geometry
trigonometry
ratio
angle bisector
trig identities
Law of Sines
Law of Cosines

[quote="frt"]The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. Compute the area of the triangle.[/quote]

This is more of an outline solution, but I believe you bisect the largest angle to get two triangles (one of which is isosceles).  You can then use Stewart's and Angle bisector to get the side lengths (which one can label x-y, x, and x+y).

Edit: Okay, label the following on the triangle:

$a = x+y, b=x-y, c=x, m=\frac{x(x+y)}{2x-y}, n = \frac{(x-y)(x+y)}{2x-y},$ and $d = \frac{x(x+y)}{2x-y}$.

After solving using Stewart's, (it states that $cnc+bmb=dad+man$), one sees the only feasible solution is $x=5y$.

This means $a=6y, b=4y, c=5y, m=\frac{10}{3}y, n=\frac{8}{3}y, d=\frac{10}{3}y$.

I used the law of sines, which gave me $\frac{8/3y}{\sin \theta} = \frac{4y}{\sin 2\theta} \Longleftrightarrow \cos \theta = 3/4$.

I suppose my method only gets you halfway there.  We know the sides are in a ratio of 4:5:6, but that's all I can figure out.  Anyone else have some ideas?

[quote="pkothari13"] I suppose my method only gets you halfway there.  We know the sides are in a ratio of 4:5:6, [/quote]
Only halfway?

[quote="frt"] The lengths of the sides of a triangle are consecutive integers [/quote]

HAHA. oops.  :blush: Wow. I didn't even notice that.  I assumed they were just in an arithmetic progression.  In that case, replace $y$ with 1.  That should make things a lottttttttttt easier.  So I guess the sides are 4,5,6 so the area, by Heron's, is $\frac{5\sqrt{63}}{4}$.

[quote="pkothari13"]So I guess the sides are 4,5,6 so the area, by Heron's, is $\frac{5\sqrt{63}}{4}$.[/quote]
Correct, but you can simplify it to $\frac{15\sqrt{7}}{4}$ ;)

Did anyone else try other ways?
(Hint for another method: Law of Cosines)

Ok, lets suppose a triangle with sides $a$ , $a-1$ and $a+1$

1. Applying the law of cosines for $a-1$, we get:

$(a-1)^2=a^2+(a+1)^2-2a(a+1)\cos\alpha$

that simplifies to:

$\cos\alpha = \frac{4+a}{2(a+1)}$ [1]

2. Applying the law of sines, we have that:

$\frac{a-1}{\sin\alpha}=\frac{a+1}{\sin2\alpha}$

from what we get that:

$\cos\alpha = \frac{a+1}{2(a-1)}$ [2]

3. From [1] and [2], we have that:

$\frac{4+a}{2(a+1)} = \frac{a+1}{2(a-1)} \Rightarrow  a=5, (a-1)=4, (a+1)=6$

4. Calling for help to our good friend Heron, we get that the area of the triangle is:

$\Delta = \sqrt{\frac{15}{2}\frac{7}{2}\frac{5}{2}\frac{3}{2}} = \frac{\sqrt{1575}}{4} = \frac{15\sqrt7}{4}$

By the way, [b]frt[/b], is this your solution? I really enjoyed solving your problem!  :)

Yeah, that's basically the same as my solution :)


Find all positive integers $n$ such that $3^{n}-1$ is divisible by $2^n$.

Let $k$ be the order of $3$ mod $2^n$. We will prove that for $n \geq 4$ there is: $k > 2^{n-3}$. In fact, we have:
$3^{2^{n-3}} \equiv 2^{n-1} + 1 \mod{2^n}$
We prove this inductively. It works for $n=4$. Suppose that it works also for $n \geq 4$, which means that we can write:
$3^{2^{n-3}} = a2^n +2^{n-1} + 1$ for $a$ integer
after squaring:
$3^{2^{n-2}} = b2^{n+1} + 2^{n} + 1$ where $b$ is some integer and we are done.
Therefore if $3^n \equiv 1 \mod{2^n}$ then $n \leq 3$ or $n > 2^{n-3}$. Simple checking yields the answer: $n \in \{1, 2, 4\}$.

It may be a known problem, however the source for this is Romania TST 2005 [ [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=198299#198299]See here[/url] ], unless someone else has knowledge of an earlier reference.

Use this Lemma:

[i]Let $ a, b, n \in \mathbb{Z}$ such that $ 2^{\alpha}\|\frac{a^{2}-b^{2}}{2}$ and $ 2^{\beta}\|n$. Then $ 2^{\alpha+\beta}\|a^{n}-b^{n}$.[/i]

[quote="Binomial-theorem"]Not sure why no one has posted a solution using Lifting the Exponent as this pretty much kills it.
[hide="Solution"]

If $n=1$ we get a true statement.  Otherwise we must have $3^n-1\equiv 0\pmod{4}$.  This results in giving us $n\equiv 0\pmod{2}$.  Let $n=2k$ to give us $9^k-1$ is divisible by $2^{2k}$.  However $v_2(9^k-1)=v_2(9-1)+v_2(k)=3+v_2(k)\ge 2k$.  However $v_2(k)\le k$ therefore $2k\le 3+v_2(k)\le 3+k$ so hence $k\le 3$.  It is now simple work to check these values and notice that $k=1,2$ satisfy this equation and combining with $n=1$ gives $n\in \{1,2,4\}$. [/hide][/quote]
More simple:
[hide="Solution"]If $n$ is odd then $v_2(3^n-1)=v_2(3-1)=1$. Thus, $n=1$.
If $n$ is even then $v_2(3^n-1)=v_2(3-1)+v_2(3+1)+v_2(n)-1=2+v_2(n) \ge n$. But $n \ge v_2(n)$ then $n-v_2(n)=0$ or $n-v_2(n)=1$. Now it is easy to do next.[/hide]

I apologize to everyone for my solution being similar to that of [b]Binomial-theorem[/b].

[b]Case1:$n=1$[/b]
$n=1$ satisfies the condition.

[b]Case2:$n\ge 2$[/b]
Since $3^n-1\equiv 0\mod{4}$,then $n$ is even.By Lifting the Exponent Lemma,
$v_2(3^n-1)=v_2(3-1)+v_2(3+1)+v_2(n)-1=v_2(n)+2$.
Thus $n\le v_2(n)+2$.Let $n=2^kt(k\in \mathbb N, t$:odd$)$.Then $2^k\le 2^kt\le k+2$.This implies that $k=1,2$.

[b]Case2-1:$k=1$[/b]
$2t\le 3\rightarrow t=1\rightarrow n=2$ which satisfies the condition.

[b]Case2-2:$k=2$[/b]
$4t\le 4\rightarrow t=1\rightarrow n=4$ which satisfies the condition.

From above cases,the answer is $\boxed{n=1,2,
4}\blacksquare$ :coolspeak:


Let $ ABC$ be a cyclic quadrilateral. Denote by $ O$ the intersection of its diagonals. The circumcircles of triangles $ ABO$ and $ CDO$ meet again at $ K$. Take point $ L$ such that triangles $ BLC$ and $ AKD$ are similar. Show that the quadrilateral $ BKCL$ has an inscribed circle.

The bisectors of two angles $ KBL,KCL$ intersect at $ I$. By simple angle chase, we have $ \Delta BKD \sim \Delta BIC$, then $ \Delta BKI \sim \Delta BDC$, so $ \angle BKI \equal{} \angle BDC \equal{} \angle BAC \equal{} 180^o \minus{} \angle BKO$. It means that $ I,K,O$ are collinear. Prove that $ KO$ is the bisector of angle $ BKC$ (it's easy by angle chase), and we have the result.

Thank you dear PDatK40SP, for your interest and your solution. It's also one of my solutions for this problem. Another one is based on Ptoleme's theorem, but it's longer and uglier.


Is it possible to add a .pdf attachment in a blog?

If not, I'm trying to post a .pdf document as an article in the WOOT blog. Any suggestions? (If I post it in the WOOT forum, then it's not avaliable to the general public, whereas the WOOT blog is.)

Hmm, that's a tricky one :)
Unfortunately you cannot attach files in blogs (and we won't probably allow this for some time). You can attach it in this post and then put the link in the blog.

That'll work. Thanks!

Here's the "Pick a Method, Any Method!" article.

Where can I find the WOOT blog?  If it has stuff like this on it, then it is definitely worth reading.


calculus
integration
geometry
calculus computations

i need to ifnd th area of an interior of a graph. so i took the integral from 0 to 2 of 4x^2 - x^4 dx , i thin i have to use u substitution but i dont know where to go from there.

So your problem is to find $\int_0^2 (4x^2-x^4)\,dx$? Just represent it as a difference of 2 integrals and recall that $\int_a^b x^p\,dx=\frac 1{p+1}(b^{p+1}-a^{p+1})$ ($p\neq -1$).


k=1.
See problem 7 from Algebraic Inequality-Old and New Method, page 300.
We find the condition $k \leq 1$ for $\frac{2b-b1}{a_{1}}=...=\frac{2b-b_{n}}{a_{n}}\geq 0$, where $b=\frac{b_{1}+...+b_{n}}{n}$.

Thank you VASC. I don't have the book. :blush:

Does your method work for the obvious generalization to $r$ rows of variables? (may you give a sketch of proof?  :) )
i.e. written somewhat floppily with $a_{i},b_{i},...,z_{i}\ge0$ $(i=1,...,n)$ and changing the condition to $\sum_{i=1}^{n}a_{i}^{2}=\sum_{i=1}^{n}b_{i}^{2}= \cdots=\sum_{i=1}^{n}z_{i}^{2}=n$ (rather than $1$) for convenience, it would state \[\frac{k}{n}\sum_{i=1}^{n}{a_{i}b_{i}\cdots z_{i}}+1\ge(k+1) \frac{\sum a_{i}}n\frac{\sum b_{i}}n\cdots\frac{\sum z_{i}}n.\] Another question: If we drop the "$\ge0$" condition, is the result still the same for $r=2$ and $r>2$  :?:


superior algebra
superior algebra unsolved

Find the group of automorphisms of group $ \mathbb{Z}_4\times\mathbb{Z}_4$.

Please, help me...:)

And the second problem... Find all groups in order of a. 2p, b. $ p^2$, where $ p\in\mathbb{P}$. I know, that all elements of the group in a. have order in {1,2,p,2p} and in b {1,p,p^2}, but what more?

Groups of order $ p^2$:

First we know, that every group of order $ p^2$ is Abelian. Now there are two options: there exists an element of order $ p^2$, in which case $ G \cong \mathbb{Z}_{p^2}$, or all elements except unity have order $ p$. Then we take elements $ a$ and $ b$, where $ b\notin \left\{a,a^2,\ldots,a^{p} \right\}$. Then there are $ p^2$ elements of the form $ a^{\alpha}b^{\beta}$, $ \alpha\in\left\{0,1,\ldots, p\minus{}1\right\}, \beta\in\left\{0,1,\ldots, p\minus{}1\right\}$, where no two are equal: $ a^{n_1}b^{m_1}\equal{}a^{n_2}b^{m_2} \Rightarrow a^{n_1\minus{}n_2}\equal{}b^{m_2\minus{}m_1},$ where we have used commutativity. Since $ A\cap B\equal{}\{e\} \Rightarrow n_1\equal{}n_2, m_1\equal{}m_2$. So $ G\cong \mathbb{Z}_{p}\times \mathbb{Z}_p.$


geometry
circumcircle
perpendicular bisector

Triangle $ABC$ is right angle at $\angle ACB$.
$CF$ is the bisector of $\angle ACB$ and crosses $AB$ at $E$.
There exists a point $D$ on $AB$ such that $FD$ is perpendicular to $AB$ and $AD=DB$.

Prove $CD=DF$

[hide="Proof"]Draw $FM$ such that $FM\perp BC$ and $M\in BC$. Now $\angle EFM=\angle ECA=45^\circ$ (alternate angles) and $\angle DFM=\angle ABC=\beta$ (perpendicular sides). Hence $\angle EFD=\angle EFM-\angle DFM=45^\circ-\beta.$

On the other hand, $AD=BD$ means that $D$ is midpoint of the hypotenuse, hence also a circumcenter, therefore \begin{eqnarray*}CD=BD\implies \angle DCB=\angle DBC=\beta\implies\angle ECD=\angle ECB-\angle DCB=45^\circ-\beta.\end{eqnarray*}

Hence, $\angle EFD=\angle ECD$, which means that $\triangle DFC$ is isosceles. Therefore, $DF=DC.$[/hide]

farenhajt

arghh i had it except for the fact i overlooked.. that D is the circumcentre.. gotta brush up on my triangle geometry..

i live in australia.. in year 10 this year.. and i have currently being in holidays for about a month.. gotta gett back to work..

thanks man^^

I shall prove that $F$ lies on the circumcircle of $\triangle{ACB}$.

