G
Topic
First Poster
Last Poster
k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
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 Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inspired by old results
sqing   7
N 23 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c> 0 $ and $ abc=1 $. Prove that
$$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$$Where $ 0<k \leq 1.$
7 replies
sqing
Yesterday at 1:42 PM
SunnyEvan
23 minutes ago
Modular Arithmetic and Integers
steven_zhang123   3
N 26 minutes ago by steven_zhang123
Integers \( n, a, b \in \mathbb{Z}^+ \) satisfies \( n + a + b = 30 \). If \( \alpha < b, \alpha \in \mathbb{Z^+} \), find the maximum possible value of $\sum_{k=1}^{\alpha} \left \lfloor \frac{kn^2 \bmod a }{b-k}  \right \rfloor $.
3 replies
steven_zhang123
Mar 28, 2025
steven_zhang123
26 minutes ago
Polynomials and their shift with all real roots and in common
Assassino9931   4
N 34 minutes ago by Assassino9931
Source: Bulgaria Spring Mathematical Competition 2025 11.4
We call two non-constant polynomials friendly if each of them has only real roots, and every root of one polynomial is also a root of the other. For two friendly polynomials \( P(x), Q(x) \) and a constant \( C \in \mathbb{R}, C \neq 0 \), it is given that \( P(x) + C \) and \( Q(x) + C \) are also friendly polynomials. Prove that \( P(x) \equiv Q(x) \).
4 replies
Assassino9931
Mar 30, 2025
Assassino9931
34 minutes ago
2025 Caucasus MO Seniors P7
BR1F1SZ   2
N 43 minutes ago by sami1618
Source: Caucasus MO
From a point $O$ lying outside the circle $\omega$, two tangents are drawn touching $\omega$ at points $M$ and $N$. A point $K$ is chosen on the segment $MN$. Let points $P$ and $Q$ be the midpoints of segments $KM$ and $OM$ respectively. The circumcircle of triangle $MPQ$ intersects $\omega$ again at point $L$ ($L \neq M$). Prove that the line $LN$ passes through the centroid of triangle $KMO$.
2 replies
BR1F1SZ
Mar 26, 2025
sami1618
43 minutes ago
Reflections of AB, AC with respect to BC and angle bisector of A
falantrng   26
N an hour ago by ehuseyinyigit
Source: BMO 2024 Problem 1
Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the
$A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points
$E$ and $F$ respectively. A line through $D$ meets $AC$ and $AB$ at $G$ and $H$ respectively such that $G$
lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of
$\triangle EDG$ and $\triangle FDH$ are tangent to each other.
26 replies
falantrng
Apr 29, 2024
ehuseyinyigit
an hour ago
configurational geometry as usual
GorgonMathDota   11
N an hour ago by ratavir
Source: Indonesia National Math Olympiad 2021 Problem 7 (INAMO 2021/7)
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
11 replies
GorgonMathDota
Nov 9, 2021
ratavir
an hour ago
kind of well known?
dotscom26   1
N an hour ago by dotscom26
Source: MBL
Let $ y_1, y_2, ..., y_{2025}$ be real numbers satisfying
$
y_1^2 + y_2^2 + \cdots + y_{2025}^2 = 1.
$
Find the maximum value of
$
|y_1 - y_2| + |y_2 - y_3| + \cdots + |y_{2025} - y_1|.
$

