is a question like this with some wierd prime numbers as factors, some other numbers that have been on the tests have included 9677954 and 3988009.  It just seems like there must be some way of doing this without just guessing and checking since that might take a while for some of these numbers.

MagnusRex, you're right.  Wow, that's pretty nice.  Must be some logic to it because it seems to work.

Nope, there is no quick way of factorising numbers like that. Its unusual to have a problem like it in a competition, either its just testing your arithmetical skills, or there is some trick to the number (like 104060401 or something). Or maybe most people don't realise its easy to work out after you get its prime factorization.

The easiest way to solve the remainder one is to know that any number to the power of (prime - 1) is 1 more than a multiple of that prime. Ie in this case 2^12 is 1 more than a multiple of 13. Or you could get that from just working your way up, 2, 4, 8, 3 (8x2 = 16, 16-13 = 3), 6, 12, etc.
And in the end 500 is so many twelves plus 8 remainder, so its the same as 2^8, which gives answer 9.

Magnus - I want to know what you have done, can you explain what your A and n are, and how you work out their units digits?

Ok. First of all let me explain that this only works when the number being expanded is one digit (unless its like 2003^2003...its worked on those before too because there are going to be lots of zeroes). Anyhow, we have 2^500. The units digits of the powers of 2 go 2, 4, 8, 6, so 2^499 would end in an 8, and 2^500 would end in a 6. I multiplied 6 by 8, and divided by the number that the problem specified (13) and got the remainder. Let me say that I'm not really sure why this works, because I'm almost positive that the last 2 digits of 2^500 are not 48. I just started using this method a while back when I did not know about mods.
So here's the thing again with algebra: When A is a one-digit number, the remainder when a^n is divided by x is
[(units digit of a^(n-1))(units digit of a^n)]/x
Maybe some of you who are really good at proofs can figure out why this method works (if, indeed, it always does :roll:  :D )

Whoa, mark, are you in Huntsville, AL? Do you go to Grissom? Who are you!

MargnusRex, I tried a few others like 2^5/13 and it doesn't seem to work.  I might have used your way wrong but anyhow I didn't get the right answer.  And yea, I do go to Grissom, what school in Birmingham do you go to?  Wow, the world is pretty small.

Using Fermat's Theorem for #2

2^12 :equiv: 1 (mod 13)
2^492  :equiv: 1^492 (mod 13)
2^500 :equiv: 2^8 :equiv: (2^4)^2  :equiv: 3^2  :equiv: 9 (mod 13)

Ah. I'm afraid your method doesn't work. If you do that, you find the remainder on division by 10 (which is what the units digit is). So you get 48 which means that 2^500 is 8 more than a multiple of 10. It can't say anything about any other number. Eg whats the remainder of dividing 2^500 by 3? 48 is divisible by 3, but the answers not 0 :)

Haha, oh well. It was fun for that problem.  :D  I have to use mods anyhow now because they ask problems like, "what is the remainder when 19^2001 is divided by 23?" Sorry I confused you guys.
Mark, I go to Briarwood. What class are you in? I'm in Alg2, but our teams sucks this year because our teacher insists on using a book (even though we did the whole thing in 8th grade).

Estimate (.9999)^7 to the nearest one-hundred thousandth. Is there some way of doing this without multiplying out.  My friend in AP Stats said something about it but he didn't explain.

0.9999 = 1 - 10-4. Using the binomial expansion, we get 1 - 7(10-4) + 21(10-8) - 35(10-12) etc. Most of the last terms are so small we don't need to take them into account. We get 0.99930.

[quote="nutz_for2.718281828"][quote="JHu"]When 2^500 is divided by 13, the remainder in the division is?  Usually the remainder repeats but I went up to 2^12 and still couldn't find anything.[/quote]

You made a mistake. Use mods. That wil take you to about 11 before you see the answer

[hide="pattern"]2^0 , 1
2^1 , 2
2^2 , 8
2^3 , 3
2^4 , 6
2^5 , 12
2^6 , 11
2^7 , 9
2^8 , 5
2^9 , 10
2^10 , 7
2^11 , 1[/hide]

By that pattern, the answer [hide="is"]7[/hide][/quote]

Please do not revive old topics (especially ones almost 3 years old  :!: )
Thanks and Locked.


methinks the eccentricity is always c/a. BTW, methinks also that one really cool defenition of a conic is that you have a focus and you have a directrix, and the distance from any point on the conic to the focus is the eccentricity times the distance from that point to the directrix. (ellipses and hyperbolas have two directrices to correspond with their two foci)

Lol... If I ever meet you, I can use this post... haha...(btw, your famous pizza is still used...)


If $-1 \le x \le-\frac{1}{\sqrt{2}}$, simplify the expression
\[\sqrt{1+2x\sqrt{1-x^{2}}}+\sqrt{1-2x\sqrt{1-x^{2}}}. \]

[quote="skimnc"] Solution[/quote]

[hide="reconsider the last step"]
$y^{2}=4x^{2}$
$\implies y= \pm 2x$

Since $x$ is negative, we take the positive solution, which is $\boxed{-2x}$.[/hide]

Thanks.  That's the one thing my teacher raves on more than anything, I can't believe I forgot that.  (Edited.)

[quote="kohjhsd"]Since $x$ is negative, we take the positive solution, which is $\boxed{-2x}$.[/quote]
Reconsider the last step.  We take the negative.  Hehe.  (I'm only kidding!)

[quote="Altheman"]let $x=\sin\theta$ where $\theta\in[\pi,\frac{5\pi}{4}]$

then $\cos\theta=-\sqrt{1-x^{2}}$ [third quadrant]

so we have $\sqrt{1-\sin2\theta}+\sqrt{1+\sin 2\theta}$
$=\sqrt{1-\cos\frac{\pi}{2}-2\theta}+\sqrt{1+\cos \frac{\pi}{2}-2\theta}=$

${\sqrt{2}(-\sin \frac{\pi}{4}-\theta}+-\cos\frac{\pi}{4}-\theta)$ [again that is in 3rd quadrant]

$=-2\sin x=-2x$[/quote]
I believe you mean
$=\sqrt{1-\cos ( \frac{\pi}{2}-2\theta )}+\sqrt{1+\cos ( \frac{\pi}{2}-2\theta)}$ ${=\sqrt{2}(-\sin ( \frac{\pi}{4}-\theta})-\cos ( \frac{\pi}{4}-\theta ) )$ $=-2 \sin \theta$ $=-2x$

[quote="Altheman"]let $x=\sin\theta$ where $\theta\in[\pi,\frac{5\pi}{4}]$

then $\cos\theta=-\sqrt{1-x^{2}}$ [third quadrant]

so we have $\sqrt{1-\sin2\theta}+\sqrt{1+\sin 2\theta}$
[/quote]
What I did from there is square it (then square root it back) to get (T is Theta):

sqrt ( 2 + 2 cos (2T) )

which simplified to:

2 cos T

which has a positive coefficient, and it ends in cosine, not sine. Is there any reason why I have a different answer by doing this? (Should the square root of the square of a number equal that number?)

[b]Compare the following formulas, where[/b] $\mathrm{sgn}x=\{\begin{array}{ccc}-x & , & x<0\\\ 0 & , & x=0\\\ x & , & x>0\end{array}$ :

$1\blacktriangleright \{\begin{array}{c}\{a,b,c\}\subset\mathcal R\\\ b\ne 0\ ,\ c>0\\\ \ a+b\sqrt c>0\\\ d>0\ ,\ d^{2}=a^{2}-b^{2}c\end{array}\|\Longrightarrow \sqrt{a+b\sqrt c}=\sqrt{\frac{a+d}{2}}+\mathrm{sgn}b\cdot\sqrt{\frac{a-d}{2}}$ ;

$2\blacktriangleright \{\begin{array}{c}\{a,b\}\subset\mathcal R\ ,\ b\ne 0\\\ r=\sqrt{a^{2}+b^{2}}\\\ z^{2}=a+b\cdot i\in\mathcal C\end{array}\|\Longrightarrow z_{1,2}=\pm(\sqrt{\frac{r+a}{2}}+i\cdot \mathrm{sgn}b\cdot \sqrt{\frac{r-a}{2}})$.

[b]Example 1.[/b] $x\in [-1,1]$ and $d=|2x^{2}-1|$ $\Longrightarrow$ $\sqrt{1\pm 2x\sqrt{1-x^{2}}}=\sqrt{\frac{1+|2x^{2}-1|}{2}}\pm \mathrm{sgn}x\cdot\sqrt{\frac{1-|2x^{2}-1|}{2}}=\{\begin{array}{ccc}\sqrt{1-x^{2}}\pm x & , & |x|\le \frac{\sqrt 2}{2}\\\\ |x|\pm \mathrm{sgn}x\cdot\sqrt{1-x^{2}}& , & \frac{\sqrt 2}{2}<x\le 1\end{array}$.

[b]Remark.[/b] $1\pm 2x\sqrt{1-x^{2}}\ge 0\Longleftrightarrow$ $1\ge \mp 2x\sqrt{1-x^{2}}$ $\Longleftrightarrow$ $1\ge |2x\sqrt{1-x^{2}}|$ $\Longleftrightarrow$ $1\ge 4x^{2}(1-x^{2})$ $\Longleftrightarrow$ $4x^{4}-4x^{2}+1\ge 0$ $\Longleftrightarrow$ $(2x^{2}-1)^{2}\ge 0$.

[b]Example 2.[/b] $z^{2}=5-12\cdot i$. Observe that $r=\sqrt{5^{2}+12^{2}}$, i.e. $r=13$. Therefore, $z_{1,2}=\pm(\sqrt{\frac{13+5}{2}}-i\cdot\sqrt{\frac{13-5}{2}})=\pm(3-2\cdot i)$.

What is $\mbox{sgn}b$?

That looks not quite High School Basic. I was not expecting this from this problem :roll:

Edit: oh thanks

[quote="p4fn2w"][quote="kohjhsd"]What is $\mbox{sgn}b$?

That looks not quite High School Basic. I was not expecting this from this problem :roll:[/quote]

I believe that means "the sign of b (positive or negative)"

Im still trying to decipher the importance of the two equations... so far I get the feeling that it has something to do with complex numbers, which have something to do with trig identities....

---
update: The main difference between the two equations, that I can see, is that in the first equation C = some real number, and the second equation C = i (imaginery), and thus squaring-and-squarerooting sqrt( a + b sqrt(c)) yields different results depending if C is real or if C is imaginery. I guess sine function uses complex numbers, somehow (and the importance of the 2 equations is that squareing-and-squarerooting doesnt work for sine functions)?

I also have one last small question for Altheman's solution -- he substituted x = sin T, and simplified the radical to -2(sin T). But when I plug in x = cos T, the simplified answer is still -2(sin T), and since sin T != cos T, you cant really say -2(sin T) = -2x if x = cos T.

Somehow though, I have the feeling there is a solid reason for this, and it too involve complex numbers...[/quote]

no you have to change the signs on the last square root thing, so it works out


calculus integration algebra polynomial geometry circumcircle inradius

[size=150][b]Selected Problems from Open Contests[/b][/size]

[b][size=150]Juniors problems[/size][/b]

[b]Problem 1.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205621#1205621
Find all positive integers N with at most 4 digits such that the number obtained by reversing the order of digits of N is divisible by N and differs from N.

[b]Problem 2.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205623#1205623
Call a scalene triangle K [i]disguisable[/i] if there exists a triangle K′ similar to K with two shorter sides precisely as long as the two longer sides of K, respectively. Call a disguisable triangle [i]integral[/i] if the lengths of all its sides are integers.
(a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter.
(b) Let K be an arbitrary integral disguisable triangle for which no smaller integral
disguisable triangle similar to it exists. Prove that at least two side lengths of K are
perfect squares.

[b]Problem 3.[/b] http://www.mathlinks.ro/viewtopic.php?t=217559
In a school tennis tournament with $ m \ge 2$ participants, each match consists of 4 sets. A player who wins more than half of all sets during a match gets 2 points for this match. A player who wins exactly half of all sets during the match gets 1 point, and a player who wins less than half of all sets gets 0 points. During the tournament, each participant plays exactly one match against each remaining player. Find the least number of participants m for which it is possible that some participant wins more sets than any other participant but obtains less points than any other participant.

[b]Problem 4.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205636#1205636
Call a k-digit positive integer a [i]hyperprime[/i] if all its segments consisting of $ 1, 2, ..., k$ consecutive digits are prime. Find all hyperprimes.

[b]Problem 5.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205638#1205638
In an exam with k questions, n students are taking part. A student fails the exam
if he answers correctly less than half of all questions. Call a question easy if more than half of all students answer it correctly. For which pairs (k, n) of positive integers is it possible that
(a) all students fail the exam although all questions are easy;
(b) no student fails the exam although no question is easy?