Let the circumcircle of triangle ABC meets CF at F'. Thus $F'D$ is perpendicular to $AB$. Thus the perpendicular bisector of $AB$ meets line $CE$ at two points, therefore $F=F'$, which means F' is on the circle with centre D and diameter AB. So FD=CD.


logarithms inequalities quadratics algebra

Solve equation $ \log(x\plus{}y)\equal{}\log_{\sqrt{x\plus{}y}}x \plus{} \frac{1}{\log_{y}{\sqrt{x\plus{}y}}}$

First change all the logs on the R.H.S to base 10. We get (all below is R.H.S)

$ \frac {\log{x}}{\log{\sqrt {x \plus{} y}}} \plus{} \frac {1}{\frac {\log{\sqrt {x \plus{} y}}}{\log{y}}}$
$ \frac {\log{xy}}{\log{\sqrt {x \plus{} y}}}$

Back to the equation:

$ \log{(x \plus{} y)} \equal{} \frac {\log{xy}}{\log{\sqrt {x \plus{} y}}}$
$ \log{\sqrt {x \plus{} y}} * \log{(x \plus{} y)} \equal{} \log{xy}$
$ \frac{1}{2}(\log{(x\plus{}y)})^2 \equal{} \log{xy}$
$ (\log{(x\plus{}y)})^2 \equal{} 2 \log{xy}$
I'm stuck..

I might be wrong, but doesn't $ \log{(x \plus{} y)}\cdot \log{\sqrt {x \plus{} y}} \not \equal{} \log{\sqrt {x \plus{} y}} \plus{} \log{(x \plus{} y)}$?

Also, the $ x \plus{} y$ and $ xy$ symmetric expressions seem like they have something to do with $ (x \plus{} y)^2 \ge 0$. (The Trivial Inequality applied to $ x \plus{} y$.)

modularmarc101        you are totally wrong again

i'll tell you how must be

$ \log{(x \plus{} y)} \equal{} \log_{\sqrt {x \plus{} y}}x \plus{} \log_{\sqrt {x \plus{} y}}y$
$ \log_{10}{(x \plus{} y)} \equal{} \log_{\sqrt {x \plus{} y}}xy$

So: $ x \plus{} y \equal{} xy ,y \equal{} \frac {x}{x \minus{} 1}$ and $ 10 \equal{} \sqrt {x \plus{} y}$
$ 10 \equal{} \sqrt {x \plus{} \frac {x}{x \minus{} 1}}$
$ x^2 \minus{} 100x \plus{} 100 \equal{} 0$
$ x_1 \equal{} 50 \plus{} 20\sqrt {6}$
$ x_2 \equal{} 50 \minus{} 20\sqrt {6}$

Edit: incomplete solution

there's also a mistake there, look at the bases.

from :

$ \log{\sqrt {x \plus{} y}}*\log{(x \plus{} y)} \equal{} \log{xy}$

we get $ 2(\log{(x \plus{} y)}) \equal{} \log{xy}$

[quote="sunehra"]How does $ \log_a{b} \equal{} \log_c{d}$ imply that $ a \equal{} c$ and $ b \equal{} d$?[/quote]

If $ a\equal{}c$ and $ b\equal{}d$ so $ \log_a{b}\equal{}\log_c{d}\equal{}\log_a{b} :$ I don't see misakes.{At least is one solution}

But they have different bases, one had base 10 and the other base $ \sqrt{x\plus{}y}$...

Obelix is wrong

   edit : not wrong but "not completed"

[quote="SHP3ND1"]Obelix is wrong and i proved you that there is no solution

  OVER   :)  :D[/quote]

I would be pleased to tell me where is my mistake  ? Thank you.

[quote="Obel1x"][quote="SHP3ND1"]Obelix is wrong and i proved you that there is no solution

  OVER   :)  :D[/quote]

I would be annoyed if you don't tell me where my mistake is ? Thank you.[/quote]

Both of you have mistakes. Obelix look at my most recent post.

Shend, it's $ (\log{(x\plus{}y)})^2$ not $ \log{(x\plus{}y)^2}$

Solution does exist.

I will continue my solution from $ \log(x \plus{} y) \equal{} \log_{\sqrt {x \plus{} y}}xy$.

[hide="Solution"]Given above condition, we have $ \sqrt {x \plus{} y} \equal{} 10^k$ and $ x \plus{} y \equal{} (xy)^k$ for some real $ k$.

$ x \plus{} y \equal{} 10^{2k} \implies xy \equal{} 100$.

Putting it in the original equation, we have $ \log10^{2k} \equal{} \log_{10^{k}}10^2\implies (10^k)^{2k} \equal{} 10^2\implies k^2 \equal{} 1\implies k \equal{} \pm 1$

If $ k \equal{} 1$, $ x \plus{} y \equal{} 100$ and $ xy \equal{} 100$.

Solving simple quadratic equation, we obtain two solutions
$ (50 \plus{} 20\sqrt {6}, 50 \minus{} 20\sqrt {6})$ and $ (50 \minus{} 20\sqrt {6}, 50 \plus{} 20\sqrt {6})$.

If $ k \equal{} \minus{} 1$, $ x \plus{} y \equal{} \frac {1}{100}$ and $ xy \equal{} 100$, in which case the solution is imaginery.

Hence, the only solutions are $ (50 \plus{} 20\sqrt {6}, 50 \minus{} 20\sqrt {6})$ and $ (50 \minus{} 20\sqrt {6}, 50 \plus{} 20\sqrt {6})$.[/hide]

@Obel1x
Your assumption is not wrong; it's just that there could be more solutions.

[quote="SHP3ND1"]to Obelix

two logarithms are equal if they are equal or they have proporcional bases and expressions above bases

so you are supposing that their base is equal and their numbers above base are equal...so you are cutting the problem just for equal and not proporcional bases[/quote]

Of course my solution is not completed because i've cut it just for equal.If there's a solution from bases equalize we can't say it's wrong :D,and we had nothing to suppose here.I just solve it when the bases are equal,which isn't wrong act.Hope you understand me and my english  :oops: .We'll disscus this @ school again.

None of the solutions so far are complete. There are an infinite number of solutions.

[hide](I'm assuming $ \log z$ means log base 10 of z)
$ \log(x \plus{} y) \equal{} \log_{\sqrt {x \plus{} y}}x \plus{} \frac {1}{\log_{y}{\sqrt {x \plus{} y}}}$
$ \log(x \plus{} y) \equal{} \log_{\sqrt {x \plus{} y}}(xy)$
$ \log\sqrt {x \plus{} y} \cdot \log(x \plus{} y) \equal{} \log(xy)$
$ \frac {1}{2}\log(x \plus{} y) \cdot \log(x \plus{} y) \equal{} \log(xy)$

Let $ \log(x \plus{} y) \equal{} k$. Then $ \log(xy) \equal{} \frac {k^2}{2}$. So $ x \plus{} y \equal{} 10^k$ and $ xy \equal{} 10^{\frac {k^2}{2}}$
Therefore x and y are the solutions to the quadratic equation $ z^2 \minus{} (10^k)z \plus{} 10^{\frac {k^2}{2}}$, which are
$ \frac {10^k \pm \sqrt {10^{2k} \minus{} 4 \cdot 10^{\frac {k^2}{2}}}}{2}$

This has real solutions for any $ k$ such that $ 10^{2k} \minus{} 4 \cdot 10^{\frac {k^2}{2}} \geq 0$
$ 10^{2k} \geq 4 \cdot 10^{\frac {k^2}{2}}$
$ 2k \geq \log 4 \plus{} \frac {k^2}{2}$
$ k^2 \minus{} 4k \plus{} 2\log 4 \leq 0$
$ (k \minus{} 2)^2 \leq 4 \minus{} 4\log 2$
$ 2 \minus{} 2\sqrt {1 \minus{} \log2} \leq k \leq 2 \plus{} 2\sqrt {1 \minus{} \log2}$

There are an infinite number of such values of $ k$, and so there are an infinite number of solutions to this equation, all of the form $ \{x,y\} \equal{} \frac {10^k \pm \sqrt {10^{2k} \minus{} 4 \cdot 10^{\frac {k^2}{2}}}}{2}$[/hide]

Nice explanation!  Two tiny notes:
[quote="gauss202"](I'm assuming $ \log z$ means log base 10 of z)

...

This has real solutions for any $ k$ such that ....[/quote]  For the first, of course what you did works no matter what the base of that logarithm is (just replace "10" in your solution everywhere with "$ b$").  For the second, actually we need not just real solutions but positive real solutions where one of them ($ y$) is not 1 and where the sum of the two solutions is not 1 -- but I don't think either of these is a problem over the range of solutions you've found.


With regards to the USNCO and Atkins, how would you prioritize reading Atkins?

I'm struggling between different factors:

(1)Doing practice tests (USNCO ones)
(2)Reading Atkins (incl. reading their example problems)
(3)Doing Atkins Practice Problems

It'd be really helpful if someone could rank these in importance. I'm fairly new to olympiad chemistry so not much experience on my end :\ HELP

2 first, then 1, and 3 is optional. So:

2
1
3

But you really should spend as much time on 1 as on 2. Again...learn some orgo as well. It will help you for the later levels.

ah ok. thanks :) it was more the practice problems of atkins that were scaring me.

also, how would you recommend I handle the masses of formulas that atkins gives? Many of them aren't all that derivable (at least, from what I've read) and mem'ing them all would be a huge pain :\

I view Atkins as additional reading that only enforces your knowledge.

The top priority should be to TAKE THE PRACTICE TESTS OVER AND OVER AND OVER AND OVER AND OVER again. Until you have mastery over every question.

Then would I read Atkins.

If you're a bonehead and don't know anything about chem (which you probably aren't =P), then I would recommend reading Atkins first: maybe skim through it not worrying so much about the problems and such first. Later on, the problems in the book become handy, ranging from the difficulty of the local exam to that of the study camp.

Oh yeah, I typed my reading notes for Atkins and other books. It was helpful...

My last piece advise is that whenever you are stuck (as in there is no way out), then look at the answer up to that point and a little past that. Then I would reason out exactly what the answer says, whether you agree with it or not ('cuz sometimes the book can be wrong, especially Atkins), and what you think got you stuck. Just skipping to the next question after copying down the answers does you harm.

haha alrite. sounds like USAMO prep except not!

thanks for the advice!

[quote]If you're a bonehead and don't know anything about chem then I would recommend reading Atkins first[/quote]

Ohhh burn

[quote]Oh yeah, I typed my reading notes for Atkins and other books.[/quote]

REAL MEN ABSORB WHILE THEY READ, AINT IN THE MOOD FOR NO SHUGAZY NOTES

@OP: Report back with how well you did on your first try (or don't). Use the first exam to see what are your strengths/weaknesses, and then go back and re-read those weak areas.


logarithms
calculus
derivative
trigonometry
calculus computations

Question: Suppose $ \ln\sqrt{x^2\plus{}y^2}\equal{}\arctan (y/x)$, find $ dy/dx$.

Of course, we can take the derivatives about $ x$ for both sides, and then we can get the answer.

However, if we put $ x\equal{}r\cos\theta$, $ y\equal{}r\sin\theta$, the original equation becomes $ \ln r\equal{}\theta$.
Then, we have $ d\theta/dr\equal{}1/r$. So, can we use this result to find out $ dy/dx$?

It is very possible (in fact I found it much more simple to do it by that approach).
[hide="Hint"]The extra piece of information that's useful is that we can claim that $ \frac {dy}{dx} \equal{} \frac {dy}{d\theta}\frac {d\theta}{dx} \equal{} \frac {\frac {dy}{d\theta}}{\frac {dx}{d\theta}}$ by the Chain Rule.[/hide]


For positive reals $ a,b,c$ if it is known that -
$ c^2 \le a^2 \plus{} b^2$ and $ c \ge b \ge a$
can u find another positive real $ t$ such that -
$ (c \minus{} t)^2 \equal{} (a \minus{} t)^2 \plus{} (b \minus{} t)^2$

[hide="Solution"]Note that the function $ f(t) \equal{} (a \minus{} t)^2 \plus{} (b \minus{} t)^2 \minus{} (c \minus{} t)^2$ is continuous for all real $ a$,$ b$,$ c$. But $ f(a) \le 0$ and $ f(0) \ge 0$, so by the Intermediate value theorem $ f$ must have a positive zero unless $ f(0) \equal{} 0$ or $ a\equal{}0$ and $ b\equal{}c$, in which case there is no positive solution. [/hide]
I know you could let $ a$,$ b$, and $ c$ be the sides of an acute triangle, and you try to shrink the sides until you get a right triangle, but in order to formalize the proof you would have to do something like I did.

thanks!
well since a "t" as above exists.. you can assume that $ a,b,c$ are sidelengths of a triangle!!

How can you say that you can keep shrinking the sides such that at one point it would be a right triangle? Is that possible without the application of intermediate value theorem etc... ?


[quote="kunny"]Welcome to this forum, idiot 1990 :)

Is your problem like this?

[quote="idiot1990"]$ a,b$ are postive real numbers and $ a \plus{} b \equal{} 2.$
Prove that
$ \frac {(a^{\frac {b}{2}} \plus{} b^{\frac {a}{2}})^2 \plus{} 3}{4}\leq a^{b} \plus{} b^{a}\leq \frac {(a^{\frac {b}{2}} \plus{} b^{\frac {a}{2}})^2 \plus{} 2}{3}$

thx :)[/quote]

Edited by kunny[/quote]
yes
thx for editing it for me :)

no idea?
is it really a hard inequality?

even you can't prove it immediately, could you provide me some direction to prove it? :(


conics
ellipse
analytic geometry
parameterization
geometry proposed
geometry

A question about circles and ellipses...

The well know property
of the ellipse: from any point on the elipse,
the sum of the distances to each focus is a constant.

