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 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
Polynomial produces perfect powers
TheUltimate123   21
N 3 minutes ago by pi271828
Source: ELMO 2023/1
Let \(m\) be a positive integer. Find, in terms of \(m\), all polynomials \(P(x)\) with integer coefficients such that for every integer \(n\), there exists an integer \(k\) such that \(P(k)=n^m\).

Proposed by Raymond Feng
21 replies
TheUltimate123
Jun 26, 2023
pi271828
3 minutes ago
geometry
srnjbr   0
5 minutes ago
the points f,n,o, t a lie in the plane such that the triangles tfo ton are similar, preserving direction and order, and fano is a parallelogram. show that of×on=oa×ot.
0 replies
srnjbr
5 minutes ago
0 replies
Problem 5
blug   0
22 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Each square on a 5×5 board contains an arrow pointing up, down, left, or right. Show that it is possible to remove exactly 20 arrows from this board so that no two of the remaining five arrows point to the same square.
0 replies
blug
22 minutes ago
0 replies
Diophantine equation with large moduli
Assassino9931   2
N 23 minutes ago by Assassino9931
Source: Bulgaria, Concours Generale Minko Balkanski 2024
Solve in positive integers $2^x - 23^y = 9$.
2 replies
Assassino9931
4 hours ago
Assassino9931
23 minutes ago
Problem 4
blug   0
23 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
In a rhombus $ABCD$, angle $\angle ABC=100^{\circ}$. Point $P$ lies on $CD$ such that $\angle PBC=20^{\circ}$. Line parallel to $AD$ passing trough $P$ intersects $AC$ at $Q$. Prove that $BP=AQ$.
0 replies
blug
23 minutes ago
0 replies
Problem 3
blug   0
26 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Find all primes $(p, q, r)$ such that
$$pq+4=r^4.$$
0 replies
blug
26 minutes ago
0 replies
Problem 2
blug   0
27 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
A party is attended by boys and girls. Each person attending the party knows exactly 3 boys and exactly 7 girls among the other people. Prove that the number of all the people attending the party is divisible by 20.
0 replies
blug
27 minutes ago
0 replies
Problem 1
blug   0
28 minutes ago
Source: Polish Junior Math Olympiad Finals 2025
Do there exists a tetrahedron, in which the lenghts of the edges are six different integers such that their sum is 25?
0 replies
blug
28 minutes ago
0 replies
A two-variable & non-homogenous inequality that seems hard to me
MyLifeMyChoice   3
N 36 minutes ago by Radin_
Source: Developing from a larger, three-variable one
For $a,b>0$, prove/disprove the following claim: :maybe:

$a^3b^3+\frac{1}{a^3}+\frac{1}{b^3}+3\stackrel{?}{\ge}a^2b+b^2a+\frac{1}{a^2b}+\frac{1}{b^2a}+\frac{a}{b}+\frac{b}{a}$
3 replies
MyLifeMyChoice
Mar 13, 2025
Radin_
36 minutes ago
exponential diophantine with factorials
skellyrah   4
N an hour ago by InftyByond
find all non negative integers (x,y) such that $$ x! + y! = 2025^x + xy$$
4 replies
skellyrah
Feb 24, 2025
InftyByond
an hour ago
Point satisfies triple property
62861   35
N an hour ago by Sanjana42
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
35 replies
62861
Jan 22, 2018
Sanjana42
an hour ago
Prove concyclic and tangency
syk0526   40
N an hour ago by Ilikeminecraft
Source: Japan Olympiad Finals 2014, #4
Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to $AB$ passing $E$ meets $l$ at $I$. Prove that there exists a circle passing four points $ F, G, H, I $ and tangent to line $ BC$.
40 replies
syk0526
May 17, 2014
Ilikeminecraft
an hour ago
p^2+3*p*q+q^2
mathbetter   0
2 hours ago
\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]
0 replies
+1 w
mathbetter
2 hours ago
0 replies
two sequences of positive integers and inequalities
rmtf1111   49
N 2 hours ago by dolphinday
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
49 replies
rmtf1111
Apr 10, 2019
dolphinday
2 hours ago
Concurrency in Parallelogram
amuthup   87
N Today at 12:40 AM by joshualiu315
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.
87 replies
amuthup
Jul 12, 2022
joshualiu315
Today at 12:40 AM
Concurrency in Parallelogram
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 ISL G1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amuthup
779 posts
#1 • 8 Y
Y by itslumi, YesToDay, Ahmed102, GeNuAlCo_pi, Mango247, deplasmanyollari, ItsBesi, ehuseyinyigit
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.
This post has been edited 1 time. Last edited by amuthup, Jul 15, 2022, 4:57 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Assassino9931
1193 posts
#3 • 5 Y
Y by PRMOisTheHardestExam, GeNuAlCo_pi, Philomath_314, Stuffybear, lomta
It turns out that only angle chasing is enough (no similarities, no radical axis etc., but only cyclic quads), but I'll let somebody else post that.

