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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Trigo relation in a right angled. ISIBS2011P10
Sayan   11
N a few seconds ago by Project_Donkey_into_M4
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
11 replies
+1 w
Sayan
Mar 31, 2013
Project_Donkey_into_M4
a few seconds ago
Expressing polynomial as product of two polynomials
Sadigly   3
N 5 minutes ago by Sadigly
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
3 replies
Sadigly
Yesterday at 9:10 PM
Sadigly
5 minutes ago
Help me this problem. Thank you
illybest   0
15 minutes ago
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
0 replies
illybest
15 minutes ago
0 replies
Product of consecutive terms divisible by a prime number
BR1F1SZ   2
N 22 minutes ago by bin_sherlo
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


2 replies
BR1F1SZ
Yesterday at 12:09 AM
bin_sherlo
22 minutes ago
No more topics!
An easy FE
oVlad   3
N Apr 21, 2025 by jasperE3
Source: Romania EGMO TST 2017 Day 1 P3
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
3 replies
oVlad
Apr 21, 2025
jasperE3
Apr 21, 2025
An easy FE
G H J
G H BBookmark kLocked kLocked NReply
Source: Romania EGMO TST 2017 Day 1 P3
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oVlad
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Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
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pco
23511 posts
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oVlad wrote:
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$
Let $c=f(1)$

If $f(0)\ne 0$, $P(x,0)$ $\implies$ $f(x)$ constant, which is never a solution. So $f(0)=0$

$P(0,0)$ $\implies$ $f(-1)=-1$
$P(1,1)$ $\implies$ $c=\pm 1$
Subtracting $P(x,1)$ from $P(-x,-1)$, we get $f(-x)=-cf(x)$

Subtracting $P(x,y)$ from $P(xy,1)$, we get new assertion $Q(x,y)$ : $f(x)f(y)=cf(xy)$
If $f(u)=0$ for some $u\ne 0$, $Q(x,u)$ implies $f(ux)=0$ $\forall x$ and so $f\equiv 0$, which is not a solution.
So $f(x)=0$ $\iff$ $x=0$

$Q(x,x)$ implies $\frac{f(x)}c$ is multiplicative and positive $\forall x>0$ and so $g(x)=\ln \frac{f(e^x)}c$ is additive

If $g(x)$ is not linear, its graph is dense in $\mathbb R^2$ and so graph of $f(x)$ is :
Either dense in $\mathbb R_{>0}\times \mathbb R_{>0}$ if $c=1$
Either dense in $\mathbb R_{>0}\times \mathbb R_{<0}$ if $c=-1$

But $P(x,x)$ $\implies$ $f(x^2-1)\le 2x^2-1$ and so contradiction with both cases
So $g(x)$ is linear and $f(x)=cx^a$ $\forall x>0$ for some real $a$
Then $P(2,1)$ implies $c+2^a=3$ and so :

If $c=1$ : $a=1$ and $f(x)=x$ $\forall x\ge 0$ and $f(-x)=-cf(x)=-f(x)$ imply $\boxed{\text{S1 : }f(x)=x\quad\forall x}$, which indeed fits

If $c=-1$ : $a=2$ and $f(x)=-x^2$ $\forall x\ge 0$ and $f(-x)=-cf(x)=f(x)$ imply $\boxed{\text{S2 : }f(x)=-x^2\quad\forall x}$, which indeed fits
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BR1F1SZ
577 posts
#3
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It is also 2015 Argentina TST P3
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jasperE3
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