We have your learning goals covered with Spring and Summer courses available. Enroll today!

G
Topic
First Poster
Last Poster
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]
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
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
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
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
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, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
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
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
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
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
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
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
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

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
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
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
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
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
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
Friday, Apr 11 - Jun 27
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
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
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

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
Monday, Mar 24 - Jun 16
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
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
H not needed
dchenmathcounts   44
N 6 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
+1 w
dchenmathcounts
May 23, 2020
Ilikeminecraft
6 minutes ago
IZHO 2017 Functional equations
user01   51
N 27 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
27 minutes ago
chat gpt
fuv870   2
N 29 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
41 minutes ago
fuv870
29 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 31 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
31 minutes ago
Segment has Length Equal to Circumradius
djmathman   72
N an hour ago by Zhaom
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
72 replies
djmathman
Apr 30, 2014
Zhaom
an hour ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   22
N an hour ago by RainbowSquirrel53B
Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
[list]
[*] Medals for the top teams
[*] Shirts, pins, stickers and certificates for all participants
[*] Additional awards provided by our wonderful sponsors!
[/list]

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
22 replies
MustangMathTournament
Mar 8, 2025
RainbowSquirrel53B
an hour ago
2025 ROSS Program
scls140511   11
N 2 hours ago by fuzimiao2013
Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
11 replies
scls140511
Today at 2:36 AM
fuzimiao2013
2 hours ago
d_k-eja Vu
ihatemath123   46
N 3 hours ago by Ilikeminecraft
Source: 2024 USAMO Problem 1
Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n!$ in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
Proposed by Luke Robitaille.
46 replies
ihatemath123
Mar 20, 2024
Ilikeminecraft
3 hours ago
average FE
KevinYang2.71   75
N 4 hours ago by Marcus_Zhang
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
75 replies
KevinYang2.71
Mar 21, 2024
Marcus_Zhang
4 hours ago
apparently circles have two intersections :'(
itised   76
N 4 hours ago by Ilikeminecraft
Source: 2020 USOJMO Problem 2
Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\ell$ be a variable line that is tangent to $\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\ell$.

Proposed by Titu Andreescu and Waldemar Pompe
76 replies
itised
Jun 21, 2020
Ilikeminecraft
4 hours ago
Too Bad I'm Lactose Intolerant
hwl0304   216
N 5 hours ago by AshAuktober
Source: 2018 USAMO Problem 1/USAJMO Problem 2
Let \(a,b,c\) be positive real numbers such that \(a+b+c=4\sqrt[3]{abc}\). Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]
216 replies
hwl0304
Apr 18, 2018
AshAuktober
5 hours ago
Easy Combinatorics
JetFire008   1
N 5 hours ago by Marcus_Zhang
Source: AMC 12 2001
How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
1 reply
JetFire008
6 hours ago
Marcus_Zhang
5 hours ago
Did this get posted yet
pog   25
N Today at 2:03 PM by santhoshn
Source: 2024 AMC 8 #1
What is the ones digit of \[222{,}222-22{,}222-2{,}222-222-22-2?\]
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
25 replies
pog
Oct 11, 2024
santhoshn
Today at 2:03 PM
Stanford Math Tournament (SMT) Online 2025
stanford-math-tournament   6
N Today at 3:49 AM by Vkmsd
[center]Register for Stanford Math Tournament (SMT) Online 2025[/center]


[center] :surf: Stanford Math Tournament (SMT) Online is happening on April 13, 2025! :surf:[/center]

[center]IMAGE[/center]

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online

When? The contest will take place April 13, 2025. The pre-contest puzzle hunt will take place on April 12, 2025 (optional, but highly encouraged!).

What? The competition features a Power, Team, Guts, General, and Subject (choose two of Algebra, Calculus, Discrete, Geometry) rounds.

Who? You!!!!! Students in high school or below, from anywhere in the world. Register in a team of 6-8 or as an individual.