Let $AQ \cap CD = T$, with the aim to show that $B$, $R$ and $T$ are collinear. Observe that $\angle QRC = \angle QAP = \angle ATC = \angle QTC$ and so $CTRQ$ is cyclic, implying $\angle CRT = \angle CQT = \angle ADC = \angle ABC$. So if we show that $\angle BRC = \angle PBC$ we shall be done. The latter is equivalent to $\triangle PBC \sim \triangle BRC$ and hence to $BC^2 = CR \cdot CP$. But since $AC = BC$, we reduce to $AC^2 = CR \cdot CP$ and hence to $\triangle ACR \sim \triangle PCA$. Now it remains to observe that $\angle ACR = \angle ACP$ and $\angle CAR = \angle CAQ + \angle QAR = \angle CDQ + \angle QPR = \angle APD + \angle DPC = \angle APC$ 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.
v_Enhance
6857 posts
#5 • 6 Y
Y by HamstPan38825, PHSH, I.owais, shujingsheng, GeNuAlCo_pi, Rounak_iitr
Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$. We wish to show $APR'Q$ is cyclic.
Let $M$ be the center of the parallelogram.
Claim: $M$, $Q$, $R'$ are collinear.
Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$. $\blacksquare$

[asy] 	size(12cm); 	pair A = dir(90); 	pair D = dir(240); 	pair C = A*A/D; 	pair Q = dir(20); 	pair B = A+C-D; 	pair P = extension(D, Q, A, B); 	pair M = midpoint(A--C); 	pair X = extension(A, Q, C, D); 	pair Rp = extension(P, C, B, X); 	filldraw(unitcircle, invisible, blue); 	filldraw(A--B--C--D--cycle, invisible, red); 	draw(A--C, red); 	draw(B--D, red); 	draw(C--X, red); 	draw(B--P, red); 	draw(A--X, lightblue); 	draw(B--X, lightblue); 	draw(C--P, lightblue); 	draw(M--Rp, deepgreen); 	draw(D--P, lightblue); 	draw(Q--C, lightblue); 	pair Z = 2*A-P; 	pair W = -D+2*foot(origin, D, Z); 	draw(Q--W, deepgreen); 	draw(D--Z--A, orange);
dot("$A$", A, dir(A)); 	dot("$D$", D, dir(D)); 	dot("$C$", C, dir(C)); 	dot("$Q$", Q, dir(72)); 	dot("$B$", B, dir(B)); 	dot("$P$", P, dir(P)); 	dot("$M$", M, dir(190)); 	dot("$X$", X, dir(X)); 	dot("$R'$", Rp, dir(Rp)); 	dot("$Z$", Z, dir(Z)); 	dot("$W$", W, dir(W));
/* TSQ Source:
!size(12cm); 	A = dir 90 	D = dir 240 	C = A*A/D 	Q = dir 20 R72 	B = A+C-D 	P = extension D Q A B 	M = midpoint A--C R190 	X = extension A Q C D 	R' = extension P C B X 	unitcircle 0.1 lightcyan / blue 	A--B--C--D--cycle 0.1 lightred / red 	A--C red 	B--D red 	C--X red 	B--P red 	A--X lightblue 	B--X lightblue 	C--P lightblue 	M--Rp deepgreen 	D--P lightblue 	Q--C lightblue 	Z = 2*A-P 	W = -D+2*foot origin D Z 	Q--W deepgreen 	D--Z--A orange
*/ [/asy]
Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$.
Claim: $\triangle ZAD \cong \triangle PAC$.
Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$. Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$, we're done. $\blacksquare$
To finish, \begin{align*} 	\measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ 	&= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*}
Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$; then the points $B$, $R'$, and $X$ may now be deleted.
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
#6 • 2 Y
Y by GeNuAlCo_pi, ehuseyinyigit
Angle chase sol sketch: Let $AQ$ and $BR$ meet at $X$, I will prove that $CX||AB$. This is equivalent to $CQRX$ being cyclic. Note that $\angle ARC= \angle AQD= \angle ACD= \angle ABC$, so $ABRC$ is cyclic. Hence $\angle CRX= \angle BAC = \angle ADC = \angle CQX$ so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Kamran011
678 posts
#7 • 2 Y
Y by PRMOisTheHardestExam, FriIzi
Let $T = AC\cap BD$. Note that $\angle QPA = \angle QDC = \angle QAC\to AC$ touches $(AQRP)$
As well as $$\angle ACQ = \angle ADQ = \angle ADC - \angle CDQ = \angle CBA - \angle CAQ = \angle CAB - \angle CAQ = \angle QAP = \angle CKQ$$
$\to AC$ touches $(CQRK).\hspace{10 cm}$
Now, we have $T\in QR$ since $QR$ bisects $AC$ and $R\in BK$ by Pappus' Theorem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DottedCaculator
7303 posts
#8 • 4 Y
Y by PROA200, s18_kanimet_bursumankulov, rafid149, Frank25
[asy]
unitsize(1cm);
pair D, C, A, B, Q, P, R, X;
D=(0,0);
A=(2,5);
C=(4,0);
B=A+C-B;
P=(10,5);
Q=intersectionpoints(D--P,circumcircle(A,D,C))[1];
R=intersectionpoints(C--P,circumcircle(A,Q,P))[0];
X=extension(A,Q,B,R);
draw(circumcircle(A,C,D));
draw(circumcircle(A,Q,P));
draw(A--D--X--C--A--P--C--Q--R--B--X--D--P--A--X);
label("$D$", D, SW);
label("$C$", C, SE);
label("$A$", A, NE);
label("$B$", B, N);
label("$Q$", Q, NNE);
label("$P$", P, SE);
label("$R$", R, ESE);
label("$X$", X, SE);
[/asy]

