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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Easy P4 combi game with nt flavour
Maths_VC   1
N 2 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
2 hours ago
Central sequences
EeEeRUT   14
N 3 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
3 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 4 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
4 hours ago
Random Points = Problem
kingu   5
N 4 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
+1 w
kingu
Apr 27, 2024
happypi31415
4 hours ago
No more topics!
Problem 2 (First Day)
Valentin Vornicu   84
N May 13, 2025 by cj13609517288
Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab+bc+ca = 0$ we have the following relations

\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]
84 replies
Valentin Vornicu
Jul 12, 2004
cj13609517288
May 13, 2025
Problem 2 (First Day)
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Valentin Vornicu
7301 posts
#1 • 9 Y
Y by Adventure10, HWenslawski, megarnie, son7, Kingsbane2139, Mango247, Rounak_iitr, cubres, MS_asdfgzxcvb
Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab+bc+ca = 0$ we have the following relations

\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]
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darij grinberg
6555 posts
#2 • 7 Y
Y by Adventure10, megarnie, son7, pluginL, Mango247, zzSpartan, cubres
straightforward solution + 2 pages of calculations = bad IMO problem.

Darij
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harazi
5526 posts
#3 • 4 Y
Y by Adventure10, son7, Mango247, cubres
In my opinion the first day characteristic was uglyness. And this problem is the best example. Of course, problem 3 is quite nice and difficult, but I bet another combinatorics problem could have been given. Again, just my opinion.
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darij grinberg
6555 posts
#4 • 9 Y
Y by TheOneYouWant, AlastorMoody, GammaBetaAlpha, Adventure10, PhysicsMonster_01, son7, Mango247, cubres, and 1 other user
darij grinberg wrote:
straightforward solution + 2 pages of calculations = bad IMO problem.

And I say this inspite of the fact that I have solved it correctly and many others have made errors :D :D :D

dg
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harazi
5526 posts
#5 • 3 Y
Y by Adventure10, son7, Mango247
Why do you say 2 pages of computations? I might be wrong, but doesn't it reduce immediately to an exponential equation which has bounded solution ( i mean, you only need to see if 6 is a solution, which yes requires computation) by 7. But probably I'm wrong. Anyway, my opinion is that this problem is very idiot. Sorry...
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harazi
5526 posts
#6 • 3 Y
Y by Adventure10, son7, Mango247
I think a much interesting problem is the following: find all polynomials f such that for any such a,b,c like in the official problem, the number $\frac{f(a-b)+f(b-c)+f(c-a)}{2} $ is in the image of f (I mean, for any such a,b,c there exists a number $ g(a,b,c) $ such that that number equals f(g(a,b,c))).
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darij grinberg
6555 posts
#7 • 3 Y
Y by Adventure10, son7, Mango247
Actually the answer is $P\left( x\right) = kx^4 + lx^4$, with constant real k and l. But in order to prove that $x^4$ works, one needs some rather messy calculations :D

Darij
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Peter Scholze
644 posts
#8 • 4 Y
Y by AlastorMoody, Adventure10, Mango247, LLL2019
you're actually right harazi(ok, i needn't tell you about mathlinks, anyway). but some idiots like me are too dumb to check that it works for n=4. hope they don't cancel too many points.
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jmerry
12096 posts
#9 • 5 Y
Y by GammaBetaAlpha, Cindy.tw, lneis1, Adventure10, Mango247
A reasonably clean solution:

We want to find all polynomials f such that $f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$ whenever $ab+bc+ca=0$. Equivalently, $ab+bc+ca$ divides $f(a-b)+f(b-c)+f(c-a)-2f(a+b+c)$.
Let $f(x)=r_nx^n+r_{n-1}x^{n-1}+...+r_0$ and $a'=ta,b'=tb,c'=tc$
Then $a'b'+b'c'+c'a'=0$ if $ab+bc+ca=0$ so $t^nr_n((a-b)^n+(b-c)^n+(c-a)^n-2(a+b+c)^n)+...+r_0(1+1+1-2)=0$
The functions of t $t^n,t^{n-1},...,1$ are linearly independent, so each coefficient $r_k((a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k)$ must be zero.
This means that either $r_k=0$ or $ab+bc+ca$ divides $(a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k$
For $k=2$, $(a-b)^2+(b-c)^2+(c-a)^2-2(a+b+c)^2=-4(ab+bc+ca)$
For $k=4$, $(a-b)^4+(b-c)^4+(c-a)^4-2(a+b+c)^4=
-6(ab+bc+ca)(2a^2+2b^2+2c^2+ab+bc+ca)$

Consider the case $a=2,b=2,c=-1$. Then $(-3)^k+(-3)^k=2*3^k$
This forces k to be even. Now take $a=6,b=3,c=-2$. This gives
$3^k+5^k+(-8)^k=2*7^k$.
Since $8^6>2*7^6$,
$3^k+5^k+(-8)^k>2*7^k$ for all even $k>4$.
By previous arguments, we must have $f(x)=dx^4+ex^2$
This post has been edited 1 time. Last edited by jmerry, Jan 2, 2017, 7:12 PM
Reason: Fixed obsolete code. No change to content.
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Michael Lipnowski
108 posts
#10 • 5 Y
Y by GammaBetaAlpha, jchang0313, Adventure10, Illuzion, Mango247
This problem becomes very simple once we get rid of all the dependencies (after milking them for all they're worth):

$a=b=c=0: f(0)=0$
$b=c=0: f(a) = f(-a)$ for all $a$. Therefore, all of $f$'s exponents have even degree. Then making the substitutions $x=a-b, y=b-c, z=c-a, w=a+b+c$ (so that $x+y+z=0$), the given condition becomes $2w^2 = x^2+y^2+z^2$. Therefore, the original problem is completely equivalent to finding all even polynomials $f$ with $f(0) = 0$ satisfying the functional equation

$f(x)+f(y)+f(x+y) = 2f(\sqrt{x^2+xy+y^2})$ for all $x,y \in R$.

