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
Simple inequality
sqing   21
N 34 minutes ago by Bexultan
Source: JBMO 2011 Shortlist A3
$\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b}$$
21 replies
sqing
May 15, 2016
Bexultan
34 minutes ago
Primes p such that p and p^2+2p-8 are primes too
mhet49   44
N 34 minutes ago by MITDragon
Source: Albanian National Math Olympiad 2012
Find all primes $p$ such that $p+2$ and $p^2+2p-8$ are also primes.
44 replies
mhet49
Apr 1, 2012
MITDragon
34 minutes ago
Integer polynomial w factorials
Solilin   1
N 44 minutes ago by Tkn
Source: 9th Thailand MO
Let $a_1, a_2, ..., a_{2012}$ be pairwise distinct integers. Show that the equation $(x -a_1)(x - a_2)...(x - a_{2012}) = (1006!)^2$ has at most one integral solution.
1 reply
Solilin
Yesterday at 2:12 PM
Tkn
44 minutes ago
Combo NT
a_507_bc   4
N an hour ago by Namura
Source: Silk Road 2024 P1
Let $n$ be a positive integer and let $p, q>n$ be odd primes. Prove that the positive integers $1, 2, \ldots, n$ can be colored in $2$ colors, such that for any $x \neq y$ of the same color, $xy-1$ is not divisible by $p$ and $q$.
4 replies
a_507_bc
Oct 20, 2024
Namura
an hour ago
No more topics!
Find the locus of all points P
orl   7
N Dec 28, 2021 by Mogmog8
Source: IMO 1992, Day 2, Problem 4
In the plane let $\,C\,$ be a circle, $\,L\,$ a line tangent to the circle $\,C,\,$ and $\,M\,$ a point on $\,L$. Find the locus of all points $\,P\,$ with the following property: there exists two points $\,Q,R\,$ on $\,L\,$ such that $\,M\,$ is the midpoint of $\,QR\,$ and $\,C\,$ is the inscribed circle of triangle $\,PQR$.
7 replies
orl
Nov 11, 2005
Mogmog8
Dec 28, 2021
Find the locus of all points P
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 1992, Day 2, Problem 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
orl
3647 posts
#1 • 5 Y
Y by nguyendangkhoa17112003, Purple_Planet, Adventure10, megarnie, Mango247
In the plane let $\,C\,$ be a circle, $\,L\,$ a line tangent to the circle $\,C,\,$ and $\,M\,$ a point on $\,L$. Find the locus of all points $\,P\,$ with the following property: there exists two points $\,Q,R\,$ on $\,L\,$ such that $\,M\,$ is the midpoint of $\,QR\,$ and $\,C\,$ is the inscribed circle of triangle $\,PQR$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
behzad
55 posts
#2 • 4 Y
Y by Purple_Planet, Adventure10, HappyMathEducation, Mango247
Let $C$ be tangent to $L$ at $X$ and $O$ be the center of $C$. Let $R$ be one of points that has the problems property. Let $P,Q$ be the vertices of triangle which has the problems property. Let $X'$ be on $L$ such that $MX=MX'$.Consider the THe circle $C'$ such that be tangent to $PQR$ externally . and let $OX$ intersects $C$ at $Y$ again. we know $P$ is center of homothety which convert $C$ to $C'$ and we know point $Y$ will conwert to $X'$ in this homothety . so $P,Y,X'$ are collinear and since $Y,X'$ are constant , so the locus of $P$ is aline passes through $Y,X'$ .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
saywhattt
14 posts
#3 • 3 Y
Y by Cipri2005, Purple_Planet, Adventure10
Let the point of tangency of $C$ with $l$ be $T$, and the reflection of $T$ across the centre of the circle be $S$, so that $ST$ is a diameter of $C$. Let the intersection of $PS$ and $QR$ be $X$. Then by a well known lemma, $QT=RX$, and therefore $TM=MX$. Since $T,M$ are fixed, $X$ must be the reflection of $T$ through $M$, and hence the locus of $P$ is the line $SX$ such that $P$ is on the other side of $S$ from $X$. It is easy to check that this is true.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wictro
119 posts
#5 • 2 Y
Y by Purple_Planet, Adventure10
Let line L be tangent to the circle at X. Point M is fixed, then so is the reflection of X over M, say Z. The circle itself is fixed then so is the point diametrically opposite X, say Y. It is well known that P,Z,Y are collinear. And since line XY is fixed, all points P lie on this line.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MP8148
888 posts
#6 • 5 Y
Y by Purple_Planet, third_one_is_jerk, Adventure10, Mango247, TheEmpress
[asy]
size(8cm);
defaultpen(fontsize(10pt));

