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

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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 upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
Hard Inequality
danilorj   5
N 16 minutes ago by danilorj
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
5 replies
danilorj
Today at 5:17 AM
danilorj
16 minutes ago
easy geo
ErTeeEs06   5
N 20 minutes ago by Adywastaken
Source: BxMO 2025 P3
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\stackrel{\frown}{BC}, \stackrel{\frown}{CA}, \stackrel{\frown}{AB}$ of $\Omega$ not containing $A, B, C$ respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$ and the midpoint $M$ of $EF$ lie on a line.
5 replies
ErTeeEs06
Apr 26, 2025
Adywastaken
20 minutes ago
Inspired by SXJX (12)2022 Q1167
sqing   4
N 21 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
4 replies
1 viewing
sqing
Yesterday at 4:01 AM
sqing
21 minutes ago
Geometry hard problem.
noneofyou34   3
N 26 minutes ago by noneofyou34
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
3 replies
1 viewing
noneofyou34
Today at 9:50 AM
noneofyou34
26 minutes ago
Function and Quadratic equations help help help
Ocean_MathGod   1
N 4 hours ago by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
4 hours ago
System of Equations
P162008   1
N 5 hours ago by alexheinis
If $a,b$ and $c$ are complex numbers such that

$\frac{ab}{b + c} + \frac{bc}{c + a} + \frac{ca}{a + b} = -9$

$\frac{ab}{c + a} + \frac{bc}{a + b} + \frac{ca}{b + c} = 10$

Compute $\frac{a}{c + a} + \frac{b}{a + b} + \frac{c}{b + c}.$
1 reply
P162008
Yesterday at 10:34 AM
alexheinis
5 hours ago
Inequalities
sqing   19
N Today at 8:40 AM by sqing
Let $ a,b,c>0 , a+b+c +abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$Let $ a,b,c>0 , ab+bc+ca+abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$
19 replies
sqing
May 15, 2025
sqing
Today at 8:40 AM
System of Equations
P162008   1
N Today at 6:33 AM by lbh_qys
If $a,b$ and $c$ are real numbers such that

$\prod_{cyc} (a + b) = abc$

$\prod_{cyc} (a^3 + b^3) = (abc)^3$

Compute the value of $abc.$
1 reply
P162008
Yesterday at 10:43 AM
lbh_qys
Today at 6:33 AM
Max min in geometry
son2007vn   0
Today at 5:33 AM
Given a triangle ABC and positive real numbers m, n, p, find the point M in the plane of the triangle such that m \cdot MA + n \cdot MB + p \cdot MC is minimized.
0 replies
son2007vn
Today at 5:33 AM
0 replies
EGMO help
mathprodigy2011   26
N Today at 5:24 AM by sunken rock
If we have a quadrilateral with 1 pair of parallel sides but the parallel sides are also equal, is that sufficient to stating the quadrilateral is a parallelogram. if it's not, please give a counter-example.
26 replies
mathprodigy2011
Sunday at 10:30 PM
sunken rock
Today at 5:24 AM
this problem pmo
derekli   2
N Today at 4:41 AM by Rabbit47
so I came across this aime #14 while practicing... i found the max value instead of min. ts pmo
2 replies
derekli
Today at 4:22 AM
Rabbit47
Today at 4:41 AM
collinear
spiralman   1
N Today at 2:32 AM by MathsII-enjoy
Given an acute triangle \( \triangle ABC \) with \( AB < AC \), inscribed in circle \( (O) \).
Let \( H \) be the orthocenter of triangle \( ABC \), and \( M \) be the midpoint of \( BC \).
A line passing through \( H \), parallel to \( AO \), intersects lines \( AB \) and \( AC \) at points \( D \) and \( E \), respectively.
Let \( K \) be the circumcenter of triangle \( ADE \). Prove that: Points \( H, K, M \) are collinear.