Where? Online - compete from anywhere!

Check out our Instagram: https://www.instagram.com/stanfordmathtournament/

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online


[center]IMAGE[/center]


[center] :surf: :surf: :surf: :surf: :surf: [/center]
6 replies
stanford-math-tournament
Mar 9, 2025
Vkmsd
Today at 3:49 AM
Easy geometry
rcorreaa   19
N Today at 4:30 AM by ihategeo_1969
Source: 2022 Brazilian National Mathematical Olympiad - Problem 2
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.
19 replies
rcorreaa
Nov 22, 2022
ihategeo_1969
Today at 4:30 AM
Easy geometry
G H J
Source: 2022 Brazilian National Mathematical Olympiad - Problem 2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rcorreaa
238 posts
#1 • 4 Y
Y by VicKmath7, itslumi, Mango247, Rounak_iitr
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MrOreoJuice
594 posts
#2
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rafinha
51 posts
#3 • 3 Y
Y by BrodyGong1027, Ubfo, Rounak_iitr
Let $I_A$ be the $A-$excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VicKmath7
1385 posts
#4
Y by
Nice config geo!
Let $D=AK \cap BC$, $I_A$ be the $A$-excenter, $M$ be the midpoint of the major arc $BC$ and $I_BI_C \cap BC=X$. We have that $AMPD$ and $BCI_BI_C$ are cyclic, so spamming PoP gives $XD.XP=XA.XM=XB.XC=XI_C.XI_B$, so $I_BI_CDP$ is cyclic. We want to prove that $Q$ lies on that circle. Indeed, we want $AI_C.AI_B=AQ.AD=AD.AK \iff$ $D$ is orthocenter for $\triangle I_BI_CK$, which follows by Brokard for the cyclic quadrilateral $IBI_AC$ (since $K$ is its center).
This post has been edited 2 times. Last edited by VicKmath7, Nov 22, 2022, 10:00 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kamatadu
466 posts
#5 • 2 Y
Y by teomihai, HoripodoKrishno
Not that hard but yeah... Nice problem for a NUB like me.

Anyways moving onto the solution,
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -28, xmax = 6, ymin = -5, ymax = 19;  /* image dimensions */

 /* draw figures */
draw((-8.132112681822244,6.103664258932486)--(-8.99,-0.77), linewidth(0.4)); 
draw((-8.99,-0.77)--(-0.31,-0.77), linewidth(0.4)); 
draw((-0.31,-0.77)--(-8.132112681822244,6.103664258932486), linewidth(0.4)); 
draw(circle((-4.65,2.1787015619620003), 5.24694586416871), linewidth(0.4)); 
draw((-0.31,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-13.3200477963969,4.134066052134884)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
draw(circle((-5.0862639235495,8.574769577234072), 9.354947603412338), linewidth(0.4) + linetype("4 4")); 
draw((-8.99,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-0.31,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
draw((-8.99,-0.77)--(-26.237391236422933,-0.77), linewidth(0.4)); 
draw((-26.237391236422933,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); 
draw((-11.614225363644488,15.275572820071684)--(-4.65,-3.0682443022067103), linewidth(0.4)); 
draw((-8.99,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); 
 /* dots and labels */
dot((-8.132112681822244,6.103664258932486),linewidth(2pt) + dotstyle); 
label("$A$", (-8.516662214016402,6.952946630403522), NE * labelscalefactor); 
dot((-8.99,-0.77),linewidth(2pt) + dotstyle); 
label("$B$", (-10.215226956958455,-2.259379272983231), NE * labelscalefactor); 
dot((-0.31,-0.77),linewidth(2pt) + dotstyle); 
label("$C$", (-0.02429676701291338,-0.9572815308149143), NE * labelscalefactor); 
dot((-4.65,-3.0682443022067103),linewidth(2pt) + dotstyle); 
label("$K$", (-4.669273367310101,-4.62494533814528), NE * labelscalefactor); 
dot((-4.65,-0.77),linewidth(2pt) + dotstyle); 
label("$P$", (-4.696326799454727,-2.2229913594429492), NE * labelscalefactor); 
dot((-13.3200477963969,4.134066052134884),linewidth(2pt) + dotstyle); 
label("$I_C$", (-15.751989828677138,4.626351465965678), NE * labelscalefactor); 
dot((4.020047796396906,10.71722880012654),linewidth(2pt) + dotstyle); 
label("$I_B$", (4.068985215404134,10.91917518067594), NE * labelscalefactor); 
dot((-11.614225363644488,15.275572820071684),linewidth(2pt) + dotstyle); 
label("$Q$", (-13.001730467854944,15.653525805985506), NE * labelscalefactor); 
dot((-5.5225278470990045,-0.77),linewidth(2pt) + dotstyle); 
label("$D$", (-6.941761667458694,-2.1959379272983231), NE * labelscalefactor); 
dot((-26.237391236422933,-0.77),linewidth(2pt) + dotstyle); 
label("$E$", (-27.09656861520515,-0.5244266165008968), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $D = AK \cap BC$, $E = I_B I_C$. Also, trivial to notice that $I_BI_CBC$ is cyclic.

