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

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
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
positive integers forming a perfect square
cielblue   1
N 6 minutes ago by aaravdodhia
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
1 reply
cielblue
Friday at 8:25 PM
aaravdodhia
6 minutes ago
Orthocenter
jayme   4
N 41 minutes ago by Sadigly
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
4 replies
jayme
Mar 25, 2015
Sadigly
41 minutes ago
Concurrency
Dadgarnia   29
N 43 minutes ago by blueprimes
Source: Iranian TST 2020, second exam day 2, problem 4
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
29 replies
Dadgarnia
Mar 12, 2020
blueprimes
43 minutes ago
Good Numbers
ilovemath04   30
N an hour ago by ihategeo_1969
Source: ISL 2019 N5
Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\tbinom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an \geq b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.
30 replies
ilovemath04
Sep 22, 2020
ihategeo_1969
an hour ago
Sequences problem
BBNoDollar   1
N an hour ago by BBNoDollar
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
Yesterday at 5:53 PM
BBNoDollar
an hour ago
square root problem
kjhgyuio   3
N an hour ago by kjhgyuio
........
3 replies
kjhgyuio
Yesterday at 4:48 AM
kjhgyuio
an hour ago
Concurrency in Parallelogram
amuthup   90
N an hour ago by Maximilian113
Source: 2021 ISL G1
Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.
90 replies
amuthup
Jul 12, 2022
Maximilian113
an hour ago
IMO 2011 Problem 4
Amir Hossein   93
N an hour ago by lpieleanu
Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0, 2^1, \cdots, 2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.

Proposed by Morteza Saghafian, Iran
93 replies
Amir Hossein
Jul 19, 2011
lpieleanu
an hour ago
Sintetic geometry problem
ICE_CNME_4   2
N 2 hours ago by ICE_CNME_4
Source: Math Gazette Contest 2025
Let there be the triangle ABC and the points E ∈ (AC), F ∈ (AB), such that BE and CF are concurrent in O.
If {L} = AO ∩ EF and K ∈ BC, such that LK ⊥ BC, show that EKL = FKL.
2 replies
ICE_CNME_4
3 hours ago
ICE_CNME_4
2 hours ago
Hard diophant equation
MuradSafarli   5
N 2 hours ago by aaravdodhia
Find all positive integers $x, y, z, t$ such that the equation

$$
2017^x + 6^y + 2^z = 2025^t
$$
is satisfied.
5 replies
MuradSafarli
Friday at 6:12 PM
aaravdodhia
2 hours ago
Geometry with orthocenter config
thdnder   4
N 3 hours ago by ohhh
Source: Own
Let $ABC$ be a triangle, and let $AD, BE, CF$ be its altitudes. Let $H$ be its orthocenter, and let $O_B$ and $O_C$ be the circumcenters of triangles $AHC$ and $AHB$. Let $G$ be the second intersection of the circumcircles of triangles $FDO_B$ and $EDO_C$. Prove that the lines $DG$, $EF$, and $A$-median of $\triangle ABC$ are concurrent.
4 replies
thdnder
Apr 29, 2025
ohhh
3 hours ago
Random modulos
m4thbl3nd3r   6
N 3 hours ago by GreekIdiot
Find all pair of integers $(x,y)$ s.t $x^2+3=y^7$
6 replies
m4thbl3nd3r
Apr 7, 2025
GreekIdiot
3 hours ago
deleting multiple or divisor in pairs from 2-50 on a blackboard
parmenides51   1
N 4 hours ago by TheBaiano
Source: 2023 May Olympiad L2 p3
The $49$ numbers $2,3,4,...,49,50$ are written on the blackboard . An allowed operation consists of choosing two different numbers $a$ and $b$ of the blackboard such that $a$ is a multiple of $b$ and delete exactly one of the two. María performs a sequence of permitted operations until she observes that it is no longer possible to perform any more. Determine the minimum number of numbers that can remain on the board at that moment.
1 reply
parmenides51
Mar 24, 2024
TheBaiano
4 hours ago
at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   9
N 4 hours ago by atdaotlohbh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
9 replies
NJAX
May 31, 2024
atdaotlohbh
4 hours ago
Problem 1
SpectralS   146
N Apr 23, 2025 by YaoAOPS
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
146 replies
SpectralS
Jul 10, 2012
YaoAOPS
Apr 23, 2025
Problem 1
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.
surpidism.
10 posts
#141
Y by
Claim 1: $FG \parallel BC$
Proof: Clearly $BJ \perp MK$ and $CJ \perp ML$, meaning $M$ is the orthocentre of $\triangle FGJ$. So $MJ \perp BC$ and $MJ \perp FG$ implies that $FG \parallel BC$.

