Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Apr 2, 2025
0 replies
Inequalities
Scientist10   0
7 minutes ago
If $x, y, z \in \mathbb{R}$, then prove that the following inequality holds:
\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]
0 replies
Scientist10
7 minutes ago
0 replies
NT from ukr contest
mshtand1   3
N 8 minutes ago by ravengsd
Source: Ukrainian TST for RMM 2021(2) and EGMO 2022 P2
Find the greatest positive integer $n$ such that there exist positive integers $a_1, a_2, ..., a_n$ for which the following holds $a_{k+2} = \dfrac{(a_{k+1}+a_k)(a_{k+1}+1)}{a_k}$ for all $1 \le k \le n-2$.
Proposed by Mykhailo Shtandenko and Oleksii Masalitin
3 replies
mshtand1
Oct 2, 2021
ravengsd
8 minutes ago
Posted before ,but no solution
Nuran2010   1
N 8 minutes ago by Nuran2010
Source: 1220 Number Theory Problems
Find all positive integers $n$ where $49n^3+42n^2+11n+1$ is a perfect cube
1 reply
Nuran2010
Apr 11, 2025
Nuran2010
8 minutes ago
The smallest of sum of elements
hlminh   0
11 minutes ago
Let $S=\{1,2,...,2014\}$ and $X\subset S$ such that for all $a,b\in X,$ if $a+b\leq 2014$ then $a+b\in X.$ Find the smallest of sum of all elements of $X.$
0 replies
+1 w
hlminh
11 minutes ago
0 replies
No more topics!
4 points defined be equal ratios in a triangle, collinearity wanted
parmenides51   2
N Oct 28, 2021 by JuanDelPan
Source: Mexican Mathematical Olympiad 1997 OMM P2
In a triangle $ABC, P$ and $P'$ are points on side $BC, Q$ on side $CA$, and $R $ on side $AB$, such that $\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP'}{P'B}$ . Let $G$ be the centroid of triangle $ABC$ and $K$ be the intersection point of $AP'$ and $RQ$. Prove that points $P,G,K$ are collinear.
2 replies
parmenides51
Jul 28, 2018
JuanDelPan
Oct 28, 2021
4 points defined be equal ratios in a triangle, collinearity wanted
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Source: Mexican Mathematical Olympiad 1997 OMM P2
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parmenides51
30630 posts
#1 • 2 Y
Y by HWenslawski, Adventure10
In a triangle $ABC, P$ and $P'$ are points on side $BC, Q$ on side $CA$, and $R $ on side $AB$, such that $\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP'}{P'B}$ . Let $G$ be the centroid of triangle $ABC$ and $K$ be the intersection point of $AP'$ and $RQ$. Prove that points $P,G,K$ are collinear.
This post has been edited 1 time. Last edited by parmenides51, Jul 28, 2018, 10:05 PM
Reason: name typo
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somebodyyouusedtoknow
255 posts
#2
Y by
P.S. This problem was also taken as Indonesia's INAMO in the Provincial Level, on the year 2010 as problem 1 of the essay section.
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JuanDelPan
122 posts
#3 • 1 Y
Y by MarioLuigi8972
Since $\frac{CQ}{QA} = \frac{CP'}{P'B} \implies AB||P'Q$. Similarly, since $\frac{CP'}{P'B} = \frac{AR}{RB} \implies RP'||AC$. This means $AQP'R$ is a paralellogram with diagonals $AP'$ and $RQ$. Therefore $RK=KQ$ and $AK=KP'$. Let $M$ be the midpoint of $BC$. Because $G$ is the centroid of $ABC$, $AG=2GM$. Let $G'$ be the centroid of triangle $APP'$. We can see $P,G',K$ are collinear, and because $\frac{CP'}{P'B}=\frac{BP}{PC}$, then $BP'=PC$, and $P'M=PM$. This means that $AM$ is also a median of triangle $APP'$, and $AG'=2G'M$. Since $G'$ and $G$ both lie on $AM$, then $G=G'$. But $P,G',K$ are collinear, so $P,G,K$ are collinear, as required. :coolspeak:
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