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
Unsymmetric FE
Lahmacuncu   0
31 minutes ago
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfies $f(x^2+xy+y)+f(x^2y)+f(xy^2)=2f(xy)+f(x)+f(y)$ for all real $(x,y)$
0 replies
Lahmacuncu
31 minutes ago
0 replies
Divisibility on 101 integers
BR1F1SZ   5
N 38 minutes ago by Grasshopper-
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
5 replies
BR1F1SZ
Aug 9, 2024
Grasshopper-
38 minutes ago
Nice number theory problem
ItsBesi   9
N 40 minutes ago by Jupiterballs
Source:  Kosovo Math Olympiad 2025, Grade 8, Problem 3
Let $m$ and $n$ be natural numbers such that $m^3-n^3$ is a prime number. What is the remainder of the number $m^3-n^3$ when divided by $6$?
9 replies
ItsBesi
Nov 17, 2024
Jupiterballs
40 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N an hour ago by wh0nix
Let \(f:[0,+\infty)\to\mathbb{R}\) be a function which is differentiable on \([0,+\infty)\) and satisfies
\[
\lim_{x\to+\infty}\bigl(f'(x)/e^x\bigr)=0.
\]Prove that
\[
\lim_{x\to+\infty}\bigl(f(x)/e^x\bigr)0.
\]
1 reply
ZeltaQN2008
2 hours ago
wh0nix
an hour ago
Find the value
sqing   8
N an hour ago by xytunghoanh
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
8 replies
sqing
Mar 17, 2025
xytunghoanh
an hour ago
Simple inequality
sqing   22
N an hour ago by ND_
Source: JBMO 2011 Shortlist A3
$\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b}$$
22 replies
sqing
May 15, 2016
ND_
an hour ago
greatest volume
hzbrl   0
an hour ago
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
0 replies
hzbrl
an hour ago
0 replies
Serbia national Olympiad Day 2 Problem 2
IgorM   19
N 2 hours ago by IndexLibrorumProhibitorum
Source: Serbia national Olympiad Day 2 Problem 2
Let $x,y,z$ be nonnegative positive integers.
Prove $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{xz+2x+1}\ge 0$
19 replies
IgorM
Mar 28, 2015
IndexLibrorumProhibitorum
2 hours ago
find angle
TBazar   2
N 2 hours ago by sunken rock
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
2 replies
TBazar
4 hours ago
sunken rock
2 hours ago
Geometry marathon
HoRI_DA_GRe8   845
N 2 hours ago by leon.tyumen
Ok so there's been no geo marathon here for more than 2 years,so lets start one,rules remain same.
1st problem.
Let $PQRS$ be a cyclic quadrilateral with $\angle PSR=90°$ and let $H$ and $K$ be the feet of altitudes from $Q$ to the lines $PR$ and $PS$,.Prove $HK$ bisects $QS$.
P.s._eeezy ,try without ss line.
845 replies
HoRI_DA_GRe8
Sep 5, 2021
leon.tyumen
2 hours ago
polonomials
Ducksohappi   0
3 hours ago
$P\in \mathbb{R}[x] $ with even-degree
Prove that there is a non-negative integer k such that
$Q_k(x)=P(x)+P(x+1)+...+P(x+k)$
has no real root
0 replies
Ducksohappi
3 hours ago
0 replies
Four tangent lines concur on the circumcircle
v_Enhance   35
N 4 hours ago by bin_sherlo
Source: USA TSTST 2018 Problem 3
Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$.

Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$.