My question is:
Circle: 1 focus (=center)(special case of the ellipse)
Ellipse: 2 focus (special case of EEE?)
EEE?: 3 focus

Is the there such a curve? A curve with 3 focus and the sum of the distances from any point on it to each focus is constant.

Can we generalize? EEEE? (4 focus)..., and so on?

Of course there is such curve, but I doubt it has any importance.

For the sake of simplicity, assume that the 3 focal points $E, F, G$ make an isosceles triangle $\triangle EFG$ with the base $EF$. Let $p$ be the constant sum of distances of the point $P$ from the points $, F, G$, i.e., $PE + PF + PG = p$. Split the distance $p$ into 2 arbitrary positive distances $q + r = p$. The locus of points $Q$ with the sum of distances $QE + QF = q$ from the points $E, F$ is an ellipse with focal points $E, F$ and major axis $a = \frac q 2$. Assuming these points have coordinates $E = [-e, 0], F = [e, 0]$, i.e., $EF = 2e$, the minor axis of the ellipse is $b = \sqrt{\left( \frac q 2 \right)^2 - e^2}$ and the ellipse equation is

$\frac{4x^2}{q^2} + \frac{4y^2}{q^2 - 4e^2} = 1$

The locus of points $R$ with the distance $RG = r$ from the point $G$ is a circle with center $G$ and radius $r$. Assuming the point $G$ has coordinates $G = [0, g]$, the circle equation is

$x^2 + (y - g)^2 = r^2 = (p - q)^2$

Eliminating the arbitrary parameter $q$ from the equations of the ellipse and circle, you can get an (ugly) equation of the desired curve. It looks like a drop of oil.


number theory
least common multiple
greatest common divisor
function
number theory unsolved

Prove, that if $[x,y]$ is the least common multiple, and $(x,y)$ is the greatest common divisor then for all natural $a,b,c$
$[(a,b),c]=([a,c],[b,c])$ and
$([a,b],c)=[(a,c),(b,c)]$

[quote="rgep"]One good way of seeing these is to translate into statements about max and min ($\wedge$ and $\vee$).[/quote]

can u explain more details [b]rgep[/b] ? It sounds interesting ! I know a sol. but very ugly!  :blush:

Whenever solving such problems of gcd and lcm involving multiple variables, we only need to remember $3$ things.

$(x_1, x_2, \ldots, x_s, x_{s+1}, \ldots, x_t) = ((x_1, x_2, \ldots, x_s), x_{s+1}, \ldots, x_t)$           Eqn (1), generalised from $(x,y,z) = ((x,y),z)$
$[x_1, x_2, \ldots, x_s, x_{s+1}, \ldots, x_t] = [[x_1, x_2, \ldots, x_s], x_{s+1}, \ldots, x_t]$           Eqn (2), generalised from $[x,y,z] = [[x,y],z]$

The last thing we need to know is the simple fact that $xy = (x,y)[x,y]$, thus giving us a relation between the gcd function and the lcm function.

For Eqn 1, a corollary would be $((a,b),(a,c)) = (a,b,c)$, and likewise, a corollary corresponding to Eqn (2) would be $[[a,b],[a,c]] = [a,b,c]$.

Now, we are equipped to tackle the two problems.

For the first problem, we start from the R.H.S.
$([a,c],[b,c]) = \frac{[a,c][b,c]}{[[a,c],[b,c]]} = \frac{[a,c][b,c]}{[a,b,c]} = \frac{[[a,c],[b,c]]([a,c],[b,c])}{[a,b,c]} = \frac{[a,b,c]([a,c],[b,c])}{[a,b,c]} = ([a,c],[b,c])$

The second problem is similar.

Let \[ a=p_1^{a_1}p_2^{a_2}...p_k^{a_k}  \] \[ b=p_1^{b_1}p_2^{b_2}...p_k^{b_k}  \] \[ c=p_1^{c_1}p_2^{c_2}...p_k^{c_k}  \]

Then \[ gcd(a,b)=p_1^{min(a_1, b_1)}p_2^{min(a_2, b_2)}...p_k^{min(a_k, b_k)}  \]
\[ lcm(a,b)=p_1^{max(a_1, b_1)}p_2^{max(a_2, b_2)}...p_k^{max(a_k, b_k)}  \]

So all we have to prove for the [b]first[/b] (of course is same for [b]2nd[/b]) problem is is that \[ max(min(a_i,b_i), c_i)=min(max(a_i,c_i), max(b_i, c_i))  \] which is trivial

I think this is very nice solution and very useful :D :D :) :D :D


How many integer values of $ n$ satisfy $ \minus{}50 < n^3 < 50$?


calculus
integration
calculus computations

$ \frac{d}{dx}(\int_{0}^{x^3 }xsin(\pi t)dt)\equal{}\frac{d}{dx}(x\int_{0}^{x^2 }sin(\pi t)dt)$

at equation above, how could the variable x at the right out of the integration sign?

n why the upper limit of integration (the right one) become $ x^2$ ?

plz help me. i really dont understand how could it be.

i'll appreciate your whole answers. tx

$ \frac{d}{dx}(\int_{0}^{x^{3}}xsin(\pi t)dt)\equal{}\frac{d}{dx}(x\int_{0}^{x^{3}}sin(\pi t)dt)$. Your equation is wrong.


limit
logarithms
real analysis
real analysis unsolved

Does there exist  $ f\in C(R)$ such that the equality $ f(f(x)) \equal{} e^{ \minus{} x}$ holds for all $ x\in R$

There is no such $ f$.

Suppose $ f$ is such a function. Then $ e^{\minus{}f(x)} \equal{} f(f(f(x))) \equal{} f(e^{\minus{}x}).$ Taking $ x \to \infty$ to both sides, we have $ \lim_{x\to\infty} e^{\minus{}f(x)} \equal{} \lim_{x\to\infty} f(e^{\minus{}x}) \equal{} f(0) \geq 0$ by continuity of $ f$.

If $ f(0) \equal{} 0$, then $ 0 \equal{} f(0) \equal{} f(f(0)) \equal{} e^0 \equal{} 1$, a contradiction. So $ f(0) > 0$. Then $ \lim_{x\to\infty} f(x) \equal{} \minus{}\ln f(0)$ is finite. So $ \lim_{x\to\minus{}\infty} e^{\minus{}f(x)} \equal{} \lim_{x\to\minus{}\infty} f(e^{\minus{}x}) \equal{} \minus{}\ln f(0) \geq 0$. This shows that $ f(0) \leq 1$.

If $ f(0) \equal{} 1$, then $ 1 \equal{} e^0 \equal{} f(f(0)) \equal{} f(1)$ and hence $ e^1 \equal{} f(f(1)) \equal{} f(1) \equal{} 1$, a contradiction. Thus $ f(0) < 1$ and $ \lim_{x\to\minus{}\infty} f(x) \equal{} \minus{}\ln(\minus{}\ln f(0))$ is finite. This shows that $ f$ is bounded for all $ x$, but this is impossible since $ e^{\minus{}x}$ is unbounded.


function
complex analysis
complex analysis unsolved

let $ f$ be analytic on $ G\equal{}\mathbb{C}\minus{}0$ st :
$ |f(z)|\leq |log(|z|)|$ for all $ z\in G$ prove that $ f$ is constant

I have tried to use the maximum modulus theorem but I cannot
any hint will be appreciated

You don't need max-mod for this. Look at $ |z|\equal{}1$.

Perhaps what was meant was that $ G$ is the punctured unit disc? If so, $ zf(z)$ is an analytic function on $ \mathbb{D}$ that is zero on the boundary.

[quote="CaptainCanada"]You don't need max-mod for this. Look at $ |z| \equal{} 1$.[/quote]

thanks so much, I got your idea  :roll:


Let $S$ be a set of integers with the property that any integer root of any non-zero polynomial with coefficients in $S$ also belongs to $S$.  If $0$ and $1000$ are elements of $S$, prove that $-2$ is also an element of $S$.

Consider 1000x + 1000. so -1 is in S

Consider $(x+2)(-x^9 + 2x^8 - 4x^7 + 8x^6 -16x^5 +31x^4 -63x^3 +125x^2 -250x + 500)$ which can be checked to have coefficents in $\{ 0, -1, 1000 \}$ when expanded.  Result follows.

As shown above, we have -1 and 1 in S.  Then, by x+1000=0, we have -1000 in S.  Then, by finding the binary expansion of 1000, which is 1111101000, we know that 2 is a root to x^9+x^8+x^7+x^6+x^5+x^3-1000=0.

I know it's not the simplest solution, but it's very intuitive.

[quote="Magnara"]As shown above, we have -1 and 1 in S.  Then, by x+1000=0, we have -1000 in S.  Then, by finding the binary expansion of 1000, which is 1111101000, we know that 2 is a root to x^9+x^8+x^7+x^6+x^5+x^3-1000=0.

[/quote]

But it asks for $-2$...

I just wanted to post a simpler solution
[hide="solution"]
As proven earlier $ 1, 10$  $ \in S$
So therefore the polynomial $ x^{3} \plus{} x \plus{} 10$  satisfies the hypothesis of the problem.
Finally checking $ x \equal{} \minus{} 2$ we see that this satisfies the polynomial and hence $ \minus{}2 \in S$ ,as desired
[/hide]


Seeing how the AMC A series of tests is coming up I thought I'd try to prepare for it. I think I know most of the theorems needed, but I still want to work a few more problems (I don't want any big disappointments). I've done most of the problems on the AMC website. What other resources (besides the already exhausted AOPS books and AMC/Harold Reiter web pages) do people use to study.

Thanks for any insight you can give.

[quote="jeffrey"]What other resources (besides the already exhausted AOPS books and AMC/Harold Reiter web pages) do people use to study.[/quote]

Several of the books on the Art of Problem [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]Book List[/url] are very helpful for learning math better for math contests, as are the books mentioned in an earlier thread about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]readable books[/url] in the Other Problem Solving Topics forum. I like anything written about math by Gelfand, Lang, or Stillwell for secondary-school-level readers, as all those books have plenty of good mathematical thinking in them. Those books have good exercises that are well worth doing.

Hope this helps! Good luck.

[quote="jeffrey"]What other resources (besides the already exhausted AOPS books and AMC/Harold Reiter web pages) do people use to study.
Thanks for any insight you can give.[/quote]

The earlier AIME problems are generally comparable with the later AMC 12 problems.
There is a full collection of the AIME problems, which I find quite useful, at the [url=http://www.kalva.demon.co.uk/aime.html]kalva site[/url].

--Dan


geometry
area of a triangle
linear algebra
linear algebra unsolved

Find the area of a triangle in the space knowing that the ortogonal projections over the planes $ xy, yz$, y $ xz$ has area $ \dfrac{1}{2}, 1, \dfrac{3}{2}$, respectively.

We may assume the triangle is in the plane $ \Pi: ux\plus{}vy\plus{}wz\equal{}0$ with $ u^2\plus{}v^2\plus{}w^2\equal{}1$. Let $ S$ be its area. If $ a$ is the angle between $ \Pi$ and $ Oxy$ then $ 1/2\equal{}S.\cos(a)$ or $ 1/2\equal{}Sw$. By the same way $ 1\equal{}Su$ and $ 3/2\equal{}Sv$. Thus $ (1/2)^2\plus{}1^2\plus{}(3/2)^2\equal{}S^2$.


abstract algebra
superior algebra
superior algebra solved

can you a example finetely generated module that is not notherian?

Any non-noetherian ring is finitely generated over itself and not noetherian. For example $\mathbb Z[X_1,\ldots,X_n,\ldots]$.


Find all pairs (x; y) of real numbers satisfying the equations

$x^{4}+y^{4}=17\left( x+y\right) ^{2}$;
$xy=2\left( x+y\right) $.

  Darij

Let $\sigma_1=x+y$ and $\sigma_2 = xy$.  We can transform the given brace of equations into $\sigma_1^4 - 4\sigma_2 \sigma_1^2 + 8\sigma_2^2 - 6\sigma_2^2 = 17\sigma_1^2$ (1) and $\sigma_2 = 2\sigma_1$ (2).
Substituting (2) into (1) gives $0 = \sigma_1^4 - 8\sigma_1^3 - 9\sigma_1^2 = \sigma_1^2 (\sigma_1-9)(\sigma_1+1)$.

If $\sigma_1 = 0$, then $\sigma_2=0$ as well giving $(x,y)=(0,0)$.

If $\sigma_1=-1$, then $\sigma_2=-2$ implying that $x$ and $y$ are roots of the quadratic $0=t^2+t-2=(t+2)(t-1)$.  This gives 2 additional solutions $(x,y)=(-2,1), (1,-2)$.

If $\sigma_1=9$, then $\sigma_2=18$ implying that $x$ and $y$ are roots of the quadratic $0=t^2-9t+18$ whose discriminant is negative thus yielding no real roots.

We conclude that $(x,y)=(0,0), (-2,1), (1,-2)$ are the only real solution to the given system of equations.

[quote="Michael Lipnowski"]that $x$ and $y$ are roots of the quadratic $0=t^2-9t+18$ whose discriminant is negative thus yielding no real roots.[/quote]

Hmm... really?  :lol:

  Darij


Hehe, I see $x=1$ works again, by inspection!

If you multiply by the conjugate of the RHS, you get

$5x^2 +10x - 15 = \sqrt{x-1}(\sqrt{6x^2+13x-10} + \sqrt{x^2+3x+5})$
I see that the LHS is $5(x-1)(x+2)$. So if we square both sides, we'll get

$25(x-1)(x+2)^2 = (\sqrt{6x^2+13x-10} + \sqrt{x^2+3x+5})^2$

I suppose I may as well check if -2 works, too. No, it looks like it doesn't.