We claim that $ABRC$ is cyclic. We will first show that $AC$ is tangent to the circumcircle of $APQ$. We have
\begin{align*}
\angle QAC&=\angle QDC\\
&=\angle QPA,
\end{align*}so $AC$ is tangent to the circumcircle of $APQ$. Now, we have
\begin{align*}
\angle ABC&=\angle CAB\\
&=180^{\circ}-\angle PRA\\
&=\angle ARC.
\end{align*}This means that $ABRC$ is cyclic. If $AQ$ and $CD$ intersect at $X$, then we have
\begin{align*}
\angle RCX&=\angle RPA\\
&=180^{\circ}-\angle AQR\\
&=\angle RQX.
\end{align*}Therefore, $RQCX$ is also cyclic. Since $AD=BC=AC$, we get $\angle CDA=\angle ACD$, so we get
\begin{align*}
\angle CRX&=\angle CQX\\
&=180^{\circ}-\angle AQC\\
&=\angle CDA\\
&=\angle ACD\\
&=\angle CAB\\
&=180^{\circ}-\angle BRC.
\end{align*}Therefore, $B$, $R$, and $X$ are collinear, so $AQ$, $BR$, and $CD$ concur at $X$.
This post has been edited 1 time. Last edited by DottedCaculator, Jul 12, 2022, 1:13 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
isaacmeng
113 posts
#9 • 1 Y
Y by Ahmed102
ISL 2021 G1. Let $ABCD$ be a parallelogram such that $AC=BC$. A point $P$ is chosen on the extension of segment $AB$ beyond $B$. Define $Q=(ACD)\cap PD$ and $R=(APQ)\cap PC$. Prove that $CD$, $AQ$, and $BR$ are concurrent.

Solution. Define $E=AQ\cap CD$. Use directed angles modulo $360^\circ$.

Claim 1. $C$, $Q$, $R$, $E$ are concyclic.

Proof. We have \begin{align*}\measuredangle CEQ&=180^\circ-\measuredangle EAD-\measuredangle ADE\\&=180^\circ-(\measuredangle QAC+\measuredangle CAD)-(\measuredangle ADQ+\measuredangle QDC)\\&=180^\circ-\measuredangle QAC-\measuredangle QDC-\measuredangle CAD-\measuredangle BAQ\\&=180^\circ-2\measuredangle QAC-(180^\circ-\measuredangle ADC-\measuredangle DCA)-\measuredangle BAQ\\&=\measuredangle ADC+\measuredangle DCA-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ+2\measuredangle QDC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ-\measuredangle BAQ\\&=\measuredangle BAQ\\&=\measuredangle CRQ.\end{align*}
Claim 2. $B$, $R$, $E$ are collinear.

