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

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
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
HCSSiM results
SurvivingInEnglish   59
N 41 minutes ago by lpieleanu
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
59 replies
SurvivingInEnglish
Apr 5, 2024
lpieleanu
41 minutes ago
[MAIN ROUND STARTS MAY 17] OMMC Year 5
DottedCaculator   35
N an hour ago by paixiao
Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/
Our Discord (6000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal.vercel.app/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]


35 replies
DottedCaculator
Apr 26, 2025
paixiao
an hour ago
USAMO Medals
YauYauFilter   32
N an hour ago by Inaaya
YauYauFilter
Apr 24, 2025
Inaaya
an hour ago
1:1 Physics Tutors
DinoDragon186   3
N 3 hours ago by talhee
I am looking for 1:1 physics tutor.
I am a beginner in physics and am in 9th grade.
I want to make it to IPhO in the coming years.
3 replies
+1 w
DinoDragon186
Dec 10, 2024
talhee
3 hours ago
No more topics!
USAJMO #5 - points on a circle
hrithikguy   208
N Apr 28, 2025 by Adywastaken
Points $A,B,C,D,E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P, A, C$ are collinear, and (iii) $DE \parallel AC$. Prove that $BE$ bisects $AC$.
208 replies
hrithikguy
Apr 28, 2011
Adywastaken
Apr 28, 2025
USAJMO #5 - points on a circle
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
InftyByond
207 posts
#207
Y by
this is an egmo problem I just did a few days ago
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
181 posts
#208
Y by
Sol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathwiz_1207
97 posts
#209
Y by
harmonic sol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
637 posts
#210
Y by
Let $F = BE \cap AC$, so we just need to prove that $OF \perp AP$. This is equivalent to $BFOP$ cyclic. Now let $\angle BED = \theta$. By the parallel condition we have $\angle BFP = \theta$ and clearly $\angle BOD = 2 \theta$ where $O$ is the center of $\omega$. Now we have $\angle POB = \frac{\angle BOD}{2} = \theta$, so $\angle POB = \angle BFP$ and thus $BFOP$ cyclic, done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Vedoral
89 posts
#211
Y by
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
endless_abyss
45 posts
#212
Y by
Noice harmonic
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YOUMAYnOTKnOWME
3 posts
#213
Y by
I don't know why but I did some messy Angle-chasing and got it done.
I am not posting my solution here as I don't have time to LaTeX it all now. I might post solution later
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3001 posts
#214 • 1 Y
Y by megahertz13
lpieleanu wrote:
Neat angle chase :)

Solution

Not so neat complex bash :)

Solution
This post has been edited 1 time. Last edited by lpieleanu, Feb 16, 2025, 11:45 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megahertz13
3183 posts
#215
Y by
Let $M$ be the intersection of $AC$ and $BE$, and $O$ be the center of $\omega$. We aim to show that $CM=MA$, or $\angle{OMA}=\angle{OMC}=90^\circ$.

Claim: The points $O, M, B, P, D$ are all concyclic.

Notice that \[ \angle{BMP} = \angle{MED} = \angle{BDP}, \]where the last part comes from the so-called Tangent Lemma. In particular, $BMDP$ is a cyclic quadrilateral, so $M$ lies on the circumcircle of $\triangle{BDP}$. Since $BODP$ is clearly cyclic ($\angle{OBP} + \angle{ODP} = 180^\circ$), the claim is proven.

