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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
an hour ago
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 16th (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
+1 w
jlacosta
an hour ago
0 replies
k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
1 viewing
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
Every rectangle is formed from a number of full squares
orl   9
N a minute ago by akliu
Source: IMO ShortList 1974, Bulgaria 1, IMO 1997, Day 2, Problem 1
Consider decompositions of an $8\times 8$ chessboard into $p$ non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If $a_i$ is the number of white squares in the $i$-th rectangle, then $a_1<a_2<\ldots <a_p$.
Find the maximum value of $p$ for which such a decomposition is possible. For this value of $p$, determine all possible sequences $a_1,a_2,\ldots ,a_p$.
9 replies
orl
Oct 29, 2005
akliu
a minute ago
Old problem :(
Drakkur   1
N 3 minutes ago by Quantum-Phantom
Let a, b, c be positive real numbers. Prove that
$$\dfrac{1}{\sqrt{a^2+bc}}+\dfrac{1}{\sqrt{b^2+ca}}+\dfrac{1}{\sqrt{c^2+ab}}\le \sqrt{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
1 reply
+1 w
Drakkur
an hour ago
Quantum-Phantom
3 minutes ago
Determinant
S_14159   0
5 minutes ago
Source: JEE adv. 2022 P1
Let $|M|$ denote the determinant of a square matrix $M$. Let $g:\left[0, \frac{\pi}{2}\right] \rightarrow R$ be the function defined by
$$
g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}
$$where $f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.

Let $p({x})$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$ and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE?

(A) $p\left(\frac{3+\sqrt{2}}{4}\right)<0$
(B) $p\left(\frac{1+3 \sqrt{2}}{4}\right)>0$
(C) $p\left(\frac{5 \sqrt{2}-1}{4}\right)>0$
(D) $p\left(\frac{5-\sqrt{2}}{4}\right)<0$
0 replies
S_14159
5 minutes ago
0 replies
Assisted perpendicular chasing
sarjinius   3
N 25 minutes ago by ZeroHero
Source: Philippine Mathematical Olympiad 2025 P7
In acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be an arbitrary point on the circumcircle of triangle $ABC$ such that $D$ does not lie on line $OB$ and that line $OD$ is not parallel to line $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$ such that $DE$ is perpendicular to $BC$, and let $F$ be the point on line $AC$ such that $FA = FE$. Let $P$ and $R$ be the points on the circumcircle of triangle $ABC$ such that $PE$ is a diameter, and $BH$ and $DR$ are parallel. Let $M$ be the midpoint of $DH$.
(a) Show that $AP$ and $BR$ are perpendicular.
(b) Show that $FM$ and $BM$ are perpendicular.
3 replies
sarjinius
Mar 9, 2025
ZeroHero
25 minutes ago
No more topics!
angle chasing in a convex ABCD <A=<C=100^o
parmenides51   1
N Sep 10, 2018 by Talmon
Source: Euler Olympiad 2017 P6 [junior]
In a convex quadrilateral $ABCD$ angles $A$ and $C$ are equal $100^\circ$. On the sides $AB$ and $BC$ points $X$ and $Y$ are selected so that $AX = CY$ . It turned out that the line $YD$ is parallel to the angle bisector of the angle $ABC$. Find the angle $AXY$.
1 reply
parmenides51
Sep 10, 2018
Talmon
Sep 10, 2018
angle chasing in a convex ABCD <A=<C=100^o
G H J
Source: Euler Olympiad 2017 P6 [junior]
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parmenides51
30629 posts
#1 • 2 Y
Y by Adventure10, Mango247
In a convex quadrilateral $ABCD$ angles $A$ and $C$ are equal $100^\circ$. On the sides $AB$ and $BC$ points $X$ and $Y$ are selected so that $AX = CY$ . It turned out that the line $YD$ is parallel to the angle bisector of the angle $ABC$. Find the angle $AXY$.
Z K Y
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Talmon
48 posts
#3 • 2 Y
Y by Adventure10, Mango247
Construct line $l$ through $Y$ parallel to $AX$. Let $l$ intersect $AD$ at $E$. Notice, that $YD$ has to be bisector of angle $Y$ and, as a consequence, of angle $D$ in $EYCD$. Thus, triangles $\triangle EYD$ and $\triangle YCD$ are equal by angles and common side. So $EY = YC = AX \implies EYCD$ is parellelogram $\implies \angle AXY = \angle XAD = 100^{\circ}$
This post has been edited 1 time. Last edited by Talmon, Sep 10, 2018, 9:51 PM
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