Proof. We have \begin{align*}\measuredangle QRB&=180^\circ-\measuredangle CRQ-\measuredangle ERC\\&=180^\circ-\measuredangle PAQ-\measuredangle EQC\\&=180^\circ-\measuredangle PAQ-\measuredangle ADC\\&=180^\circ-\measuredangle ACQ-\measuredangle DCA\\&=\measuredangle QCE\\&=180^\circ-\measuredangle ERQ.\end{align*}
Now, Claim 2 implies the result.

[asy]
        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 = -4.195830202854993, xmax = 18.824455296769358, ymin = -6.88018031555221, ymax = 5.561592787377907;  /* image dimensions */
        
         /* draw figures */
        draw((4.37,-1.48)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); 
        draw((5.65,1.88)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); 
        draw((9.429564356435643,-1.4836435643564359)--(5.65,1.88), linewidth(0.8)); 
        draw((5.65,1.88)--(3.733168316831683,-3.151683168316832), linewidth(0.8)); 
        draw((3.733168316831683,-3.151683168316832)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); 
        draw(circle((8.179089667867803,0.9165372693836932), 2.7063915055664625), linewidth(0.8)); 
        draw(circle((3.7352424196126757,-0.2715209217572098), 2.8801629933754995), linewidth(0.8)); 
        draw((3.733168316831683,-3.151683168316832)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); 
        draw((5.65,1.88)--(7.737054951313305,-5.926480752802572), linewidth(0.8)); 
        draw((7.737054951313305,-5.926480752802572)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); 
        draw((7.737054951313305,-5.926480752802572)--(4.37,-1.48), linewidth(0.8) + linetype("2 2")); 
         /* dots and labels */
        dot((5.65,1.88),dotstyle); 
        label("$A$", (5.706498873027804,2.0304132231404948), NE * labelscalefactor); 
        dot((4.37,-1.48),dotstyle); 
        label("$B$", (4.429263711495121,-1.335477084898571), NE * labelscalefactor); 
        dot((9.429564356435643,-1.4836435643564359),dotstyle); 
        label("$C$", (9.493125469571757,-1.335477084898571), NE * labelscalefactor); 
        dot((10.709564356435642,1.8763564356435638),linewidth(4pt) + dotstyle); 
        label("$D$", (10.77036063110444,2.0003606311044315), NE * labelscalefactor); 
        dot((3.733168316831683,-3.151683168316832),dotstyle); 
        label("$P$", (3.798159278737796,-3.0033959429000725), NE * labelscalefactor); 
        dot((6.468215171114042,-1.1804757104902723),linewidth(4pt) + dotstyle); 
        label("$Q$", (6.53294515401954,-1.0650037565740034), NE * labelscalefactor); 
        dot((5.290538122451863,-2.6956484235895024),linewidth(4pt) + dotstyle); 
        label("$R$", (5.345867768595046,-2.582659654395189), NE * labelscalefactor); 
        dot((7.737054951313305,-5.926480752802572),linewidth(4pt) + dotstyle); 
        label("$E$", (7.79515401953419,-5.813313298271971), NE * labelscalefactor); 
        clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
This post has been edited 1 time. Last edited by isaacmeng, Jul 13, 2022, 9:15 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hopeooooo
819 posts
#10 • 4 Y
Y by Mango247, Upwgs_2008, H_Taken, Math_.only.
The circumcircle $DCR$ is $w$ and $BR$ Intersects $w$ at $F$.
Claim:$D,A,F$ are collinear.
$\angle{PAF}=\angle{PRF}=\angle{CDF}$ and $CD||AB$ implies the result.
Radical axis theorem on $(CDFR),(AQRP),(AQCD)$
This post has been edited 4 times. Last edited by Hopeooooo, Jul 12, 2022, 1:38 PM
Reason: 1
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
#11 • 1 Y
Y by PHSH
Cute
$$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC$$meaning $ABRC$ is cyclic. Let $S= \overline{AD} \cap \overline{BR}$.
$$\measuredangle SAP = \measuredangle ABC  = \measuredangle CAB = \measuredangle CRB = \measuredangle SRP$$meaning $SAQRP$ is cyclic.
$$\measuredangle SRC = \measuredangle BRC = \measuredangle BAC = \measuredangle ADC = \measuredangle SDC$$meaning $SRCD$ is cyclic. Finish by radical axis theorem on $(SAQR), (AQCD) , (SRCD)$.
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
#12
Y by
Let $S=AD\cap BR.$ Note that $\measuredangle ABC=\measuredangle ACD=\measuredangle AQD=\measuredangle ARC\implies R\in \odot (ABC).$ Next $\measuredangle SDP=$ $=\measuredangle ADC=$ $\measuredangle SRP\implies S\in \odot (APQ)\cap \odot (CDR).$Radical axis on $\odot (APQRS),\odot (ACDQ),\odot (CDRS)$ finish.
This post has been edited 2 times. Last edited by JAnatolGT_00, Jul 12, 2022, 2:02 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ppanther
160 posts
#13
Y by
Angles are directed. We have
\[
\angle ARC = \angle ARP = \angle AQP = \angle AQD = \angle ACD = \angle ABC,
\]so $BRCA$ is cyclic. Let ray $\overline{RB}$ intersect $(APQ)$ at $E$. We claim that $E$ lies on $\overline{AB}$. Indeed,
\[
\angle EAP = \angle ERP = \angle BRC = \angle BAC = \angle DAB.
\]Finally, as
\[
\angle ERC = \angle BRC = \angle BAC = \angle EDC,
\]quadrilateral $ERCD$ is cyclic. Considering the radical center of $(ERQA)$, $(AQCD)$, and $(ERCD)$ establishes the concurrency.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lneis1
243 posts
#14 • 1 Y
Y by Mango247
Let $\angle ABC=x$
Note that, $AB$ is tangent to $(ACD)$ since $\angle BAC=\angle ADC$.
Similarly $AC$ is tangent to $(AQRP)$ since $\angle PQA=180-\angle AQD=180-\angle ACD=180-\angle CAP=180-x$

