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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
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
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Any nice way to do this?
NamelyOrange   0
18 minutes ago
Source: Taichung P.S.1 math program tryouts

How many ordered pairs $(a,b,c)\in\mathbb{N}^3$ are there such that $c=ab$ and $1\le a\le b\le c\le60$?
0 replies
NamelyOrange
18 minutes ago
0 replies
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Polynomial optimization problem
ReticulatedPython   2
N an hour ago by Mathzeus1024
Let $$p(x)=-ax^4+x^3$$, where $a$ is a real number. Prove that for all positive $a$, $$p(x) \le \frac{27}{256a^3}.$$
2 replies
ReticulatedPython
Monday at 2:51 PM
Mathzeus1024
an hour ago
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a+b+c=3 inequality
JK1603JK   1
N an hour ago by KhuongTrang
Let $a,b,c\ge 0: a+b+c=3$ then prove
$$\color{black}{\sqrt{a+b+2c^{2}}+\sqrt{b+c+2a^{2}}+\sqrt{c+a+2b^{2}}\le 3\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}+3}.}$$
1 reply
JK1603JK
2 hours ago
KhuongTrang
an hour ago
J
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Complex + Radical Evaluation
Saucepan_man02   4
N 2 hours ago by Mathzeus1024
Evaluate: (without calculators)
$$ (\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$$
4 replies
Saucepan_man02
Mar 17, 2025
Mathzeus1024
2 hours ago
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Inequality
JK1603JK   1
N 4 hours ago by lbh_qys
Let $a,b,c\ge 0: a+b+c=3$ then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
1 reply
JK1603JK
4 hours ago
lbh_qys
4 hours ago
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Inequality
JK1603JK   1
N 5 hours ago by lbh_qys
Prove that 9ab\left(a-b+c\right)+9bc\left(b-c+a\right)+9ca\left(c-a+b\right)\ge \left(a+b+c\right)^{3},\ \ a\ge 0\ge b\ge c: a+b+c\le 0.
1 reply
JK1603JK
5 hours ago
lbh_qys
5 hours ago
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Solve the equetion
yt12   4
N 5 hours ago by lgx57
Solve the equetion:$\sin 2x+\tan x=2$
4 replies
yt12
Mar 31, 2025
lgx57
5 hours ago
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Inequalities
sqing   2
N 6 hours ago by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
6 hours ago
J
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geometry incentre config
Tony_stark0094   1
N 6 hours ago by Tony_stark0094
In a triangle $\Delta ABC$ $I$ is the incentre and point $F$ is defined such that $F \in AC$ and $F \in \odot BIC$
prove that $AI$ is the perpendicular bisector of $BF$
1 reply
Tony_stark0094
Yesterday at 4:09 PM
Tony_stark0094
6 hours ago
J
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geometry
Tony_stark0094   1
N 6 hours ago by Tony_stark0094
Consider $\Delta ABC$ let $\omega_1$ and $\omega_2$ be the circles passing through $A,B$ and $A,C$ respectively such that $BC$ is tangent to $\omega_1$ and $\omega_2$ define $R$ to be a point such that it lies on both the circles $\omega_1$ and $\omega_2$ prove that $HR$ and $AR$ are perpendicular.
1 reply
Tony_stark0094
Today at 7:04 AM
Tony_stark0094
6 hours ago
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one very nice!
MihaiT   1
N Today at 5:28 AM by MihaiT
Given $m_1$ weights, each weighing $k_1$ and another $m_2$ weights with $k_2$ each. Write a algorithm that determines the ways in which a scale can be balanced with a weight $X$ on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers $m_1,k_1,m_2,k_2,X$. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?