Evan Chen and Yannick Yao
35 replies
v_Enhance
Jun 26, 2018
bin_sherlo
4 hours ago
tangent and tangent again
ItzsleepyXD   0
4 hours ago
Source: holder send me this
Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of $\overarc{BC}$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX \parallel CY \parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN \parallel AM, \omega_b$ be a circle thaat passes through $B$, $X$ and tangents to $BC$ and $\omega_C$ be a circle that passes through $C,Y$ and tangents to $BC$. Prove that circle with center $K$ and radias $KP$ is tangent to circles $\omega_B,\omega_C$ and $\Gamma$.
0 replies
ItzsleepyXD
4 hours ago
0 replies
Geometry
Lukariman   10
N 4 hours ago by Captainscrubz
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
10 replies
Lukariman
Tuesday at 12:43 PM
Captainscrubz
4 hours ago
IMO Problem 4
iandrei   105
N Apr 13, 2025 by cj13609517288
Source: IMO ShortList 2003, geometry problem 1
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
105 replies
iandrei
Jul 14, 2003
cj13609517288
Apr 13, 2025
Source: IMO ShortList 2003, geometry problem 1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iandrei
138 posts
#1 • 13 Y
Y by Davi-8191, nguyendangkhoa17112003, TurtleKing123, HWenslawski, Adventure10, centslordm, megarnie, proxima1681, Mahmood.sy, Mango247, Rounak_iitr, and 2 other users
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sebadollahi
1 post
#2 • 4 Y
Y by Adventure10, centslordm, Mango247, and 1 other user
in two cyclic quadrilatrals APRD & DRCQ ,AD & CD are diameters respectively and we have:
RQ/sin(<RCQ)=CD/2 or RQ/sin(<BCA)=CD/2
PR/sin(<PAR)=AD/2 or PR/sin(<BAC)=AD/2
By dividing:
CD/AD= sin(<BAC)/sin(<BCA)=BC/AB
Since CD/AD=BC/AB thus bisectors of <CBA and <ADC intersect AC in the same poin.
S.Ebadollahi
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
galois
400 posts
#3 • 2 Y
Y by Adventure10, Mango247
i must admit that this problem was quite easy by imo standards.my solution is based on a standard trick using pedal triangles and projections which is damn easy and pretty similar to the proof posted here.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lagrangia
1326 posts
#4 • 5 Y
Y by Inconsistent, Adventure10, Mango247, and 2 other users
here is an interesting thing about this problem and other 2 IMO problems in the past! it's about the so called Pedal Triangle Trick!

this was posted on a forum by: fritue2000 and I thought it would be interesting to post it here also!

"The previous two IMO problems solved by the Pedal Triangle Trick.
(1996 IMO) Let P be a point inside the triangle ABC such that angle
APB - angle ACB = angle APC - angle ABC. Let D, E be the incenters of
triangles APB, APC respectively. Show that AP, BD, CE meet at a
point.

As I know, at the jury meeting at IMO 1996, there is a discussion
for this problem becasuse its solution is very similar to problem
2 of the 34th IMO. The techniques include 'inversion' and 'PTT', etc.
For example, http://home1.pacific.net.sg/~slwee/imo96/imo96op.htm )

(1993 IMO) Let D be a point inside the acute-angled triangle ABC such
that angle ADB = angle ACB + 90 degrees, and AC*BD = AD*BC.

(a) Calculate the ratio AB*CD/(AC*BD).

The well-known 'Pedal Triangle Trick' is "For any point D, let
X, Y, Z be feet of the altitudes from D to AB, BC and CA. Then,
YZ = (DA*BC)/2r, etc, where r is the circumradius of ABC."
The proof is very easy, since D, A, Y, Z lies on a circle with
diameter DA, by the law of sines, YZ = DA sin A = DA*(BC/2r).
(2003 IMO) Given is a cyclic quadrilateral ABCD and let P, Q, R
be feet of the altitudes from D to AB, BC and CA respectively.

Prove that if PR = RQ then the interior angle bisectors of the
angles <ABC and <ADC are concurrent on AC.
Solution) By PTT, PR=RQ implies (DA*BC)/2r = (DC/AB)/2r, so,
CD/DA=BC/AB implies the results. q.e.d.

The most noticeable thing is in the above solution, we did not
used the condtion 'ABCD is cyclic'. And as an IMO problem, it
is not so intersting because all three problems 1993, 1996,
2003 solved by exactly the same TWO ways (PTT and inversion).
Well, the PTT is really well-known, for example, it appears
at "Geometry Revisited" by Coexter. It also used for a proof
of Ptolemy's Theorem.

In a cyclic qudrilateral ABCD, the three points are on the
simson line, we have PR+RQ=RP, with the same notation of 2003
imo. Then, (DA*BC)/2r + (DC*AB)/2r = (DB*AC)/2r or
DA*BC + DC*AB = DB*AC. "
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anonymous
334 posts
#5 • 2 Y
Y by Adventure10, Mango247
A bit of standard angle chasing shows that triangles DPR and DBC are similar and so are DQR and DBA. Thus PR=QR gives DC/CB = DR/RP = DR/RQ = DA/AB, and the result follows from the angle bisector theorem.
Z K
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Peter
3615 posts
#6 • 4 Y
Y by Adventure10, Mango247, Dream6068., ehuseyinyigit
Yeah, I had something like that too, was quite easy :D solved in about 30 minutes (which is extremely few for me)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Philip_Leszczynski
327 posts
#7 • 3 Y
Y by Adventure10, Adventure10, Mango247
Lemma: Let $ABCD$ be a cyclic quadrilateral. Let $DR$ and $DP$ be the altitudes from $D$ to $AB$ and $DC$, respectively. Let $RP$ intersect $AC$ at $Q$. Then $\angle AQD$ is a right angle.