I have seen many problems with the same structure, Id really appreciate if someone could explain which approach is suitable here
1 reply
dotscom26
Today at 4:11 AM
dotscom26
an hour ago
sum(ab/4a^2+b^2) <= 3/5
truongphatt2668   2
N 2 hours ago by arqady
Source: I remember I read it somewhere
Let $a,b,c>0$. Prove that:
$$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$$
2 replies
1 viewing
truongphatt2668
Yesterday at 1:23 PM
arqady
2 hours ago
April Fools Geometry
awesomeming327.   3
N 2 hours ago by awesomeming327.
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ be the projection from $A$ onto $BC$. Let $E$ be a point on the extension of $AD$ past $D$ such that $\angle BAC+\angle BEC=90^\circ$. Let $L$ be on the perpendicular bisector of $AE$ such that $L$ and $C$ are on the same side of $AE$ and
\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]Let the reflection of $D$ across $AB$ and $AC$ be $W$ and $Y$, respectively. Let $X\in AW$ and $Z\in AY$ such that $\angle XBE=\angle ZCE=90^\circ$. Let $EX$ and $EZ$ intersect the circumcircles of $EBD$ and $ECD$ at $J$ and $K$, respectively. Let $LB$ and $LC$ intersect $WJ$ and $YK$ at $P$ and $Q$. Let $PQ$ intersect $BC$ at $F$. Prove that $FB/FC=DB/DC$.
3 replies
awesomeming327.
Today at 2:52 PM
awesomeming327.
2 hours ago
hard problem
pennypc123456789   2
N 2 hours ago by aaravdodhia
Let $\triangle ABC$ be an acute triangle inscribed in a circle $(O)$ with orthocenter $H$ and altitude $AD$. The line passing through $D$ perpendicular to $OD$ intersects $AB$ at $E$. The perpendicular bisector of $AC$ intersects $DE$ at $F$. Let $OB$ intersect $DE$ at $K$. Let $L$ be the reflection of $O$ across $EF$. The circumcircle of triangle $BDE$ intersects $(O)$ at $G$ different from $B$. Prove that $GF$ and $KL$ intersect on the circumcircle of triangle $DEH$.
2 replies
pennypc123456789
Mar 26, 2025
aaravdodhia
2 hours ago
Geometry
Emirhan   1
N 2 hours ago by ehuseyinyigit
Let $ABC$ be an equilateral triangle with side lenght is $1$ $cm$.Let $D \in [AB]$ is a point. Perpendiculars from $D$ to $[AC]$ and $[BC]$ intersects with $[AC]$ and $[BC]$ at points $E$ and $F$ respectively. Perpendiculars from $E$ and $F$ to $[AB]$ intersects with $[AB]$ at points $E_1$ and $F_1$. Prove that
$$[E_1F_1]=\frac{3}{4}$$
1 reply
Emirhan
Jan 30, 2016
ehuseyinyigit
2 hours ago
Polynomials
Pao_de_sal   2
N 2 hours ago by ektorasmiliotis
find all natural numbers n such that the polynomial x²ⁿ + xⁿ + 1 is divisible by x² + x + 1
2 replies
Pao_de_sal
3 hours ago
ektorasmiliotis
2 hours ago
inequalities
Cobedangiu   2
N 3 hours ago by ehuseyinyigit
Source: own
$a,b>0$ and $a+b=1$. Find min P:
$P=\sqrt{\frac{1-a}{1+7a}}+\sqrt{\frac{1-b}{1+7b}}$
2 replies
Cobedangiu
5 hours ago
ehuseyinyigit
3 hours ago
very cute geo
rafaello   3
N 3 hours ago by bin_sherlo
Source: MODSMO 2021 July Contest P7
Consider a triangle $ABC$ with incircle $\omega$. Let $S$ be the point on $\omega$ such that the circumcircle of $BSC$ is tangent to $\omega$ and let the $A$-excircle be tangent to $BC$ at $A_1$. Prove that the tangent from $S$ to $\omega$ and the tangent from $A_1$ to $\omega$ (distinct from $BC$) meet on the line parallel to $BC$ and passing through $A$.
3 replies
rafaello
Oct 26, 2021
bin_sherlo
3 hours ago
Isotomic point of the height foot
silouan   19
N Jun 3, 2024 by buratinogigle
Source: All Russian MO 2015, grade 10, problem 7
In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$
Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$.
The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
M. Didin
19 replies
silouan
Aug 7, 2015
buratinogigle
Jun 3, 2024
Isotomic point of the height foot
G H J
G H BBookmark kLocked kLocked NReply
Source: All Russian MO 2015, grade 10, problem 7
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
silouan
3952 posts
#1 • 3 Y
Y by agimog, Adventure10, Mango247
In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$
Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$.
The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
M. Didin
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TelvCohl
2312 posts
#2 • 8 Y
Y by A_Math_Lover, baladin, jty123, enhanced, LoloChen, Adventure10, Mango247, Tastymooncake2
My solution :