Now,$$AQ \cdot AD = AK \cdot AD \stackrel{\dagger}{=} (\sqrt{bc})^2 = AB \cdot AC \stackrel{\ddagger}{=} AI_B \cdot AI_C \implies I_B I_C DQ\text{ is cyclic}.$$
$\dagger$ follows straight from $\sqrt{bc}$ Inversion, and $\ddagger$ follows from here.

And, $$EI_B \cdot EI_C = EB \cdot EC \stackrel{\star}{=} ED \cdot EP \implies I_B  I_C PD \text{ is cyclic}.$$
$\star$ is just EGMO Lemma 9.17 as it's well known $(E, D; B, C) = -1$.

Both these together imply that $I_B I_C PDQ$ is cyclic.$\blacksquare$

$\textbf{Remark:}$ This might've been well known for $\sqrt{bc}$ Inversion that $I_B \xleftrightarrow{} I_C$, and due to not being in touch with geo for so long I had forgotten the result and had to prove it myself :).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
teomihai
2944 posts
#7
Y by
Rafinha wrote:
Let $I_A$ be the $A-$excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic.

nice solution!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
geometry6
304 posts
#8
Y by
My solution is also posted above I just wanted to post this since I think it is a very cool one!
2022 Brazilian National Mathematical Olympiad - Problem 2 wrote:
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.
By some angle chasing we get:
$$\angle I_AII_B=\angle BCI_B\land\angle II_AI_B=\angle CBIB_B\implies \Delta II_AI_B\cup\{ K\}\sim\Delta CBI_B\cup\{P\}\implies\angle IKI_B=\angle CPI_B$$By a similar argument, we can get $\angle IKI_C=\angle BPI_C$.Thus, we have $$\angle I_CQI_B=\angle I_CKI_B=\angle CPI_B+\angle BPI_C=180^\circ-\angle I_CPI_B\implies P,Q,I_B,I_C \text{ are concyclic.}\blacksquare$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bruckner
106 posts
#9
Y by
Straightforward by complex numbers
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
naman12
1358 posts
#10 • 3 Y
Y by Mango247, Mango247, Mango247
Let $L$ be the midpoint of arc $BAC$ and $X=I_BI_C\cap BC$. Let $K'$ be the reflection of $K$ over $L$. Then, $\angle DAL=\angle DPL=90^\circ$, so $ADLP$ is cyclic. Thus, by say Power of a Point at $K$, $PDQK'$ is also cyclic. In addition, Power of a Point gives $XD\cdot XP=XA\cdot XL=XB\cdot XC=XI_C\cdot XI_B$ so $I_BI_CPD$ is cyclic.