Note that $\angle ARP=180- \angle ARP=180- \angle AQP=x$, hence $R \in (ABC)$.

Let $AQ \cap CD=E$
Then since $AB || CD \implies \angle RCE=\angle RPA=\angle RQE$ where the last equality holds because $Q \in (ARP)$
Hence $E \in (RCQ)$

Note that $\angle ERC=x$, since $\angle ERC=\angle CQE=180-\angle AQC=180-(180-x)$
But since $R \in (ABC)$ we have $\angle BRC=180-x$ Hence we get $B-R-E$
This post has been edited 2 times. Last edited by lneis1, Jul 12, 2022, 2:58 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BVKRB-
322 posts
#15 • 1 Y
Y by adorefunctionalequation
Yay I can solve something now :D
Let $RB \cap \odot(APQ) = X$
$$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC \implies \odot(ABRC)$$This implies that $D-A-X$, Now $$\measuredangle DXR = \measuredangle AXR = \measuredangle APR = \measuredangle DCR \implies \odot(CDXR)$$Now radical centre on $\odot(AQDC),\odot(AQRP),\odot(CDXR)$ finishes the problem :D $\blacksquare$
v_Enhance wrote:
Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$. We wish to show $APR'Q$ is cyclic.
Let $M$ be the center of the parallelogram.
Claim: $M$, $Q$, $R'$ are collinear.
Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$. $\blacksquare$