[b][size=150]Seniors problems[/size][/b]

[b]Problem 1.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205641#1205641
Let $ a_n \equal{} 1\plus{}2\plus{}...\plus{}n$ for every $ n \ge 1$; the numbers an are called triangular. Prove that if $ 2a_m \equal{} a_n$ then $ a_{2m\minus{}n}$ is a perfect square.

[b]Problem 2.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205643#1205643
Three circles with centres A, B, C touch each other pairwise externally, and touch circle c from inside. Prove that if the centre of c coincideswith the orthocentre of triangle ABC, then ABC is equilateral.

[b]Problem 3.[/b] http://www.mathlinks.ro/viewtopic.php?t=217566
Let $ b$ be an even positive integer for which there exists a natural number n such
that $ n>1$ and $ \frac{b^n\minus{}1}{b\minus{}1}$ is a perfect square. Prove that $ b$ is divisible by 8.

[b]Problem 4.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205653#1205653
The Fibonacci sequence is determined by conditions $ F_0 \equal{} 0, F1 \equal{} 1$, and $ F_k\equal{}F_{k\minus{}1}\plus{}F_{k\minus{}2}$ for all $ k \ge 2$. Let $ n$ be a positive integer and let $ P(x) \equal{} a_mx^m \plus{}. . .\plus{} a_1x\plus{} a_0$ be a polynomial that satisfies the following two conditions:
(1) $ P(F_n) \equal{} F_{n}^{2}$ ;
(2) $ P(F_k) \equal{} P(F_{k\minus{}1}) \plus{} P(F_{k\minus{}2}$ for all $ k \ge 2$.
Find the sum of the coefficients of P.

[b]Problem 5.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205655#1205655
Does there exist a natural number $ n$ such that $ n>2$ and the sum of squares of some $ n$ consecutive integers is a perfect square?

[b]Problem 6.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205656#1205656
Tangents $ l_1$ and $ l_2$ common to circles $ c_1$ and $ c_2$ intersect at point $ P$, whereby tangent points remain to different sides from $ P$ on both tangent lines. Through some point $ T$, tangents $ p_1$ and $ p_2$ to circle $ c_1$ and tangents $ p_3$ and $ p_4$ to circle $ c_2$ are drawn. The intersection points of $ l_1$ with lines $ p_1, p_2, p_3, p_4$ are $ A_1, B_1, C_1, D_1$, respectively, whereby the order of points on $ l_1$ is: $ A_1, B_1, P, C_1, D_1$. Analogously, the intersection points of $ l_2$ with lines $ p_1, p_2, p_3, p_4$ are $ A_2, B_2, C_2, D_2$, respectively. Prove that if both quadrangles $ A_1A_2D_1D_2$ and $ B_1B_2C_1C_2$ are cyclic then radii of $ c_1$ and $ c_2$ are equal.

[b]Problem 7.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205659#1205659
Find all positive integers n such that one can write an integer 1 to $ n^2$ into each unit square of a $ n^2 \times n^2$ table in such a way that, in each row, each column and each $ n \times n$ block of unit squares, each number 1 to $ n^2$ occurs exactly once.

[b]Problem 8.[/b] http://www.mathlinks.ro/viewtopic.php?p=1205661#1205661
Consider triangles whose each side length squared is a rational number. Is it true
that
(a) the square of the circumradius of every such triangle is rational;
(b) the square of the inradius of every such triangle is rational?

[b][size=150]Selected Problems from the Final Round of National Olympiad[/size][/b]

[b][size=150]Grade 9[/size][/b]

[b]Problem 1[/b] http://www.mathlinks.ro/viewtopic.php?p=1205666#1205666
Two medians drawn from vertices A and B of triangle ABC are perpendicular. Prove that side AB is the shortest side of ABC.

[b]Problem 2[/b]  http://www.mathlinks.ro/viewtopic.php?p=1205665#1205665
Juhan wants to order by weight five balls of pairwise different weight, using only a balance scale. First, he labels the balls with numbers 1 to 5 and creates a list of weigh-ings, such that each element in the list is a pair of two balls. Then, for every pair in the list, he weighs the two balls against each other. Can Juhan sort the balls by weight, using a list with less than 10 pairs?

[b][size=150]Grade 10[/size][/b]

[b]Problem 3[/b] http://www.mathlinks.ro/viewtopic.php?p=1205667#1205667
Two radii OA and OB of a circle c with midpoint O are perpendicular. Another circle touches c in point Q and the radii in points C and D, respectively. Determine $ \angle{AQC}$.

[b]Problem 4[/b] http://www.mathlinks.ro/viewtopic.php?p=1205669#1205669
Prove that the sum of the squares of any three pairwise different positive odd integers can be represented as the sum of the squares of six (not necessarily  different) positive integers.

[b]Problem 5[/b] http://www.mathlinks.ro/viewtopic.php?p=1205671#1205671
Two triangles are drawn on a plane in such a way that the area covered by their union is an n-gon (not necessarily convex). Find all possible values of the number of vertices n.

[b]Problem 6[/b] http://www.mathlinks.ro/viewtopic.php?p=1205673#1205673

The identifier of a book is an n-tuple of numbers 0, 1, .... , 9, followed by a checksum. The checksum is computed by a fixed rule that satisfies the following property: whenever one increases a single number in the n-tuple (without modifying the other numbers), the checksum also increases. Find the smallest possible number of required checksums if all possible n-tuples are in use.

[b][size=150]Grade 11[/size][/b]

[b]Problem 7[/b] http://www.mathlinks.ro/viewtopic.php?t=217585
Find all real numbers a such that all solutions to the quadratic equation $ x^2 \minus{} ax \plus{} a \equal{} 0$ are integers.

[b]Problem 8[/b] http://www.mathlinks.ro/viewtopic.php?p=1205676#1205676
A 3-dimensional chess board consists of $ 4 \times 4 \times 4$ unit cubes. A rook can step from any unit cube K to any other unit cube that has a common face with K. A bishop can step from any unit cube K to any other unit cube that has a common edge with K, but does not have a common face. One move of both a rook and a bishop consists of an arbitrary positive number of consecutive steps in the same direction. Find the average number of possible moves for either piece, where the average is taken over all possible starting cubes K.

[b]Problem 9[/b] http://www.mathlinks.ro/viewtopic.php?p=1205677#1205677
A circle passing through the endpoints of the leg AB of an isosceles triangle ABC
intersects the base BC in point P. A line tangent to the circle in point B intersects the circumcircle of ABC in point Q. Prove that P lies on line AQ if and only if AQ and BC are perpendicular.

[b]Problem 10[/b] http://www.mathlinks.ro/viewtopic.php?p=1205680#1205680
Find all pairs $ (m, n)$ of positive integers such that $ m^n \minus{} n^m \equal{} 3$.

[b]Problem 11[/b] http://www.mathlinks.ro/viewtopic.php?p=1205681#1205681
Some circles of radius 2 are drawn on the plane. Prove that the numerical value of
the total area covered by these circles is at least as big as the total length of arcs bounding the area.

[b][size=150]Grade 12[/size][/b]

[b]Problem 12[/b] http://www.mathlinks.ro/viewtopic.php?p=1205682#1205682
Consider a cylinder and a cone with a common base such that the volume of the
part of the cylinder enclosed in the cone equals the volume of the part of the cylinder outside the cone. Find the ratio of the height of the cone to the height of the cylinder.

[b]Problem 13[/b] http://www.mathlinks.ro/viewtopic.php?p=1205685#1205685
Let $ x, y, z$ be positive real numbers such that $ x^n, y^n$ and $ z^n$ are side lengths of some triangle for all positive integers $ n$. Prove that at least two of x, y and z are equal.

[b]Problem 14[/b] http://www.mathlinks.ro/viewtopic.php?p=1205688#1205688
Does there exist an equilateral triangle
(a) on a plane; (b) in a 3-dimensional space;
such that all its three vertices have integral coordinates?

The end !!!!!!!!!!!!!!!!!!!!!!!!!!!

the above posts collections have been created, here they are
[url=https://artofproblemsolving.com/community/c1090810_2007_estonia_math_open_junior_contests]2007 Estonia Math Open Junior Contests[/url],
  [url=https://artofproblemsolving.com/community/c1090811_2007_estonia_math_open_senior_contests]2007 Estonia Math Open Junior Contests[/url],
 [url=https://artofproblemsolving.com/community/c1090813_2007_estonia_national_olympiad]2007 Estonia National Olympiad[/url]

PS. I have asked to create the corresponding folder for these problems in Aops Contests Collections [url=https://artofproblemsolving.com/community/c40244h1064480p14240613]here[/url]


How many positive integer roots does the equation $\frac1{9a}+\frac1{223b}= \frac1{2007}$ have?

[quote="EUCLA"]
Then $223b+9a=ab$

 $a=\frac{223b}{b-9}$

$a=\frac{223b-2007+2007}{b-9}$

 $a=223+\frac{2007}{b-9}$
[/quote]

Sorry, but could you be more elaborate? I don't understand how did you reach those conclusions.

EDIT: Oh, I see it:

$223b+9a=ab$

$223b = ab-9a$

$223b = a(b-9)$

$\frac{223b}{b-9}= a$

$a=\frac{223b-2007+2007}{b-9}$

$a=\frac{223(b-9)+2007}{b-9}$

:wink:

by the way, welocome on Mathlinks  :lol:

EDIT: I didn't realised you'd answered yourself

I don't believe this thread belongs in the [b]Pre-Olympiad[/b] forum.  :)

[hide="Another approach"]
Multiply out denominators to obtain $223b+9a=ab$. Then, we have that \[(a-223)(b-9)=2007 \cdot 1 = 3 \cdot 669 = 9 \cdot 223 \] and we proceed as in the solution above.[/hide]

Notice $2007=9*223$, we have $\frac{1}{9a}+\frac{1}{223b}=\frac{1}{2007}\Longleftrightarrow \frac{223}{a}+\frac{9}{b}=1$, which shows that the line

$\frac{x}{a}+\frac{y}{b}=1$ passes through the point $(223,\ 9)$.

We are to find the $x,y$ intersept $a,\ b$ respectively satisfying this condition.


probability
geometry
geometric transformation
LaTeX
complementary counting
Pascal\u0027s Triangle

We flip a fair coin 10 times.  What is the probability that we get heads in at least 6 of the 10 flips?

[quote="Kouichi Nakagawa"]$ \binom{10}{6}\left(\frac {1}{2}\right)^6\left(\frac {1}{2}\right)^{10 \minus{} 6} \equal{} \frac {210}{1024} \equal{} \boxed{\frac {105}{512}}$.[/quote]
I think it's wrong. The question asks about the probability of appearance of at least 6 heads in 10 flips.So using complement rule, we need to find the probability of appearance of at most 5 heads in 10 flips, then get 1-this probability. The ans is: $ 0.37695$. Your result only shows the probability of 6 heads in 10 flips, which is not enough.

It is just as easy to count the cases themselves, rather than complementary counting them. The total number of ways is 2^10=1024.
10 Heads- 1 way
9 Heads- 10C9= 10
8 Heads- 10C8= 45
7 Heads- 10C7= 120
6 Heads- 10C6= 210

Adding these yields 396, so the answer is $ 384/1024\equal{} 193/512$

Another solution follows-
We know we need 10C10+10C9+....10C6
Through Pascal's triangle, we know 10C10+10C9+...10C0= 2^10=1024. This makes sense, as this would be all the possibilities for 1024/1024=1. To find what we need, we first subtract the term in the middle, 10C5. Then we take 1/2 the remaining number for our answer.

1024-252= 772
772/2= 386
and we get the same answer 193/512. This way is probably more elegant.

I overlooked "least".

When Japanese translates this problem, are you as follows?
[quote="Japanese translates"]私たちはコインを $ 10$ 回はじきます. $ 10$ 回はじいたとき少なくとも $ 6$ 回表が出る確率は何ですか?[/quote]
If translation is right...

$ \sum_{i\equal{}6}^{10}\binom{10}{i}\left(\frac{1}{2}\right)^i\left(\frac{1}{2}\right)^{10\minus{}i}\equal{}\frac{1}{1024}\sum_{i\equal{}6}^{10}\binom{10}{i}\equal{}\frac{1}{1024}\cdot\frac{2^{10}\minus{}\binom{10}{5}}{2}\equal{}\frac{386}{1024}\equal{}\boxed{\frac{193}{512}}$.

