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a My Retirement & New Leadership at AoPS
rrusczyk   1619
N a few seconds ago by Matthew_B
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
1619 replies
+8 w
rrusczyk
Mar 24, 2025
Matthew_B
a few seconds ago
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
Partition set with equal sum and differnt cardinality
psi241   73
N 44 minutes ago by mananaban
Source: IMO Shortlist 2018 C1
Let $n\geqslant 3$ be an integer. Prove that there exists a set $S$ of $2n$ positive integers satisfying the following property: For every $m=2,3,...,n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$.
73 replies
psi241
Jul 17, 2019
mananaban
44 minutes ago
IMO 2018 Problem 5
orthocentre   75
N an hour ago by VideoCake
Source: IMO 2018
Let $a_1$, $a_2$, $\ldots$ be an infinite sequence of positive integers. Suppose that there is an integer $N > 1$ such that, for each $n \geq N$, the number
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$$is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \geq M$.

Proposed by Bayarmagnai Gombodorj, Mongolia
75 replies
orthocentre
Jul 10, 2018
VideoCake
an hour ago
Ornaments and Christmas trees
Morskow   29
N 2 hours ago by gladIasked
Source: Slovenia IMO TST 2018, Day 1, Problem 1
Let $n$ be a positive integer. On the table, we have $n^2$ ornaments in $n$ different colours, not necessarily $n$ of each colour. Prove that we can hang the ornaments on $n$ Christmas trees in such a way that there are exactly $n$ ornaments on each tree and the ornaments on every tree are of at most $2$ different colours.
29 replies
Morskow
Dec 17, 2017
gladIasked
2 hours ago
Another square grid :D
MathLuis   42
N 2 hours ago by gladIasked
Source: USEMO 2021 P1
Let $n$ be a fixed positive integer and consider an $n\times n$ grid of real numbers. Determine the greatest possible number of cells $c$ in the grid such that the entry in $c$ is both strictly greater than the average of $c$'s column and strictly less than the average of $c$'s row.

Proposed by Holden Mui
42 replies
MathLuis
Oct 30, 2021
gladIasked
2 hours ago
Find the midpoint of the chord of a conic
Hunter87   5
N 3 hours ago by vanstraelen
From P(4,5), the chord of contact to the conic 3x² + 4y² = 1 is AB, we are to find the midpoint of this chord.

I used T(4,5)=0 to get eqn. of AB, then assuming (h,k) to be the midpoint, T(h,k)=S1(h,k) should give the equation of AB again. But comparing both equations to get h,k does not give me the correct answer.

What am I doing wrong?
5 replies
Hunter87
Today at 8:15 AM
vanstraelen
3 hours ago
An easy inequality
jokehim   3
N 3 hours ago by no_room_for_error
Let $a,b,c\ge 0: a+b+c=3$ then prove that $$\color{black}{\frac{a+bc}{b+c+2}+\frac{b+ca}{c+a+2}+\frac{c+ab}{a+b+2}\le \frac{3}{2}.}$$
3 replies
jokehim
Today at 1:22 PM
no_room_for_error
3 hours ago
2^a+7^b=c^2+4 (Puerto Rico TST 2024.3)
Equinox8   2
N 4 hours ago by MuradSafarli
Find all positive integers $a,b,c$ such that: $$2^a+7^b=c^2+4$$
2 replies
Equinox8
Mar 12, 2025
MuradSafarli
4 hours ago
Inequalities
sqing   7
N 6 hours ago by DAVROS
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 $a=b=\frac{25}{66},c=\frac{8}{33}.$
7 replies
sqing
Today at 3:07 AM
DAVROS
6 hours ago
3D Geometry Problem
ReticulatedPython   1
N Today at 3:09 PM by ReticulatedPython
Three mutually tangent non-degenerate spheres rest on a plane. Let their centers be $C_1, C_2$, and $C_3$. The spheres with centers $C_1, C_2$, and $C_3$ touch the plane at $P_1, P_2$, and $P_3$, respectively. Prove that $$\frac{(P_1P_2)(P_2P_3)(P_1P_3)}{(C_1P_1)(C_2P_2)(C_3P_3)}=8.$$
Source: Own
1 reply
ReticulatedPython
Yesterday at 8:12 PM
ReticulatedPython
Today at 3:09 PM
Inequalities
sqing   13
N Today at 12:18 PM by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that$$a^3b+b^3c+c^3a+\frac{473}{256}abc\le\frac{27}{256}$$Equality holds when $ a=b=c=\frac{1}{3} $ or $ a=0,b=\frac{3}{4},c=\frac{1}{4} $ or $ a=\frac{1}{4} ,b=0,c=\frac{3}{4} $
or $ a=\frac{3}{4} ,b=\frac{1}{4},c=0. $
13 replies
sqing
Mar 22, 2025
sqing
Today at 12:18 PM
Euclid 2022 Question 10
mockingjay11   1
N Today at 12:08 PM by grey_blue_sky7
10. At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping’s semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizontal is selected uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping.

(a) For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings.

(b) For a 3-topping pizza, determine the probability that some region of the pizza with non-zero area is covered by all 3 toppings.