Claim 2: $A$, $G$, $L$, $J$, $K$, $F$ are concylic
Proof: $\angle GFJ = \angle MBJ = \angle MKJ = \angle GKJ \implies  JKFG$ is cyclic. Similarly, $JLGF$ is cyclic. Thus with $AKJL$ cyclic as well, $A$, $G$, $L$, $J$, $K$, $F$ are concylic.

Claim 3: $ AT \parallel FM$
Proof: $ \angle GFM = \angle CML = \angle CLM = \angle ALF = \angle AGF$.

Claim 4: $F$ is the midpoint of $\overline{AS} $
Proof: $ \angle AKJ = \angle AFJ = \angle AFB = 90^{\circ} = \angle SFB$ and $ \angle FBS = \angle MBJ = \angle KBJ = \angle FBA $. So, $\triangle FBS \cong \triangle FBA $, which implies $ SF = FA$.

Hence, claim 3 and 4 follows that $M$ is the midpoint of $ST$. $\square$
This post has been edited 1 time. Last edited by surpidism., Jun 3, 2024, 7:16 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PEKKA
1848 posts
#142
Y by
By right angles, notice that $AKJL$ is a cyclic quad. Let the circumcircle of this quadrilateral be $\omega.$
Then we can angle chase to get $\angle JFM= \frac{A}{2}$ and $\angle LKJ=\frac{A}{2}.$ This implies that $F$ lies on $\omega.$
By symmetry $G$ lies on $\omega$ too.
Then by cyclic quads, $\frac{B}{2}=\angle BJK=\angle BAF.$
By definition then vertical angles, $\angle KBJ=\angle FBA=90-\frac{B}{2}.$
Therefore $ABF$ is a right triangle and therefore $FJ \perp AS.$
By definition then vertical angles again, $\angle SBF=90-\frac{B}{2}.$
Therefore by ASA, triangles $SBF$ and $ABF$ are congruent.
By symmetry, triangles $ACG$ and $TCG$ are congruent too.
By tangents excircle properties, $SM=SB+SM=AB+BK=s$ and $AL=AC+CL=CM+CT=MT=s$ where $s$ is the semiperimeter of ABC.
Therefore $SM=MT$ and we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
G0d_0f_D34th_h3r3
22 posts
#143
Y by
Let's solve this easy angle chase problem.

First we notice that $\angle SMJ = \angle SFJ$. So, quadrilateral $FSJM$ is cyclic.

Now, Since $\angle CML = 180^{\circ} - \frac{180^{\circ}-\angle C}{2}$ and
$$\angle FSJ = 90^{\circ}- \angle SJF = 90^{\circ}-\angle SMF = 90^{\circ}-\angle CML = 90^{\circ} - \frac{\angle C}{2}$$
Also,
$$\angle ACJ = \angle C + \frac{180^{\circ}-\angle C}{2}= 90^{\circ} +\frac{\angle C}{2}$$
Therefore, quadrilateral $ACJS$ is cyclic. Since $A$, $I$ and $J$ are collinear, we get,
$$\angle JSM = \angle JSC = \angle JAC = \angle IAC = \frac{\angle C}{2}$$
Similarly, $\angle JTM = \frac{\angle C}{2}$. Hence by congruency, we get that $M$ is the midpoint of $ST$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1262 posts
#144
Y by
bad formatting because i copied my solution from math dash

let JC meet ML at X, then JXM is right, so JFL = 90 - BJX = 90 - BJC = a/2. since JAL is also a/2, we see that (AFJL) is cyclic, and since (AKJL) is also cyclic so is (AFJKL). now AFJ is right, then so is SFJ, so (SFJM) is cyclic. since JFM = JFL = JAL = a/2, we have JSM = JFM = a/2. repeating this argument gives JTM = a/2. since JMS = JMT = 90, by AAS congruency we have JSM and JTM are congruent, so SM = TM.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1002 posts
#146
Y by
We present three claims, all of which may be proven by angle chasing with respect to the angles of the triangle $ABC$.
Claim 1: $A, F, G, J, K, L$ are concylic.
Claim 2: $AFMG$ is a parallelogram.
Claim 3: $FG \parallel ST$.