[b]Remark[/b]
Let $ l\equal{}\left\{l\in\mathbb{Z}|l\le\frac{m}{2}\right\}$.
$ 2^m\equal{}\sum_{i\equal{}0}^{m}\binom{m}{i}\equal{}\sum_{i\equal{}0}^{l}\binom{m}{i}\plus{}\sum_{i\equal{}l\plus{}1}^{m\minus{}(l\plus{}1)}\binom{m}{i}\plus{}\sum_{i\equal{}m\minus{}l}^{m}\binom{m}{i}$.
and
$ \sum_{i\equal{}0}^{l}\binom{m}{i}\equal{}\sum_{i\equal{}m\minus{}l}^{m}\binom{m}{i}$.
So, $ \sum_{i\equal{}0}^{l}\binom{m}{i}\equal{}\sum_{i\equal{}m\minus{}l}^{m}\binom{m}{i}\equal{}\frac{1}{2}\left(2^m\minus{}\sum_{i\equal{}l\plus{}1}^{m\minus{}(l\plus{}1)}\binom{m}{i}\right)$.

[quote="akhil0422"]Dude, what the heck are you talking about?[/quote]
I talked about generalization in Remark.

I don't want to spam, but I just have to say that's pretty awesome. It took me a while to start understanding though. Nice $ LaTex$.

Two problems before this, I get the problem of:

[quote=Alcumus]What is the probability that Fatima gets fewer heads than tails if she flips 10 coins? [/quote]

Compare to this problem. Basically the same thing. :P (exactly the same numbers etc)

Anyway, [hide=sketch]Just find the probability of getting 5 and 5, then you know it's symmetrical in if there are more heads or tails, so just cut $1-P(5,5)$ in half.[/hide]

Here's my solution to this problem:
Since there has to be at least 6 heads, we need P(Heads>Tails)=1/2(1-P(Heads=Tails)), the only way we can achieve this is 5 heads and 5 tails so this can be done in 10C5 ways so it can be done in 252 ways, this has to be over 2^10 ways, so P(Heads=Tails)= 252/2^10. If we plug this in, we have 193/512 after calculation.

[quote=shresthahari]hey rubixsolver, don't post direct alcumus solutions, please. The AoPS police reported me for doing this.[/quote]

Also happened to me before :(

There are $2^{10} = 1024$ possible outcomes of the 10 coin flips. The number of ways to get 8, 9, or 10 heads is $\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=45+10+1=56$. So the probability is $\dfrac{56}{1024}=\boxed{\dfrac{7}{128}}$.

We could solve this problem using casework, but using a little bit of symmetry and complementary probability gives us a more elegant solution. Since each coin flips heads or tails with equal probability, by the principle of symmetry the probability of getting fewer heads than tails is the equal to the probability of getting fewer tails than heads. Additionally, there are only three possible outcomes: getting fewer heads than tails, getting fewer tails than heads, or getting the same number of both. If we let $x$ represent the probability of the first outcome (which is the same as the probability of the second outcome) and $y$ represent the probability of the third outcome, we get the equation $2x + y = 1 \Rightarrow x=\dfrac{1-y}{2}$. So all we need to do is calculate the probability of getting the same number of heads and tails and we can then easily solve for what we want using the principle of complementary probability. Since there are two equally likely outcomes for each flip, there are a total of $2^{10}$ equally likely possible outcomes for flipping the 10 coins. We will have the same number of both heads and tails if we have exactly 5 of each, which we can count by selecting 5 out of the 10 flips to be heads which can occur in $\binom{10}{5}=252$ ways. So $y=\dfrac{252}{1024}=\dfrac{63}{256}$, and substituting that back into our first equation gives us the probability that we want: $x=\boxed{\dfrac{193}{512}}$. $THE ONE BEFORE IS WRONG$

[hide=A very simple solution]It is just the same as getting less than 4 tails. That is $\frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4}}{2^{10}}$ which simplifies to $\boxed{\frac{193}{512}}$[/hide]

Phew that was tricky!

[quote=shresthahari]Yes it is wrong...as far as I can see, it is just a calculation error.[/quote]

no, it was wrong because that only counted the number of ways if you need 8 heads

[hide=my way]
[hide=Part 1: Finding the total number of ways to flip 10 coins]First, we will find the total number of ways to flip $10$ coins. There are $2$ ways for the coin to land, heads or tails, and there are $10$ coins, so there are a total of $$\underbrace{2\cdot2\cdot2\cdot\cdots\cdot2}_{\text{10 twos}}=2^{10}=1024.$$[/hide]

[hide=Part 2: Finding the number of ways to get at least 6 heads]For flipping at least $6$ heads, we have:

\begin{align*}
\text{number of ways to flip at least 6 heads} & = & \text{number of ways to flip 6 heads}\\
& + & \text{number of ways to flip 7 heads}\\
& + & \text{number of ways to flip 8 heads}\\
& + & \text{number of ways to flip 9 heads}\\
& + & \text{number of ways to flip 10 heads.}
\end{align*}

To flip $6$ heads, we must first choose $6$ of the $10$ coins that will be heads. This can be done in $\dbinom{10}{6}$ ways. Similarly, to flip $7$ heads, there are $\dbinom{10}{7}$ ways. Continuing this way, we have $$\text{number of ways to flip at least 6 heads} = \sum_{n=6}^{10}\dbinom{10}{n}.$$ (That means $\dbinom{10}{6}+\dbinom{10}{7}+\dbinom{10}{8}+\dbinom{10}{9}+\dbinom{10}{10}.$) Simplifying, we have $$\sum_{n=6}^{10}\dbinom{10}{n}=210+120+45+10+1=386.$$[/hide]

[hide=Part 3: Answer]Combining these two parts, we have $$\frac{386}{1024}=\boxed{\frac{193}{512}}.$$

[quote=Kouichi Nakagawa]$ \binom{10}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^{10\minus{}6}\equal{}\frac{210}{1024}\equal{}\boxed{\frac{105}{512}}$.[/quote]

I did this first :oops:


Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $ 8/13$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $ \pi$ radians is 180 degrees).

[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/diagram_circle.PNG[/img]


In how many ways can we make ten pounds out of five-pence, ten-pence and twenty-pence coins?

I think that this can be done recursively.

There are $ 51$ cases: using $ 50; 49; ...; 0$ twenty-pence coins. Then for, say,  $ 49$ twenty-pence coins we could make up the other $ 20$ pence with tens and fives, again using $ 2, 1$ or $ 0$ tens.

But it seems that this will take some time.

Perhaps there is an easier way.


calculus
derivative
trigonometry
complex analysis

Note that f_k(x) = sin kx has unbounded derivative at 0 but f_k(z) = sin kz on the unit disc, appealing to the Cauchy estimates, has bounded partial derivative w.r.t z. Why is this not contradictory?

Because $ \sin kz$ is not bounded by $ 1$ in the unit disk ;)


I select a $4$ Digit number at random .

What is the Probability that it is not divisible by any pair of Primes between $5$ and $23$ both inclusive?

[quote="hydro"]I found around $82 \%$...[/quote]

how do you get this? direct calculation or otherwise?

I got it by ugly calculation... I got $p=\frac{7403}{9000}=0.822 \bar{5}$


[quote="guile"][hide="question"]Ms. Perez wanted to cancel the field trip the scheduled the next so she had to inform all her students. However, Ms. Perez only chose to call one student who will call three students. Now each of these three students would again call three students. How many student phone calls is needed to call 51 students completely?
[b]Note: [/b]A student is required to call three students unless there is less than three left.

What is the rule for any total number of students if a student can call 3 other students? If a student can call [b]n[/b] students?

Please forgive my English. I had hard time constructing this.[/hide][/quote]

[hide="solution"]
Okay.

We just list the number of students called at each point of time:
$ 1,3,9,27,11$
[11 students are called last because you can't have more students appearing out of nowhere!]

Are you asking for the number of times [the people called call 3 students for each person?] [b]If so, the answer is 4.[/b] [i]Or are you asking for the number of students who called other students? The answer is then $ 1 \plus{} 3 \plus{} 9 \plus{} 27 \plus{} \lfloor{11/3}\rfloor$[/i]. Please clarify this matter.

[/hide]


A*B = A^3-B^3
and
A~B = (A+B)^3

So what does (2*3)~9 = ?

I figured (8-27)~9 = -15~9 = (-15+9)^3 = (-6)^3 = -216... but that was not the correct answer, any help?

$ 9\minus{}27 \equal{} \minus{}19 \text{ not} \minus{}15$


http://www.bored.com/games/puzzle/1/one_impossible_level/fullscreen/

I hate it

I beat it!!! I don't know what the "now vote 5" is about

omg i cant get past the part where its like floors

Video deleted for showing the "trick" about the last part.

What is "now vote 5"?

I think "now vote 5" means "ok, you beat the game, now rate it 5 out of 5."

[hide="Spoiler?"]I don't understand the last part either... I thought ANYTHING you touched killed you...[/hide]


calculus
vector
geometry
3D geometry
prism
linear algebra
matrix

Three edges ota triangular prism are defined by vectors a=[1,3,2],vector b=[2,2,4], and vector c=[12,0,-6]

a)draw a diagram of the prism identifying  which edge of the prism is defined by each vector
******b)Determine the VOLUME of the prism.

general principle: the volume of the solid in $ n$-space spanned by $ n$ vectors $ v_1, ..., v_n$ is the absolute value of the determinant of the $ n\times n$ matrix whose columns are the $ v_i$'s.

Note: where pleurestique says "solid", they're referring to the parallelipiped. A triangular prism cuts off part of that solid, for a fraction of the volume.

There are several possible choices for (a), based on choosing which two vectors are edges of a triangle. All of them lead to the same volume in (b).

lol, i was looking for this question to for the test tmrw. sufiya, - and i found this off google

didnt mrs tytanek take this up


Find all functions $ f$ defined on $ \mathbb R$ with the property :

   $ f(xf(y)\plus{}x) \equal{} xy \plus{} f(x) , \forall x ,y \in \mathbb R$.

[color=red][i] I'm seeking for a  simple and short solution.[/i][/color]

For $ x\equal{}1$, we have $ f(f(y)\plus{}1)\equal{}y\plus{}f(1)$.
It is now easy to deduce that $ f$ is bijective from $ \mathbb{R}$ onto itself.

For $ y\equal{}0$, we have $ f(xf(0)\plus{}x) \equal{} f(x)$ so that $ xf(0)\plus{}x \equal{} x$ for all $ x$, which gives $ f(0)\equal{}0$.

Now, there exists $ a$ such that $ f(a)\equal{}\minus{}1$. For $ y\equal{}a$, it gives $ 0 \equal{} ax\plus{}f(x)$, so that $ f$ is linear.

Conversely, it is easy to verify that the only linear solutions are $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$.

Hope, it is sufficiently short and simple...

Pierre.

[quote="pbornsztein"]For $ x \equal{} 1$, we have $ f(f(y) \plus{} 1) \equal{} y \plus{} f(1)$.
It is now easy to deduce that $ f$ is bijective from $ \mathbb{R}$ onto itself.

For $ y \equal{} 0$, we have $ f(xf(0) \plus{} x) \equal{} f(x)$ so that $ xf(0) \plus{} x \equal{} x$ for all $ x$, which gives $ f(0) \equal{} 0$.

Now, there exists $ a$ such that $ f(a) \equal{} \minus{} 1$. For $ y \equal{} a$, it gives $ 0 \equal{} ax \plus{} f(x)$, so that $ f$ is linear.

Conversely, it is easy to verify that the only linear solutions are $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{} x$.

Hope, it is sufficiently short and simple...

Pierre.[/quote]

 Yes , it is a simple and nice solution. Thanks. As I can see , we can not skip the fact that $ f$ is onto . How ever , after your solution ,  I can make my solution essential better and this is important :) .

[color=red][b] I wish a happy new year! [/b][/color]
 Babis


Can someone explain the concept of the LOCUS.

My textbook states that there are 5 basic Loci.

I have no idea what to do with this locus stuff.

Locus is the set of something that satisfies a certain condition. the locus of the points equidistant from a certain point is a circle.

Great but what are the other conditions and locus per condition?

The locus of the points equidistant from two points is its perpendicular bisector.

A locus is the set of points that satisfy given criteria.  For example, the locus of points that are in a plane and four units from a given point is a circle with the given point as the center and a radius of four.

Also, the locus of points a set distance from a line is two parallel lines, one above and one below the original line.


function
algebra
domain
real analysis
real analysis theorems

Show directly from the definition of uniform continuity (without using the Boundedness Theorem)
that a function f uniformly continuous on a closed, finite interval is necessarily bounded there.

Definition: A function f with domain D is called uniformly continuous on the domain D if for every  :varepsilon:  > 0 there exists a  :gamma:   > 0 such that: if s, t is in D and | x - t | <  :gamma:  then | f(x) - f(t) | <  :varepsilon:

I don't know what's your proof of Boundness Theorem...
The following proof seems to be a repeat of it...