I've got to go, and I'm not sure what to do with this now anyway. Have fun, folks!

btw, for these type of questions, do you guys find the actual solutions or just leave it in this form?

*Usually* we try to crack out the real roots, unless otherwise stated.  But its good to at least have a factored expression/equation to work with.

[quote="Jimmy"]Square both sides to get
$7x^2+16x-5-2\sqrt{(x^2+3x+5)(6x^2+13x-10)}= 9x-9 \Rightarrow 7x^2+7x+4$$=2\sqrt{(x^2+3x+5)(6x^2+13x-10)}$.

Then square both sides again, move everything to the left, and factor to get

$(x-1)(25x^3-x^2-132x-216)=0$.[/quote]
Do you prove the equation $25x^3-x^2-132x-216=0$ no solutions?
 :?:I think this problem didn't need to solved by a muscle method


IMO Shortlist
functional equation
algebra

Yay! My first (first) IMOSL solution!

First we let $f(0) = \alpha$.

$f(f(x)+y) = 2x + f(f(y)-x)$ $(1)$

Plugging in $x=0$ into (1):

$f(y+\alpha) = f(f(y))$ $(2)$

Plugging in $x=f(y)$ into (1):

$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$

Now plugging in $y=-\alpha$ into (3):

$f(-\alpha) = 2f(-\alpha) + \alpha \Rightarrow f(-\alpha) = -\alpha$

Plugging in $y=-\alpha$ into (2):

$\alpha = f(f(-\alpha)) \Rightarrow \alpha = -\alpha \Rightarrow \alpha = 0$

So (2) can be rewritten as: $f(f(x)) = x$ $(4)$

Plugging in $y=0$ into (1):

$f(f(x)) = 2x + f(-x) \Rightarrow f(-x) = -x$

Plugging in $x=-x$ for the last equation gives $f(x)=x$ for all $x$.

It is easy to see that this function indeed satisfies the initial conditions given in the problem.

[quote="Philip_Leszczynski"]$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$ [/quote]

You are wrong here sorry you miscalculate I think.
The correct is

$f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(f(y+\alpha)+y)=2f(y)+\alpha$

from where we can find  (plugging where $y=-a$)  that  $f(-\alpha)=0$  and finally
the solutions in not $f(x)=x$  but $f(x)=x+c$

[hide="Solution"]
$ f$ is [b]surjective[/b]:
We first choose $ y\equal{}\minus{}f(x)$, from where $ f(0)\equal{}2x\plus{}f(f(\minus{}f(x))\minus{}x)$. From this it is obvious that $ f$ is surjective.

$ f$ is [b]injective[/b]:
Suppose $ f(a)\equal{}f(b)$, then we choose $ f(y)\equal{}a\plus{}b$ (because $ f$ is surjective), and $ x\equal{}a,b$. We have $ f(f(a)\plus{}a\plus{}b))\equal{}f(f(b)\plus{}a\plus{}b)$ and $ f(f(y)\minus{}a)\equal{}f(a\plus{}b\minus{}a)\equal{}f(a\plus{}b\minus{}b)\equal{}f(f(y)\minus{}b)$, so $ 2a\equal{}2b\Leftrightarrow a\equal{}b$, and $ f$ is injective.

Now we choose $ x\equal{}0$, and get $ f(y\plus{}f(0))\equal{}f(f(y))\Rightarrow f(y)\equal{}y\plus{}f(0)$, from injectivity.

Finally $ f(x)\equal{}x\plus{}c$, $ c\in\mathbb{R}$ satisfies the equation, since $ f(f(x)\plus{}y)\equal{}x\plus{}y\plus{}2c\equal{}2x\plus{}f(f(y)\minus{}x)$
[/hide]

Setting $ y\equal{}\minus{}f(x)$, we have $ f(0)\equal{}2x\plus{}f(f(\minus{}x)\minus{}x))$, so $ f(0)\minus{}2x\equal{}f(f(\minus{}x)\minus{}x)$, and $ f(0)\minus{}2x$ is a surjective function of $ x$, so $ f$ is surjective as well.

Then, by surjectivity, we can let $ c$ be a number for which $ f(c)\equal{}0$. Setting $ x\equal{}c$, we have $ f(y)\equal{}2c\plus{}f(f(y)\minus{}c)$. Since $ f(y)$ can take on all real values $ r$, $ r\equal{}2c\plus{}f(r\minus{}c)$ for all real numbers $ r$, implying $ f(r\minus{}c)\equal{}r\minus{}2c$. Letting $ s\equal{}r\minus{}c$, we find $ \boxed{f(s)\equal{}s\minus{}c}$ for some constant $ c$ for all real numbers $ s$. Also, it can be easily verified that all functions of this form satisfy this equation.

Let $P(x,y)$ denote the assertion $f(f(x)+y) = 2x + f(f(y)-x)$.
Now,
\[P(0,x) \implies f(x+f(0)) = f(f(x))\]
\[P(f(y),y) \implies f(f(f(y)) + y ) = 2f(y)+f(0) \implies f(f(y+f(0))+y) = 2f(y) + f(0)\]
Taking $y=-f(0)$ here, implies $f(0) = 2f(-f(0)) +f(0) \implies f(-f(0)) = 0$.
Next,
\[P(-\frac x2, -f(-\frac x2) - f(0)) \implies x = f(f(-f(-\frac x2) - f(0)) + \frac x2)\]
Therefore, $f$ is onto i.e. covers all values of $x$.
Now,let $f(0) = c$, and define a function $g : \mathbb{R} \rightarrow \mathbb{R}$ as $g(x) = f(x - c)$.
Therefore, our original assertion $P(x,y)$ for $f$ changes into the equivalent assertion $Q(x,y)$ for $g$, as being
\[g(g(x+c) + y + c) = 2x + g(g(y+c) - x +c)\]
As $f$ is onto, therefore $g$ is also onto. Also, $g(0) = 0$. And then finally,
\[Q(-c,x-c) \implies g(x) = -2c + g(g(x) + 2c) \implies g(g(x) + 2c) = g(x) + 2c\]
As $g$ is onto, therefore, $g(x)$ covers all values of $x$, and therefore, $g(x) + 2c$ also covers all values of $x$,
Thus, $g(x) = x$ for every $x \in \mathbb{R}$.
And therefore, $f(x) = x+f(0)$ for every $x \in \mathbb{R}$ and this is indeed a solution.

Set $y=-f(x)$ then $-2x=f(f(-f(x))-x)$ so $f$ is $surjective$.
So there exist $a,b$ such that $f(a)=0$ and $f(b)=a$. Then set $x=a$ and $y=b$ $\Rightarrow$ $a=2a+f(0)$ so $f(-f(0))=0$.
Now let $x=-f(0)$ $\Rightarrow$ $f(y)=-2f(0)+f(f(y)+f(0))$
let $f(0)=c$ $\Rightarrow$ $(f(y)+c)+c=f(f(y)+c)$. Since $f$ is $surjective$ we get that $f(y)+c$ is also $surjective$ which implies $f(x)=x+c$ $\forall$ $x$ $\in$ $\mathbb{R}$	where $c=f(0)$.

[quote="orl"]Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.[/quote]
This is my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3376958#p3376958

Notice that we can take $x$ such that $x=\frac{1}{2}\cdot f(y)$.Also it is impossible to be $f(x)>0$ because we can always take $x$ so small to make that absurd.Putting $y=y_1$,($y_1$ is such that $f(y_1)=0$),and $x->x_1=\frac{1}{2}\cdot f(y)$ we easily get $f(f(x_1)+y)=2x_1+f(\frac{1}{2}\cdot f(y))$.Lets assume that there is $x_2$ such that $x_1\neq x_2$ and $f(x_1)=f(x_2)$.From that it follows: $2x_1+f(\frac{1}{2}\cdot f(y))=f(f(x_1)+y)=f(f(x_2)+y)=2x_2+f(\frac{1}{2}\cdot f(y))$,then we get $2x_1=2x_2$.Clearly the function is an injection.
Putting $x=0$ we get $f(f(0)+y)=f(f(y))$ which is equivalent to $f(x)=x+f(0)$,(lets take $f(0)=c$).
Now it remains to check if that is a solution:$f(f(x)+y)=f(x+c+y)=x+y+2c$ and $RHS=f(f(y)-x)+2x=f(y-x+c)+2x=y-x+2c+2x=y+x+2c$.Obviously $LHS=RHS$.Hence the only solution is $f(x)=x+c$

Let $f(0) = a$. We will first prove that $f$ is surjective i.e. the range of $f$ is $\mathbb{R}$. Let $y = -f(x)$ and we get
\[a - 2x = f(f(-f(x)) - x)\]
and because $a-2x$ can range over all the reals, there must exist some real $z$ such that $f(z) = a-2x$.

Now let $r$ be some real constant such that $f(r) = 0$. Let $x = r$ and we have $f(y) = 2r + f(f(y) - r)$. If we let $y$ be some number such that $f(y) = r$, then we have $r = 2r + a$ so $r = -a$, so $f(-a) = 0$. Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$, and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$. Letting $x = 0$ and $y = z$ in the original FE gives us
\[f(z+a) = f(f(z))\]
so we have $f(f(z)) = z+2a$. Now we show that $f$ is injective. Suppose we have $f(x) = f(y)$ for some $x, y$, so then $f(f(x)) = x+2a = f(f(y)) = y+2a$ which implies that $x = y$. So now, since $f(z+a) = f(f(z))$, we have $f(z) = z+a$.

Therefore, our solutions are $f(x) = x+a$ for some $a \in \mathbb{R}$. Checking that these work is trivial.

[quote="fclvbfm934"]Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$, and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$. .[/quote]

I think you could have stopped here, as $a$ is constant and $z$ can be chosen arbitrarily, so for any $x$ , we may let $z=x-a$, and we get the solution .. Am I messing something ?

By putting $P(x;0)$ we have $f(f(x))=2x+f(f(0)-x)$ which begets $f$ is injective and by $P(0,y)$ we have $f(f(0)+y)=f(f(y))$. Obviously  $f(y)=y+f(0)$. It is easy to verify that it is indeed a solution.

[quote=awesomeming327.]Not quite. $\tfrac{y_0-|x|}{2}$ does not run through every value, and $|x|$ is also not constant.[/quote]

That's not what I meant. From the last line doesn't it imply that the range of $g$ is a singleton set? I might be wrong.
@below counter example?

We first prove that $f$ is a surjective function. Specifically, substituting $y=-f(x)$ into our original relation $f(f(x)+y)=2x+f(f(y)-x)$ gives $$f(f(x)-f(x))=2x+f(f(-f(x))-x),$$ or equivalently $$f(f(-f(x))-x)=-2x+f(0).$$ As $x$ varies across $\mathbb{R}$, the value of $-2x+f(0)$ sweeps across $\mathbb{R}$ too. Thus, the range of $g(x)=f(f(-f(x))-x)$ is simply $\mathbb{R}$. Since the range of $f$ must contain that of $g$, it follows that $f$ has range $\mathbb{R}$ as well, implying that $f$ is surjective. It follows that there must exist some $c$ such that $f(c)=0.$

Now let $z$ be an arbitrary real number; again, surjectivity implies that there must exist some $y$ such that $f(y)=c+z.$ Setting $x=c$ in our given relationship gives $$f(f(c)+y)=2c+f(f(y)-c).$$ Since $f(c)=0$, it follows that $f(y)=2c+f(f(y)-c)$, so that $$f(f(y)-c)=(f(y)-c)-c.$$ However, since we have defined $y$ such that $f(y)-c=z$, it follows from the above equation that $$f(z)=z-c.$$ Thus $f$ must be a linear function of slope $1.$

Substituting the above into the original equation, we see that
\begin{align*}
f(f(x)+y)&=f((x-c)+y)\\
&=((x-c)+y)-c\\
&=2x+((y-c)-x)-c\\
&=2x+f(f(y)-x),
\end{align*}
which works. Switching dummy variables from $z$ to $x$, it follows that the solutions to the original functional equation are exactly those of the form $$f(x)=x-c,$$ where $c$ is any real number. $\blacksquare$

$P(x,-f(x))\Rightarrow f(0)=2x+f(f(-f(x))-x)\Rightarrow f$ is surjective $\Rightarrow \exists c\in \mathbb{R}$ such that $f(c)=0$

$P(c,y)\Rightarrow f(y)=2c+f(f(y)-c)\Rightarrow f(x-c)=x-2c\Rightarrow f(x)=x-c$

Conclusion: $\boxed{f(x)=x-c,\forall x\in \mathbb{R}}$

My first functional equation  :)
Let $P(x,y)$ denote the assertion of $f(f(x)+y)=2x+f(f(y)-x)$
I'll prove $f$ is surjective
$P(x,-f(x))\implies f(f(x)-f(x))=2x+f(f(-f(x))-x)\implies -2x+f(0)=f(f(-f(x))-x+c)\implies f$ is surjective because all real numbers can be expressed as $-2x+f(0)$
Let $f(a)=0$ (we can do this because $f$ is surjective)
$P(a,x)\implies f(x)=f(f(x)-a)+2a$
Let $f(x)=z$
$\implies f(z-a)+2a=z$
Now let $t=z-a\implies f(t)=t-a$ where a is an arbitrary real number
Final answer $f(x)=x+c,\forall c\in\mathbb{R}$ are the solutions of the functional equation

[hide=length from mahaler][quote=mahaler][quote=mahaler][quote=mahaler]This will be my first ISL solved if this is right! :jump: I would greatly appreciate it if someone could check this solution.
---------------------------------
Let $P(x, y)$ denote the given assertion.

