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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
A sharp one with 3 var
mihaig   3
N 14 minutes ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
3 replies
+1 w
mihaig
May 13, 2025
mihaig
14 minutes ago
Acute triangle, equality of areas
mruczek   5
N 15 minutes ago by LeYohan
Source: XIII Polish Junior MO 2018 Second Round - Problem 2
Let $ABC$ be an acute traingle with $AC \neq BC$. Point $K$ is a foot of altitude through vertex $C$. Point $O$ is a circumcenter of $ABC$. Prove that areas of quadrilaterals $AKOC$ and $BKOC$ are equal.
5 replies
mruczek
Apr 24, 2018
LeYohan
15 minutes ago
Gives typical russian combinatorics vibes
Sadigly   3
N 43 minutes ago by AL1296
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
3 replies
Sadigly
May 8, 2025
AL1296
43 minutes ago
Triangular Numbers in action
integrated_JRC   29
N 2 hours ago by Aiden-1089
Source: RMO 2018 P5
Find all natural numbers $n$ such that $1+[\sqrt{2n}]~$ divides $2n$.

( For any real number $x$ , $[x]$ denotes the largest integer not exceeding $x$. )
29 replies
integrated_JRC
Oct 7, 2018
Aiden-1089
2 hours ago
No more topics!
concurent lines related to midpoints of chords and midpoints of arcs
parmenides51   2
N Apr 20, 2023 by yofro
Source: 2011 Sharygin Geometry Olympiad Correspondence Round P21
On a circle with diameter $AC$, let $B$ be an arbitrary point distinct from $A$ and $C$. Points $M, N$ are the midpoints of chords $AB, BC$, and points $P, Q$ are the midpoints of smaller arcs restricted by these chords. Lines $AQ$ and $BC$ meet at point $K$, and lines $CP$ and $AB$ meet at point $L$. Prove that lines $MQ, NP$ and $KL$ concur.
2 replies
parmenides51
Apr 7, 2019
yofro
Apr 20, 2023
concurent lines related to midpoints of chords and midpoints of arcs
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Source: 2011 Sharygin Geometry Olympiad Correspondence Round P21
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parmenides51
30652 posts
#1 • 1 Y
Y by Adventure10
On a circle with diameter $AC$, let $B$ be an arbitrary point distinct from $A$ and $C$. Points $M, N$ are the midpoints of chords $AB, BC$, and points $P, Q$ are the midpoints of smaller arcs restricted by these chords. Lines $AQ$ and $BC$ meet at point $K$, and lines $CP$ and $AB$ meet at point $L$. Prove that lines $MQ, NP$ and $KL$ concur.
Z K Y
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aa1024
1882 posts
#2
Y by
We use barycentric coordinates with $\triangle{ABC}$ as the reference triangle. We have,
$a^2 + c^2 = b^2.$ Since $Q$ is a arc midpoint, it follows that $AQ$ passes through the incenter. In other words, we parameterize $Q = (t : b : c).$ Plugging this into the circumcircle equation gives,
$$a^2bc + b^2ct + c^2bt = 0 \rightarrow t = \left(-\dfrac{a^2}{b+c} : b : c \right) .$$In a similar fashion we have
$$P = \left(a : b : -\dfrac{c^2}{a+b} \right).$$Trivially,
$$K = (0 : b : c) \quad\,\,\text{and}\quad\,\, L=(a : b : 0)$$Also,
$$M = (1 : 1 : 0) \quad\,\,\text{and}\quad\,\, N = (0 : 1 : 1).$$Therefore, the equation for line $KL$ is
$$
\begin{vmatrix}
0 & b & c \\
a & b & 0 \\
x & y & z
\end{vmatrix}
 = x(-bc)+y(ac)+z(-ab)=0.
$$The equation for line $MQ$ is
$$
\begin{vmatrix}
1 & 1 & 0 \\
-\dfrac{a^2}{b+c} & b & c \\
x & y & z
\end{vmatrix}
= x(c) + y(-c) + z\left(\dfrac{a^2}{b+c} + b \right) =0.
$$The equation for line $NP$ is
$$
\begin{vmatrix}
0 & 1 & 1 \\
a & b & -\dfrac{c^2}{a+b} \\
x & y & z
\end{vmatrix}
 = x \left(-\dfrac{c^2}{a+b}-b \right) + y(a)+z(-a)= 0.
$$By the concurrency lemma, it suffices to show that
$$
\begin{vmatrix}
-bc & ac & -ab \\
c & -c & \dfrac{a^2}{b+c} + b \\
-\dfrac{c^2}{a+b} - b & a & -a
\end{vmatrix}
=0.
$$However, the determinant in question turns out to be
$$ab^2c(b^2 - a^2 - c^2) = 0 $$by the Pythagorean Theorem.
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yofro
3151 posts
#3
Y by
It’s true by Desargues on $\triangle PLM$ and $\triangle NQK$.
This post has been edited 2 times. Last edited by yofro, Apr 20, 2023, 9:28 PM
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