Suppose a uniformly continuous function $f$ is not bounded on $[a,b]$. Then there exist $\{x_n\}$ such that $|f|\to\infty$. (We can choose $x_n$ s.t. $|f(x_n)|>n$)

Notice that $\{x_n\}\subset [a,b]$ then we know that $\{x_n\}$ has at least a limit point $\xi$ from Bolzano-Weierstrass Theorem. Suppose $x_{n_k} \to \xi$ as $k\to\infty$. Since $F$ is uniformly continuous, for any $\epsilon>0$, there exists $\delta>0$ s.t. $|f(x)-f(y)|<\varepsilon$ for all $|x-y|<\epsilon$. And from the definition of limit, for the $\delta$ above, there exists $K$ such that $|x_{n_j} - \xi|<delta$ for all $j>K$. Hence $|f(x_{n_j})-f(\xi)|<\epsilon$. All $f(x_{n_j}) (j>K)$ are bounded, so are the whole sequence of $\{f(x_{n_k})\}$. This contradicts with our choose of $x_n$...

There's a much easier argument, since we already know that the function is uniformly continuous. Let $\epsilon=1.$ Find $\delta$ such that if $|x-y|<\delta,\, |f(x)-f(y)|<1.$ Cut the interval $[a,b]$ into $n$ equal nonoverlapping pieces of width $\frac{b-a}n$, where $\frac{b-a}n<\delta.$ Choose a point $x_i$ in each subinterval. Then for any $x,$ $|f(x)-f(x_i)|<1$ for some $i$ and thus

$|f(x)| < \max_i|f(x_i)|+1.$


quadratics
number theory proposed
number theory

Solve equation, $ x;y\in Z$
$ x^{6}\plus{}x^{3}y\equal{}y^{3}\plus{}2y^{2}$

note that $ y\mid x\implies x \equal{} ay$.
\[ a^6y^6 \plus{} a^3y^3.y \equal{} y^3 \plus{} 2y^2\]

\[ a^6y^4 \plus{} a^3y^2 \equal{} y \plus{} 2\]

\[ \implies y\mid 2\implies y \equal{} 1,2\]
if $ y \equal{} 1$,
\[ a^6 \plus{} a^3 \equal{} 3\]
otherwise
\[ 16a^3 \plus{} 4a^3 \equal{} 4\implies 4a^6 \plus{} a^3 \equal{} 1\]
solve quadratic in $ a^3$. done.

I think $ y|x^{6}$ , we can't tell $ y|x$ directly

[quote="uNc"]Solve equation, $ x;y\in Z$
$ x^{6} \plus{} x^{3}y \equal{} y^{3} \plus{} 2y^{2}$[/quote]

$ x^3(x^3\plus{}y)\equal{}y^2(y\plus{}2)$ and so $ y\equal{}2u$ is even (else LHS even while RHS odd) and so $ x\equal{}2v$ is even and whe got $ 2v^3(4v^3\plus{}u)\equal{}u^2(u\plus{}1)$

Then $ w\equal{}gcd(u,v)\equal{}1$ else $ w^4|LHS$ while $ w^2|RHS$ then $ u|8v^6$ and so $ u|8$ and $ u\in\{1,2,4,8\}$

$ u\equal{}1$ $ \implies$ $ v^3(4v^3\plus{}1)\equal{}1$ and no solution
$ u\equal{}2$ $ \implies$ $ v^3(2v^3\plus{}1)\equal{}3$ and the solution $ (u,v)\equal{}(2,1)$ and so $ \boxed{(x,y)\equal{}(2,4)}$
$ u\equal{}4$ $ \implies$ $ v^3(v^3\plus{}1)\equal{}10$ and no solution
$ u\equal{}8$ $ \implies$ $ v^3(v^3\plus{}2)\equal{}72$ and no solution


Two squares on a $8*8$ chessboard are called touching if they have at least one common vertex. Determine if its possible for a king to begin in some square and visit all squares exacly once in such a manner that all moves except the first are made into squares touching an even number of squares already visited.

It is not possible.
It can be easily shown by contradiction on counting a summation in two ways.
Let $S(i,j)$ represent the no. of squares consecutive to $(i,j)$ already visited on reaching $(i,j)$ where $1\leq i,j \leq 8$.
Suppose that there exists a path satisfying given conditions.
we find the sum $\sum \sum_{1\leq i,j \leq 8}S(i.j)$
(i) as it is $1$ for first move and even for others(given) hence it must come out to be odd.
(ii) Now, consider any consecutive pair $A,B$. Only one of them is counted in the summation as one is reached before the other.
Hence the sum $\sum \sum_{1\leq i,j \leq 8}S(i.j)$
is the no. of possible consecutive squares which is even by symmetry(horizontal no. of consecutive squares= vertical and right diagonal =left diagonal. In fact it comes out to be $2*56+2*49 = 210$)
This gives a contradiction.


Give [tex] a,b,c,d \geq 0 [/tex].Prove that:
a)If [tex] a+b+c=3 [/tex] then [tex] a^3+b^3+c^3-3abc \geq \frac94 (a^2+b^2-2ab) [/tex].
b)If [tex] a+b+c+d=4 [/tex] then [tex] a^4+b^4+c^4+d^4 -4abcd \geq \frac{11}{12}(a^3+b^3+c^3-3abc) [/tex]  :D
The second one is not really strong.I hope that someone will help me :)

for a) (it's not very nice )
With the indentity $\displaystyle\sum a^3-3abc=\frac12(\sum a)(\sum(a-b)^2)$ we have to prove that $\displaystyle\sum(a-b)^2\ge \frac 32(a-b)^2$

which is equivalent to $\displaystyle(b-c)^2+(c-a)^2\ge\frac12(a-b)^2$
and by the hypothesis we get the equivalence $\displaystyle(2b+a-3)^2+(2a+b-3)\ge\frac12(a-b)^2\hfill{(1)}\\ $
WLOG we can say that $a\ge b$.
We will use the inequality $\displaystyle x^2+y^2\ge\frac12(x+y)^2\hfill{(*)}$
[b]case I[/b]: $b\ge c$ results that $2b+a-3\ge 0$ and $2a+b-3\ge 0$ and by (*) we get that $\displaystyle(2b+a-3)^2+(2a+b-3)^2\ge\frac12(3(a+b-2))^2$. Now we should prove that $(3(a+b-2))^2\ge(a-b)^2$ and because $a\ge b\ge c$ we have that $a+b-2\ge 0$ so the inequality becomes $2a+4b-6\ge 0\Leftrightarrow a+2b-3\ge 0$ whic is true.

[b]case II[/b]: $a\ge c\ge b$ results that $2a+b-3\ge 0$ and $3-2b-a\ge 0$ and by (*) we get that $(2a+b-3)^2+(3-a-2b)^2\ge\frac12(a-b)^2$

[b]case III[/b]: Analoguous with [b]case I[/b]
hope it's ok

$\displaystyle x^2+y^2\ge\frac12(x+y)^2$ implies that $\displaystyle(b-c)^2+(c-a)^2\ge\frac12(a-b)^2$ immediately, does it?

ouppps.... i'm dumb.. you'r right.. i don't know why I thoug that the inequality holds only when $x,y\ge 0$...  :blush:

Very nice solution :) ,try the second easy one too.Will that method work? :?

[quote="math_sipo"]notice that a^3+b^3+c^3=(a+b+c)(a+b+c-ab-bc-ca)=3((a-b)+(b-c)+(c-a))/2
the rest is simple (prove that ((a-c)+(b-c)>=(a-b)/2)
bye[/quote]

a^3+b^3+c^3-3abc=(a+b+c)(a+b+c-ab-bc-ca)


inequalities
function
inequalities theorems

Could anyone say to me what is Vornicu-Shur ineq?

[quote="darij grinberg"]It's actually called Vornicu-S[b]c[/b]hur. It was first introduced in http://www.mathlinks.ro/Forum/viewtopic.php?t=1166 . Later, I gave a more detailed proof of the case of a monotonous function f in http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9 (I showed only the case of a monotonous function f, since I have never applied the case of a convex function f on ML). Some extensions can be found in http://www.mathlinks.ro/Forum/viewtopic.php?t=41284 .

  darij[/quote]

A sharper version of Theorem 3 from the last link, as well as an upper bound, is given at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45902 .

The core idea of Vornicu-Schur is rearrangement, where if $a \ge b \ge c$, then $\min\{(a-b)(a-c),\, (c-a)(c-b)\} \ge \left|(b-c)(b-a)\right| \ge 0$. Suppose $x \ge y \ge z$ or $x \le y \le z$. If $f(x),f(y),f(z) \ge 0$ and $\max\{f(x),f(z)\} \ge f(y)$ (*), then

\[
\sum_{cyc} f(x) (a-b)(a-c) \ge 0
\]

is quite clear. Indeed, the two mentioned conditions $f$ convex and positive, or $f$ monotic and positive, are sufficient to guarantee (*).

Excuse my sloppy writing in previous posts here. I will write my even more general version here in precise language.

[color=blue][b]Theorem 1 (Generalized Schur.)[/b] Let $a, b, c$ be arbitrary reals such that $a \ge b \ge c$, and $x, y, z$ nonnegative reals such that $\sqrt{x} + \sqrt{z} \ge \sqrt{y}$. Then
\[
\sum_{cyc} x(a-b)(a-c) \ge 0
\]
With equality if and only if $a=b=c$, or $a = b, z = 0$, or $b=c, x = 0$, or $x(a-b)^2 = z(b-c)^2$ and $\sqrt{x} + \sqrt{z} = \sqrt{y}$.[/color]

Proof: For $a=b=c$, $a=b>c$, and $a>b=c$, the claim is obvious. Now suppose $a > b > c$. We have

\[
x(a-b)(a-c) + y(b-c)(b-a) + z(c-a)(c-b) \ge 0 \\
\iff x \cdot \frac{a-c}{b-c} - y + z \cdot \frac{a-c}{a-b} \ge 0 \\
\iff x + z + \left(x \cdot \left(\frac{a-b}{b-c}\right) + z \cdot \left(\frac{b-c}{a-b}\right)\right) \ge y \\
\Longleftarrow x + z + 2\sqrt{xy} = \left(\sqrt{x} + \sqrt{z}\right)^2 \ge y
\]

Where the last step used AM-GM inside the parenthesis. $\blacksquare$

[color=blue][b]Corollary to Generalized Schur.[/b] Let $a, b, c, x, y, z$ be arbitrary reals such that $x+y+z = 0$. Then

\[
\sum_{cyc} x^2(a-b)(a-c) \ge 0
\]

with equality if and only if $x(a-b)=z(b-c)$ (or, equivalently, $ax+by+cz=0$.)[/color]

Proof 1. Suppose WLOG $a \ge b \ge c$. Since $\sqrt{x^2} + \sqrt{z^2} = |x| + |z| \ge |x + z| = |y| = \sqrt{y^2}$, the desired holds by Theorem 1.

Proof 2. Write $y = -x-z$ and observe that the inequality in question reduces to $(ax-bx-bz+cz)^2 \ge 0$.

It is my opinion that the corollary is considerably easier to apply than its original theorem!

However, Theorem 1 implies the Vornicu-Schur theorem for both the monotonic and convex cases. This also allows us to introduce a third, really weird class of acceptable $f$: All $f : \mathbb{R} \rightarrow [k,4k]$ where $k$ is a nonnegative real. Examples include $f(x) = \frac{x^2+4}{x^2+1}$ and $f(x) = \frac{5}{3} + \sin(x)$. However, just how to make use of this new class in a serious inequality is unclear.

Which brings me to my point.

[quote="perfect_radio"]Is there on ML any application of Vornicu-Schur for convex functions?

I've seen it been used only for monotonically increasing/decreasing functions, and I'm wondering how it is used for convex functions[/quote]

I am attempting to develop that idea, but it seems pretty hard to do. I should've realized that my first try, [url]http://www.mathlinks.ro/Forum/topic-49482.html[/url], is easily substituted to one of Darij's generalizations of Schur. I have now posed more problems (see [url]http://www.mathlinks.ro/Forum/topic-49606.html[/url]) for which I know no solution besides this (convex $f$) extension of Schur.

[quote="me@home"]After such an amazing post, I feel stupid posting so redundantly... but what do you mean by $f : \mathbb{R}\rightarrow [k,4k]$?[/quote]
It means $f$ is a function that takes a real value and returns something in the interval $[k, 4k]$.

but the shur' inequaliti for n number is true or false

Just a remark concerning the last line of the proof of the first theorem :
Note that the it is actually 2\sqrt{xz} and not 2\sqrt{xy} ;)
Nice topic, by the way!


vector
linear algebra
linear algebra unsolved

The first part can be brute-forced:  First, say that wlog ||a|| = 1.  Then write x = alpha.a + beta.b, for some alpha, beta in C and some b in H, satisfying (b,a) = 0 and (b,b) = 1, and proceed from there.