1 reply
spiralman
Yesterday at 5:50 PM
MathsII-enjoy
Today at 2:32 AM
shadow of a cylinder, shadow of a cone
vanstraelen   4
N Yesterday at 10:59 PM by mathafou

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
4 replies
vanstraelen
May 9, 2025
mathafou
Yesterday at 10:59 PM
Hard Problem help?
7168   5
N Yesterday at 10:22 PM by Shan3t
Suppose that a,b < or = 2010 are positive integers with b * Lcm(a,b) < or = a * gcd(a,b). Let N be the smallest positive integer value of n such that either nb/a and na/b are integers. What, over all a,b, is the greatest possible value of N?

THx!
5 replies
7168
Feb 25, 2011
Shan3t
Yesterday at 10:22 PM
Really easy "divide the cube in cubes" problem
cyshine   4
N Nov 8, 2009 by davicoelhoamorim
Source: Brazilian Math Olympiad, Problem 4
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
4 replies
cyshine
Oct 22, 2009
davicoelhoamorim
Nov 8, 2009
Really easy "divide the cube in cubes" problem
G H J
Source: Brazilian Math Olympiad, Problem 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cyshine
236 posts
#1 • 3 Y
Y by Davi-8191, Adventure10, Mango247
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathias_DK
1312 posts
#2 • 2 Y
Y by Adventure10, Mango247
cyshine wrote:
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
We can partion any cube into $ 8$ smaller cubes or $ 27$ smaller cubes easily.($ 2^3=8,3^3=27$)

So if we can split a cube into $ m$ smaller cubes we can also split it into $ m+7$ and $ m+26$ smaller cubes. Since we can "split" it into $ 1$ cube we can split it into $ 1+7x+26y$ for any $ x,y \in \mathbb{N}_0$. $ 1+7x+26y = n \iff x = \frac{n-1-26y}{7}$. Let $ 0 \le y \le 6$ be such that $ 26y \equiv n-1 \bmod 7 \iff y \equiv 3n-3 \bmod 7$. Let $ n_0 = 1 + 26 \cdot 6$, then $ x \ge 0$ and we can split it into $ n$ smaller cubes whenever $ n \ge n_0$. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mszew
1045 posts
#3 • 1 Y
Y by Adventure10
check http://www.research.att.com/~njas/sequences/A014544

for minimum $ n_0=48$[/hide]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jmerry
12096 posts
#4 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Mathias' argument gives us that $ n_0=1+(7-1)(26-1)+1=152$ works; $ 150$ is the largest positive integer that can't be written as a sum of sevens and 26s. Obviously, we can do better.

First, that $ 26$ could be $ 19$; rather than replacing a $ 1\times 1\times 1$ cube with $ 27$ $ \frac13\times\frac13\times\frac13$ cubes, we replace it with one $ \frac23\times\frac23\times\frac23$ and $ 19$ $ \frac13\times\frac13\times\frac13$ cubes. Since adding $ 26$ is possible by doing the above and then splitting the large cube into eight pieces, we don't need to include it separately. From sevens and nineteens, we get a new estimate of $ n_0\ge 6\cdot 18+2=110$.
For the next improvement, consider breaking a $ 1\times 1\times 1$ into a $ \frac34\times\frac34\times\frac34$ and $ 37$ $ \frac14\times\frac14\times\frac14$ cubes; this allows us to add $ 37$. This is very useful, since $ 19\equiv -2\mod 7$ and $ 37\equiv 2\mod 7$. The multiples $ 0,19,38,57,76,37,74$ form a complete system of residues mod $ 7$, and $ 69$ is the last number we can't get as a sum of copies of $ 7,19,37$. The new estimate is $ n_0\ge 71$.

Finally, there are ways to dissect a cube into $ 49$, $ 51$, or $ 54$ smaller cubes, which effectively replace my dissections into $ 77,$ $ 58,$ and $ 75$ cubes respectively. You can look them up at mszew's link, or try to find them yourself if you're feeling clever. That gets $ n_0$ down to $ 48$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
davicoelhoamorim
2 posts
#5 • 1 Y
Y by Adventure10
We can partion any cube into 8 and 27 cubes. So, we can partion any cube into 8+7k, 8k+26, 27+7K and 27+7k.
8+26k and 27+26k form a complete set of residue classes modulo 7. So, we form all numbers greater than 152 (n0=152).. If we can partion any cube into x cubes, we can partion into x+7k
Z K Y
N Quick Reply
G
H
=
a