Let the tangent of $ \odot (ABC) $ passing through $ B, C $ meets each other at $ T $ .
Let $ S $ be the isogonal conjugate of $ T $ WRT $ \triangle ABC $ ($ SB \parallel CA, SC \parallel AB $) .

From $ PM \perp AB, TM \perp BC \Longrightarrow \angle TMP=\angle CBA=\angle TCP $ ,
so $ C, P, T, M $ lie on a circle with diameter $ CT $ $ \Longrightarrow $ $ P $ is the projection of $ T $ on $ CA $ .
Similarly, $ Q $ is the projection of $ T $ on $ AB $ $ \Longrightarrow \odot (PMQ) $ is the pedal circle of $ T $ WRT $ \triangle ABC $ ,
so $ X $ is the projection of $ S $ on $ BC $ due to $ T, S $ share the same pedal circle (WRT $ \triangle ABC $) $ \Longrightarrow BH=CX $ .

Q.E.D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Luis González
4145 posts
#3 • 5 Y
Y by A_Math_Lover, baladin, Adventure10, Mango247, Tastymooncake2
Let $U \equiv QM \cap AC,$ $V \equiv PM \cap AB.$ $T$ is the midpoint of $PQ$ and $AM$ cuts $\odot(APQ)$ again at $N$ (reflection of orthocenter $M$ of $\triangle APQ$ on $PQ$). By symmetry, reflections $Y$ and $Z$ of $X$ on $PQ$ and $T$ lie on $\odot(APQ)$ and $PQ \parallel YZ$ $\Longrightarrow$ $\angle (XZ,XM)=\angle (ZN,ZX)=\angle ANZ=\angle AYZ$ $\Longrightarrow$ $AHY \perp BC.$ But by Butterfly theorem for the cyclic $PQVU,$ it follows that $TM \perp BC$ $\Longrightarrow$ $TM$ is X-midline of $\triangle XHY$ $\Longrightarrow$ $MX=MH$ or $BH=CX.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
livetolove212
859 posts
#4 • 7 Y
Y by baladin, naw.ngs, karitoshi, mijail, IAmTheHazard, Adventure10, Tastymooncake2
Let $I$ be the midpoint of $PQ$. Since $M$ is the orthocenter of triangle $APQ$ and $MB=MC$ then applying butterfly theorem for the circle with diameter $PQ$, $IM\perp BC$. Let $J$ be the reflection of $A$ wrt $M, O$ be the circumcenter of triangle $APQ, N$ is the center of $(MPQ)$, $L$ be the antipode of $M$ wrt $(MPQ)$, $Y$ be the midpoint of $LJ.$
We have $NY\parallel =\frac{1}{2}MJ\parallel =\frac{1}{2}AM\parallel =IN$ hence $N$ is the midpoint of $IY$. This means $IY\parallel =MJ$ or $JY\parallel IM$. Therefore $LJ\perp BC$ at $X$. But $A$ and $J$ are symmetric wrt $M$ hence $MH=MX$ or $BH=CX.$
Attachments:
This post has been edited 2 times. Last edited by livetolove212, Nov 17, 2015, 5:06 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NguyenHungQuangKhai
16 posts
#5 • 2 Y
Y by Adventure10, Mango247
This question is a little bit silly but how your guys can thinking like that ? ANY SECRET :-D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kapilpavase
595 posts
#6 • 5 Y
Y by Ankoganit, guptaamitu1, Adventure10, Mango247, Tastymooncake2
Let $PM\cap AB=D, QM\cap AP=E$ performing inversion wrt $M$ and ratio $\sqrt{-MD.MP}$ it amounts to showing that $H'=DE\cap BC$ & $X'=PQ\cap BC$ are equidistant from $M$.