But, of course, it'd be a whole lot better to get some idea why the thing's true, rather than merely 'bash it out'.


function
vector
real analysis
real analysis unsolved

Let $ R$ be a bounded convex set in $ \mathbb{R}^3$.  Show that for any $ C^2$ vecotor field $ H$ on $ \bar{R}$ there exists a $ C^2$ function $ f$ and a $ C^2$ function vector field $ G$ such that $ H=\nabla f+\nabla\times G$

(you may assume the equation $ \nabla^2 f=\nabla\cdot H$ has a solution).

oops, the last line should be $ \mathbf{H} \equal{} \nabla f \plus{} \nabla\times \mathbf{G}$.

any ideas on how to solve this problem? i've been thinking about this for the past day, and cant seem to find a place to start!

Place to start? Well, $ \operatorname{div}\operatorname{curl}\equal{}0$, so taking the $ \operatorname{div}$ of both sides, you get an equation for $ f$. Moreover, if $ \operatorname{div} F\equal{}0$, than $ F$ is a curl of something (if I remember right, it is called Poincare lemma; more generally, it states that every closed differential form in an open convex set is exact). That's all you need to know.


function
logarithms
calculus
calculus computations

Let $ g(x) \equal{} \log f(x)$ where $ f(x)$ is a twice differentiable positive function on $ (0, \infty)$ such that $ f(x \plus{} 1) \equal{} f(x)$. Then, for $ N \equal{} 1, 2, 3, \ldots$,

$ g''(N \plus{} \frac {1}{2}) \minus{} g''(\frac {1}{2}) \equal{} ?$
[b]Select the correct option[/b]:

$ A)$ $ \minus{} 4[1 \plus{} \frac {1}{9} \plus{} \frac {1}{25} \plus{} \ldots \plus{} \frac {1}{(2N \minus{} 1)^2}]$
$ B)$ $ 4[1 \plus{} \frac {1}{9} \plus{} \frac {1}{25} \plus{} \ldots \plus{} \frac {1}{(2N \minus{} 1)^2}]$
$ C)$ $ \minus{} 4[1 \plus{} \frac {1}{9} \plus{} \frac {1}{25} \plus{} \ldots \plus{} \frac {1}{(2N \plus{} 1)^2}]$
$ D)$ $ 4[1 \plus{} \frac {1}{9} \plus{} \frac {1}{25} \plus{} \ldots \plus{} \frac {1}{(2N \plus{} 1)^2}]$

$ f(x\plus{}1)\equal{}f(x) \to f'(x\plus{}1)\equal{}f'(x), f''(x\plus{}1)\equal{}f''(x), g(x\plus{}1)\equal{}g(x)$
$ g'(x)\equal{}\frac{f'(x)}{f(x)} \to g'(x\plus{}1)\equal{}g'(x)$

$ g''(x)\equal{}\frac{f''(x)f(x)\minus{}f'(x)^2}{f(x)^2}\to g''(x\plus{}1)\equal{}g''(x)$

so shouldnt we just get zero? unless i misunderstand the question

You should conform to the fact $ f(x)\equal{}f(x\plus{}1)$ only if $ x\in \mathbb{N}$.


Isn't that a programming competition?  Should be moved out of MathCounts.

No its not! its for mathcounts problems!

Oh, whoops.  Should read the second google entry.

My parents don't know about it, but they'll probly sign me up.  Yay!

1.Can u get nats problems on OPLET?
2. Do you have OPLET
3. If you do, can u give me the answers to 2006 and 2007 National Sprints? (or just help me solve the problems i posted)

i don't have oplet, its like subscribtion (250 bucks a year or something like that)

I don't know. My coach hasn't even done anything productive for this MATHCOUNTS year yet... :(

Thats alot of money just for some game...

btw, heres a great $ \text{\LaTeX}$ thing!

$ \sum_{i=1}^{\infty}\frac{1}{i}$

1. its not a game
2. You have absolutely no clue what you're talking about when it comes to OPLET
3. You copied that latex thing off the latex guide on AoPsWiki

[quote="math154"]OPLET can't be that useful...the key is having challenging problems, which MC doesn't do as well as some other competitions.[/quote]
i agree but i think mc is just trying to get ppl interested in math


If $S,n\in\mathbb{N}$ , $S$ is odd and $S>2^n$ . Is it possible to find a neccesary or sufficient condition for $S$ such that for the set of number $\{S-2,S-2^2,S-2^3,\cdots ,S-2^n\}$ , there are at most one prime number in the set .

It'd be very hard (impossible?) to find such a condition for every $n$ - look for example at $n=2$ - your problem then asks for finding conditions for which two numbers are twin primes. Maybe for some special $n$ it is easier but it doesn't seem so to me. Anyway it is an interesting problem  :)

Sufficient conditions are easy: something like $S$ being even works.
Also $S$ being at the end of an intervall without primes is sufficent.

[quote="ZetaX"] something like $S$ being even works.
Also $S$ being at the end of an intervall without primes is sufficent.[/quote]

$S$ is supposed to be odd in the statement. Secondly being at the end of interval without primes means nothing because you can take large enough $n$ so $S-2^j$ for $n\geq j \geq k$ for some $k$ will be out of your interval without primes (if I understand correctly what you mean by interval without primes)

Sorry, I overread that odd part :blush:
I understood it in the way that $n$ is given and then we shall give $S$. Then using a large intervall works. If $S-2^k$ shall not be a positive prime for any $k$, I agree with you.

Ah, that's okay - it seems that I misunderstood the problem however the way I get it is not so bad :P - let me state clearly what I meant:

[b]Problem:[/b] Let $S$ be an arbitrary odd natural number and let $m$ be a natural number satisfying $2^m < S < 2^{m+1}$ . Find all $k \leq m$ for which among the numbers $S-2$,$S-2^2$,...,$S-2^k$ there is at most one prime.

$S=5^{n!}$

$5^{n!}-2^{k}$ is divisable by $5^{\frac{n!}{k}}-2$, for every $k$ less than $n$

I didnt state very clear in my question  :blush:

Should be for $2^{n+1}>S>2^n$ ... actually this is the problem I faced

Example : $S=27$ , then we have the set of integer to be $\{27-2,27-4,27-8,27-16\}=\{25,21,19,11\}$ . Here we can see that there are two primes inside the set . So I would like to know $S$ must have what condition for it such that the set contains at most one prime number . Does such "condition(s)" exists ?


geometry perimeter MATHCOUNTS trigonometry inequalities

A slightly more challanging , yet, IMO, still  not too advance for this form:

Q. 2:

[b]Find all right angle triangles (all sides positve integers) which have their area and perimeter same (numerical value) [/b]

[size=75][Split from parent thread by rcv.  I believe this problem is above the level you would find on a National MATHCOUNTS competition.  (And far above the level you would find on a Chapter or State competition.)  If proposed solutions do not use MATHCOUNTS-level techniques, I will bump this to a higher forum.][/size]

I think this problem should be moved.  It's been here a while and no one has gotten it.  I have a way to solve it with trig, but I don't want to post it until it's out of the Mathcounts forum. :theta:  :theta:  :theta:  :theta:  :theta:

Hint : The problem is hard but does not need anything higher in terms of tools   than  used in  mathcounts . FWIW  out of four mathcounts students, I gave this problem a few yeas ago  (national level) everyone solved it fairy fast..(under a minute) . The  tool in the  toolbox (or factoring trick) given  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25476]here[/url]  can be used (see, for example  the "factoring tricks"  )

Here are some hints:   (Go ahead peek even if you have gotten a solution )

 [hide]Hint 1: Assume the  legs are $a$ and $b$
Hint 2:  What is the area? What is the perimeter?
Hint 3: Equate the two
Hint 4: If you see a square root: square the side (which has the root) it to get rid of it...(try to move terms so that square root is only on    one side  of the equatuion)
Hint  5: Simplify ... (Look out for  the nice  mathount fairy ,, lot of terms will simply go away ...)
Hint  5: Use the old trick [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25476]here[/url]
[/hide]
Good luck.

Anoter Set of Hints/Tricks   ( And other set of tricks which no mathcounts student should leave home without )

Go ahead and peek ..

[hide]Can you try some simple cases .. say small triangles ...

What happens if the one side is too big, can you see/prove there are no  solutions ..For example, say both legs are >8  Can you see/prove there are no solutions .. In fact it is  "easy"  to "see" (if you happen to "see" at the right place) that  one of the sides has to be less than 8, actually less than 7 .. see if you can prove that..

Are there only a few cases left which you can try  one by one .. ..
[/hide]
[hide]In anycase , if you try a few cases like (3,4,5),(5,12,13) etc and saw what you are looking for  and can show nothing more is left.. you are done..
[/hide]
Hope you have fun trying ...

Another set of Hints .. tools .. fun items:..... (Peek at them , if you already have a set of  solutions .. or have tried for a while and won't mind lookling at hints . and some new ideas ...If you  do not have a solution yet .. try on your own  before peeking ...)

Some items may be interesting ..

[hide]Can you   find the radius of inscribed circle ?
Does knowing this help?
For example can you  "see"  (right away) that one side(leg)  has to be less than 7, (and thus  less than or equal to $6$ if it is an integer?  (And the other - or the larger -  leg  has to greater than or equal to 8 - by the similar logic )
[hide]Area of the traingle is  $(1/2)$  times $perimeter$  times $radius$
Radius is also given as $\frac {a+b-c} {2} $

Does that suggest something? [/hide] [/hide]

[hide="Beyond MATHCOUNTS level.  Peek at your own risk."]By the condition of the problem we have:
\[
\sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}  = 2p
\]
Let us denote p-a = x.......we have

\[
\sqrt {\left( {x + y + z} \right)xyz}  = 2\left( {x + y + z} \right) \Rightarrow xyz = 4\left( {x + y + z} \right)
\]
x, y, z are integer. Assume:
\[
x \ge y \ge z
\]
and then
\[
\begin{array}{l}
 x = \frac{{4y + 4z}}{{yz - 4}} \\

\frac{{4y + 4z}}{{yz - 4}} \ge y \\
 \end{array}
\]
The last inequality brings to

\[
y \le \frac{{4 + \sqrt {16 + 4z^2 } }}{z}
\]
Thus we have
\[
yz \le 4 + \sqrt {16 + 4z^2 }
\]
and we can write
\[
z^2  - 4 \le \sqrt {16 + 4z^2 }  \Rightarrow z \le 3
\]
 Now let z = 1 , z =2 , z =3 and  we find five solutions (6,25,29) ; (7,15,20); (6,8,10); (5,12,13); (9,10,17).[/hide]
[size=75][Solution hidden by moderator rcv.][/size]

Nene - Nice work but this is mathcounts level,  and the problem is much more simple.. The triangle is "right angle" . The  equation  is simpler and easier for area etc ($\frac{ab}{2}$ ) etc.. .(Your answer  gives [b]all [/b] triangles including right angle  ones and thus.  is more  work than  than the simpler problem...

(I think mods also want it to be left for middle schoolers .. others can give helpful hints etc..but ..  hide the solutions so other can tryl
Thanks)


$\triangle ABC,\ M\in (BC),\ N\in (CA),\ P\in (AB),\ \frac {MB}{MC}=\frac{NC}{NA}=\frac{PA}{PB}=r>0\Longrightarrow$
\[\frac{AM}{a}\cdot \frac{BN}{b}+\frac{BN}{b}\cdot \frac{CP}{c}+\frac{CP}{c}\cdot \frac{AM}{a}\ge \frac{3(r^2+r+1)}{(r+1)^2}\ge \frac 94\ \ (*)\]When is happen the equality in the inequality (*) ?. Particular case: $r=1\Longrightarrow \sum \frac{m_a\cdot m_b}{a\cdot b}\ge \frac 94.$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=46861

My God Levi, your inequality is so beautiful.Well,I haven't tried a proof yet,but it seems...... :coolspeak:


[quote="Jimmy"]Let $f : \mathbb{Z}^{+}\to\mathbb{Z}^{+}$ be a function such that $f(n+1)>f(f(n)) \ \forall \ n\in\mathbb{Z}^{+}$.  Prove that $f(n) = n$.[/quote]

[hide]I don't know if this is the right logic.

If $f : \mathbb{Z}^{+}\to\mathbb{Z}^{+}$, then $f(n)\ge n$

$f(n+1)>f(f(n))$
$n+1>f(n)$

So $n\le f(x)<n+1 \Rightarrow f(n)=n$

Is that right?[/hide]

[quote="SplashD"]If $f : \mathbb{Z}^{+}\to\mathbb{Z}^{+}$, then $f(n)\ge n$[/quote]

Try $f: \mathbb{Z}^{+}\to\mathbb{Z}^{+}: x\mapsto (x-6)^{2}$.