$P(0, k)$: $f(c+k) = f(f(k))$, where $f(0) = c$.
$P(k, 0)$: $f(f(k)) = 2k + f(c-k) \Rightarrow f(c+k) = 2k + f(c-k)$.
By the above equation, $f$ is in the form $mx+b$ since it has a linear rate of change.
Hence, $mc + mk + b = 2k + mc - mk + b \Rightarrow 2mk = 2k \Rightarrow m=1$.
Therefore, $\boxed{f(x) = x+b}$ for all real numbers $b$ satisfy the given functional equation. It is easy to see that all functions of this form work.[/quote]

[b]Remark:[/b] @fuzimiao2013 suggested that I prove why $f(c+k) = 2k + f(c-k)$ implies that the $f(x) = mx +b$.

[i]Proof:[/i]
Plug in $k+t$ for $k$: $f(c+k+t) = 2k + 2t + f(c-k-t)$. Now, notice that if you subtract the $x$-values of our related $f$-values in this equation, we get $c+k+t - (c-k-t) = 2(k+t) = 2k + 2t$. Also, if you subtract the $f$-values themselves, you get $2k + 2t$. Hence, we get that the rate of change in this function is constant, or more specifically, $\frac{2k+2t}{2k+2t} = 1$. So, in fact, we can directly get the conclusion that $\boxed{f(x) = x+b}$.[/quote]

[b]Second Remark:[/b] Ok, well seems like $f(c+k) = 2k + f(c-k)$ doesn't imply $mx +b$, it actually implies $f(x) = x +g(x)$, where $g(x)$ is an even function.

Let $Q(x, y)$ denote the given assertion.
By $Q(0, y)$: $f(x + y + g(x)) = 2x + f(y-x + g(y)) \Rightarrow g(x) + g(x + y + g(x)) = g(y) + g(y-x+g(y)) \Rightarrow g(0) + g(y+g(0)) = g(y) + g(y + g(y))$, notice that this is true when $g(y) = g(0)$, so the said even function is a constant. Which gets us to the answer of $\boxed{f(x) = x+b}$.[/quote][/hide]
[quote]Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.[/quote]
I know I'm late and also you've probably changed that habit by now, but even though it was your first ISL, okay you're happy make two posts but you don't need to make three: 1. dont quote your own post since its consecutive and 2. theres an edit button also 3. you should try to get it fixed as john pointed out

Anyways, here's what I hope is a motivated solution to this beautiful problem :)
Our preliminary checks of the form f(x)=mx+b yields that f(x)=x+c for some constant c is all the solutions, which clearly works.
Claim 1: f is surjective.
Proof. Whenever we see an isolated 2x term, "DURR WE WANT STUFF TO CANCEL" to try to prove surjectivity by getting something of the form f(...)=2x since x can take on all real values. Let $P(x,y)$ denote the assertion of plugging in x and y. $P(x,-f(x))\implies f(f(-f(x))-x)=f(0)-2x,$ which is a constant, meaning f is surjective. $\square$
Claim 2: f is injective.
Proof. Suppose $$f(a)=f(b): (P(a,y),P(b,y))\implies 2a+f(f(y)-a)=2b+f(f(y)-b),$$ whence to prove a=b we try to show that $f(f(y)-a)=f(f(y)-b).$ Indeed, this becomes evident after choosing some y such that $f(y)=a+b. \square$
Finally, setting the common x=0 yields $$f(f(0)+y)=f(f(y))\iff f(y)=f(0)+y$$ for any real y, hence for f(0)=c we have as claimed. $\blacksquare$

[b]S[/b]olved with [b]NTGuy, everythingpi3141592[/b]

[hide = solution]
We're going to show that all solutions to the functional equation are given by $f(x) = x+c$ for some $c \in \mathbb{R}$. It is easy to see that these work. Let us show these are the only ones.

We first show that $f$ is injective. Suppose that $u, v \in \mathbb{R}$ satisfy $f(u) = f(v)$. Plugging $x = u, v$ we obtain that $2u+f(f(y)-u) = 2v+f(f(y)-v)$ for all real $y$. Plugging $y = u, v$ we obtain that $f(f(x)+u) = f(f(x)+v)$ for all real $x$. Now suppose we managed to find $z_1, z_2 \in \mathbb{R}$ such that $f(z_2)-f(z_1) = u+v$. We would have by setting $y = z_2$ in the first equation: $2u+f(f(z_1)+v) = 2v+f(f(z_1)+u)$. But we also have $f(f(z_1)+u) = f(f(z_1)+v)$ by setting $x = z_1$ in the second equation. This would imply finally that $u = v$. To find such $z_1, z_2$ we simply set $x = \frac{u+v}{2}$ in the original equation with $y$ arbitrary. Thus $f$ is injective. Now we simply set $x = 0$ in the original equation and obtain $f(y+f(0)) = f(f(y))$ for all $y \in \mathbb{R}$. Thus $f(x) = x+c$ for some $c \in \mathbb{R}$ and we're done.
[/hide]

[hide=Oh gosh]
Let $f(0) = k$
$P(x,-f(x)):$\\
$x = \frac{k-f(f(f(x)-x))}{2}$
So $f$ is surjective.
Now FTSOC suppose $\exists a,b$ $a \neq b$ st $f(a)=f(b)$.
$P(a,y): a = \frac{f(f(a)+y)-f(f(y)-a)}{2}$
$P(b,y): b= \frac{f(f(b)+y)-f(f(y)-b)}{2}$
Setting $f(y) = a+b$ we get a contradiction so $f$ is injective.
$P(x,0):$
$f(k+y)=f(f(y))$
and we get $f(y) = k + y$ which clearly works
[/hide]

We will prove that $f$ is injective. If $f(a)=f(b)$, we have

\begin{align*}
f(f(a)+a+b)&=f(f(b)+a+b) \\
\implies 2a+f(f(a+b)-a)&=2b+f(f(a+b)-b)
\end{align*}

so $a=b$.

Then, plugging in $(x,y)=(0,y)$ gives

\[f(f(0)+y)=f(f(y))\]

implying $f(y)=y+f(0)$. Thus, our answer is $\boxed{f(x)=x+c, \ c \in \mathbb{R}}$.

Denote the assertion as $P(x, y)$. Note that
\[ P(x, -f(x)) \implies f(0) = 2x + f(f(-f(x))- x) \]
which implies that $f$ is surjective.
Now, suppose that we have a $u$ such that $f(u) = 0$. Then
\[ P(u, y) \implies f(y) = 2u + f(f(y) - u) \]
Let $v = f(y) - u$, noting that $v$ runs through all the real numbers. Then we must have that
\[ v+u = 2u + f(v) \implies f(v) = v - u \]
Thus, we find that $f(x) = x + c$ for some real constant $c$. This obviously works. $\blacksquare$

[color=#00f][b]Claim 1:[/b][/color] $f$ is surjective.

Substitute $y=-f(x)$ to get
\[f(f(x)-f(x)) = 2x+f(f(-f(x))-x) \implies f(f(-f(x))-x) = -2x+f(0). \quad {\color{blue} \Box}\]

[color=#00f][b]Claim 2:[/b][/color] $f$ is injective.

Suppose we have real numbers $a$, $b$, and $t$ such that $f(a)=f(b)$ and $f(t)=a+b$. Substituting $(a,t)$ and $(b,t)$, we have
\[2a+f(b)=f(f(a)+t)=f(f(b)+t)=2b+f(a) \implies a=b. \quad {\color{blue} \Box}\]

We finish by substituting $(0,y)$ to get
\[f(f(0)+y)=f(f(y)) \implies f(y)=y+f(0),\]

so our only possible solution is $\boxed{f(x)=x+c, \quad c \in \mathbb{R}}$, which clearly works. $\blacksquare$

The answer is $f(x)=x+c$ for any $c$ which works.

[b][color=red]Claim:[/color][/b] $f$ is surjective
[i]Proof.[/i] Rewrite as $f\left(f(x)+y\right)-2x=f\left(f(y)-x\right)$. The claim follows by varying $x$ ut choosing $y$ suh that $f(x)+y$ is constant. $\blacksquare$

Now, setting $x$ such that $f(x)=0$ gives $f(y)=2x+f(f(y)-x)$. As $f(y)$ can be anything we conclude $f(x)=x+c$ for some $c$.

We claim $f(x) = x + c$ for some constant $c$, which clearly works.

Substitute $y = -f(x)$. This yields \[f(0) - 2x = f(f(-f(x)) - x)\text{.}\] This implies $f$ is surjective.

Let $x$ satisfy $f(x) = 0$ and $y$ satisfy $f(y) = a + x$ for a fixed $a$. Directly substituting yields
\begin{align*}
    f(y) &= 2x + f(a + x - x) \\
    a + x - 2x &= f(a) \\
    f(a) &= a - x\text{.}
\end{align*}
Since $a$ can be anything, $f$ must be linear with slope $1$. (Note that this also implies $x$ is unique).

[hide=Solution] $f(x)=x \ \forall x\in \mathbb{R}$ [/hide]

Let $P(x,y)$ denote the given assertion.

[color=#00f]Claim:[/color] $f$-is surjective

[color=#00f]Proof:[/color] [hide]$P(x,-f(x)) \implies f(-x-f(x))=f(0)-2x \implies f$- is surjective  [/hide]

Since $f$ is surjective we get that  $\exists \alpha : f(\alpha)=0$
$P(\alpha,\alpha) \implies f(-\alpha)=-2\alpha$ $...(1)$
$P(\alpha,f(\alpha)) \implies f(-\alpha)=f(0)-2\alpha$ Combining with $(1)$ we get: $f(0)=0$

[color=#f00]Claim:[/color] $f$-is injective

[color=#f00]Proof:[/color] [hide] $\exists a,b : f(a)=f(b)$
$P(a,x) \implies f(f(a)+x)=2a+f(f(x)-a) \implies f(f(a)+x)-f(f(x-a)-a)=2a$
$P(b,x) \implies f(f(b)+x)=2b+f(f(x)-b)\implies f(f(b)+x)-f(f(x-b)-b)=2b$
We know that $f(f(a)+x)-f(f(x-a)-a)= f(f(b)+x)-f(f(x-b)-b) \implies 2a=2b \implies a=b \implies f$-is injective[/hide]

$P(0,x) \implies f(f(0)+x)=f(f(x))$ Combining this with $f(0)=0$ we get : $f(x)=f(f(x))$ Combining with $f$-is injective we get: $f(x)=x$

Remind me if I'm wrong at injectivity part  :roll:[/quote][/quote]

The answer is $f(x) = x + c$ for any real $c$, which works. Denote the original assertion as $P(x, y)$.

[color=#f00][b]Claim:[/b][/color] $f$ is surjective

[color=#f00][b]Proof:[/b][/color] From $P(x, -f(x))$ we have $f(0) = 2x + f(f(-f(x)) - x)$ or $f(f(-f(x)) - x) = f(0) - 2x$. Since $f(0)$ is constant, from varying $x$ we see that $f$ can take on any real output.

Now, from surjectivity, we can choose $t$ such that $f(t) = 0$. Then $P(t, x)$ gives $f(x) = 2t + f(f(x) - t)$. Letting $x_1 = f(x) - t$, this simplifies to $x_1 + t = 2t + f(x_1)$, or $f(x_1) = x_1 - t$. Since $x_1$ can take on any real value by surjectivity, this implies the solution set.


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I'm in 9th grade in Poland, but I'd really like to attend MIT after high school... Unfortunately there are some problems for which I'd like to have some help from you :)
First, I've read that there are about 2,000 international students applying to MIT, from which only about 100 are admitted (moreover at MIT's website it was written that most of them succeeded in international contests). Is all of that true? If so, do you know any statistics what SAT, TOEFL, contests etc.  scores and results they had?
Second... is Polish high school diploma worth anything? The problem with polish education system is that you have no choice in taking courses (dispite of (every) science courses, we have to "take" hisotory, polish language, arts and so on, no matter if we want it or no... :/ ).
That's the reason why I'm considering going abroad. There are a few firms that offer studying abroad. There's something like "a year abroad program in USA or UK", and IB school in UK (2 years). Unfortunately private schools, both in US and UK are too expensive :( (maybe my parents would have money for private school, but university will also cost a lot :/ ). So the only way for me to study abroad is to study in public school ...
Public school in US for me is only for 1 year :/ because of visa problems and so on.... maybe you know some schoolarships for high school international students? Have you got any knowledge about it? (neither me nor my parents have "permanent residency" or whatever).
Fortunately schools in UK look much more interesting :) Poland and UK are in EU so it's a huge advantage :) One of the schools offered by those firms is "Impington Sixth Form" - they offer IB, and courses which I'd like to take (maths & science)... here are links:
[url=http://www.ibsca.org.uk/ibscamember-impington.htm]www.ibsca.org.uk/ibscamember-impington.htm[/url]
and their official website:
[url=http://cgu.e2bn.net/e2bn/leas/c99/schools/cgu/]cgu.e2bn.net/e2bn/leas/c99/schools/cgu/[/url]

Can somebody read a little about them, or maybe somebody knows this school, and tell if it's a "good" school. You know what I mean, the education level and so on... (especially when it comes to math)...