Note $D$ is the orthocenter of $I_BI_CK$. Note clearly $KA\perp I_BI_C$ (note $I_B,I_C,A,L$ are collinear) and $\measuredangle DI_CI_B=\measuredangle DPI_B=\measuredangle CPI_B$. In addition, if $I_A$ is the $A$-excenter, $II_AI_B\sim CBI_B$, so the midpoints are corresponding. In particular, $\measuredangle CPI_B=\measuredangle I_BKI=90^\circ-\measuredangle I_CI_BK$. Thus, $I_CD\perp I_BK$ and vice versa, so $D$ is the orthocenter. This implies that the reflection of $(I_BI_CK)$ is $(I_BI_CD)$, so the reflection of $K$ over $I_BI_C$ lies on $(I_BI_CD)$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
v_Enhance
6857 posts
#13 • 4 Y
Y by HamstPan38825, mod_x, Rounak_iitr, ehuseyinyigit
Solution from Twitch Solves ISL:

Let $I_A$ be the $A$-excenter, and let $H = \overline{AIK} \cap \overline{BC}$.
[asy] /*   Converted from GeoGebra by User:Azjps using Evan's magic cleaner   https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-3.81664,2.88806); pair B = (-3.5,-0.5); pair C = (1.5,-0.5); pair I = (-2.45079,0.65183); pair K = (-1.,-1.72347); pair P = (-1.,-0.5); pair Q = (-6.63328,7.49959); pair I_B = (3.85362,7.57292); pair I_C = (-5.85362,1.64391); pair I_A = (0.45079,-4.09877); pair H = (-1.74727,-0.5); pair T = (-9.36373,-0.5);
import graph; size(9cm); pen zzttqq = rgb(0.6,0.2,0.); pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(I_A--I_B--I_C--cycle, linewidth(0.6) + qqffff);
draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((-1.,1.44247), 3.16594), linewidth(0.6)); draw(circle((-1.37363,5.22015), 5.73234), linewidth(0.6)); draw(I_A--I_B, linewidth(0.6) + qqffff); draw(I_B--I_C, linewidth(0.6) + qqffff); draw(I_C--I_A, linewidth(0.6) + qqffff); draw(Q--I_A, linewidth(0.6)); draw(B--I_B, linewidth(0.6) + blue); draw(C--I_C, linewidth(0.6) + blue); draw(I_B--T, linewidth(0.6) + qqwuqq); draw(T--B, linewidth(0.6) + qqwuqq);
dot("$A$", A, dir((-14.232, 52.186))); dot("$B$", B, dir((-41.152, -54.958))); dot("$C$", C, dir((13.913, -43.888))); dot("$I$", I, dir((5.740, 13.299))); dot("$K$", K, dir((6.148, 12.041))); dot("$P$", P, dir(110)); dot("$Q$", Q, dir((-50.427, 29.076))); dot("$I_B$", I_B, dir((6.270, 12.255))); dot("$I_C$", I_C, dir((-9.789, 50.090))); dot("$I_A$", I_A, dir((-12.421, -42.985))); dot("$H$", H, dir(225)); dot("$T$", T, dir((-60.449, -26.493))); [/asy]
  • Then by Brokard's theorem on cyclic quadrilateral $BICI_A$, it follows that $\triangle I_B I_C H$ is self-polar to this circle.
  • In particular $K$ is the orthocenter. Equivalently, $H$ is the orthocenter of $\triangle I_B K I_C$.
  • Define $T = \overline{I_B I_C} \cap \overline{BC}$. Then $(TH;BC) = -1$.
  • I claim that $I_B I_C P H$ is cyclic. Indeed, $TH \cdot TP = TB \cdot TC = TI_B \cdot I_C$.
  • The reflection of the orthocenter $K$ of $\triangle I_BI_CH$ over a side $\overline{I_B I_C}$ then coincides with $Q$, which now lies on $(PHI_BI_C)$.
This post has been edited 1 time. Last edited by v_Enhance, Nov 26, 2022, 3:53 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Quidditch
815 posts
#14 • 1 Y
Y by teomihai
rcorreaa wrote:
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.