[quote="SplashD"]$f(n+1)>f(f(n))$
$n+1>f(n)$[/quote]

This is only true if $f$ is increasing.

Hint. For what $x$ is $f(x)$ the smallest?

[hide]
I will assume $\mathbb{Z}^{+}$ to mean the set $\{0, 1, 2, \ldots\}$. Let $n_{0}\in \mathbb{Z}^{+}$ be the nonnegative integer that minimizes $f(n)$. Then

$f(n_{0}) > f(f(n_{0}-1))$ if $n_{0}-1 \in \mathbb{Z}^{+}$.

But this is clearly a contradiction since $f(n_{0})$ is the minimum. Hence $n_{0}-1 \not\in \mathbb{Z}^{+}$ but $n_{0}\in \mathbb{Z}^{+}$ so $n_{0}= 0$. Knowing that $f(0)$ is the minimum, we pick the next minimum of the elements remaining and we call it $n_{1}$. Then

$f(n_{1}) > f(f(n_{1}-1))$,

but since it's the smallest we know $f(n_{1}-1)$ is not one of the remaining numbers, which means it is $0$. Consequently, $f(n_{1}-1)$ is the global minimum and $n_{1}-1 = 0 \Rightarrow n_{1}= 1$. We thus obtain $f(0) = 0$ as well.

Continuing this argument by induction I believe solves the problem.[/hide]


probability geometry ratio Alcumus rectangle

Someone told me to do a problem by graphing it and looking at the area, but I forgot everything they told me. Can someone give me an example of when I'd do this and how to do it? Much appreciated.

Suppose for example, we're throwing darts at a circular target.  Suppose that the target has radius $ R$, with a small circular bulls-eye in the center, of radius $ r$.  If you throw a dart, what's the probability of getting a bulls-eye?

Well area you want is $ \pi r^{2}$.  The total area is $ \pi R^{2}$.  So the probability is just the ratio of areas, $ \frac{r^{2}}{R^{2}}$.

Well in JRav's example, graphing is not needed because the shapes are exactly circular. However, in irregular figures, graphing, along with [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]the Shoelace Theorem[/url] may be employed.

A slightly modified Alcumus problem:

Ernie needs to take a train to New York for a secret meeting with JRav (:D).  The train will come at a random time between $ 1: 00$ and $ 1: 30$, and will stay there for $ 15$ minutes.  Ernie will show up at a random time between $ 1: 00$ and $ 1: 30$.  What is the probability that Ernie will arrive when the train is there?

A secret rendezvous? :wink:   That problem is really similar to one in AoPS Volume 2, Chapter 19.
[hide="Hint"]Draw a rectangle[/hide]

Hmmmm... really?

I don't have any AoPS books. :)

Is it a rectangle with side 30 with a rectangle with side 15 inside it and find ratio of areas?

About the train problem, why can't I just use a number line and plot it?

Because you have two factors to account for, me and the train.

So the train wouldn't be a line segment length 15 and the guy a point that drops on the larger line segment 0-30? Doesn't make sense to me.

There are two variables, ernie and the train.  Just like in the x-y plane, you need two axes, the "ernie-train" axes.


If ${  \sqrt[3]{7}= a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\cdots}}}}$ where $ a,b,c,d, \cdots$ are positive integers, find $ b-a$.

[hide]
First, $ 1 < \sqrt[3]7 < 2 \implies a=1$.

Now, $ b+\frac{1}{c+\frac{1}{\ldots}}= \frac{1}{\sqrt[3]7-1}= \frac{ \sqrt[3]{49}+\sqrt[3]7+1}{6}\approx \frac{3.5+2+1}{6}\implies b=1$
Thus, $ b-a = \boxed{0}$.

(Note that the rough bound on $ \sqrt[3]{49}$ is a lower bound, as $ \left( \frac{7}{2}\right)^{3}= \frac{343}{8}< 43$. But we don't need to worry; the problem turns out that only an extremely rough approximation would change anything in the final answer.)[/hide]


calculus
derivative
function
real analysis
real analysis unsolved

Let [i]f:R-> R[/i] be smooth, such that for any (fixed) [i]x[/i] there exists natural [i]n[/i] with [i]n[/i]-th derivative of [i]f [/i]evaluated at [i]x[/i] being zero. Prove f is actually a polynomial.

Thank you.
Can smn explain me the part of the sol I don't get there though? :)


[quote="scarface"]ABC is a triangle that is not isosceles.external bisectors of angle A; internal bisectors of angles B,C are meeting the sides(or its extensions) at points D,E,F respectively.prove that intersecting points of line pairs {AB,DE}, {AC,DF}, {BC,EF} are collinear.[/quote]

Easy Menelaus gives $D$, $E$, $F$ are collinear. Thus $AB$ and $DE$ intersect at $F$; $AC$, $DF$ intersect at $E$; and $BC$ and $EF$ intersect at $D$...and we're done

thank you for your solution,sunchips. I made a mistake.I'm sorry.now, I'm writing correct problem.

PROBLEM: ABC is a triangle that is not isosceles.internal bisectors of angle A; external bisectors of angles B,C are meeting the sides(or its extensions) at points D,E,F respectively.prove that intersecting points of line pairs {AB,DE}, {AC,DF}, {BC,EF} are collinear.

This problem is also true in general configuration of three arbitrary cevians, in a given triangle. So, let $AP,$ $BP,$ $CP$ be, three cevians which intersect the sideline $BC$ and the extensions of the sidelines $AC,$ $AB,$ at points $D,$ $E,$ $F,$ respectivey.

If we consider an arbitrary point $D,$ on the segment line $AP$ and we denote as $B',$ $C',$ the intersection points of $AB,$ $AC,$ from the segment lines $DE,$ $DF$ respectively, we have the result that the segment lines $B'C',$ $BC,$ $EF,$ are concurrent at one point $($ easy to prove from complete quadrilaterals $ABPCEF$ and $AB'DC'EF$ $).$

Kostas Vittas.


Woahh no more middle school math and high school math! :o  What do Illinoisans think of this new change?

Agrees with above poster. I wonder what would happen if AoPS converted had an option of making the forum look like an e-mail inbox, where new topics would go in the "inbox" folder and old messages be sorted in folders arranged in the current layout. Interesting idea, but probably ineffective.

Just had to spam.  :lol:

[quote="ThinkFlow"]Interesting idea, but probably ineffective.[/quote]
No. It's a very BAD idea.

Thought so, just thinking out loud (and sort of spamming)! :lol:


Hi all,

I'm on the hunt for a graduate school in mathematics are for various reasons am very keen on going to California. For personal reasons, a graduate program that has options for doing cross disciplinary research in Biologi is reasonably important. I know that both Berkeley and UCLA have strong math programs (at least by rankings), and have people doing research in mathematical biology (which is a very small field at this point).

What do you guys know about the graduate program at Berkeley and UCLA? Are both reputable programs? How about professors, student body, atmosphere in general? Quality of life? Intensity of study?

One particular thing: I am coming from a background in biology and am studying on my own make up for my lack of upper division courses. Thus, I am not sure that going to a competitive program might be a good idea because I might not be ready.

Thanks for the advice!
Joshua

Berkeley, along with Princeton, is generally regarded as the best graduate program in the country.  Stanford is also considered very, very good.

[quote="JRav"]Berkeley, along with Princeton, is generally regarded as the best graduate program in the country.[/quote]

wait what?!?!?!

where do you even hear these things?!

Berkeley and UCLA are both very good programs with Berkeley generally being slightly better regarded.  But it really depends on what you're interested in.  Since you want to do mathematical biology you should talk to someone you know who works in the field and ask for their opinion of various schools.  This is particularly important since math bio people tend to be somewhat isolated from the rest of the math graduate students (at least they are where i am) and hence the quality of the math bio program is a lot more important than the quality of the institution as a whole.

I'm referring to math department rankings, obviously.  From US News 2008:

1-Princeton
2-Harvard, MIT, Stanford, Berkeley
12-UCLA

See here for the list: http://grad-schools.usnews.rankingsandreviews.com/grad/mat/search

Or, in other words, you are providing no information whatsoever that the O.P. hasn't already claimed access to.  Here's seconding blahblahblah's suggestion that you go talk to some professors in mathematical biology and see what they have to say.  The perspective of people in the field is likely to be much more valuable than that of the rest of us.


If angle A equals angle B, then this triangle must be isosceles. Thus $ 2x \minus{} 2 \equal{} 3x \minus{} 1$ and $ x \equal{} \boxed{1}$.

yea but the answer it said was 3 and dont you mean -1?

Oops, I did mean -1, sorry.

I don't see how the answer's 3 though...it has to be isosceles...

Probably error. And if x=3 then x-4=-1 Now the sides are negative!


real analysis
limit
inequalities
advanced fields
advanced fields unsolved

I need help proving

If liminf Sn = m then for all epsilon>0, Sn < m+epsilon for infinitely many terms[/quote]

That is more or less the definition of lim inf.

To say that lim inf $s_n = a$  means that $a$ is the optimal value for which you can assert that "$s_n \geq a$ for large $n$", i.e. the statement becomes false if $a$ is replaced by any $a + \epsilon$.

That's not quite right, fleeting_guest. Suppose $s_n$ is a strictly increasing sequence with limit $a$. Even though $\liminf s_n=a$, it is never true that $s_n>a.$

To help out mathdork:

Define $b_n= \inf_{k\ge n}s_k.$

We are trying to show this:

If $\lim b_n=L$ then $\forall\,\epsilon>0,\,\forall\,N\,\exists\,n>N$ such that $ S_n<L+\epsilon.$

Proof: Start with $\epsilon>0.$ Find $N$ such that for $m > n,\,|b_m-L|<\frac{\epsilon}2.$

For any $m > N,\,b_m+\frac{\epsilon}2$ is not the infinum of tail-end set of $s_k$, so there exists $k>m$ such that $s_k<b_m+\frac{\epsilon}2.$

Then for that $k,\,s_k<b_m+\frac{\epsilon}2<L+\frac{\epsilon}2+ \frac{\epsilon}2=L+\epsilon.$

Since we can find such a $k>m$ for any $m>N$, we have proved "infinitely often."

Two very useful terms in discussing sequences are "eventually" and "infinitely often. Let $P(n)$ be a statement which depends on the positive integer $n.$

$P(n)$ is [b]eventually[/b] true iff $\exists\,N$ such that $P(n)$ is true $\forall\,n>N.$

$P(n)$ is true [b]infinitely often[/b] iff $\forall\,N\,\exists\,n>N$ such that $P(n)$ is true.

$\lim s_n=L\iff \forall\,\epsilon>0$, eventually, $|s_n-L|<\epsilon.$

(Or to summarize it more briskly, $s_n\toL$ iff $s_n$ is eventually arbitrarily close to $L$.)

$\lim s_n\ne L$ (which includes the limit not exising) iff $\exists\,\epsilon>0$ such that $|s_n-L|\ge\epsilon$ infinitely often.

$\limsup s_n=L$ iff $\forall\epsilon>0$,
(1) $ s_n<L+\epsilon$ eventually, and
(2) $s_n>L-\epsilon$ infinitely often.

$\liminf s_n=L$ iff $\forall\epsilon>0$,
(1) $ s_n<L+\epsilon$ infinitely often, and
(2) $s_n>L-\epsilon$ eventually.

[quote="Kent Merryfield"]That's not quite right, fleeting_guest. Suppose $s_n$ is a strictly increasing sequence with limit $a$. Even though $\liminf s_n=a$, it is never true that $s_n>a.$ [/quote]

Yes, you are right.  For finite $a$ one should weaken the inequality to $s_n \geq a - o(1)$ (with optimality of $a$ being the meaning of "lim inf = $a$").  I hope that doesn't introduce any new mistakes; I find the optimal-asymptotic-inequality formulation a useful way to think.


What kind of student are you and what grades do you get?

The poll answer for a is my current grade for every class.

i'm in all honors to let you know

I get all a's, but I don't have to work very hard to get them (especially in math).

The general philosophy is - the poll is probably going to show most A's because why would anybody waste their time sharing the fact that they aren't 'A' students? Just a thought... on psychology.

I get mostly A's. Just 2 B's in Social Studies and Language Arts because I'm in a gifted program and it's a lot harder.

[quote="yif man12"]I'm not voting because there's no choice for F.[/quote]

lol and la mafioso does not want to share his grades 4 1st quarter i think :roll:

Geez for pete's sake slam that was 3D ART!

Of course I woulda gotten a B+...

