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k a My Retirement & New Leadership at AoPS
rrusczyk   1573
N Yesterday at 11:40 PM 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
1573 replies
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rrusczyk
Mar 24, 2025
SmartGroot
Yesterday at 11:40 PM
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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.
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0 replies
jlacosta
Mar 2, 2025
0 replies
J
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a+b+c=3 inequality
JK1603JK   0
7 minutes ago
Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}+\frac{a+b+c}{ab+bc+ca}\ge \frac{12}{a+b+c}.}
0 replies
1 viewing
JK1603JK
7 minutes ago
0 replies
J
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An inequality
jokehim   4
N 8 minutes ago by JK1603JK
Let $a,b,c \in \mathbb{R}: a+b+c=3$ then prove $$\color{black}{\frac{a^2}{a^{2}-2a+3}+\frac{b^2}{b^{2}-2b+3}+\frac{c^2}{c^{2}-2c+3}\ge \frac{3}{2}.}$$
4 replies
jokehim
Mar 21, 2025
JK1603JK
8 minutes ago
J
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Inequalities
sqing   0
13 minutes ago
Let $ a,b,c\geq 1 $ and $ abc-\frac{1}{3}( ab+bc+ca)\leq 4. $ Prove that
$$20\geq 4(a+b+c) - (a b+b c+c a)-a b c \geq 4$$
0 replies
sqing
13 minutes ago
0 replies
J
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Inequalities
sqing   9
N 22 minutes ago by MathRook7817
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +21abc\leq\frac{512}{441}$$Equality holds when $a=b=\frac{38}{21},c=\frac{5}{214}.$
$$a^2+b^2+ ab +19abc\leq\frac{10648}{9747}$$Equality holds when $a=b=\frac{22}{57},c=\frac{13}{57}.$
$$a^2+b^2+ ab +22abc\leq\frac{15625}{13068}$$Equality holds when $a=b=\frac{25}{66},c=\frac{8}{33}.$
9 replies
sqing
Yesterday at 3:07 AM
MathRook7817
22 minutes ago
J
No more topics!
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2011 AMC12B
Quaratinium   2
N Mar 24, 2025 by mpcnotnpc
[quote=2012 AMC12B]Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?

$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$[/quote]

How come the answer isn't,

The area of the rhombus is $\frac{2\sqrt{3}\cdot2}{2}=2\sqrt{3}$

Since all of the points have to be closest to one of the vertexes of the rhombus, the answer is $\frac{2\sqrt{3}}{4}=\frac{\sqrt3}{2}$
2 replies
Quaratinium
Jul 29, 2017
mpcnotnpc
Mar 24, 2025
J
2011 AMC12B
G H J
Reveal topic tags
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AMC 12
AMC 12 B
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rhombus
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Quaratinium
1030 posts
Quaratinium
#1 h  Jul 29, 2017, 3:08 AM • 2 Y
Y by Adventure10, Mango247
2012 AMC12B wrote:
Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?

$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$

How come the answer isn't,

The area of the rhombus is $\frac{2\sqrt{3}\cdot2}{2}=2\sqrt{3}$

Since all of the points have to be closest to one of the vertexes of the rhombus, the answer is $\frac{2\sqrt{3}}{4}=\frac{\sqrt3}{2}$
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vanstraelen
8941 posts
vanstraelen
#3 h  Mar 24, 2025, 7:25 PM
Y by
Given the rhombus $ABCD\ :\ A(\sqrt{3},0),B(0,1),C(-\sqrt{3},0),D(0,-1)$.

Area region $R$ is the area of $BEFHG\ =\ \frac{2\sqrt{3}}{3}$.
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mpcnotnpc
39 posts
mpcnotnpc
#4 h  Mar 24, 2025, 8:01 PM
Y by
buh, answer is 8 years late :skull:
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