Can't believe no one has posted this sol yet.

We bary with respective triangle $ABC$ and set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$.

First, let's find $K$. Note that $K$ lies on $AI$ where $I$ is the incenter so $K=(x:b:c)$. Moreover, it lies on $(ABC)$ so
$$a^2bc+b^2cx+c^2xb=0\implies x=-\frac{a^2}{b+c}\implies K=(-\frac{a^2}{b+c}:b:c).$$Now, let's find $Q$. We know $K=(-\frac{a^2}{b+c}:b:c)$ and $A=(b+c-\frac{a^2}{b+c}:0:0)$. Thus,
$$Q=2A-K=(2(b+c)-\frac{a^2}{b+c}:-b:-c)=(2(b+c)^2-a^2:-b(b+c):-c(b+c)).$$And lastly, we know $P=(0:1:1),I_B=(a:-b:c),I_C=(a:b:-c)$. Let's find the equation of $(PI_BI_C)$.
$$(x,y,z)\mapsto (0:1:1)\implies -a^2+2(v+w)=0\implies v+w=\frac{a^2}{2}$$$$(x,y,z)\mapsto (a:-b:c)\implies a^2bc-b^2ca+c^2ab+(ua-vb+wc)(a-b+c)=0\implies abc+ua-vb+wc=0$$$$(x,y,z)\mapsto (a:b:-c)\implies a^2bc+b^2ca-c^2ab+(ua+vb-wc)(a+b-c)=0\implies abc+ua+vb-wc=0$$The second plus the third equation gives $2abc+2ua=0\implies u=-bc$. Plug this back to the second equation to get $vb=wc$, so $w=\frac{vb}{c}$. Plug this into the first equation, and we'll have $v+\frac{vb}{c}=\frac{a^2}{2}\implies v=\frac{a^2c}{2(b+c)}$, so $w=\frac{a^2b}{2(b+c)}$.
$$(PI_BI_C): -a^2yz-b^2zx-c^2xy+\left(-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z\right)(x+y+z)=0.$$Finally, we plug $(x,y,z)\mapsto (2(b+c)^2-a^2:-b(b+c):-c(b+c))=Q$.
$$-a^2yz-b^2zx-c^2xy=-a^2bc(b+c)^2+b^2c(b+c)(2(b+c)^2-a^2)+bc^2(b+c)(2(b+c)^2-a^2)=2bc(b+c)^2((b+c)^2-a^2)$$$$-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z=(-2bc(b+c)^2+a^2bc)-\frac{a^2bc}{2}-\frac{a^2bc}{2}=-2bc(b+c)^2$$$$x+y+z=(2(b+c)^2-a^2)-b(b+c)-c(b+c)=(b+c)^2-a^2$$Hence, we're done. :D
This post has been edited 1 time. Last edited by Quidditch, Nov 27, 2022, 9:30 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5000 posts
#15 • 1 Y
Y by samrocksnature
Rename $K$ to $M_A$. Suppose $(ABC)$ is the unit circle, and let $A=x^2$, etc., so $M_A=-yz$, etc., and $I=-xy-yz-zx$. By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$. Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$. We then wish to prove that
$$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$$We can simplify the expression to
$$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$$which of course is real, so we are done. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Mar 8, 2023, 3:00 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_comb01
659 posts
#16 • 1 Y
Y by teomihai
Beautiful problem! I liked the configuration
Let $M$ be the midpoint of $I_BI_C$, then $M$ is on $(ABC)$ because it is nine-point circle of $I_AI_BI_C$
Let $K'$ be the reflection of $K$ with respect to $A$

Claim 1:$I_CQK'I_B$ is cyclic.
Proof: Notice $\angle I_CKI_B= \angle I_CQI_B$ because of the reflection and $KA \perp I_BI_C$, then we observe $I_CKI_BK'$ is parallelogram as diagonals bisect each other, hence $\angle I_CKI_B= \angle I_CK'I_B$, hence the claim.