One answer is possible:
a)10;5;0;
b)20;7;0;
c)20;7;1;
d)10;10;0;
e)10;7;0;
f)20;5;0,
1 reply
MihaiT
Mar 31, 2025
MihaiT
Today at 5:28 AM
J
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Geo Mock #4
Bluesoul   1
N Today at 4:44 AM by Sedro
Consider acute triangle $ABC$ with orthocenter $H$. Extend $AH$ to meet $BC$ at $D$. The angle bisector of $\angle{ABH}$ meets the midpoint of $AD$. If $AB=10, BH=6$, compute the area of $\triangle{ABC}$.
1 reply
Bluesoul
Yesterday at 7:03 AM
Sedro
Today at 4:44 AM
J
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An inequality
JK1603JK   2
N Today at 3:24 AM by lbh_qys
Let a,b,c\ge 0: a+b+c=3 then prove \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{2}\cdot\frac{1}{2ab+2bc+2ca+3}.
2 replies
JK1603JK
Today at 3:11 AM
lbh_qys
Today at 3:24 AM
J
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Geo Mock #3
Bluesoul   2
N Yesterday at 11:31 PM by mathprodigy2011
Consider square $ABCD$ with side length of $5$. The point $P$ is selected on the diagonal $AC$ such that $\angle{BPD}=135^{\circ}$. Denote the circumcenters of $\triangle{BPA}, \triangle{APD}$ as $O_1,O_2$. Find the length of $O_1O_2$
2 replies
Bluesoul
Yesterday at 7:02 AM
mathprodigy2011
Yesterday at 11:31 PM
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J
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AC = BC + BM, angle bisector of ABCD (Zhautykov City MO VII 2019 P7)
parmenides51   3
N Jul 31, 2019 by mufree
The bisector of angle $ \angle BAC $ of $ ABCD, $ intersects $ BC $ at $ M. $ Prove that $ AC = BC + BM. $
3 replies
parmenides51
Jul 29, 2019
mufree
Jul 31, 2019
J
AC = BC + BM, angle bisector of ABCD (Zhautykov City MO VII 2019 P7)
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Reveal topic tags
geometry
angle bisector
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parmenides51
30629 posts
parmenides51
#1 h  Jul 29, 2019, 7:39 PM • 2 Y
Y by Adventure10, Mango247
The bisector of angle $ \angle BAC $ of $ ABCD, $ intersects $ BC $ at $ M. $ Prove that $ AC = BC + BM. $
This post has been edited 1 time. Last edited by parmenides51, Dec 19, 2022, 12:07 PM
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kittenwarrior
2351 posts
kittenwarrior
#2 h  Jul 29, 2019, 11:48 PM • 2 Y
Y by Adventure10, Mango247
Can we WLOG $ABCD$ as a square? Or would that invalidate the proof?
This post has been edited 1 time. Last edited by kittenwarrior, Jul 29, 2019, 11:49 PM
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parmenides51
30629 posts
parmenides51
#3 h  Jul 29, 2019, 11:59 PM • 2 Y
Y by Adventure10, Mango247
no way you may choose a random case in geometry in order to prove the general case

as fas as I know this may be true using a transformation .e.g. parallel projection

(these one sends lines to lines, intersections to intersections, preserves ratios of segments and areas,
and any problem involving lines, intersections and ratios can be solved any specific case and is trus for the general one)

suggest we want to prove that in a triangle all medians are concurrent
since it involves only lines, intersections and ratio 1/2 (midpoint) it is enough to prove it in any triangle
let's use the equilateral one, in an equilateral we take as granted that perpendicular bisectors are concurrent
and because in any equilateral, every perpendicular bisector is also a median, we conclude that in the equilateral all medians are concurrent
and as it is true in this specific triangle, due to parallel projection, it is true for any triangle
so in any triangle all medians are concurrent

related pdf
 for further reading
This post has been edited 1 time. Last edited by parmenides51, Jul 30, 2019, 12:00 AM
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mufree
792 posts
mufree
#4 h  Jul 31, 2019, 5:57 PM • 2 Y
Y by Adventure10, Mango247
This is actually a not so famous theorem (Van Schooten’s theorem).
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