[asy] 	size(12cm); 	pair A = dir(90); 	pair D = dir(240); 	pair C = A*A/D; 	pair Q = dir(20); 	pair B = A+C-D; 	pair P = extension(D, Q, A, B); 	pair M = midpoint(A--C); 	pair X = extension(A, Q, C, D); 	pair Rp = extension(P, C, B, X); 	filldraw(unitcircle, invisible, blue); 	filldraw(A--B--C--D--cycle, invisible, red); 	draw(A--C, red); 	draw(B--D, red); 	draw(C--X, red); 	draw(B--P, red); 	draw(A--X, lightblue); 	draw(B--X, lightblue); 	draw(C--P, lightblue); 	draw(M--Rp, deepgreen); 	draw(D--P, lightblue); 	draw(Q--C, lightblue); 	pair Z = 2*A-P; 	pair W = -D+2*foot(origin, D, Z); 	draw(Q--W, deepgreen); 	draw(D--Z--A, orange);
dot("$A$", A, dir(A)); 	dot("$D$", D, dir(D)); 	dot("$C$", C, dir(C)); 	dot("$Q$", Q, dir(72)); 	dot("$B$", B, dir(B)); 	dot("$P$", P, dir(P)); 	dot("$M$", M, dir(190)); 	dot("$X$", X, dir(X)); 	dot("$R'$", Rp, dir(Rp)); 	dot("$Z$", Z, dir(Z)); 	dot("$W$", W, dir(W));
/* TSQ Source:
!size(12cm); 	A = dir 90 	D = dir 240 	C = A*A/D 	Q = dir 20 R72 	B = A+C-D 	P = extension D Q A B 	M = midpoint A--C R190 	X = extension A Q C D 	R' = extension P C B X 	unitcircle 0.1 lightcyan / blue 	A--B--C--D--cycle 0.1 lightred / red 	A--C red 	B--D red 	C--X red 	B--P red 	A--X lightblue 	B--X lightblue 	C--P lightblue 	M--Rp deepgreen 	D--P lightblue 	Q--C lightblue 	Z = 2*A-P 	W = -D+2*foot origin D Z 	Q--W deepgreen 	D--Z--A orange
*/ [/asy]
Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$.
Claim: $\triangle ZAD \cong \triangle PAC$.
Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$. Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$, we're done. $\blacksquare$
To finish, \begin{align*} 	\measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ 	&= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*}
Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$; then the points $B$, $R'$, and $X$ may now be deleted.

It seems that vEnhance has forgotten simple thinking :rotfl:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Leartia
93 posts
#16
Y by
Let $E\neq R$ be such that $E=BR\cap (APQ)$ and let $F=BC\cap PE.$
By parallelism we have $AD=AC=BC.$
Since $\angle BAC=\angle ADC$ and $\angle PAC=180^{\circ}-\angle AQP=\angle AQD=\angle ACD.$ we have that $AP$ is tangent with $(ADC)$ and $AC$ is tangent with $(APQ).$
Let $\angle BAC=\alpha.$
We have $\angle CDA=\angle CBA=\angle CAB=\angle ACD$, $\angle CAB=\angle CAP=\angle AEP,$ and $\angle CBA=\angle FBP.$
Since $\angle ABF=180^{\circ}- \alpha$ we have that $A,B,F,E$ are concyclic.
Therefore, $\angle FAP=\angle FAB=\angle FEB=\angle PER=\angle PAR.$
We also have $\angle CPA=\angle CAR=\angle AER=\angle AFC$ hence $A,C,P,F$ are concyclic.
So we get $\angle BAE=\angle CFP=\angle CAP=\alpha$ so $P, A,$ and $E$ are collinear.
Since $\angle PRE=\angle PAE=\alpha$ we get that $\angle EDC+\angle CRE=180^{\circ}$ therefore $D,C,R,$ and $E$ are concyclic.
Applying the radical axis theorem on $(AQCD), (AQRE)$ and $(DCRE)$ we get that $DC, AQ,$ and $RE$ concur. \qed
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mahmood.sy
26 posts
#17 • 1 Y
Y by Muaaz.SY
let $L$ is the intersection of $(CD)$ & $(BR)$
claim_1:$ACRB$ is cyclic
proof :$\angle AQD=\angle ACD=\angle ADC=\angle ABC$
$\rightarrow \angle AQP=180^o -\angle ABC$
$angle ARP=\angle AQP$ $\rightarrow \angle ARC=\angle ABC$ so $ACRB$ is cyclic
also $\angle LCB=\angle ABC=\angle CAB$ so $LC $ is tangent to the circle of $ACRB$
$\rightarrow LC^2=LR.LB$
let $T$ is the intersection of $BR$ and the circum circle of traingle $APQ$
claim_2: traingles $ABT$ & $LRC$ are similiarity
proof: $\angle CLR=\angle BTA$ by paralle
$\angle LRC=\angle PRT=\angle BAT$ so we proof our claim , this mean that $LR.BT=LC.AB=LC.CD$
$\rightarrow LC^2+LC.CD=LR.LB+LR.BT \rightarrow LC.LD=LR.LT$
this mean that $L\in(AQ)$, so we done
This post has been edited 2 times. Last edited by Mahmood.sy, Jul 12, 2022, 3:16 PM
Z K Y
G
H
=
a