Claim 2:The problem itself
Proof: Note that by Claim 1 it suffices to prove $I_CK'I_BP$ is cyclic, $$\angle I_CK'P = \angle I_CI_BP$$$$\angle K'KI_B=I_CI_BP$$$$MP*MK = MI_B^2$$$$MP^2+MP*PK=MI_B^2$$$$MP^2+PC^2=MI_B^2$$$$MC=MI_B$$which is true since $I_BCI_C=90^\circ$ and $M$ is midpoint.



To make more readable: here's diagram
Attachments:
This post has been edited 3 times. Last edited by math_comb01, May 20, 2023, 7:39 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
L567
1184 posts
#17
Y by
Solved with mueller.25, starchan, Siddharth03, AdhityaMV

Let $AK$ intersect segment $BC$ at $R$ and let $I_BI_C$ intersect line $BC$ at $T$. Let $N$ be the midpoint of major arc $BC$. Then note that $(N,K;B,C) = -1$ so projecting through $A$, we get that $(T,R;B,C) = -1$ as well. This means that since $P$ is the midpoint of $BC$ and $(BCI_BI_C)$, we have $TR \cdot TP = TB \cdot TC = TI_B \cdot TI_C$ so points $P,R,I_B,I_C$ are concyclic.

Let $I$ be the incenter of $\triangle ABC$. Note that $(A,R;I,I_A) = -1$ and $K$ is the midpoint of $II_A$. So, $AQ \cdot AR = AK \cdot AR = AI \cdot AI_A = AI_B \cdot AI_C$, implying that points $R,Q,I_B,I_C$ are concyclic.

Together, these imply that points $P,Q,I_B,I_C$ are concyclic, as desired. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5000 posts
#18 • 1 Y
Y by L13832
IAmTheHazard wrote:
Rename $K$ to $M_A$. Suppose $(ABC)$ is the unit circle, and let $A=x^2$, etc., so $M_A=-yz$, etc., and $I=-xy-yz-zx$. By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$. Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$. We then wish to prove that
$$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$$We can simplify the expression to
$$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$$which of course is real, so we are done. $\blacksquare$

oh

Let $D$ and $E$ be the intersections of the interior and exterior $\angle A$-bisectors with $\overline{BC}$ respectively. By $\sqrt{bc}$ inversion we have
$$AI_B\cdot AI_C=AB\cdot AC=AD\cdot AK=AD\cdot AQ,$$hence $I_BI_CQD$ is cyclic. On the other hand, it is well-known (extraversion of incenter/excenter) that $I_BI_CBC$ is cyclic, so $EI_B\cdot EI_C=EB\cdot EC$. Furthermore, since $(B,C;D,E)=-1$, we have $EB\cdot EC=ED\cdot EP$, hence $I_BI_CPD$ is cyclic by power of a point, so we're done. $\blacksquare$
This post has been edited 2 times. Last edited by IAmTheHazard, Oct 26, 2023, 1:54 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Thapakazi
53 posts
#19 • 1 Y
Y by surpidism.
Let $H$ be the intersection of $AK$ and $BC$. We first show that $QI_CHI_B$ is cyclic. First as $\triangle I_CAC \sim \triangle I_BAB$, we see that

\[\frac{I_CA}{AC} = \frac{AB}{I_BA} \implies I_CA \cdot I_BA = AB \cdot AC.\]
Secondly, as $\triangle ABK \sim \triangle ACH$, we again see