Cyclic quadrilateral $BMOP$ tells us that $\angle{OBP}=\angle{OMP}=90^\circ$, so we are done.
This post has been edited 1 time. Last edited by megahertz13, Feb 17, 2025, 1:34 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joebub56
14 posts
#216 • 1 Y
Y by KnowingAnt
Note that $ABCD$ is harmonic, so $BD$ is a symmedian. Also, $ACDE$ is cyclic with $AC\mid\mid DE$, so $ACDE$ is an isosceles trapezoid. Then $AE = CD$ implies $\angle DBC = \angle EBA$, so $BE$ and $BD$ are isogonal, and $BE$ is a median.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
795 posts
#217
Y by
Let $M = AC \cap BE$. Notice that $\angle BMP = \angle BED = \angle BDP$, so $MBPD$ is cyclic. But since $(BPD)$ has diameter $OP$, we have $\angle OMP = \angle OBP = 90$ and hence $M$ is the midpoint of $AC$.
This post has been edited 2 times. Last edited by shendrew7, Feb 19, 2025, 10:16 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cappucher
94 posts
#218
Y by
Let $X = AC \cap BC$. We wish to show that $\overline{OX} \perp \overline{AC}$; if these two segments are perpendicular, then $X$ must be the midpoint of $\overline{AC}$, as $\triangle{AOC}$ is isosceles.

We claim $BXOP$ is cyclic. We can do so by proving $\angle{BXO} = \angle{BOP}$:

\[\angle{BXO} = \angle{BED} = \frac{\overarc{BD}}{2} = \frac{\angle{BOD}}{2} = \angle{BOP}\]
This is because $\overline{AC} \parallel \overline{ED}$, and $P$ bisects $\angle{BOD}$ (this can be shown by proving $\triangle{BOP} \cong \triangle{DOP}$, which is a simple right angle SSA congruence).

Because $BXOP$ is cyclic, we have $\angle{OXP} = \angle{OBP}$, and thus $\angle{OXP} = 90^{\circ}$ (as $\angle{OBP} = 90^{\circ}$ by the tangent condition). Since $A$, $C$, and $P$ are collinear, we have that $\overline{OX} \perp \overline{AC}$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
616 posts
#219
Y by
Note that $(AC;BD) = -1$ and then project through $E$ onto $AC$ to finish.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LeYohan
43 posts
#220
Y by
It's easy even without projective geo....

Let $O$ be the center of $\omega$ and $F = BE \cap AC$.

Due to tangency, $OBPD$ is cyclic and by the $SAS$ criterion, $\triangle OBP \cong \triangle ODP \implies \angle BOP = \angle POD$. Now we angle chase using the fact that $DE \parallel AC$, basic tangency properties and the congruency we just found:

$\angle BFP = \angle BED = \angle PBD = \angle POD = \angle BOP \implies OFBP$ is cyclic. This means $\angle OFP = \angle OBP = 90^{\circ}$, so $O$ lies in the perpendicular bisector of $AC \implies AF = AC$ and we're done. $\square$
'
This post has been edited 1 time. Last edited by LeYohan, Mar 31, 2025, 2:23 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Adywastaken
21 posts
#221
Y by
Just complex bash.
$p=\frac{2bd}{b+d}$
$C=AP\cap\left(ABD\right)=\frac{a-p}{a\overline{p}-1}$
$=\frac{ab+ad-2bd}{2a-b-d}$
$e=\frac{ac}{d}=\left(\frac{a}{d}\right)\left(\frac{ab+ad-2bd}{2a-b-d}\right)$
$\frac{a+c}{2}=\frac{a^2-bd}{2a-b-d}$
So, we need
$\frac{a+c}{2}=\frac{be\left(a+c\right)-ac\left(b+e\right)}{be-ac}$
$b+e=\left(\frac{b+d}{d}\right)\left(\frac{a^2-bd}{2a-b-d}\right)$
So, we need $\frac{a^2-bd}{2a-b-d}=\frac{be\left(a+c\right)-ac\left(b+e\right)}{be-ac}$
$\Longleftrightarrow 1=\frac{2be-ac\left(\frac{b+d}{d}\right)}{be-ac}$
$\Longleftrightarrow ac=ac\left(1+\frac{b}{d}\right)-be$
$\Longleftrightarrow 0=\frac{bde}{d}-be$
$\Longleftrightarrow be=be$, true.
This post has been edited 2 times. Last edited by Adywastaken, Apr 29, 2025, 11:28 AM
Z K Y
N Quick Reply
G
H
=
a