Suppose $deg f = 2n$. Equating terms of degree $2n$, we find that

$x^{2n} + y^{2n} + (x+y)^{2n} = 2(x^2+xy+y^2)^n$ or letting $y=tx$, $1+t^{2n}+(1+t)^{2n} = 2(1+t+t^2)^n$.

Assume that $n \geq 3$. Then the root $\omega = \frac{-1+\sqrt{3} i}{2}$ appears with multiplicity at least $3$ on the $RHS$, and hence on the $LHS$ as well.
However, letting $g(t)=LHS$, $0=g''(\omega) =  2n(2n-1)(\omega^{2n-2}+(1+w)^{2n-2}) = 2n(2n-1)(\omega^{2n-2}+(-\omega^2)^{2n-2}) = 2n(2n-1)\omega^{2n-2}(1+w^{2n-2}) \not= 0$, a contradiction.

It follows that $deg f \leq 4$ so that $f(x) = kx^4 + lx^2$ (since $f(0)=0$) for $k,l \in R$ are the only possibly solutions. A quick check shows that these solutions are indeed valid.

The fact that I found this problem simple supports your opinions, namely that this problem is relatively simple.
However, I disagree that this problem is ugly. I think the concepts behind it are reasonably nice, and appropriate. But then again, this is all a matter of perspective. If you, Darij, spent 2 pages tediously computing this answer, I understand why you're bitter. But it's really not that tedious, if you check things the right way. When you reduce this to 2 variables instead of 3, it's very simple to check manually (or doing a root check the way I did above...the above method (with slightly more explanation) shows necessity AND sufficiency).
Finally, I'm not sure how they mark these, but wouldn't you be safe just saying that "plugging this back into the functional equation, we find that these solutions indeed satisfy it"? That way, you avoid the hassle of computing the thing over 2 pages, saving lots of time and effort.

~Mike
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Charlydif
72 posts
#11 • 2 Y
Y by Adventure10, Mango247
If a=b=c you get f(0)=0. With a=b=0 we get f(c)=f(-c), then f(x)=P(x^2) for some P(x).(P(0)=0). Then put x=a-b=b-c (you can easy check that the system have real solution for infinite values of x) then the identity becames 2P(x^2)+P(4x^2)=2P(3x^2)..... looking at the leading coefficient we find that 2+4^n=2*3^n. But 4^n>2*3^n for n>2 hence f(x)= ax^2+bx^4 as we want.
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vinoth_90_2004
301 posts
#12 • 2 Y
Y by Adventure10, Mango247
A much less elegant solution :( :( :
(1) we have f(0)=0, f is even by trivial substituions
(2) Define g(x1,x2,k), where x1,x2 are reals and k is an even integer:
g(x1,x2,k) = (x1-x2)^k + (x2+x3)^k + (x1+x3)^k - (x1+x2-x3)^k,
where x3 = - (x1*x2)/(x1 + x2).
since f is even, let the coefficients of x^2,x^4 ... x^(2n) be a(2),a(4) ... a(2n).
it follows from the given condition that, for all reals x1,x2:
sum, from k=1 to n of a(2k)*g(x1,x2,2k) = 0 ... (*)
now g(x1,-3*x1,2k) = x1^2k ( 4^(2k) + (3/2)^(2k) + (5/2)^(2k) - 2*(7/2)^2k) (where x1 is not 0). now after some computations we have
b(2k) = 4^2k + (3/2)^2k + (5/2)^2k - 2*(7/2)^2k
is strictly positive for k>2, and zero for k=1,2. (for k=1,2 we just verify. for k>=3,
8^k > 2*7^k)
using that in (*), and setting for convenience c(2k)=b(2k)*a(2k),
sum of, from k=3 to n, x^(2k) * c(2k) = 0 for all x.
this implies c(2k) = 0 for all k>3, and since b(2k) is non-zero for thos values, a(2k)=0 for all k>3 -- > f is of the form a*x^2 + b*x^4. now we verify that it works ...
i totally agree that this problem isn't a very good IMO problem, despite some of the nicer solutions that it does have. It doesn't really use any clever ideas at all, and is a bit too simple for q2/5 level problems.
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darij grinberg
6555 posts
#13 • 3 Y
Y by Adventure10, son7, Mango247
Indeed my solution was quite similar to all of your ones.
Michael Lipnowski wrote:
Finally, I'm not sure how they mark these, but wouldn't you be safe just saying that "plugging this back into the functional equation, we find that these solutions indeed satisfy it"? That way, you avoid the hassle of computing the thing over 2 pages, saving lots of time and effort.

Well, maybe they wouldn't like this... the calculation is really heavy if you do not have the right idea, so I think one would lose some points with this.

darij
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harazi
5526 posts
#14 • 5 Y
Y by Adventure10, son7, Mango247, and 2 other users
Anyway, a really cool solution is the following: just take instead of a,b,c the numbers 6x, 3x, -2x. Then identify the coefficients and you will find a simple exponential equation. That's all!
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Namdung
585 posts
#15 • 2 Y
Y by Adventure10, Mango247
It is really nice and natural idea. In sln of jmery it was mentioned in other form. I think it is shortest way to find that $degf \le 4$. The rest is so trivial.

Namdung
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