(c) Suppose that $N$ is a positive integer. For an $N$-topping pizza, determine the probability, in terms of $N$, that some region of the pizza with non-zero area is covered by all $N$ toppings.

I already solved (a), but I don't seem to be able to do (b). Is there any trick to tackle this problem?
1 reply
mockingjay11
Sep 3, 2022
grey_blue_sky7
Today at 12:08 PM
Inequalities
sqing   1
N Today at 11:58 AM by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a(b+c+ 5bc +1)\leq\frac{676}{675}$$$$a(b+c+6bc +1)\leq\frac{245}{243}$$
1 reply
sqing
Today at 11:34 AM
sqing
Today at 11:58 AM
Serving some good tea
smartvong   1
N Today at 10:43 AM by Chanome
Consider a teapot that holds $1$ liter and is initially filled with tea of $m$% concentration. I want to serve tea to my guests, but they insist that the tea must have at least a $n$% tea concentration to be considered good, such that $0<n<m\le100$. I can add as much water as needed, as long as the teapot’s capacity isn’t exceeded, and I can pour out tea in arbitrarily small amounts—allowing me to continuously adjust the concentration by topping it off with water. Using an optimal strategy, what is the maximum total volume of good tea (i.e. tea with at least $n$% concentration) that I can serve to my guests?
1 reply
smartvong
Today at 8:42 AM
Chanome
Today at 10:43 AM
computational in an isosceles triangle (Singapore Junior 2018)
parmenides51   2
N Today at 3:46 AM by lightsynth123
In $\vartriangle ABC, AB=AC=14 \sqrt2 , D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\vartriangle CDE$ is similar to $\vartriangle ABC$. Find the length of $BD$.
2 replies
parmenides51
Jul 10, 2019
lightsynth123
Today at 3:46 AM
integral points
jhz   2
N Today at 2:50 PM by DottedCaculator
Source: 2025 CTST P17
Prove: there exist integer $x_1,x_2,\cdots x_{10},y_1,y_2,\cdots y_{10}$ satisfying the following conditions:
$(1)$ $|x_i|,|y_i|\le 10^{10} $ for all $1\le i \le 10$
$(2)$ Define the set \[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]then \(|S| = 1024\)and any rectangular strip of width 1 covers at most two points of S.
2 replies
jhz
Today at 1:14 AM
DottedCaculator
Today at 2:50 PM
integral points
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G H BBookmark kLocked kLocked NReply
Source: 2025 CTST P17
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jhz
10 posts
#1 • 1 Y
Y by MS_asdfgzxcvb
Prove: there exist integer $x_1,x_2,\cdots x_{10},y_1,y_2,\cdots y_{10}$ satisfying the following conditions:
$(1)$ $|x_i|,|y_i|\le 10^{10} $ for all $1\le i \le 10$
$(2)$ Define the set \[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]then \(|S| = 1024\)and any rectangular strip of width 1 covers at most two points of S.
This post has been edited 1 time. Last edited by jhz, Today at 1:17 AM
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gaussiemann144
71 posts
#2
Y by
well the only thing i could find is-
The integers $$x_i = 3^{i-1}
, y_i = 3^{i+9}$$satisfy for $i \in {1,2,3, \dots 10}$
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DottedCaculator
7316 posts
#3
Y by
Induct on the length of the sequence. Consider the geometric interpretation. For the first $2^1$ points, pick points at an axis-aligned $1\times 10$ rectangle $(0,1)$ and $(10,0)$. Now, for the inductive step, assume that there is a $10^{n-1}\times 10^n$ rectangle that contains all $2^n$ points. Then, shift this rectangle by $(9\times 10^n,K)$. Consider the angles of rectangular strips of width $1$ that cover two points in the $10^{n-1}\times 10^n$ rectangle. Suppose the maximum angle is $\theta$. Then, we can take $K=10^{n-1}+10^{n+1}\tan\theta+1$. The maximum angle of a rectangular strip of width $1$ that covers two points of this new set of $2^{n+1}$ points is at most $\arctan\left(\frac1{40}+\frac54\tan\theta+\frac1{8\cdot10^n}\right)+\arctan\frac1{8\cdot10^n}$. For $n=1$, $\theta$ starts at $0$, so the maximum angle for the first $2^2$ points is at most $\arctan\frac3{80}+\arctan\frac1{80}=\arctan\frac{320}{6397}\leq\arctan0.0501$. Otherwise, for $n\geq2$, the tangent is at most
$\frac{\frac{11}{400}+\frac54\tan\theta}{\frac{31999}{32000}-\frac1{640}\tan\theta}\leq\frac1{38}+\frac{500}{399}\tan\theta\leq0.03+1.3\tan\theta$. Then, the maximum value of $\tan\theta$ is at most $0.09513$ for $2^3$ points, $0.1537$ for $2^4$ points, $0.23$ for $2^5$ points, $0.329$ for $2^6$ points, $0.459$ for $2^7$ points, $0.628$ for $2^8$ points, and $0.836$ for $2^9$ points. Then, the last $2^9$ points are placed in a rectangle with $y$-coordinate at most $10^8+10^{10}\cdot0.836+10^8+1<10^{10}$. Therefore, all $1024$ points fit inside the $10^{10}\times 10^{10}$ square, and no rectangular strip of width $1$ covers at most two points of $S$.
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