\[\frac{AK}{AC} = \frac{AB}{AH} \implies AK \cdot AH = AB \cdot AC.\]
So,

$$QA \cdot AH = KA \cdot AH = BA \cdot AC = I_CA \cdot AI_B$$
implying the claim. Now let $J$ be the midpoint of $I_BI_C$. Note that $J$ lies on $(ABC)$ as it is the nine point circle of $\triangle I_AI_BI_C$. Furthermore $J$ is the arc midpoint of arc $BC$ containing $A$. That means $PHAJ$ is cyclic. Now, letting $I_CI_B$ intersect $BC$ at $X$, we see

\[HX \cdot XP = AX \cdot XJ = BX \cdot XC = I_CX \cdot XI_B\]
finishing the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
L13832
248 posts
#20
Y by
[asy]
        /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -5.803255101716253, xmax = 20.94508689023541, ymin = -2.1002703047388525, ymax = 22.319316398917017;  /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); 

draw((7.575733531948671,10.863916969703546)--(5,5)--(12,5)--cycle, linewidth(2) + zzttqq); 
 /* draw figures */
draw((7.575733531948671,10.863916969703546)--(5,5), linewidth(2)); 
draw((5,5)--(12,5), linewidth(2)); 
draw((12,5)--(7.575733531948671,10.863916969703546), linewidth(2)); 
draw(circle((8.5,6.960275942712022), 4.011568492694037), linewidth(2)); 
draw((1.6247981902196311,10.169020646842817)--(8.970521844693746,-1.0807354791321093), linewidth(2)); 
draw((8.970521844693746,-1.0807354791321093)--(15.375201809780371,11.7746682239693), linewidth(2)); 
draw((1.6247981902196311,10.169020646842817)--(15.375201809780371,11.7746682239693), linewidth(2)); 
draw(circle((8.380234318788622,11.99749071039707), 6.998515561209342), linewidth(2)); 
draw((7.575733531948671,10.863916969703546)--(8.970521844693746,-1.0807354791321093), linewidth(2)); 
draw((1.6247981902196311,10.169020646842817)--(12,5), linewidth(2)); 
draw((5,5)--(15.375201809780371,11.7746682239693), linewidth(2)); 
draw((15.375201809780371,11.7746682239693)--(8.5,5), linewidth(2) + linetype("2 2") + qqwuqq); 
draw((8.5,2.9487074500179844)--(15.375201809780371,11.7746682239693), linewidth(2) + linetype("2 2") + zzttqq); 
draw((8.5,5)--(1.6247981902196311,10.169020646842817), linewidth(2) + linetype("2 2") + qqwuqq); 
draw((1.6247981902196311,10.169020646842817)--(8.5,2.9487074500179844), linewidth(2) + linetype("2 2") + zzttqq); 
 /* dots and labels */
dot((7.575733531948671,10.863916969703546),dotstyle); 
label("$A$", (7.716463099778057,11.193041132615535), NE * labelscalefactor); 
dot((5,5),dotstyle); 
label("$B$", (5.128957223893979,5.338809088427837), NE * labelscalefactor); 
dot((12,5),dotstyle); 
label("$C$", (12.11522308878099,5.338809088427837), NE * labelscalefactor); 
dot((8.5,5),linewidth(4pt) + dotstyle); 
label("$P$", (8.622090156337485,5.274121441530735), NE * labelscalefactor); 
dot((8.5,2.9487074500179844),linewidth(4pt) + dotstyle); 
label("$K$", (8.622090156337485,3.2041167408234825), NE * labelscalefactor); 
dot((1.6247981902196311,10.169020646842817),linewidth(4pt) + dotstyle); 
label("$I_C$", (1.7651995852446765,10.416789369850315), NE * labelscalefactor); 
dot((8.970521844693746,-1.0807354791321093),linewidth(4pt) + dotstyle); 
label("$I_A$", (9.107247508065749,-0.8065173667968194), NE * labelscalefactor); 
dot((15.375201809780371,11.7746682239693),linewidth(4pt) + dotstyle); 
label("$I_B$", (15.511324550878843,12.033980542277858), NE * labelscalefactor); 
dot((6.651467063897343,18.779126489389107),linewidth(4pt) + dotstyle); 
label("$Q$", (6.7784922197700785,19.052590230613387), NE * labelscalefactor); 
dot((8.029478155306254,6.978150379168081),linewidth(4pt) + dotstyle); 
label("$I$", (8.169276628057771,7.247094671892335), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
 [/asy]
Let $I_A$ be the $A-$excenter. Note that $\angle I_CKI_B=\angle I_CQI_B=90^{\circ}$ this is because $Q$ is the reflection of $K$ over $A$, so we have to show that $90^{\circ}=\angle I_CDI_B$, so we need to show that $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$. Now observe that $K$ is midpoint of the $II_a$ because the circumcircle of the orthic triangle is nine point circle of the reference triangle. Note that $II_CBA$ and $II_ABC$ are cyclic, this gives us $\triangle I_BI_AI\sim \triangle I_BCI$ and $\triangle I_CI_AI \sim \triangle I_CCB$ because of this similarity we have $\angle I_BPC=\angle I_BKA$ and $\angle I_CPB=\angle I_CKA$ so $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$ so $P,Q,I_B,I_C$ are concyclic.
This post has been edited 1 time. Last edited by L13832, Jun 28, 2024, 5:10 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1979 posts
#21
Y by
Cute! :)
rcorreaa wrote:
Let $ABC$ be an acute triangle, with $AB<AC$. Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$. Let $I_B,I_C$ be the $B$-excenter and $C$-excenter of $ABC$, respectively. Let $Q$ be the reflection of $K$ with respect to $A$. Prove that the points $P,Q,I_B,I_C$ are concyclic.

