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
Miklos Schweitzer 1971_7
ehsan2004   1
N 16 minutes ago by pi_quadrat_sechstel
Let $ n \geq 2$ be an integer, let $ S$ be a set of $ n$ elements, and let $ A_i , \; 1\leq i \leq m$, be distinct subsets of $ S$ of size at least $ 2$ such that \[ A_i \cap A_j \not= \emptyset, A_i \cap A_k \not= \emptyset, A_j \cap A_k \not= \emptyset, \;\textrm{imply}\ \;A_i \cap A_j \cap A_k \not= \emptyset \ .\] Show that $ m \leq 2^{n-1}-1$.

P. Erdos
1 reply
ehsan2004
Oct 29, 2008
pi_quadrat_sechstel
16 minutes ago
Functional equation with a twist (it's number theory)
Davdav1232   0
25 minutes ago
Source: Israel TST 8 2025 p2
Prove that for all primes \( p \) such that \( p \equiv 3 \pmod{4} \) or \( p \equiv 5 \pmod{8} \), there exist integers
\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]such that
\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]
0 replies
Davdav1232
25 minutes ago
0 replies
Grid combi with T-tetrominos
Davdav1232   0
28 minutes ago
Source: Israel TST 8 2025 p1
Let \( f(N) \) denote the maximum number of \( T \)-tetrominoes that can be placed on an \( N \times N \) board such that each \( T \)-tetromino covers at least one cell that is not covered by any other \( T \)-tetromino.

Find the smallest real number \( c \) such that
\[
f(N) \leq cN^2
\]for all positive integers \( N \).
0 replies
Davdav1232
28 minutes ago
0 replies
forced vertices in graphs
Davdav1232   0
30 minutes ago
Source: Israel TST 7 2025 p2
Let \( G \) be a graph colored using \( k \) colors. We say that a vertex is forced if it has neighbors in all the other \( k - 1 \) colors.

Prove that for any \( 2024 \)-regular graph \( G \), there exists a coloring using \( 2025 \) colors such that at least \( 1013 \) of the colors have a forced vertex of that color.

Note: The graph coloring must be valid, this means no \( 2 \) vertices of the same color may be adjacent.
0 replies
Davdav1232
30 minutes ago
0 replies
four point lie on circle
Kizaruno   0
5 hours ago
Let triangle ABC be inscribed in a circle with center O. A line d intersects sides AB and AC at points E and D, respectively. Let M, N, and P be the midpoints of segments BD, CE, and DE, respectively. Let Q be the foot of the perpendicular from O to line DE. Prove that the points M, N, P, and Q lie on a circle.
0 replies
Kizaruno
5 hours ago
0 replies
Inequalities
sqing   0
Today at 2:42 PM
Let $ a,b>0, a^2+ab+b^2 \geq 6  $. Prove that
$$a^4+ab+b^4\geq 10$$Let $ a,b>0, a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$a^4+ab+b^4  \leq 10$$Let $ a,b>0,  a^2+ab+b^2 \geq \frac{15}{2}  $. Prove that
$$ a^4-ab+b^4\geq 10$$Let $ a,b>0,  a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$-\frac{1}{8}\leq  a^4-ab+b^4\leq 10$$
0 replies
sqing
Today at 2:42 PM
0 replies
Range if \omega for No Inscribed Right Triangle y = \sin(\omega x)
ThisIsJoe   0
Today at 2:02 PM
For a positive number \omega , determine the range of \omega for which the curve y = \sin(\omega x) has no inscribed right triangle.
Could someone help me figure out how to approach this?
0 replies
ThisIsJoe
Today at 2:02 PM
0 replies
Interesting question from Al-Khwarezmi olympiad 2024 P3, day1
Adventure1000   1
N Today at 1:10 PM by pooh123
Find all $x, y, z \in \left (0, \frac{1}{2}\right )$ such that
$$
\begin{cases}
(3 x^{2}+y^{2}) \sqrt{1-4 z^{2}} \geq z; \\
(3 y^{2}+z^{2}) \sqrt{1-4 x^{2}} \geq x; \\
(3 z^{2}+x^{2}) \sqrt{1-4 y^{2}} \geq y.
\end{cases}
$$Proposed by Ngo Van Trang, Vietnam
1 reply
Adventure1000
Yesterday at 4:10 PM
pooh123
Today at 1:10 PM
one nice!
MihaiT   3
N Today at 12:53 PM by Pin123
Find positiv integer numbers $(a,b) $ s.t. $\frac{a}{b-2}  $ and $\frac{3b-6}{a-3}$ be positiv integer numbers.
3 replies
MihaiT
Jan 14, 2025
Pin123
Today at 12:53 PM
Acute Angle Altitudes... say that ten times fast
Math-lover1   1
N Today at 12:29 PM by pooh123
In acute triangle $ABC$, points $D$ and $E$ are the feet of the angle bisector and altitude from $A$, respectively. Suppose that $AC-AB=36$ and $DC-DB=24$. Compute $EC-EB$.
1 reply
Math-lover1
Yesterday at 11:30 PM
pooh123
Today at 12:29 PM
Find a and b such that a^2 = (a-b)^3 + b and a and b are coprimes
picysm   2
N Today at 8:28 AM by picysm
it is given that a and b are coprime to each other and a, b belong to N*
2 replies
picysm
Apr 25, 2025
picysm
Today at 8:28 AM
Algebra problem
Deomad123   1
N Today at 8:28 AM by lbh_qys
Let $n$ be a positive integer.Prove that there is a polynomial $P$ with integer coefficients so that $a+b+c=0$,then$$a^{2n+1}+b^{2n+1}+c^{2n+1}=abc[P(a,b)+P(b,c)+P(a,c)]$$.
1 reply
Deomad123
May 3, 2025
lbh_qys
Today at 8:28 AM
Palindrome
Darealzolt   1
N Today at 8:01 AM by ehz2701
Find the number of six-digit palindromic numbers that are divisible by \( 37 \).
1 reply
Darealzolt
Today at 4:13 AM
ehz2701
Today at 8:01 AM
Geometry Proof
strongstephen   17
N Today at 3:59 AM by ohiorizzler1434
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
17 replies
strongstephen
May 6, 2025
ohiorizzler1434
Today at 3:59 AM
Constructing triangles holding many similarities
WakeUp   4
N Apr 15, 2025 by Nari_Tom
Source: Baltic Way 2011
Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
\[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]
4 replies
WakeUp
Nov 6, 2011
Nari_Tom
Apr 15, 2025
Constructing triangles holding many similarities
G H J
Source: Baltic Way 2011
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WakeUp
1347 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
\[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
skytin
418 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hints :
Let E' is isogonal to E wrt ABC
E'BCD ~ AFEI , AFBE , IAED , EBGC , DECH are cycic
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KRIS17
134 posts
#3 • 1 Y
Y by Adventure10
Based on the given similarities of the triangles, it is easy to see that quadrilaterals $AFBE$ and $DECH$ are similar.
Thus, we have:
$AB/DC$ = $FE/EH$
or $AB.EH$ = $FE.DC$