Note that from Claim 2, $AM$ bisects $FG$. But now taking the homothety at $A$ sending $FG$ to $ST$ (which exists by Claim 3), $AM$ bisects $ST$, so $M$ is the midpoint of $ST$. $\square$

Remark(due to Rijul Saini): The above claim 2 and 3 may essentially be summarised as Iran Lemma on the $A-$ excircle.
This post has been edited 1 time. Last edited by AshAuktober, Sep 18, 2024, 8:34 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
794 posts
#147
Y by
Using angle chase, notice that
\[\angle LFJ = \angle XBJ - \angle CXL = 90 - \frac B2 - \frac C2 = \frac A2 = \angle LAJ,\]
so $AFJL$ is cyclic. Notice that $\angle ALJ = \angle AKJ = 90$ implies $AFKJL$ is also cyclic. Instead exploiting symmetry like a normal person, we can instead notice that the converse of Pascal's on hexagon $AKGJFL$ implies these six points lie on a conic, which must be a circle as we know five of the points define a circle.

Hence we have $\angle AFB = \angle AGC = 90$, giving us isosceles trapezoids $AXKS$ and $AXLT$, so
\[XS = AK = AL = XT. \quad \blacksquare\]
[asy]
size(330);

pair A, B, C, J, K, L, F, G, S, T, X;
A = dir(120);
B = dir(210);
C = dir(330);
J = extension(A, incenter(A, B, C), B, rotate(90, B)*incenter(A, B, C));
K = foot(J, A, B);
L = foot(J, A, C);
X = foot(J, B, C);
F = extension(B, J, X, L);
G = extension(C, J, X, K);
S = extension(A, F, B, C);
T = extension(A, G, B, C);

draw(A--K--J--L--A--J--X^^K--G--J--F--L^^A--S--T--cycle);
draw(circumcircle(A, K, L), dashed);
draw(circumcircle(X, K, L), dotted);
draw(A--K^^S--X, red+linewidth(1.5));
draw(A--L^^T--X, blue+linewidth(1.5));

dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$K$", K, W);
dot("$L$", L, E);
dot("$J$", J, SE);
dot("$X$", X, N);
dot("$F$", F, NW);
dot("$G$", G, NE);
dot("$S$", S, W);
dot("$T$", T, E);
[/asy]
This post has been edited 1 time. Last edited by shendrew7, Oct 22, 2024, 1:35 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bonime
36 posts
#148 • 1 Y
Y by Jorge_144p
Claim 1: $AF\bot BF$ and $AG\bot CG$
Prove: Let $P=AC\cap KM$ and $Q=AB\cap LM$. Clearly, $BJ$ is a perpendicular bissector of $KM$ e $CJ$ of $LM$, so $\angle KFB=\angle BFQ$ and $\angle LGC=\angle CGP$. Also, by ceva and then menelaus with the points $K$, $M$ and $L$ and $Q$, $M$ and $L$ likewise $P$, $M$ and $P$, we get that $(A, B; K, Q)=(A, C; P, L)=-1$, so we´re done $\blacksquare$

Claim 2: $FG//BC$
It´s easy to see that $GK\bot FJ$ and $FL\bot GJ$, therefore $M$ is the orthocenter of $\triangle FJG$ so $JM\bot FG$, but $JM\bot BC$. $\blacksquare$.