I'm an a student

my bro...he's a c/b student...

well generally the grades tend to vary depending on competition difficulty and age level so you cannot compare it like this effectively

I can get all A's without even really trying. :roll: (I try and get mostly A+'s :D )My school is WAY too easy. They don't have any honors classes but I get to take math with the 8th graders.

B overall
I get
Math-A easy
Spanish-A easy
History-B I don't try hard enough
English-A my teacher doesn't believe in homework :)  :)  :)  :)
Science-B I hate doing labs and thats all we do
Computers-F my teacher looks like a rat and acts like one too :D
Art-F I hate art
PE-F I always forget my uniform
Band A+ Not too special so does everyone else

I want to be in your english class :) ...I mean who gives the majority of kids single digit scores on a timed writing honestly. I think they enjoy torture. They also make you read like 3 books at the same time. We don't even need half the stuff in real life :mad:

This poll is poorly worded, is it the general trend of your grades(mean, median, mode), your lowest grade, your gpa...  Currently my average (mean) grade is an A-, the median and mode are both A's, and im not telling you my low.  My weighted gpa is 5.4(this may be a little off) something form the first quarter.

How is 5.4 a possible gpa? I thought they had to be on a scale of 0.0 to 4.0.

Some schools have different gpa rating systems.  Some schools weight their courses differently, mostly pending on the difficulty of the course (honors vs non-honors).

That's a weighted GPA. Unweighted is on a scale of 0 - 4.0.

Straight A's, easy.

Too bad I'm only taking 2 APs. High schools don't have faith in 9th graders.  :(

you would think people who go to MATH FORUMS for FUN would get good grades :lol:

i currently get A's (improving)

To get an A in my school is like trying to drink chocolate milk: it's a breeze. Teachers offer so much extra credit and corrections that it's nearly impossible [i]not[/i] to get an A.

I have a question, is this a high school/math competition/university grade poll

Only in the latter can I participate

Also, in high schools here in Belgium, we don't have A B or C, every subject is evaluated with percents : less than 50 means flunked, below sixty isn't good, like seventy is the average, eighty is good, ninety is better,when you are lucky you can get a 92/93 sometimes

So far I've been a straigh A student.  This semester I may be looking at a B in English, though.  We'll see.

[quote="fredbel6"]

Also, in high schools here in Belgium, we don't have A B or C, every subject is evaluated with percents : less than 50 means flunked, below sixty isn't good, like seventy is the average, eighty is good, ninety is better,when you are lucky you can get a 92/93 sometimes[/quote]
the same in Italy :)

Straight A's, except that I'm getting a B on English.
But only 2 years has past since I came to the U.S. and I'm taking regular English, so I'm fine with that :roll:

D Overall.

B in Pure Maths
C in Physics
D in English, Chemistry
F in Spanish

Everything over 90 and biology 88. :mad: Bah, I never liked the stupid subject anyway...(No offense to those who like it)


Two cards are chosen at random from a standard 52-card deck.  What is the probability that both are face cards (Jacks, Queens, or Kings)?

There are 12 face cards total in a standard card deck. Therefore the probability is:
\[ \frac{12}{52}\cdot\frac{11}{51}\equal{}\boxed{\frac{11}{221}}
\]

[hide=combinations solution]There are 12 different face cars, so there are $\binom{12}{2}$ different ways to choose 2 face cards. There are $\binom{52}{2}$ to choose 2 different cards, so the answer is $\dfrac{\binom{12}{2}}{\binom{52}{2}} = \boxed{\dfrac{11}{221}.}$

[quote=awesomeguy856][hide=combinations solution]There are 12 different face car[size=200][b][color=#0f0]d[/color][/b][/size]s, so there are $\binom{12}{2}$ different ways to choose 2 face cards. There are $\binom{52}{2}$ to choose 2 different cards, so the answer is $\dfrac{\binom{12}{2}}{\binom{52}{2}} = \boxed{\dfrac{11}{221}.}$[/quote]

I thinks u meant cards not cars lol

[hide=Sol]12 Cards you pick on the first try. 11 cards you can pick on the second try. Ans = 12*11/52/51 = 11/221[/hide]

[quote=hansenhe]posting the answer doesn't contribute to the discussion. please post a full solution or helpful hints.[/quote]

[hide]12/52 * 11/51= 11/221[/hide]

[hide = @above & @2above Solutions using LaTeX]$\frac{12 \times 11}{52 \times 51} = 11/221$[/hide]


calculus
integration
number theory unsolved
number theory

Find integral solutions to the equation \[(m^{2}-n^{2})^{2}=16n+1.\]

This is problem from Russian mathematical olympiad 1997. It can be found in book MATHEMATICAL OLYMPIADS FROM AROUND A WORLD 1997-1998 ublished my MAA It is problem 32 Russia 1977 page 97 or 98.
Here is link so you can download this book.

[url]http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf[/url]

After looking at the elegant proof in the link above, I decided to destroy my rough sketch...

My solution:

If $|m|>n$ $\Longrightarrow$ $16n+1=(m^2-n^2)^2\geq (2n+1)^2$ $\Longrightarrow$ $16n+1\geq 4n^2+4n+1$ $\Longrightarrow$ $3\geq n$
$\Longrightarrow$ if $n=0$ then $m=1,-1$ $\Longrightarrow$ $(m,n)=(1,0);(-1,0)$ are solutions
if $n=1,2$ it has no solutions since $17,33$ are not perfect squares
if $n=3$ then $m=4,-4$ $\Longrightarrow$ $(m,n)=(4,3);(-4,3)$ are solutions

If $n>|m|$ $\Longrightarrow$ $16n+1=(m^2-n^2)^2\geq (2n-1)^2$ $\Longrightarrow$ $16n+1\geq 4n^2-4n+1$ $\Longrightarrow$ $5\geq n$
If $n=1,2,4$ it has no solutions since $17,33,65$ are no perfect squares
If $n=5$ then $m=4,-4$ $\Longrightarrow$ $(m,n)=(4,5);(-4,5)$
If $n=0,3$ solutions have been found in the previous case

Then all solutions are: $(m,n)=(4,3);(-4,3);(4,5);(-4,5);(1,0);(-1,0)$... :laugh:


${f=(x y z)^{3}+(x y+z y+x z)^{3}}$

${x+y+z=3 \theta }$
${(x,y,z,\theta )>0}$

Proove that

${0\leq \text{ }f\text{ }\leq \theta^{6}\left(3^{3}+\theta^{3}\right)}$

$\theta$  it's just a positive constant

[quote="Thales418"]${f=(x y z)^{3}+(x y+z y+x z)^{3}}$

${x+y+z=3 \theta }$
${(x|y|z|\theta )>0}$

Proove that

${0\leq \text{ }f\text{ }\leq \theta^{6}\left(3^{3}+\theta^{3}\right)}$

$\theta$  it's just a positive constant[/quote]
What is ${(x|y|z|\theta )>0}$?  :maybe:

In mathematica  when we do

(x|y|z) -> k

means that x-> k , y->k, z -> k

sorry for the busy with notation

x>0
y>0
z>0

etc...

N.T.TUAN

good proof

:)

congratulations

but could you do the double sum of square roots?!

i know that is rigth, but i cant proove


Hey, what is recursion? I've seen people used it left and right, so I was wondering if anyone can explain it.

btw, good luck on the aime tomorrow everyone!

Basically a recursion is a way of getting one term of a sequence from previous term(s) of the sequence. The most commonly know recursion is the Fibonacci Sequence, in which one term is the sum of the two previous terms (the first two terms are 1). In mathematical notation, that would be:

$F_{1}=1$
$F_{2}=2$
$F_{n+2}=F_{n+1}+F_{n}$ for $n\ge 1$

The first few terms of this sequence are:
1,1,2,3,5,8,13,21,34,...

Usually the key in recursion problems is finding the sequence and then finding the recursion.
...if that makes sense


trigonometry
function
algebra proposed
algebra

For $ x,y\in [0, \frac{\pi}{3}]$ prove that $ \cos x+\cos y\leq 1+\cos xy.$

[quote="natural123"]With $ x,y\in [0, \frac {\pi}{3}]$, prove that $ \cos x \plus{} \cos y\leq 1 \plus{} \cos xy$[/quote]

For $ x\in[0,\frac{\pi}2]$, the function $ g(x)\equal{}x\sin(x)$ for $ x\in[0,\frac{\pi}3]$ is increasing and so $ x\sin(x)>xy\sin(xy)$ $ \forall y\in[0,1]$ and $ x\sin(x)<xy\sin(xy)$ $ \forall y\in[1,\frac{\pi}3]$

So $ \sin(x)\minus{} y\sin(xy)\geq 0$ $ \forall x\in[0,\frac{\pi}3],y\in[0,1]$ and $ \sin(x)\minus{} y\sin(xy)\leq 0$ $ \forall x\in[0,\frac{\pi}3],y\in[1,\frac{\pi}3]$

So $ f(x)\equal{}1\plus{}\cos(xy)\minus{}\cos(x)\minus{}\cos(y)$ is an increasing function $ \forall x\in[0,\frac{\pi}3],y\in[0,1]$ and is a decreasing function $ \forall x\in[0,\frac{\pi}3],y\in[1,\frac{\pi}3]$ (since $ f'(x)\equal{}\sin(x)\minus{} y\sin(xy)$)

So :
If $ y\in[0,1]$ $ f(x)\geq f(0)\equal{}1\minus{}\cos(y)\geq 0$
If $ y\in[1,\frac{\pi}3]$ $ f(x)\geq f(\frac{\pi}3)\equal{}\frac 12\plus{}\cos(y\frac{\pi}3)\minus{}\cos (y)$ $ >\frac 12\plus{}\cos((\frac{\pi}3)^2)\minus{}\cos (1)$ $ \equal{}0.416>0$

So $ f(x)\geq 0$ $ \forall x,y\in [0, \frac {\pi}{3}]$ and so $ 1\plus{}\cos(xy)\geq \cos(x)\plus{}\cos(y)$ $ \forall x,y\in[0,\frac{\pi}3]$

Q.E.D.


Someone please help with this problem, its killing me!

Determine whether the following language is regular or not, and prove your answer:

{[i]x[/i] ∈ ∑⎮there exists a non empty [i]y[/i] such that [i]x[/i] contains [i]yyy[/i] as a substring}

You may assume that there are arbitrarily long strings in the complement of this language—that is, there are arbitrarily long strings in ∑* that do not contain a non-empty substring of the form [i]yyy[/i].

thanks so much

Hm, yeah, looks pretty hard. I couldn't solve it yet but I've been thinking about using the pumping lemma on some specific word from your language L, for example word yyy where y is some word from L complement of length greater than p (p is the constant from pumping lemma), or something like that.... of course this probably won't work, it's just an idea. Oh, and yes, I think that the answer is almost surely that L is not regular.


Thanks to Olena Bormashenko
http://www.kalva.demon.co.uk/apmo/asoln/asol034.html

Just prove if$0<x\leq y$,then $\sqrt[n]{x^n+y^n}\leq(\sqrt[n]{2}-1)x+y$
Then it's easy.