Same way, since quadrilaterals $DIAE$ and $CEBG$ are similar, we get:
$EG.DA$ = $EI.BC$

It remains to be proven that:
$EG.DA$ = $AB.EH$ as the fourth eqiality follows directly form these 3 eualities.

Since quadrilateral $PQRS$ is cyclic, we get that:
$\angle RSE$ = $\angle RDE$ = $\angle BAF$ (from similarities given)
and $\angle ESP$ = $\angle EAP$
Thus, we have $\angle RSP$ = $\angle EAF$.

In the same manner we can derive that $\angle PQR$ = $\angle FBE$.

Since, $\angle RSP$ and $\angle PQR$ are supplementry, so are $\angle EAF$ and $\angle FBE$.
Thus, quadrilateral $AFBE$ is cyclic. Since quadrilaterals $AFBE$ and $DECH$ are similar, $DECH$ is cyclic too.
In the same manner, we can derive that, quadrilaterals $DIAE$ and $CEBG$ are cyclic too.

Applying Ptolemy's theorem to $AFBE$, we get:
$AB.EF$ = $AF.EB$ + $FB.AE$
or $AB$ = $(AF/EF).EB + (FB/EF).AE$ = $(DE/EH).EB + (EC/EH).AE$ (from similarity of quadrilaterals $AFBE$ and $DECH$)
or $AB.EH$ = $DE.EB$ + $EC.AE ---> (1)$

In a similar manner, by applying Ptolemy's theorem to $DIAE$, we get:
$EG.AD$ = $DE.EB$ + $EC.AE ---> (2)$

From $(1)$ and $(2)$, we have $EG.DA$ = $AB.EH$ which completes the proof.
This post has been edited 1 time. Last edited by KRIS17, Aug 18, 2019, 10:03 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rafaello
1079 posts
#4
Y by
Claim. $ABEF$, $BECG$, $CEDH$ and $AEDI$ are cyclic.
Proof. We have
\begin{align*}\measuredangle IAE&=\measuredangle SAE+\measuredangle EBQ\\&=\measuredangle SPE+\measuredangle EPQ\\&=\measuredangle SPQ\\&=\measuredangle SRQ\\&=\measuredangle SRE+\measuredangle ERQ\\&=\measuredangle SDE+\measuredangle ECQ\\&=\measuredangle SDE+\measuredangle IDA\\&=\measuredangle IDE,
\end{align*}we do the same for others, the claim follows.

Claim.
There exists $T$, the isogonal conjugate of $E$ wrt $ABCD$.
Proof. By six-point circle theorem, we conclude that there exists $T$, the isogonal conjugate of $E$.

Now, note that angle chasing implies that $\triangle GBE\sim\triangle CTD$ and $\triangle FBE\sim\triangle ATD$, thus
$$\frac{EF}{EG}=\frac{AD}{DC}.$$Similarly, $\dfrac{EG}{EH}=\dfrac{AB}{AD}$, $\dfrac{EH}{EI}=\dfrac{BC}{AB}$ and $\dfrac{EI}{EF}=\dfrac{CD}{BC}$.
This yields that
$$EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Nari_Tom
117 posts
#5
Y by
Here is a solution using spiral similarity and avoiding isogonal conjugates.

$EF \cdot CD= EH \cdot AB$ is clear since there is similar quadrilaterals. So let's prove that $EF \cdot CD=EG \cdot DA$.

We have $\angle FEG=\angle FEB+\angle BEG=\angle FAB+\angle BCG= \angle EDC+\angle EDA=\angle ADC$. Now let's consider $S$, the center of the spiral similarity that takes $AB$ to $CD$. So it takes $F$ to $E$. We have $\frac{EF}{ES}=\frac{AD}{DS}$.
Since $S$ is also the center of the similarity that takes $AD$ to $BC$, we have $\frac{EG}{ES}=\frac{CD}{DS}$.
Combining these two relations we get $\frac{AD}{CD}=\frac{EF}{EG}$, So $\triangle FEG \sim \triangle ADC$. Conclusion follows.
Z K Y
N Quick Reply
G
H
=
a