But by converse of butterfly thm on $DQPE$ with $DQ$ and $EP$ as 'wings', it follows that $OM\perp BC$($O$ is centre of $\odot DQPE$. Now we see that the above thing that we wanted to prove is nothing but butterfly on $DQPE$ with $DE$ and $QP$ as wings. ;)
This post has been edited 2 times. Last edited by kapilpavase, Jun 10, 2016, 12:45 PM
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
#7 • 2 Y
Y by kapilpavase, Adventure10
Here is my solution.

Reflecting $P,Q$ in $M$ the problem is equivalent to the following:

In triangle $ABC$ the feet from $A$ onto $BC$ is $H$ and $M$ is the midpoint of $BC$. Points $K,N$ are on $AB,AC$ such that $\angle AKM=\angle ANM=90^{\circ}$. Let rays $MK,MN$ meet the lines through $B,C$ parallel to $AC,AB$ respectively at $P,Q$. Prove that $P,Q,H,M$ are concyclic.

For this newer statement, the proof is as follows: It suffices to prove that $H$ is the centre of a spiral similarity sending $KP$ to $NQ$ since we already know that $A,H,M,K,N$ are concyclic. Now, for this it suffices that $\frac{HK}{HN}=\frac{KP}{NQ}$.

Consider the following equalities: \begin{align*} \frac{HK}{HN}=\frac{\sin HNK}{\sin HKN}=\frac{\sin HAK}{\sin HAN}=\frac{\cos B}{\cos C} \end{align*}
And observe that $BM\cdot \cos B=BK$ and so $KP=BK\cdot \tan A$. Similarly, we observe that $CM\cdot \cos C=CN$ and $NQ=CN\cdot \tan A$.

Therefore, we conclude that \begin{align*} \frac{HK}{HN}=\frac{\cos B}{\cos C}=\frac{BM\cdot \tan A}{CM\cdot \tan A}\cdot \frac{\cos B}{\cos C}=\frac{KP}{NQ} \end{align*}
Thus, our claim holds. Now, it follows that points $P,Q,H,M$ are concyclic.
This post has been edited 1 time. Last edited by anantmudgal09, Jun 10, 2016, 4:32 PM
Reason: Latex error
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
v_Enhance
6870 posts
#8 • 10 Y
Y by baladin, CaptainCuong, anantmudgal09, MeowX2, v4913, megarnie, IAmTheHazard, Adventure10, Mango247, Tastymooncake2
Consider triangle $APQ$, which has orthocenter $M$. Extend line $BC$ to intersect the circle with diameter $PQ$ at $Z$, $Y$. Then $YZ$ also has midpoint $M$ by Butterfly theorem. By inversion at $M$, Butterfly theorem then implies $MH = MX$ too.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WizardMath
2487 posts
#9 • 2 Y
Y by Adventure10, Mango247
Let $N$ be the midpoint of $PQ$. By the converse of butterfly theorem, $MN \perp BC$. Also $M$ is the orthocenter of $APQ$.
Now let the intersection of $(AM),(APQ)$ be $L \ne A$.
Then we claim that $AHML$ is a rectangle.
Proof:
Make $APQ$ our "reference triangle". Then the antipode of $A$ in $(APQ)$ is reflection of $M$ over $N$, and so as $LA \perp LM$ , $A', N' M; L$ are collinear.
So the conclusion follows.
So $HM = AL$.
Consider the translation with vector $AM$.It sends $AQP)$ to $(MQP$ as these circles are congruent and are reflections over $PQ$ and their radical axis is perpendicular to $AM$.
As $MX \parallel AL$, and $X \in (MQP)$, $L$ is mapped to $M$, so $MX = AL = HM$ and so $BH = CX$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9774 posts
#10 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
according a converse of the butterfly theorem and
https://www.artofproblemsolving.com/community/c6h1387927_a_nice_lemma
we can have a nice proof of the problem...