Hence, from the Claim 1, we get that $AGMF$ is a paralelogram. We want to prove that $(S,T;M, P_{\infty, BC})=-1$ but $(S, T; M,  P_{\infty, BC})\stackrel{A}{=}(F, G; AM\cap FG,  P_{\infty, FG})=-1$ so we´re done. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lnzhonglp
120 posts
#149
Y by
Note that $$\angle JFL = \angle JAL = \angle KAJ = \angle KGJ,$$so $AKJG$ and $AFJL$ are cyclic. Then $$\angle AGM = \angle AJK = \angle AJL = \angle AFL = 90^\circ - \frac{\angle A}{2}$$and $$\angle FAG = \angle FAJ + \angle GAJ = \angle FLJ + \angle GKJ = \angle FMG = 90^\circ + \frac{\angle{A}}{2},$$so $AFMG$ is a parallelogram. We also have $AS \perp FJ$ and $AT \perp JG$, so since $\angle ABF = \angle FBS$ and $\angle ACG = \angle GCT$, we get $SF = AF = MG$ and $FM = AG = GT$. Then $$\triangle SFM \cong \triangle MGT,$$so $SM = MT$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cj13609517288
1900 posts
#150
Y by
This is really a problem about triangle $MKL$. We claim that $BA=BS$ and $CA=CT$, which will clearly finish.

Note that if we show that $\triangle BKM\sim\triangle BSA$, we will win. Thus it suffices to show that $KM\parallel AF$, or $\angle AFJ=90^{\circ}$, or $F\in(AKJL)$.

Now let's invert around the excircle. Then $F'\in (JLM)$, so
\[\angle JF'L=\angle JML=90^{\circ}-\angle K,\]so since $F'$ lies on $BJ$, it is exactly $BJ\cap KL$, 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.
Maximilian113
575 posts
#152
Y by
Notice that $BJ$ is the perpendicular bisector of $MK$ so $FK=FM.$ Therefore $\angle MFK = 2 \angle JFM = 2(90^\circ - \angle BJC) = \angle A.$ Thus $KFAL$ is cyclic. Similarly $LGAK$ is cyclic and it follows that pentagon $FAGLK$ is cyclic.

Thus,$$\angle FAG = 180^\circ - \angle FLG = 90^\circ +\frac12 \angle JGL = 90^\circ + \frac{\angle A}{2}.$$Also, $\angle AGK = \angle ALK = 90^\circ - \frac{\angle A}{2},$ so $\angle FAG + \angle AGK = 180^\circ \implies AS \parallel KG.$ But $BK = BM$ so by homothety $BS=BA.$ Therefore $MS = AK = s$ where $s$ denotes the semiperimeter.

Similarly $MT=s$ too, and we are done. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KevinYang2.71
421 posts
#153
Y by
We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$. Then
\begin{align*}
J&=(-a:b:c)\\
M&=(0:s-b:s-c)\\
K&=(c-s:s:0)\\
L&=(b-s:0:s).
\end{align*}$F$ is given by $(-a:t:c)$ for some $t$ where
\[
\begin{vmatrix}
-a & t & c\\
0 & s-b & s-c\\
b-s & 0 & s
\end{vmatrix}=0.
\]It follows that $-as(s-b)+t(b-s)(s-c)-c(b-s)(s-b)=0$ so $t=\frac{cs-bc-as}{s-c}=-a-c$. Hence
\[
S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)
\]and similarly
\[
T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).
\]Thus $\frac{1}{2}(S+T)=\left(0,\,\frac{s-b}{a},\,\frac{s-c}{a}\right)=M$, as desired. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
imagien_bad
57 posts
#154
Y by
KevinYang2.71 wrote:
We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$. Then
\begin{align*}
J&=(-a:b:c)\\
M&=(0:s-b:s-c)\\
K&=(c-s:s:0)\\
L&=(b-s:0:s).
\end{align*}$F$ is given by $(-a:t:c)$ for some $t$ where
\[
\begin{vmatrix}
-a & t & c\\
0 & s-b & s-c\\
b-s & 0 & s
\end{vmatrix}=0.
\]It follows that $-as(s-b)+t(b-s)(s-c)-c(b-s)(s-b)=0$ so $t=\frac{cs-bc-as}{s-c}=-a-c$. Hence
\[
S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)
\]and similarly
\[
T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).
\]Thus $\frac{1}{2}(S+T)=\left(0,\,\frac{s-b}{a},\,\frac{s-c}{a}\right)=M$, as desired. $\square$

bh bary is banned
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sadigly
158 posts
#155
Y by
Honestly it is easy with syntethic but I'm trying to improve my complex bash