Do you know any schoolarships (EU or UK), which would help me with studying in UK private school that offer IB?

Will IB diploma be worth anything when I'll be applying for MIT?
What would you recommend me to do to increase my chances to get to MIT? Which [maths & programming] contests should I start in?
I already participate in usaco and polish olympiad in informatics.

Hi, if you have not checked, please check earlier discussion,  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=MIT+dream&t=42985]here[/url]

It may be helpful.

It is true that it is very difficult for international students to get into MIT--because most international applicants are very highly accomplished students. But it is worth trying. It shouldn't be necessary to move out of your country to make a strong application, because MIT repeatedly mentions that it evaluates applicants according to their personal contexts, and if your country has a particular curriculum, MIT will be aware of that.

[quote="tokenadult"]it is very difficult for international students to get into MIT (...) It shouldn't be necessary to move out of your country to make a strong application, [/quote]

This raises the question: what is the definition of "international student" used by MIT and other selective universities?   Is it based on place of legal residence, physical residence, country of citizenship, or case-by-case determinations?  Are some countries more equal than others for purposes of admissions, all other things being equal?  Depending on the answers to these questions, one could apparently influence the chances of admission by moving from one country to another.

[quote] if your country has a particular curriculum, MIT will be aware of that.[/quote]

MIT is probably aware of certain generalities, such as the rough comparison between A-levels and AP or IB certificates.  Beyond that, my impression (from knowing some admissions officers as well as common sense) is that the admissions officers' knowledge decreases rapidly for the non-English-speaking systems and the smaller countries that send relatively few students.  For those, they rely heavily on alumni interviewers "on the ground" and familiar recommenders at known foreign institutions (e.g. the top 1-2 high schools of a given country, or IMO coaches), and there is a pipeline effect as a result.

[url=http://matt.mitblogs.com/mt/mt-search.cgi?IncludeBlogs=1&search=international]Matt McGann's blog[/url] has, I think, a statement somewhere of what MIT's definition of "international" is. If I remember correctly, the basic idea is that eligible for United States federally sponsored student aid --> domestic, while not eligible for such aid (ordinarily because one's citizenship is of another country, without United States permanent residence status) --> international. And once one is in the "international" pool of applicants, which is very competitive at MIT, there isn't any further quota or any reason to suppose that one country is favored over any other. I think I have seen MIT statistics on countries of origin (this, I presume, means countries of students' mailing address at time of admission, or might mean countries of students' secondary schooling) and there are very many countries represented in MIT's current class.

To be sure, one way to stand out from the general international applicant pool to a university with a strong math program is to win a gold medal at the IMO. But even that doesn't guarantee admission, and some admittees have never participated in IMO-related contests at all.

I don't know the fine details about MIT and certainly not for 2005-6, but there are reasons to believe that not all countries are created equal for purposes of admission, or at least reason enough to ask whether this is the case, even with most other things being equal (comparable curricula, letters of recommendation, etc).  A country with more alumni may have more alumni-donated scholarships to defray tuition for its students, for example, and some places have generous burses for students travelling abroad (e.g. NSERC).  MIT explicitly considers finances for international students.
There is probably no explicit quota differentiating between different countries but there may well be de facto differences.

tokenadult can you give me a link to those MIT statistics on coutries of origin? I've been looking for them but no results :/

In my opinion not all countries are equal.... and i think that most of the "english speaking" coutries (eg. Canada, GB, Ireland, Australia) are in (little) favour than others (especially after terroris attacks, for students in Arabic coutries it may be harder) :/

But that's just my opinion, and I hope it's not true at all ;]

Hey Victor,

just go to this URL: http://ivysuccess.com/index.html

Here are the MIT Admissions Stats for the Class of 2008: http://ivysuccess.com/mit_2008.html

"...27% are Asian American, 35 % are Caucasian/White, 17 % are members of underrepresented minority groups (African-American, Mexican-American, Puerto Rican, Native American, and other Hispanic groups). A few years ago, underrepresented minorities comprised of only 11 percent of the admitted..."

And go here: http://www-tech.mit.edu/V125/PDF/N14.pdf

And take a look here:

[url=http://admissions.mit.edu/AdmissionsWeb/appmanager/AdmissionsWeb/Main?_nfpb=true&_pageLabel=pageWhosAtMIT]MIT class of 2008[/url]

"The MIT class of 2008

1087 students

Male     57%
Female     43%

African American     6%
Asian American     26%
Hispanic     12%
Native American     2%
White/Caucasian     35%
Other/no response     13%
International students     7%

New England     12.4%
Mid Atlantic     19.3%
Southeast     15.0%
Midwest     13.5%
South     11.3%
West     19.9%
Abroad     8.2%

Speak a language other than English at home     39%
First generation college students     17%
Receiving some sort of financial aid (including a job on campus)     90%"

Orlando

Thanks orl :)
Quite interesting facts :> but it's still not a list of countries.... anyway it's enough too ;)

orl - i see you're from England... can you tell me what you think about school (and  their education level) I mentioned earlier?
Thanks ;]

[quote="_victor_"]Thanks orl :)[/quote]

You are welcome!

[quote="_victor_"]
orl - i see you're from England... can you tell me what you think about school (and  their education level) I mentioned earlier?
Thanks ;][/quote]

Well, I didn't go to school in UK. Last year I was an exchange student at the University of Edinburgh. And I am a bit familiar with good universities in UK. So are you really talking about school?

Fleeting guest wrote:

[quote] MIT explicitly considers finances for international students. [/quote]

MIT admissions officer Matt McGann says that MIT is need-blind for international students, meaning that any given international student will be admitted without consideration of his/her ability-to-pay.

At the same time, Matt also says that MIT has an overall limit on the total number of international students it can accept.  Presumably this overall limit is set based on a statistical projection that will allow MIT to break even.

http://matt.mitblogs.com/archives/2005/01/international_m.html

Fleeting guest also wrote:

[quote] A country with more alumni may have more alumni-donated scholarships to defray tuition for its students, for example, and some places have generous burses for students travelling abroad (e.g. NSERC). MIT explicitly considers finances for international students.  There is probably no explicit quota differentiating between different countries but there may well be de facto differences. [/quote]

So, fleeting guest, are you suggesting that MIT will be more likely to admit a student who comes from a country that has generous scholarship funding available for that student?

Such a policy seems radically at variance with what Matt states in his blog above.

Token adult, as moderator of this forum, would you be willing to ask Matt to comment in this forum to clarify whether Fleeting Guest's speculations reflect actual MIT admissions practice?

My impression is that MIT goes to exceptional lengths to communicate clear and accurate information about its admissions process (far moreso than many other schools), so I would be very surprised to learn that MIT somehow gives secret preference in undergraduate admissions to internatiional students from countries that make generous funding available.

[quote="sophia"]Token adult, as moderator of this forum, would you be willing to ask Matt to comment in this forum to clarify whether Fleeting Guest's speculations reflect actual MIT admissions practice?[/quote]

I would like to follow up on this, because I am sure that MIT's admission practice for international students in recent years is NOT as fleeting_guest suggests. As you note, MIT's explicit statement about international student admission in recent years has been that admission is "need blind," in accord with the policies at several peer institutions.

AoPS member orl has kindly pointed to one source of summary statistics about a recent MIT entering class. It's maddening that MIT's Web site does NOT have the kind of search capabilities that allow one to conveniently search out the page I have previously seen detailing countries of origin of MIT students in a particular year. (I can't remember offhand whether or not those figures lumped together undergraduates and graduates.) In any event, students are admitted each year to MIT from countries all over the world, and I think anyone with _victor_'s ability in English as a ninth grader doesn't need to complete his secondary education in an English-speaking country to have an excellent chance of admission at MIT--as long as everything else about his application is competitive, which I acknowledge is not easy. English-language ability is, of course, a criterion for admission to a university at which the langauge of instruction is English. But [url=http://matt.mitblogs.com/]Matt[/url] and [url=http://ben.mitblogs.com/]Ben[/url] and their colleagues in MIT's admission office have been quite consistent in urging international applicants to take the TOEFL test, OR the SAT I test, whichever makes the applicant look better. For applicants who have a sufficient level of English to study at MIT, there is no further advantage in having had a secondary education in an English-speaking country. All the students who attend MIT get the best diversity advantage from having classmates with varied backgrounds.

To briefly respond to some issues brought up in this thread:

For the purposes of MIT admissions, a domestic student is a student who is a US citizen or permanent resident, and an international student is a student without US citizenship or permanent resident status.

MIT considers all applications, domestic and international, in a need blind process, and we meet full financial need for all students, international and domestic.  MIT does have a quota on the number of international citizens we can admit, currently about 8% of the enrolling class.

Last year, we received ~2300 applications from international students.  We admitted 113 students representing ~65 countries, with 96 of those students enrolling in the MIT Class of 2009.

We do not have preferences for any individual or group of countries.  It is not true that we favor countries with more alumni or that it is harder for Arabic students to be admitted.

You can view the current international enrollment by country here:
[url]http://web.mit.edu/registrar/www/stats/geofinal.html[/url]
Remember that this is enrolling students; I presently don't have statistics on numbers of applications or admitted students by country.

I hope this is helpful!
-[url=http://matt.mitblogs.com]Matt[/url]

Thank you, Matt, for the link to the document I remembered seeing (probably with the previous year's figures the last time I saw it).

http://web.mit.edu/registrar/www/stats/geofinal.html

As I remembered, there are undergraduates from Poland and from several nearby countries, which is the concern of the questioner who opened this thread.

As always, it is tough for students and parents who have ambitious college dreams to decide what is a good plan for high school. There are young people who go to high school in another country (e.g., at Phillips Exeter Academy in Exeter, New Hampshire, U.S.A.) to gain what they think is better preparation for making a successful application to a famous college. That may be a good idea in a particular case (my family has been thinking about the school I just mentioned ourselves), but it shouldn't be strictly necessary to move just to get a good education that gets you ready for college. MIT, and all colleges, recognize that people are born and grow up in places all over the world. Just be sure to maximize your learning in the environment you are in, and do both

1) what is required by your country's schools,

and

2) what the most self-motivated learners in your country do to go BEYOND the school curriculum

and then apply and don't worry about it. Apply to the schools you admire, and make sure your list of schools you apply to includes schools that offer very high odds of admission ("safety schools"). While you are still of high school age, enjoy learning and gaining deeper understanding for the value those activities have whether or not you go to college.

Best wishes on your application a few years from now. Thanks to Matt for clearing up the informational questions.

[quote="MattMcGannMIT"] MIT considers all applications, domestic and international, in a need blind process, and we meet full financial need for all students, international and domestic.  MIT does have a quota on the number of international citizens we can admit, currently about 8% of the enrolling class. [/quote]

Well most of the excellent universities keep a quota of below 10 % for international students, at least for the undergraduate degree. Why are they keeping this protectionist policy? Thus they should implement a country-blind admission process as well! But I guess American students are more likely to keep financially connected with their former almer mater and support them with a lot of money once they took hold of an influencial position. I guess more excellent students will increase the top-level even more. Therefore I wonder whether they have to solve the excellence - patriotism trade-off issue. Opinions wanted!

Actually, historically most countries have been MORE restrictive of foreign students attending the top national universities than the United States has. This is partly because in the United States, unlike most countries in the world, the "top" universities are mostly privately operated, and they can decide for themselves whom to admit. But all countries around the world are gradually figuring out that their own domestic students get a better education if they are exposed to the differing backgrounds of foreign students, and some countries have long, for foreign policy reasons, invited students from particular countries. (Thus the former Soviet Union was very welcoming of foreign students, the better to influence young leaders in other countries to be favorable to the Soviet Union.)

I think MIT (and also Harvard, Yale, and other top universities in the United States that have adopted similar policies) are doing pretty well: they make admission "need blind" for everyone in the whole world, meaning that they admit applicants without regard to their ability to pay full list price. They do keep the percentage of foreign students from rising to what might be its "natural" level (100 percent?) because they do, after all, have a constituency of United States students to serve. They know the foreign students have other universities in their home countries (in most cases) that they could attend if they don't get into a United States university. But they recognize that everyone attending the university gets a better education if they have classmates from all over the world (and, incidentally, from all states of the United States).