Who want to start in this adventure?

Sincerely
Jean-Louis
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9774 posts
#11 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

see

http://jl.ayme.pagesperso-orange.fr/Docs/Simplicity%201.pdf p. 19.

Sincerely
Jean-Louis
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
#12 • 7 Y
Y by RudraRockstar, RAMUGAUSS, karitoshi, p_square, L567, Adventure10, Mango247
No need for summoning butterflies :)
silouan wrote:
In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$
Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$.
The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
M. Didin

Draw parallelogram $ABDC$ and let $T$ be the intersection of tangents to $\odot(ABC)$ at $B$ and $C$. Notice that $$\angle MQB=90^{\circ}-A=\angle MTB$$so $BMTQ$ is cyclic. Likewise, $CMTP$ is cyclic. It follows that $\odot(PMQ)$ is the pedal circle of point $T$ wrt $\triangle ABC$. Since $T$ and $D$ are isogonal conjugates in $\triangle ABC$, and $\overline{DX} \perp \overline{BC}$, we see that $X \in \odot(PMQ)$, 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.
TheUltimate123
1740 posts
#13 • 6 Y
Y by ETS1331, PROA200, VulcanForge, tree_3, khina, IAmTheHazard
Solved with Christopher Qiu, Elliott Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, Rey Li, and Ryan Yang.

We present two comedies.

First comedy (inversion) Invert at \(M\); butterfly.

Second comedy (wild goose chase) Let \(APA'Q\) be a parallelogram. By Reim's theorem, \(\overline{A'Q}\) is tangent to \((QME)\) and \(\overline{A'P}\) is tangent to \((PMF)\).

Now \(\angle A'QM=\angle A'PM=90^\circ\), so it will suffice to show \(\angle A'XM=90^\circ\). In other words, if \(O\) is the midpoint of \(\overline{PQ}\), it will suffice to show \(\overline{OM}\perp\overline{BC}\).

Let \(B'\) and \(C'\) be the reflections of \(B\) and \(C\) over \(Q\) and \(P\), and let \(N\) be the midpoint of \(\overline{B'C'}\). By homothety \((B,2)\), we have \(\overline{B'C}\perp\overline{AC'}\), and analogously \(\overline{C'B}\perp\overline{AB'}\), so points \(B\), \(C\), \(B'\), \(C'\) lie on a circle centered at \(N\).

Now \(NQMP\) is a Varignon parallelogram, and since \(NB=NC\), we have \(\overline{NOM}\perp\overline{BC}\). The end.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Cindy.tw
54 posts
#14
Y by
Let $U$ be the reflection of $A$ over $MQ$. Note that $\measuredangle QUP = \measuredangle PAQ = \measuredangle QMP$. Hence $P, Q, M, U$ are concyclic. Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$. Since $QM$ is the perpendicular bisector of $BF$, $AF = BU$. And note that $\triangle BPM \sim \triangle HAC$, hence
$$ BU \times BP = AF \times (CH \times \frac{BM}{AH}) = BM \times (CH \times \frac{AF}{AH}) = BM \times CD = BM \times BX $$Hence $M, X, U, P$ are concyclic, which means $X \in \odot(MPQ)$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JAnatolGT_00
559 posts
#15
Y by
Define $MP\cap AB=R,MQ\cap AC=S.$ Since $H\in \odot (ARMS),$ inversion wrt $M,$ which fixes $\odot (PQRS),$ maps $H,X$ onto points $QP\cap BC, RS\cap BC$ respectively. These points are symmetric wrt $M$ by DIT, thus the conclusion.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
starchan
1602 posts
#17
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.
IAmTheHazard
5000 posts
#18 • 2 Y
Y by centslordm, NO_SQUARES
synthetic skill issue.

