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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)!

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[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
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Factorials divide
va2010   37
N 3 minutes ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
va2010
Jul 7, 2016
ND_
3 minutes ago
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IMO Shortlist 2011, Number Theory 2
orl   24
N 6 minutes ago by ezpotd
Source: IMO Shortlist 2011, Number Theory 2
Consider a polynomial $P(x) = \prod^9_{j=1}(x+d_j),$ where $d_1, d_2, \ldots d_9$ are nine distinct integers. Prove that there exists an integer $N,$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20.

Proposed by Luxembourg
24 replies
orl
Jul 11, 2012
ezpotd
6 minutes ago
J
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Inequality in triangle
Nguyenhuyen_AG   3
N 14 minutes ago by Nguyenhuyen_AG
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
3 replies
Nguyenhuyen_AG
Today at 6:17 AM
Nguyenhuyen_AG
14 minutes ago
J
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Problem 1
randomusername   73
N 26 minutes ago by ND_
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
73 replies
randomusername
Jul 10, 2015
ND_
26 minutes ago
J
No more topics!
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convex ABCD, AB=CB, <ABC +2<CDA = \pi, AE=EC, prove <CDE=<BDA
parmenides51   1
N Sep 26, 2018 by little-fermat
Source: IMAR 2009 p3
Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$.

Paolo Leonetti
1 reply
parmenides51
Sep 26, 2018
little-fermat
Sep 26, 2018
J
convex ABCD, AB=CB, <ABC +2<CDA = \pi, AE=EC, prove <CDE=<BDA
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Reveal topic tags
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Angle Chasing
convex quadrilateral
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equal angles
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Source: IMAR 2009 p3
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parmenides51
30653 posts
parmenides51
#1 h  Sep 26, 2018, 7:33 PM • 1 Y
Y by Adventure10
Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$.

Paolo Leonetti
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little-fermat
147 posts
little-fermat
#2 h  Sep 26, 2018, 8:29 PM • 2 Y
Y by Adventure10, Mango247
Note that $B$ is the intersection of the tangents from $A,C$ to the circle $(ADC)$
So $DB$ is a symmedian in $\triangle ADC$ which implies the result .
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