[quote="tokenadult"]Actually, historically most countries have been MORE restrictive of foreign students attending the top national universities than the United States has... They do keep the percentage of foreign students from rising to what might be its "natural" level (100 percent?) because they do, after all, have a constituency of United States students to serve. They know the foreign students have other universities in their home countries (in most cases) that they could attend if they don't get into a United States university. But they recognize that everyone attending the university gets a better education if they have classmates from all over the world (and, incidentally, from all states of the United States).[/quote]

Well, I guess your 100 % are a little bit ironic unless you violate the constituency of United States and consider American applicants international. I don't say that Americans are admitted based on their citizenship only. You do have a lot of excellent students.
Thus we have to discuss what's an appropriate percentage! And what does the constituency of United States enforces in this accord in detail? And do these top universities benefit increasing the number of international students? Bearing in mind that the US has most of the top universities in the world and wants to keep/improve this status quo your argument of being restrictive cannot be handled as in other countries! So do you think that the upper half according to performance at MIT benefits more from the other students (lower half of American/International students) or do they benefit more from a higer percentage of excellent international students? Doesn't the constituency of United States assume that the latter idea is better for MIT's top American/International students? Don't you agree that most nations have a limited number of the very top students and that Harvard for instance dilutes this argument when admitting mostly Americans? Don't you think it is a fine idea to have the best students world-wide united at a few universities? Think about these issues as well.
I guess a proportion of 30 - 50 % of international students sounds like a fair compromise. Ideas and different evaluations requested.


inequalities
function
inequalities unsolved

For $x_i (i=1,..,n)$ positive real numbers such that

 $\displaystyle \frac{1}{x_1}+...+\frac{1}{x_n}=n$

prove that

$\displaystyle \displaystyle \frac{x_2^2+\cdots+x_n^2}{n-1}\cdot \frac {x_1^2+x_3^2+\cdots +x_n^2} {n-1} \cdots \frac{x_1^2+\cdots+x_{n-1}^2}{n-1}\geq \left(\frac{x_1^2+...+x_n^2}{n}\right)^{n-1}$

You forgot the main condition!!!
\[\frac{1}{x_1}+...+\frac{1}{x_n}=n.\]
And positive numbers are $x_1$, $x_2$, and so on ;)

I think $\sum_{i=1}^{n}{\frac{1}{x_{i}}=n}$ , am I right?

Manlio, you can start the research using a substitution and then the Suranyi inequality. I would be very pleased to see a very simple solution to this problem/

[quote="harazi"]... the Suranyi inequality...[/quote]

what is the Suranyi inequality ?  :?

I think I have posted the statement and a proof, but I do not remember where/

It is here

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Surany&t=14008

Which is the magic Harazi substitution ???????

Does anyone see it ?

I don't  :blush:  :blush:

Manlio, using the substituion $ x_2^2+...+x_n^2=(n-1)a_1^2$ I think we get a problem that I have already posted in the forum and that has a solution in Old and New Inequalities. Do you see the correct way to arrive at that inequality?

I still don't know.because I don't know how to use $\frac{1}{x_1}+...+\frac{1}{x_n}=n$ .BTW:I think if $n=2$ .it should be $ \leq$

I also have a lot of problems with this

$\displaystyle \frac{1}{x_1}+...+\frac{1}{x_n}=n $

I don't know how to treat it :?   to arrive at Suranyi inequality

Can you please detail better your proof, Harazi

Thank you very much ;)

Harazi, can you please  post your proof?

Many thanks

[quote="harazi"]Wait a few days, Manlio. I will post it, now it is a little difficult, since I don't have a computer...[/quote]
 :? much surprise.but how can you post it?

Ok, guys, sorry for the delay, I just wanted to let you think more on it. Let's go. The idea is to use the inequality that we have already discusses (but to which noone gave a solution):
    The minimum of the expression $\sum {\sqrt\frac{a_1...a_n}{1-(n-1)a_i}}$ when $a_i$ add up to 1 is attained when the variables are equal. Then, substitute here $ 1-(n-1)a_i=x_i^2$, homogenize the inequality you obtain and then use the given condition. You find exactly the desired inequality. For the inequality I used, you can find a solution in Old and New Inequalities. If you have time, you can post it, I do not remember the solution. ;)

[quote="harazi"]Just use the old trick of dehomogenization-homogenization.[/quote]
forgive my stupidness,can you show me clearly?

[quote="harazi"]
    The minimum of the expression $\sum {\sqrt\frac{a_1...a_n}{1-(n-1)a_i}}$ when $a_i$ add up to 1 is attained when the variables are equal.[/quote]
 Can anybody be so kind to post solution for it here?

[quote="harazi"]Just use the old trick of dehomogenization-homogenization.[/quote]

what is exactly this trick?

Ok, sorry, I simply forgot about this problem. Thanks to zhaobin for remembering about it. So, we agree that we only have to show now that $\sum{\sqrt{\frac{a_1...a_n}{1-(n-1)a_1}}}$ attains minimal value when all variables are $1/n$.
  Now, remember Suranyi's inequality: $(n-1)(x_1^n+...+x_n^n)+nx_1...x_n\geq (x_1+...+x_n)(x_1^{n-1}+...+x_n^{n-1})$. It follows that $ na_1...a_n\geq\sum{a_1^{n-1}(1-(n-1)a_1)}$. Now, suppose that $n$ is at least 4. Write $\sum{a_1^{n-1}(1-(n-1)a_1)}=\sum{\frac{(a_1^{\frac{n-1}{3}})^3}{(\sqrt{\frac{1}{1-(n-1)a_1}})^2}}$ and apply Holder. You will find that it is at least equal to $\frac{(\sum{a_1^{\frac{n-1}{3}}})^3}}{n(\sum{\sqrt{\frac{1}{1-(n-1)a_1}})^2}$. Next, apply Power-Mean inequality (or Jensen) to $\sum{a_1^{\frac{n-1}{3}}}$ and find a minoration for the last one in function of n. Insert this minoration in the last inequality and combine with the fact that $ na_1...a_n\geq\sum{a_1^{n-1}(1-(n-1)a_1)}$. This will show that $\sum{\sqrt{\frac{a_1...a_n}{1-(n-1)a_1}}}$ attains minimal value when all variables are $1/n$. I will leave the pleasure for $n=3$. Hint: use Schur inequality $ a^4+b^4+c^4+abc(a+b+c)\geq a^3(b+c)+b^3(c+a)+c^3(a+b)$ in this case.

sorry,harazi.
I still don't know your meaning  :(
(1)if you mean we just have do a work when $\sum_{i=1}^n x_i^2=n$,but why?
(2)if (1) is right.where would $\sum {\sqrt\frac{a_1...a_n}{1-(n-1)a_i} }$ come from?
sorry again

Zhaobin! Are you following me? Here are the steps (I already explained this!):
  1. Homogenize the inequality and bring it to the form
   $\frac{x_2^2+...+x_n^2}{n-1}\cdot...\cdot\frac{x_1^2+...+x_{n-1}^2}{n-1}\cdot (1/x_1+...+1/x_n)^2\geq n^2\cdot (\frac{x_1^2+...+x_n^2}{n})^{n-1}$ for any $x_1,...,x_n>0$ with absolutely no other condition on them.
  2. Since the inequality in 1. is homogenous, we can take $ x_1^2+...+x_n^2=1$. Consider the numbers $a_1=\frac{x_2^2+...+x_n^2}{n-1}$ and so on. Of course, they add to 1 since $x_1^2+...+x_n^2=1$. But $a_1=\frac{1-x_1^2}{n-1}$ thus $ x_1=\sqrt{1-(n-1)a_1}$. The left hand side is nothing else than $ (\sum{\sqrt{\frac{a_1...a_n}{1-(n-1)a_1}}})^2$. Is it clear now?

thank you of your kindness.it is clear now,thank you :)

[quote="harazi"]I think the best example of application is the hard improvement of Walther Janous's inequality:
  FOr all $a,b,c,x,y,z$ positive real numbers such that $ xy+yz+zx=3$ we have
    $\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq 3$
  Here is the trick: we homogenize with respect to $x,y,z$ and we "reduce" it to
   $\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq\sqrt{3(xy+yz+zx)}$ for all $a,b,c,x,y,z$, with no conditions now. Therefore, we have freedom to "dehomogenize", that is to choose some convenable conditions for $x,y,z$ that make our life easier. In this case take $x+y+z=1$ and you will see how nice the life becomes. I will let you now find a proof with these hints.[/quote]

can anyone show me how to finish it? i really can't continue from where harazi left it


eigentlich an darij, aber wenn mir noch jemand diesen bew erklären kann wäre ich ihm sehr verbunden:

Def: Die Maximalanzahl linear unabhängiger Spalten nennt man den $\textit{Spaltenrang}$ von $A$, die Maximalanzahl linear unabhängiger Zeilen den $\textit{Zeilenrang}$ von $A$.

Satz: Spaltenrang $A$ = Zeilenrang $A$

Beweis:
(ich werde nur den schwierigen Teil auf schreiben, da es sonst zu viel Text würde...)
Wir wollen eine Spalte oder Zeile linear überflüssig nenen, wenn sie aus den übrigen Spalten bzw. Zeilen linearkombiniert werden kann. Verkleinert man eine Matrix durch Weglassen einer linear überflüssigen Spalte, so ändert sich natürlich der Spaltenrang nicht. --- Angenommen, in einer Matrix $A$ sei die $j$-te Spalte linear überflüssig. Dann ist auch für jede Zeile und jede Linearkombination von Zeilen die $j$-te Komponente überflüssig (im eindimensionalen Vektorraum $\mathbb R$ bzw. $\mathbb C$). Das ist klar: Mit denselben Koeffizienten, mit denen man die $j$-te Spalte aus den übrigen Elementen. --- Daraus folgt, dass eine Linearkombination von Zeilen von von $A$ $\textit{ genau dann}$ Null ist, wenn die entsprechende Zeilenkombination der durch Weglassen der $j$-ten Spalte verkleinerten Matrix Null ist. Deshalb haben $A$ und die durch weglassen einer linear überflüssigen Spalte entstehende Matrix denselben Zeilenrang.
(...)

danke, peter

danke darij.

hat sich alles geklärt (das kursive "genau dann, wenn" wurde mir nicht ganz klar), aber jetzt


Okay. I wonder why this idea never cam to anyone (including me). I just saw it in a forum on age of empires age of king.

[u][size=150]Rules:[/size][/u]
[b]1. Every post consists of one word in italics for the stiry.
2. Consecutive posts are not allowed. Even if you post two consecutive posts, only can contribute a word to the story.
3. Besides the word, comments are welcome. Indeed, making the post large enough using thiings like asdfsdfsa or !!!!!!!!!!!!!!!!! is strongly discouraged.
4. Separate your comments from the main word by a series of underscores like ___________.
5. After certain intervals, someone compiles all the words before to make the story so far.[/b]

This game has a nonsense capacity matched onlt by - among topics I have seen - The Never Ending Story, The conversation, Spontaneous prose and to some extent The Never Beginning Story.

I hope you will not disappoint me. Make it a cracking story! And maintain the high level of activity this requires to get anywhere.
_____________________________________________________________

[hide="Hers is the first word:"][i]Google[/i][/hide]

[i]the[/i]

_______________

I'm making it a rhyming story.

[i]spoogle[/i]

words that connect so far:

Google the spoogle

Sorry, but this is going to get as junky as the other spam topics in this forum.


We can calculate the determinant of $ A$ w.r.t. the first column and get:

${{ \det A=a_{1}x_{2}\cdots x_n \prod_{
2 \leq j <k \leq n}(x_{k}-x_j)
-\ldots+
(-1)^{i+1}
a_{i} \prod_{
1 \leq l \leq n
}^{l \neq i}
}x_{l} \prod_{
1 \leq j <k \leq n}
^{j,k \neq i}}(x_k-x_j)
+(-1)^{n+1}a_{n}x_{1}\cdots x_{n-1} \prod_{
1 \leq j <k \leq n-1
}(x_k-x_j)$

Hence
\begin{eqnarray*}
\det{A}&=& \prod_{
i=1
}^{n}
(-1)^{i+1}
a_{i}\prod_{
1 \leq l \leq n
}^{l \neq i}
x_{l} \prod_{
1 \leq j <k \leq n}
^{j,k \neq i}(x_k-x_j)=\\
&=&
\prod_{
i=1
}^{n}
\prod_{
1 \leq l \leq n
}^{l \neq i}
\prod_{
1 \leq j <k \leq n}
^{j,k \neq i}
(-1)^{i+1}
a_{i}x_{l}(x_k-x_j)
\end{eqnarray*}

you need to repair $ \prod_{i\equal{}1}^{n}$ by $ \sum_{i\equal{}1}^{n}$ for correct


Well, nothing much to be proud of but still an improvement :D

S.Shreyas Total-198. 96.7 Percentile.

Best scores
Maths - 45 (Paper II)
Physics - 47 (Paper I)
Chemistry - 22 (Paper II)

82.9  :rotfl:
all score worst....much less than calculated marks taht is seeing their solutions...
seems they borrowed software from IAPT :P   :huh:

95.5 GOT 187 AND GOT WHAT I EXPECTED  :lol:  :D

BEST SCORES
MATHS : 49
PHY : 31
CHEM 37

DID YOU SEE THE TOPPER GOT 357.
AND THERE WERE 2 FULL SCORES IN CHEM AND MATH  :o

[b][i][size=200][color=orange]PATHETIC EVALUVATION :furious:  :mad:  :bomb:  :censored: [/color][/size][/i][/b]
I got just 6 in the paperII of physics when I expected 42 in the second my best in physics.
As for the others I got 81-chem and 85-maths

Total-204 (35 less than expected)

i gor about 45 less(but in terms of percentile must be around 15)

I may have fot 97 and odd percentile but I am highly disappointed by the way
my mark has come out.

It is damaging my image in physics especially when I thought I had done so well.
Even if I had got 235 odd I would have got 1 percentile above and it doesnt matter
much for me but scoring 6/81 has made me really really really upset :(  :(  :(

anyway anit congrats but u r certainly those lucky ones to get extra marks.