Since $KA \perp I_BI_C$ and $KQ$ bisects $I_BI_C$, we just need to prove $Q$ is the $K$-HM point in $\triangle KI_BI_C$. This is equivalent to showing that $L\overset{\text{def}}{:=} KA \cap BC$ is the orthocentre of $\triangle KI_BI_C$, as $LQ \perp KQ$. Finally, this last part follows as $L$ lies on ray $AK$ and $AL \cdot AK = AB \cdot AC = AI_B \cdot AI_C$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bin_sherlo
661 posts
#22 • 2 Y
Y by ehuseyinyigit, teomihai
If we rephrase the problem according to $\triangle I_AI_BI_C$ where $I_A$ is the $A-$excenter, we get the following problem.
New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ and $K=AD\cap (DEF), \ M$ is the midpoint of $EF$. Prove that reflection of $K$ with respect to $BC,B,C,M$ are concyclic.
Proof: Let $N$ be the midpoint of $BC$ and $Q,K'$ be the reflections of $K$ to $BC,N$. $P=EF\cap AD,R=EF\cap BC$. Note that $K,M,N$ are collinear. $NB.NC=NM.NK=NM.NK'$ hence $B,C,M,K'$ are concyclic. Also $RB.RC=RE.RF=RD.RN=RP.RM$ which implies $B,C,M,P$ are cyclic. Since $P,M,K,Q$ lie on the circle with diameter $PK',$ we get that $P,K'\in (BCM)$ and $Q\in (PMK)$ thus, $B,C,M,P,K',Q$ are concyclic as desired.$\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihategeo_1969
161 posts
#23
Y by
One of my quickest headsolves.

If we let $D=\overline{AI} \cap \overline{BC}$ then see that $(I_BI_CQD)$ is cyclic since $AI_B \cdot AI_C=AD \cdot AK=AD \cdot AQ$ (first two quantities are both equal to $AB \cdot AC$ by say $\sqrt{bc}$ inversion).

Now let $X$ be $I_A$-Ex point of $\triangle I_AI_BI_C$ And now $(XD;BC)=-1$ (by Cevs Menelaus or whatever) and hence by Mclaurin's we get $XI_C \cdot XI_B=XB \cdot XC=XD \cdot XP$ and done by PoP.
Z K Y
N Quick Reply
G
H
=
a