Rename $H$ to $D$; let $D'$ be the reflection of $D$ over $M$ and $A'$ be the reflection of $A$ over the midpoint of $\overline{PQ}$. I claim that $\overline{A'D'} \perp \overline{BC}$. This is by the sophisticated method of brainlessly coordinate bashing.

Let $A=(a,b),B=(-1,0),C=(1,0)$, so $M=(0,0)$ and $D=(a,0),D'=(-a,0)$. Then $\overline{MQ}$ has equation $y=-\tfrac{a-1}{b}x$ by computing the slope of $\overline{AC}$, and $\overline{AB}$ has equation $y=\tfrac{b}{a+1}(x+1)$, so their intersection $Q$ has $x$-coordinate $-\tfrac{b^2}{a^2+b^2+1}$. Likewise, $\overline{MP}$ has equation $y=-\tfrac{a+1}{b}x$ and $\overline{AC}$ has equation $y=\tfrac{b}{a-1}(x-1)$, so their intersection $P$ has $x$-coordinate $\tfrac{b^2}{a^2+b^2+1}$, so the midpoint of $\overline{PQ}$ lies on the $y$-axis and hence $A'$ has $x$-coordinate $-a$.

To finish, note that this means $\angle A'PM=\angle A'QM=\angle AD'M=90^\circ$, hence $A'PQMD'$ is cyclic which finishes. $\blacksquare$

Remark: This finish seems much cleaner than all the other solutions that prove the perpendicularity synthetically (imagine butterfly)
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 23, 2023, 2:07 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
buratinogigle
2320 posts
#19 • 1 Y
Y by GeoKing
This is very nice problem and it is my favorite problem. I would like to introduce an extension to this problem.

Given a cyclic quadrilateral $ABCD$ and $ABCD$ is not a trapezoid. Let $M$ be the midpoint of $AD$. Points $P$ and $Q$ lie on lines $CD$ and $AB$ respectively such that $MP \parallel BD$ and $MQ \parallel AC$. The circumcircle of triangle $MPQ$ intersects $AD$ again at $N$ different from $M$. Let $AC$ intersect $BD$ at $E$. $F$ is the projection of $E$ onto $AD$.

a) Prove that $AF = ND$.

b) Take points $I$ and $J$ on $MQ$ and $MP$ respectively such that $AI \parallel CD$ and $DJ \parallel AB$. Prove that $M, N, J, I$ lie on the same circle.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
buratinogigle
2320 posts
#20 • 1 Y
Y by dungnguyentien
Another generalization of this beautiful problem

Given an acute triangle $ABC$. Two points $M$ and $N$ lie on side $BC$ such that $M$ and $N$ are different from $B$ and $C$, and $BM = CN$. The line through $M$ perpendicular to $CA$ intersects line $AB$ at $F$. The line through $N$ perpendicular to $AB$ intersects line $AC$ at $E$. $MF$ intersects $NE$ at $P$. On the circumcircle of triangle $PEF$, take $Q$ such that $PQ \parallel BC$. Let $R$ be the point symmetric to $Q$ with respect to the midpoint of $BC$. Prove that $AR \perp BC$.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
buratinogigle
2320 posts
#21 • 2 Y
Y by dungnguyentien, nguyenducmanh2705
And now this is the combination of both two extesions above

Given a cyclic quadrilateral $ABCD$ that is not a trapezoid. On $CD$, take points $M$ and $N$ that are symmetric with respect to the midpoint of $CD$. On lines $BC$ and $AD$, take points $E$ and $F$ respectively such that $ME \parallel AC$ and $NF \parallel BD$. Let $P$ be the intersection of $ME$ and $NF$. On the circle $(PEF)$, take $Q$ such that $PQ \parallel CD$. Let $R$ be the point symmetric to $Q$ with respect to the midpoint of $CD$. Prove that $IR \perp CD$ where $I$ is the intersection of $AC$ and $BD$.
Attachments:
Z K Y
N Quick Reply
G
H
=
a