[quote="anitsahu"]i dont seeem to understand fiitjee people at all :mad:
in pt2 i was expecting 265 got 246  :o
in pt3 expecting 223 but i got 246 :)
percentile 99.3
i believe they have assigned me 246 as my marks[/quote]

That's great  :D
What's ur split up???

anyway i had done really bad in physics ....actually what did he want he want to test :)

phy 92 I50 II42
chem 72 I40 II32
maths 82 I40 II42

anit how did u get the assertion reasoning's right...
and i didn't seem to understand the ist comprehension(stress...strain) para
all the results hould be half of what they have been

no i mean individual in paper one and 2 in each subject  :D

What is chapli's split up btw????????? :huh:  :huh:

hey for your info
shuddhodhan i got all my assertion reasons wrong in phy 1st paper
in 2nd paper i managed to get 1 of them right
really pathetic answers
 :wallbash_red:

no iassure u it was a real typo as i had already tried ur name in google that day itself(when i pmed you)  :D

friends of mine r crazy about 99 percentile in fiitjee
i have lots of them in 98-99 range
but i know something a guy in my city was projected by fiitjee of a 183 rank in iitjee
but he managed only 1976

yes as they say it all depends on that day  :)  :D  :)

but why fiitjee claims all this stuff
its just rubbish
and these 99 or 100 percentiles dont matter at all
its the confidence level u have on that day

And someone try to answer the 10th question of the IITJEE Sticky  :D  :D  :D

I'm not blaming FIITJEE for wrong correction cause I got what I expected exactly. Expected 199, got 198.... (well nothings wrong in 1 mark) :D

neither did i last tiem...i got 104 in physics...expected 102 :P

i got 82 in physics 65 in chemistry and 52 in maths.i expected to get 221 but had to settle finally for 199 :bomb:

[quote="anitsahu"]
 was expecting
in pt3 expecting 223 but i got 246 :)
percentile 99.3
[/quote]
@physici olympiaci
seems anit took away ur marks :P

i too am not blaming FIITJEE bcos i got wat i expected totalwise, subjectwise and each testwise so no probs. Even though i am inclined to think i should get 4 more marks bcos i perfectly proved one Binomial question (that n>2 question with some binomial expansion) that For n>3 only it is $ a^2$ and for n = 3 it is $ a^2 \minus{} 2$  :mad:  :mad:   :huh:

Rohit: No da your ans is wrong as usual.. (okay okay just kidding!)

Anyways I just recieved that Fiitjee slip for PTII. Doing basic addition I see that not even 10K have written the test. This is a pathetic and a far cry from ppl's claims of 50K. :(

I got it the day before yest and basic counting tells not more than 9635 could have written pt 2.. N i am blamin fiitjee dearly for the correction.. Chemrock wont differ.. I lost 20 in physics paper 2 from my expected mark, he lost 36..  :lol:

Even srini lost 20 marks and sriki lost 25 so it is just pathetic work by them

i, well, um........... in short, got humped.......

129........ 80 percentile...... i mean wtf?


A semipro baseball league has teams with $ 21$ players each. League rules state that a player must be paid at least $ \$15,000$, and that the total of all players' salaries for each team cannot exceed $ \$700,000$. What is the maximum possiblle salary, in dollars, for a single player?

$ \textbf{(A)}\ 270,000 \qquad
\textbf{(B)}\ 385,000 \qquad
\textbf{(C)}\ 400,000 \qquad
\textbf{(D)}\ 430,000 \qquad
\textbf{(E)}\ 700,000$

[hide]A player makes a maximum salary if all other players make minimum salary. Thus, the salary of the 20 players making a minimum salary will be 15,000 * 20 = 300,000. The resulting salary of the other player will be 700,000 - 300,000 = 400,000.

Answer: C[/hide]

For such a probelm, I think Aznwithabrain doesn't need to post the solution seriously.

He's done it several other times. Even with a relatively simple problem, you shouldn't just post an answer. Give solutions.

[quote="perfect628"][quote="Aznwithabrain"][hide]
I got 400000
so C
[/hide][/quote]

Seriously, if you're going to post, post your solution. Not, "I got ___. So ___.[/quote]

Well the questions were just so easy...
and some ppl don't have the solutions packet from their schools which means they don't know the answers
so just thought I might just write the answer for those that desperately need to know..

[quote="mathcountsfreak"]Wasn't this on the AMC 10 too?[/quote]
Yes, it was number 4.


This is my second inequality, and I have made it as strong as possible. But still it is not strong enough. :(   Here goes;

[hide="My 2nd inequality"]Prove that for $ x, y, z\in \mathbb{R}^{\plus{}}$ such that $ x\plus{}y\plus{}z\equal{}1$,

                       $ \frac{1\minus{}x}{2yz}\plus{}\frac{1\minus{}y}{2zx}\plus{}\frac{1\minus{}z}{2xy}\ge \frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}\plus{}\frac{1}{\sqrt{y^2\plus{}z^2\minus{}yz}}\plus{}\frac{1}{\sqrt{z^2\plus{}x^2\minus{}zx}}\ge \frac{1}{1\minus{}x}\plus{}\frac{1}{1\minus{}y}\plus{}\frac{1}{1\minus{}z}\ge \frac{9}{2}$                     !!! :D

[/hide]
Hope you will enjoy it (of course, it's easy!)

[hide]
From AM-GM
$ \sum \frac{1\minus{}z}{2yx} \equal{}\sum\frac{x\plus{}y}{2xy} \ge \sum \frac{1}{\sqrt{xy}} \ge \frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}$   because $ x^2\plus{}y^2\ge 2xy \qquad(1)$

$ \sum \frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}} > \sum \frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy\plus{}3xy}} \equal{} \sum \frac{1}{x\plus{}y} \equal{} \sum \frac{1}{1\minus{}z} \qquad(2)$

Now from Jensen's Inequality for $ f(x) \equal{} \frac{1}{1\minus{}x}$

$ \sum \frac{1}{1\minus{}x} \ge (1\plus{}1\plus{}1)\frac{1}{1\minus{}\frac{x\plus{}y\plus{}x}{3}} \equal{} \frac{9}{2}\qquad(3)$

 :)
Nice inequality Polta, how did you arrive at the second one?
[/hide]

I arrived here through this line:

[hide]
$ (\frac {1}{xy}(\frac {1}{x} \plus{} \frac {1}{y}))^2\ge \frac {4}{x^3y^3}$
and using $ \frac {4}{x^3y^3}\ge \frac {16}{(x^3 \plus{} y^3)^2}$(AM GM) I got
$ \frac {x \plus{} y}{x^2y^2}\ge \frac {4}{x^3 \plus{} y^3}$ which eventually simplified to
$ \frac {1 \minus{} z}{xy}\ge \frac {2}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}$

Summing up gives:
$ \sum \frac {1 \minus{} z}{2xy}\ge \sum\frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}$.....(1)
Observe that:
$ x^2 \plus{} y^2 \minus{} xy \equal{} (x \plus{} y)^2 \minus{} 3xy\le (x \plus{} y)^2$
Using this,
$ \sum\frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}\ge \sum \frac {1}{1 \minus{} x}$ as x+y+z=1. Now,
from CS it follows that:$ 2\sum \frac {1}{1 \minus{} x}\ge 9$ as x+y+z=1.
Or $ \sum \frac {1}{1 \minus{} x}\ge 9/2$........(2)

Comparing 1 & 2 gives the result.
[/hide]

However, your solution is nicer and shorter, Ocha!! :lol:
Mine is too long for words..........

DID YOU ENJOY IT??? :D

I found out some interesting fact here:

$ \frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}} \plus{} \frac {1}{\sqrt {y^2 \plus{} z^2 \minus{} yz}} \plus{} \frac {1}{\sqrt {z^2 \plus{} x^2 \minus{} zx}}\le \frac {2}{1 \minus{} x} \plus{} \frac {2}{1 \minus{} y} \plus{} \frac {2}{1 \minus{} z}$

[hide="my solution"]

We havet to prove
$ \sum\frac{1\minus{}x}{2yz}\ge\sum\frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}$ (1)
$ \sum\frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}\ge\sum\frac{1}{1\minus{}x}$ (2)
$ \sum\frac{1}{1\minus{}x}\ge\frac{9}{2}$ (3)
The first one, we have to prove

$ \frac{1\minus{}x}{2yz}\equal{}\frac{y\plus{}z}{2yz}\ge\frac{1}{\sqrt{y^2\plus{}z^2\minus{}yz}}$
So, we need to prove
$ (y\plus{}z)\sqrt{y^2\plus{}z^2\minus{}yz}\ge2yz$
By $ AM\minus{}GM$, $ LHS\ge2\sqrt{yz}.\sqrt{yz}\equal{}2yz$ , proved

So, we get $ \frac{1\minus{}x}{2yz}\equal{}\frac{y\plus{}z}{2yz}\ge\frac{1}{\sqrt{y^2\plus{}z^2\minus{}yz}}$
Then,from this we get $ \frac{1\minus{}y}{2zx}\ge\frac{1}{\sqrt{z^2\plus{}x^2\minus{}xz}}$ and $ \frac{1\minus{}z}{2xy}\ge\frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}$, summing this three, we have proved (1)

The second one, we have to prove
$ \frac{1}{\sqrt{x^2\plus{}y^2\minus{}xy}}\ge\frac{1}{1\minus{}z}\equal{}\frac{1}{x\plus{}y}$ or $ x\plus{}y\ge\sqrt{x^2\plus{}y^2\minus{}xy}$
This true since $ x^2\plus{}y^2\plus{}2xy\ge x^2\plus{}y^2\minus{}xy$, then from this we also get $ \frac{1}{\sqrt{y^2\plus{}z^2\minus{}yz}}\ge\frac{1}{1\minus{}x}$ and $ \frac{1}{\sqrt{z^2\plus{}x^2\minus{}zx}}\ge\frac{1}{1\minus{}y}$, summing this three, we have proved (2)

The third one, from Cauchy-Schwarz
$ [(y\plus{}z)\plus{}(x\plus{}z)\plus{}(x\plus{}y)][\frac{1}{1\minus{}x}\plus{}\frac{1}{1\minus{}y}\plus{}\frac{1}{1\minus{}z}]\equal{}[(y\plus{}z)\plus{}(x\plus{}z)\plus{}(x\plus{}y)][\frac{1}{y\plus{}z}\plus{}\frac{1}{x\plus{}z}\plus{}\frac{1}{x\plus{}y}]\ge 9$
$ \frac{1}{y\plus{}z}\plus{}\frac{1}{x\plus{}z}\plus{}\frac{1}{x\plus{}y}\equal{}\frac{1}{1\minus{}x}\plus{}\frac{1}{1\minus{}y}\plus{}\frac{1}{1\minus{}z}\ge \frac{9}{2}$, we haved proved (3)

$ QED$

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My solution is too long.... :D

[quote="IW@IT"][hide="my solution"]

We havet to prove
$ \sum\frac {1 \minus{} x}{2yz}\ge\sum\frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}$ (1)
$ \sum\frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}\ge\sum\frac {1}{1 \minus{} x}$ (2)
$ \sum\frac {1}{1 \minus{} x}\ge\frac {9}{2}$ (3)
The first one, we have to prove

$ \frac {1 \minus{} x}{2yz} \equal{} \frac {y \plus{} z}{2yz}\ge\frac {1}{\sqrt {y^2 \plus{} z^2 \minus{} yz}}$
So, we need to prove
$ (y \plus{} z)\sqrt {y^2 \plus{} z^2 \minus{} yz}\ge2yz$
By $ AM \minus{} GM$, $ LHS\ge2\sqrt {yz}.\sqrt {yz} \equal{} 2yz$ , proved

So, we get $ \frac {1 \minus{} x}{2yz} \equal{} \frac {y \plus{} z}{2yz}\ge\frac {1}{\sqrt {y^2 \plus{} z^2 \minus{} yz}}$
Then,from this we get $ \frac {1 \minus{} y}{2zx}\ge\frac {1}{\sqrt {z^2 \plus{} x^2 \minus{} xz}}$ and $ \frac {1 \minus{} z}{2xy}\ge\frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}$, summing this three, we have proved (1)

The second one, we have to prove
$ \frac {1}{\sqrt {x^2 \plus{} y^2 \minus{} xy}}\ge\frac {1}{1 \minus{} z} \equal{} \frac {1}{x \plus{} y}$ or $ x \plus{} y\ge\sqrt {x^2 \plus{} y^2 \minus{} xy}$
This true since $ x^2 \plus{} y^2 \plus{} 2xy\ge x^2 \plus{} y^2 \minus{} xy$, then from this we also get $ \frac {1}{\sqrt {y^2 \plus{} z^2 \minus{} yz}}\ge\frac {1}{1 \minus{} x}$ and $ \frac {1}{\sqrt {z^2 \plus{} x^2 \minus{} zx}}\ge\frac {1}{1 \minus{} y}$, summing this three, we have proved (2)

The third one, from Cauchy-Schwarz
$ [(y \plus{} z) \plus{} (x \plus{} z) \plus{} (x \plus{} y)][\frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z}] \equal{} [(y \plus{} z) \plus{} (x \plus{} z) \plus{} (x \plus{} y)][\frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} z} \plus{} \frac {1}{x \plus{} y}]\ge 9$
$ \frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} z} \plus{} \frac {1}{x \plus{} y} \equal{} \frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z}\ge \frac {9}{2}$, we haved proved (3)

$ QED$

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My solution is too long.... :D[/quote]

Nope, your solution is nice and don't you think it resembles mine? :D