$ x\equal{}\dfrac{\minus{}a\plus{}b\plus{}c}{2}$ ; $ y\equal{}\dfrac{a\minus{}b\plus{}c}{2}$ ; $ z\equal{}\dfrac{a\plus{}b\minus{}c}{2}$

By Minkowsky:

$ \sqrt[n]{(y\plus{}z)^n\plus{}(x\plus{}z)^n}\plus{}\sqrt[n]{(z\plus{}x)^n\plus{}(y\plus{}x)^n}\plus{}\sqrt[n]{(x\plus{}y)^n\plus{}(z\plus{}y)^n}\le (\sqrt[n]{y^n\plus{}x^n}\plus{}\sqrt[n]{z^n\plus{}z^n})\plus{}(\sqrt[n]{z^n\plus{}y^n}\plus{}\sqrt[n]{x^n\plus{}x^n})\plus{}(\sqrt[n]{x^n\plus{}z^n}\plus{}\sqrt[n]{y^n\plus{}y^n})$

$ \sqrt[n]{a^n\plus{}b^n}\plus{}\sqrt[n]{b^n\plus{}c^n}\plus{}\sqrt[n]{c^n\plus{}a^n}\le \sqrt[n]{x^n\plus{}y^n}\plus{}\sqrt[n]{y^n\plus{}z^n}\plus{}\sqrt[n]{z^n\plus{}x^n}\plus{}\sqrt[n]{2}(x\plus{}y\plus{}z)$

$ \sqrt[n]{a^n\plus{}b^n}\plus{}\sqrt[n]{b^n\plus{}c^n}\plus{}\sqrt[n]{c^n\plus{}a^n}\le \sqrt[n]{x^n\plus{}y^n}\plus{}\sqrt[n]{y^n\plus{}z^n}\plus{}\sqrt[n]{z^n\plus{}x^n}\plus{}\dfrac{\sqrt[n]{2}}{2}$

And because:

$ x^n\plus{}y^n<(x\plus{}y)^n$ $ \Rightarrow$ $ \sqrt[n]{x^n\plus{}y^n}<x\plus{}y$

$ y^n\plus{}z^n<(y\plus{}z)^n$ $ \Rightarrow$ $ \sqrt[n]{y^n\plus{}z^n}<y\plus{}z$

$ z^n\plus{}x^n<(z\plus{}x)^n$ $ \Rightarrow$ $ \sqrt[n]{z^n\plus{}x^n}<z\plus{}x$

Then:

$ \sqrt[n]{a^n\plus{}b^n}\plus{}\sqrt[n]{b^n\plus{}c^n}\plus{}\sqrt[n]{c^n\plus{}a^n}<(x\plus{}y)\plus{}(y\plus{}z)\plus{}(z\plus{}x)\plus{}\dfrac{\sqrt[n]{2}}{2}$

$ \sqrt[n]{a^n\plus{}b^n}\plus{}\sqrt[n]{b^n\plus{}c^n}\plus{}\sqrt[n]{c^n\plus{}a^n}<2(x\plus{}y\plus{}z)\plus{}\dfrac{\sqrt[n]{2}}{2}$

$ \boxed{\sqrt[n]{a^n\plus{}b^n}\plus{}\sqrt[n]{b^n\plus{}c^n}\plus{}\sqrt[n]{c^n\plus{}a^n}<1\plus{}\dfrac{\sqrt[n]{2}}{2}}$

$ \mbox{ My solutions of the Jin's inequality, in attachment: }$

[quote="shobber"]Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that
\[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \][/quote]
We may assume that $a \le b \le c < \frac 12$. Then \begin{align*}
		\sqrt[n]{b^n+a^n} &< b + \frac1n \frac{a^n}{b^{n-1}} < b + \frac1n a \\
		\sqrt[n]{c^n+a^n} &< c + \frac1n \frac{a^n}{c^{n-1}} < c + \frac1n a \\
		\sqrt[n]{b^n+c^n} &< \sqrt[n]{\left( \frac12 \right)^n + \left( \frac12 \right)^n} = \frac{\sqrt[n]{2}}{2}.
	\end{align*} Summing and using the fact $n \ge 2$ yields the conclusion.

[quote=shobber]Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that
\[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \][/quote]
[url=http://www.artofproblemsolving.com/community/c6h1306943p6988594http://www.artofproblemsolving.com/community/c6h1306943p6988594]n=2,Romania 1983[/url]

Let $a$, $b$ and $c$ be sides-lengths of triangle such that $a+b+c=1$. [url=http://www.artofproblemsolving.com/community/c6h1306943p6989154]Prove that:[/url]$$\sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{{{b}^{2}}+{{c}^{2}}}+\sqrt{{{c}^{2}}+{{a}^{2}}}+\frac{abc}{ab+bc+ca}\le 1+\frac{1}{\sqrt{2}}$$

[quote=shobber]Let $a,b,c$ be the sides of a triangle, with $a+b+c=1$, and let $n\ge 2$ be an integer. Show that
\[ \sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]{2}}{2}. \][/quote]

WLOG. $c\le b\le a<\frac{1}{2}$.
Claim. $b^n+a^n<\left(b+\frac{c}{2}\right)^n$.
Proof.
$b^n+c^n<b^n+b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}+\cdots+\left(\frac{c}{2}\right)^n$
$c^n<b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}+\cdots+\left(\frac{c}{2}\right)^n$
See $n\ge 2$ such that,
$b^{n-1}\times\frac{c}{2}\times{{n}\choose{n-1}}=b^{n-1}\times\frac{c}{2}\times n>b^{n-1}\times c$
$c^n<b^{n-1}\times c$
QED.

Next, we get :
$\sqrt[n]{a^n+b^n}\le\sqrt[n]{a^n+a^n}=a\sqrt[n]{2}<\frac{\sqrt[n]{2}}{2}$
$\sqrt[n]{b^n+a^n}<b+\frac{c}{2}$
$\sqrt[n]{a^n+c^n}<a+\frac{c}{2}$
Thus, we get $\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<a+b+c+\frac{\sqrt[n]{2}}{2}=1+\frac{\sqrt[n]{2}}{2}$.
QED

[quote=strong_boy]Nice problem , But hard  :roll:

Lemma : if $x \geq 1$ then $\sqrt[n]{x^n+1} \leq x+\sqrt[n]{2}-1$ .

Prove : We can see $f^{'}(x) = \frac{x^{n-1}}{(x^n+1)^{\frac{n-1}{n}}}-1 <0$

So $f$ is descending function . Then we can see :
$$f(x) \leq f(1)=\sqrt[n]{2}-1$$

Now if we let $m \geq j >0$ , $x=\frac{m}{j}$ . Then $\sqrt[n]{m^n+j^n} \leq m+(\sqrt[n]{2}-1)j$ . ( :w00t: )

Now let $a \geq b \geq c$ . Now we can see :

$$\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n} \leq \sqrt[n]{2}(a+b+c) =  \sqrt[n]{2} < 1+\frac{ \sqrt[n]{2}}{2}$$[/quote]

nice problem-beautiful solution!


I thought you guys would be interested in giving these a shot. I'll post the senior division questions if I can get them.

1.) Matthew is choosing from the dimensions of tiles he will use. He can have three tiles. One with length 12, another with 10, and another with 15. The widths of the tiles will be 9,8,7. He can pair any length with any width. Compute the maximum total area of the tiles he chooses.
2.) Each student in a group of 12 math students is to be assigned into one of three teams. One team will have 3 students, another will have 4, and one will have 5. Find the number of such assignments.

3.) Find all ordered pairs (x,y) of positive integers such that a number, x, squared, minus a number, y, squared is equal to 55.
4.) Compute the square root of 20 plus the square root of 20, plus the square root of 20... and so on.

5.) The sum of 5 consecutive integers, all of which are greater than 1000, is divisible by 15. Compute the least possible sum of these 5 integers.
6.) In trapezoid ABCD with bases AB and CD, E is on AD, and F is on BC so that EF and AB are parallel. The area of ABFE and EFCD are equal. IF AB= 6, BC=9, CD=10, and AD=7, compute EF.

Which division are these from (Soph-Frosh, Junior, Senior B, Senior A)?  Are they from this school year?  If so, which round?  Thanks.

[hide="Problem 1"]$ 15\times 9 \plus{} 12 \times 8 \plus{} 10 \times 7 \equal{} 135 \plus{} 96 \plus{} 70 \equal{} \boxed{301}$[/hide]

[hide="Problem 3"]$ x^2 \minus{} y^2 \equal{} 55$ implies $ (x\plus{}y)(x\minus{}y) \equal{} 55$. There are two cases, $ x\plus{}y \equal{} 55$ with $ x\minus{}y \equal{} 1$ and $ x\plus{}y \equal{} 11$ with $ x \minus{} y \equal{} 5$. The two solutions are $ \boxed{(27, 26)}$ and $ \boxed{(8, 3)}$[/hide]

[hide="Problem 4"]Let $ x$ be the given sum. Note that $ x^2 \minus{} 20 \equal{} x$. Equivalently, $ x^2 \minus{} x \minus{} 20 \equal{} 0$, or $ (x\minus{}5)(x\plus{}4) \equal{} 0$. Since $ x > 0$, $ x \equal{} \boxed{5}$.[/hide]

[hide="Problem 5"]Let the smallest integer be $ a$. We want $ 5a \plus{} 10$ to be divisible by $ 15$. Therefore, $ a \equiv 1$ modulo $ 3$. Since $ a > 1000$, $ a \equal{} 1003$. The smallest possible sum: $ \boxed{5025}$[/hide]

Yes they are from this school year. These are the jr. division problems.

[hide] Problem 2
From 12 students, for the first team we can make 12C3 arrangements. For the second team we can make 9C4 arrangements. And for the last team we can make 4C4 arrangements. 12C3= 220. 9C4= 126. And 4C4= 1. 220*126*1= 27,720 ways.[/hide]

If anyone could provide a non-trig solution for number 6, that would be greatly appreciated.

[hide="Hint for Problem 6"]Extend the two legs of the trapezoid to complete a triangle.  Then compare the area of similar triangles.[/hide]

Ravi can you post your solution? I get to the part with the area ratio. But, I cant figure out how we can use that to find EF.  :(

[hide="Start of Solution for Problem 6"]Let $ G$ be the point at which $ DA$ and $ CB$ would meet when extended.  Let $ x \equal{} EF$.  The areas of $ ABG$, $ EFG$, and $ CDG$ are in the ratio $ 6^2 : x^2 : 10^2$.  So the areas of $ ABFE$ and $ EFCD$ are in the ratio $ x^2 \minus{} 36$ to $ 100 \minus{} {x}^2$.[/hide]

I read somewhere a while ago that EF can be represented by the adding the squares of the bases, dividing by two, and taking the square root.

So for this question the bases are 6 and 10. 6²+10²= 136. 136/2= 68. Square root of 68 is 2√17. So EF is 2√17.

If someone can give me the name of the theorem that would be a great help.  :D


calculus
integration
function
calculus computations

Any ideas on how to obtain a closed form solution for the integral in the range [0,1] of the following function?

1/(sqrt[(1 - w^2)*(4 - w^2)*(9 - w^2)*(16 - w^2)])

Thanks in advance for any help!

My first guess is that there is no closed form solution; it's possible, but there's no good reason to expect one.

Hi.

Thanks for your response.

If you are interested in some hints I received about this integral
see here

http://groups.google.gr/group/sci.math.symbolic/browse_thread/thread
/417ec8df07895e60/7997a49ba3e72a06?hl=el#7997a49ba3e72a06


number theory
greatest common divisor
number theory proposed

Is there exists infinitive couples of natural numbers $(m,n)$ such that $gcd(m!+1,n!+1) >1$ and  $gcd(m!-1,n!-1) >1$

[mod.: lets add that $m \neq n$ ;)]

Of coures  :oops: , Another problem form this kind is:

 Is there exists many infinitive tribles of positive integers $(m,n,l)$ such that $(m>n>l)$and $gcd(m!+1,l!+1)>1;gcd(l!+1,n!+1)>1,gcd(n!+1,m!+1)>1$.

Try to kill all of them.  :)

[quote="ZetaX"]What about taking any triple of even numbers in that order  :huh:[/quote] U know what i was talking about, it was just the type mistake


advanced fields
advanced fields theorems

Let $ X$ normed space, and $ Y$ Banach space. Then normed space $ B(X,Y)$ is complete, that is every C-sequence converges.

can be found in every book about functional analysis and also in a dozen of threads here.


[quote="levi"][b]Let ABC be an acute triangle. Prove that there is the following inequality: [/b]

\[\underline {\overline {\left|\ \overline {\underline {\left| \ a^2+b^2+c^2\ge 2R^2+s^2+r^2\ \right| }}\ \right| }}.\] [b][u]Remark.[/u][/b]$2R(R-2r)\le a^2+b^2+c^2-(ab+bc+ca).$[/quote]

[b]Remark. [/b]$\underline {\overline {\left| \frac{a^3+b^3+c^3}{abc}\ge 1+\frac Rr\right| }}\ge 3$. [b]This is a beautiful inequality !.[/b]


Prove or disprove: For any set of integers ${a_1,a_2,...,a_n}$, there exists integers $b_1,b_2,...,b_n$ such that $a_1b_1+a_2b_2+...+a_nb_n = \gcd (a_1,a_2,...,a_n)$

proof by induction:
true for $n=2$.
suppose it is true for $n-1$
Because $gcg(a,b,c)=gcd(gcd(a,b),c)$ it is true then for $n$.

What if I include the condition that the $b_i$'s are non-zero integers.

[quote="K.W.M.A.N"]
Because $gcg(a,b,c)=gcd(gcd(a,b),c)$ it is true then for $n$.[/quote]

Hmm... why is it true for $n$?

there are numbers so that $x\cdot gcd(a_1,...,a_{n-1})+b_na_n=gcd(a_1,...,a_n)$, and by induction there are numbers so that  $gcd(a_1,...,a_{n-1})=b_1'a_1+...+b_{n-1}'a_{n-1}$, so if you chose $b_1 = xb_1', ..., b_{n-1}=xb_{n-1}'$ then the numbers $b_i$ will be you solution

And is it true that
$a_1b_1+a_2b_2+...+a_nb_n = k$ has a solution in integers $b_i$'s $\leftrightarrow \gcd (a_1,a_2,...,a_n) |k$

I think so: suppose not, then the left hand can be divided by $gcd(a_i)$, but the right hand can't, a contradiction