Let $(KLM)\in\mathbb{S}^1\hspace{5mm}K=k\hspace{2mm}L=l\hspace{2mm}M=m\hspace{2mm}J=0$

$A=\frac{2kl}{k+l}\hspace{2mm}B=\frac{2lm}{l+m}\hspace{2mm}C=\frac{2mk}{m+k}$

$F=LM\cap BJ=\frac{k(m+l)}{k+l}\hspace{5mm}G=KM\cap CJ=\frac{l(m+k)}{k+l}$

$S=AF\cap BC=\frac{2km}{k+l}\hspace{5mm}T=AG\cap BC=\frac{2lm}{k+l}$

$\frac{\frac{2km}{k+l}+\frac{2lm}{k+l}}{2}=m$, so midpoint of $ST$ is $M$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IndexLibrorumProhibitorum
8 posts
#156
Y by
Claim I. $ A , F , J , L $ are concyclic.
This is because $\angle AJF = \pi-\angle BAI-\angle ABI-\angle IBJ=\pi-\frac{A}{2}-\frac{B}{2}-\frac{\pi}{2}=\frac{C}{2}=\angle FLA$
$\implies  A , F , J , L $ are concyclic.$\quad \square$

Claim II. $ FA \parallel BI \parallel KM$.
At first, we have $\measuredangle JBI =\measuredangle (JB,KM)=\frac{\pi}{2} \implies BI \parallel KM. $
Then by Claim I, we have $\angle JLM = \angle JFB= \frac{\pi}{2}=\angle JBI \implies BI \parallel FA.\quad \square$

Claim III. Quadrilateral $FGMS$ is a parallelogram
By Claim II, we have $-1=( K ,  M ; BJ \cap KM , \infty )= ( A ,  S ; F , \infty ) \implies AF=SF$.
The same is true of $AG=GT$, so $FG \parallel ST$.
Hence, Quadrilateral $FGMS$ is a parallelogram. $\quad \square$

Thus now it's clear to know $ FG = SM $ by Claim III. The same is true of $ FG = TM $ , therefore, $SM=TM$.

Q.E.D.

[asy]
import olympiad;
import cse5;
size(11cm);

pair A = dir(75);
pair B = dir(210);
pair C = dir(-30);
pair X= dir((B+C)/2);
pair I = incenter(A,B,C);
pair J = (X-I)*2+I;
pair L= foot(J,A,C);
pair K= foot(J,A,B);
pair M= foot(J,B,C);
pair F =IP(L(L,M,5,5),L(J,B,5,5));
pair G =IP(L(K,M,5,5),L(J,C,5,5));
pair S =IP(L(A,F,5,5),L(B,C,5,5));
pair T =IP(L(A,G,5,5),L(B,C,5,5));


draw(A -- C--B--cycle);
draw(unitcircle);
draw(circumcircle(K,L,M),blue);
draw(circumcircle(A,J,F),red + dotted);
draw(L(A,C,0,1.5));
draw(L(A,B,0,1.5));
draw(L--F);
draw(J--F);
draw(J--G);
draw(K--G);
draw(S--T);
draw(A--T);
draw(S--A);


dot("$A$", A, dir(A));
dot("$B$", B,dir(145)*2);
dot("$C$", C, dir(33)*2);
dot("$I$",I,dir(45));
dot("$J$",J,dir(270));
dot("$K$", K, dir(180));
dot("$L$",L,dir(45));
dot("$M$",M,N);
dot("$F$",F,dir(135));
dot("$G$", G, dir(45));
dot("$S$",S,dir(180));
dot("$T$",T,E);
[/asy]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YaoAOPS
1537 posts
#157
Y by
We claim that $JG \perp AG$. To prove this, note that if $K'$ is the reflection of $K$ over $JG$, then by Brokard's, $G' \coloneqq KL \cap MK' \cap GJ$ is the polar of $AG$ since $A$ is the polar of $KL$. Similarly, $AF \perp FM$. Since $AL$ is the reflection of $BC$ over $GJ$, $G$ is the midpoint of $AT$, and similarly $F$ is the midpoint of $AS$. To show that $M$ is the midpoint of $ST$, it thus remains to show that $AFMG$ is a parallelogram which follows by the perpendicularities.
Z K Y
N Quick Reply
G
H
=
a