pair P = dir(120), Q = dir(210), R = dir(330), I = incenter(P,Q,R), T = foot(I,Q,R), M = (Q+R)/2, S = Q+R-T, Tt = 2I-T;

dot("$P$", P, dir(45));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(270));
dot("$T$", T, dir(270));
dot("$M$", M, dir(270));
dot("$S$", S, dir(270));
dot("$T'$", Tt, dir(240));

draw(Tt--Tt+dir(S--Tt)*1.2, red, Arrow);
draw(Q--P--R, blue);
draw(2Q-T--2R-S, Arrows);
draw(incircle(P,Q,R));
draw(T--Tt--S, blue+dashed);
[/asy]

Let $T$ be the point where $C$ touches $L$, $T'$ be the point diametrically opposite $T$ on $C$, and $S$ be the reflection of $T$ over $M$. We claim that the locus is a ray with endpoint $T'$ pointing toward the direction from $S$ to $T'$, not including $T'$.

It is well known that in order to satisfy the given conditions, we must have $P$, $T'$, and $S$ collinear (consider a homothety at $P$ that sends $C$ to the $P$-excircle, where $S$ is the point where the excircle touches $\overline{QR}$). Thus $P$ must lie on line $\overline{T'S}$. However, if $P$ is closer to $S$ than $T'$ or $P = T'$, it is impossible to draw the tangents $\overline{PQ}$ and $\overline{PR}$ such that $C$ is the incircle.

To show that all points in the locus works, pick any $P$ and let its tangents to $C$ meet $L$ at $Q$ and $R$. Since the midpoint $M'$ of $\overline{PQ}$ is the unique point such that $T$ is the reflection of $S$ over $M'$, we must have $M'=M$, completing the proof. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AwesomeYRY
579 posts
#7
Y by
Let the point $T$ be the tangency point, and $E$ be the reflection of $T$ over $M$. Then, let $D$ be the antipode of $T$.

We claim that the locus is the points beyond $D$ on the ray $ED$. We first show such points work. It is well known that $PD\cap L$ is the excenter of $\triangle PQR$ (proof by homothety). Then, the midpoint of QR is the midpoint of $T$ and $PD\cap L = E$ which is $M$, so we're done.

We now show that only points on this ray work. Firstly, points on the other side of $\overline{DE}$ fail since $D$ must be in the interior of $\triangle PQR$. Then, for $P\notin DE$, note that $PD\cap L= E'$ is still the excenter of $\triangle PQR$. Then, the midpoint of $QR$ is the midpoint of $TE'$ which is not equal to the midpoint of $TE$ since $E\neq E'$, so these points don't work.

Thus, we 've shown that all points on the ray work, and all points not on the ray don't work, so we're done. $\blacksquare$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Jalil_Huseynov
439 posts
#8
Y by
Seems easy for even P1.
Let $A$ be center of $C$ and $B=L\cap C$ and $B'$ be reflection of $B$ over $M$ and $B_1$ be antipode of $B$ . So, because of " Diameter of incircle Lemma" we know that $P$ should be on line $MB_1$. So, locus of $P$ is $\overrightarrow{\rm B_1X}$ where $X$ is any point on $\overrightarrow{\rm MB_1}$ such that $B_1$ lies between $M$ and $X$. The question "Why all points on this ray work?" can be easily answered like this : Take any point $P$ on this ray and draw tangents from $P$ to $C$ and intersect them with $L$ at $Q,R$. Then again because of "Diameter of Incircle Lemma" $M$ is midpoint of $QR$. So we are done!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#9 • 1 Y
Y by centslordm
Let $X=\mathcal{C}\cap\ell,$ $X'$ the antipode of $X$ with respect to $\mathcal{C},$ and $Y$ the reflection of $X$ over $M.$ We claim that $P$ lies on $\overline{X'Y}$ beyond $X'.$ Since $\mathcal{C}$ is the incircle of $\triangle PQR,$ we know $P$ must lie on $\overline{X'Y}$ by the Diameter of an Incircle lemma. Moreover, with this lemma, we can draw the tangents from $P$ (on $\overline{X'Y}$ beyond $X'$) to $\mathcal{C}$ to intersect $\ell$ at $Q$ and $R,$ noting that $MQ=MR.$ $\square$
This post has been edited 1 time. Last edited by Mogmog8, Dec 28, 2021, 1:56 AM
Z K Y
N Quick Reply
G
H
=
a