Proof of Lemma:

$\angle DRB = \angle DPB = \pi / 2$, so $DRPB$ is cyclic. Then $\angle DRP = \angle DBP$. From cyclic quadrilateral $ABCD$, $\angle DBP = \angle DAP$. $\angle DRP = \angle DAP$, so $RAQD$ is cyclic. Thus $\angle AQD = \pi - \angle ARD = \pi / 2$.

Proof:

Let the bisector of $\angle ABC$ meet $AC$ at S. Let the bisector of $\angle ADC$ meet $AC$ at T.
By the Lemma, $R,Q,P$ are collinear.
Let $\angle DRQ = \alpha$. $\angle APD = \pi - \alpha - \angle RDP = \pi - \alpha - (\pi - \angle ABC) = \angle ABC - \alpha$.
$\frac{RQ}{\sin \angle RDQ} = \frac{DQ}{\sin \alpha}$, $\frac{PQ}{\sin \angle PDQ} = \frac{DQ}{\sin (\angle ABC - \alpha)}$.
$\frac{RQ}{PQ} \cdot \frac{\sin PDQ}{\sin RDQ} = \frac{sin(\angle ABC - \alpha)}{\sin \alpha}$.
$\angle RDQ = \pi - \angle RAQ = \angle BAC$. Also $\angle PDQ = \angle BCA$.
$\frac{RQ}{PQ} \cdot \frac{\sin \angle BCA}{\sin \angle BAC} = \frac{\sin (\angle ABC - \alpha)}{\sin \alpha}$.
$\frac{RQ}{PQ} \cdot \frac{AB}{BC} = \frac{\sin \angle QCD}{\sin \angle QAD} = \frac{AD}{DC}$.
So $RQ=PQ$ if and only if $\frac{AB}{BC} = \frac{AD}{DC}$.

By the Angle Bisector Theorem,
$\frac{AS}{SC} = \frac{AB}{BC}$ and $\frac{AT}{TC} = \frac{AD}{DC}$.
So $\frac{AB}{BC} = \frac{AD}{DC}$ if and only if $\frac{AS}{SC} = \frac{AT}{TC}$.
This can happen if and only if $S=T$.
So $PQ=QR$ if and only if the angle bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$. QED.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#8 • 2 Y
Y by Adventure10, Mango247
Yay I exploded this problem with like 20 cyclic quadrilaterals.

Okay first well-know fact: Simson's Line, so we know P, Q, R are collinear.

i'll go with the only if direction
Let X be the intersection of the angle bisectors. Extend BX, DX to Y, Z, where Y and Z lies on the circumcircle of ABCD. Y, Z lies on the perpendicular bisector of AC, hence YZ is a diameter, and it goes through midpoint M of AC. angle AMY = 90, angle ZDY = 90 because YZ is a diameter, so MYDX is cyclic, so angle ZYB = MDX = ADB. DX bisects ADC hence ADM = BDC = BAC = PDQ.
angle QAD=QPD. Hence triangle PQD is similar to triangle AMD, and triangle ADC is simlar to PDR. DM is a median hence DQ is a median => Q is the midpoint of PR as desired.

If direction:
Define X to be on AC such that DX bisects ADC, DX intersect the circumcircle at Z, M is the midpoint of AC, Y is the intersection of MZ and the circumcircle. triangle ADC is simlar to PDR. DM is a median and DQ is a median, so triangle PQD is similar to triangle AMD, angle ADM = PDQ=BAC = BDC. By definition XD bisects ADC hence XD bisects MDB so MDX = XDQ. QMY=90, YZ is still diameter, MYDX is cyclic, ZYX = MDX = ZDB. Now let BX intersect the circumcircle at Y'. XY'Z = XYZ, now Y' is unique(consider the locus of points Y' such that XY'Z = XYZ, which is a circle, and Y' lies on the cirucmcircle of ABCD. Two circle intersects in two points, one of which is Z, hence there's only one such point left and thus it's unique), hence Y = Y', thus B, X, Y are collinear. Y lies on the perpendicular bisector of AC so YA=YC, thus angle ABY = YBC.

QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
FelixD
588 posts
#9 • 2 Y
Y by Adventure10, Mango247
The following will be useful: Let $ \triangle ABC$ be a triangle, $ D$ an arbitrary point and $ X$, $ Y$, $ Z$ the pedal points wrt the sides $ BC$, $ CA$, $ AB$. Then $ YZ= \frac{AD \cdot BC}{2R}$, where $ R$ denotes the circrumradius of $ \triangle ABC$. That's very easy to prove, I won't write it down now :) .
Using the lemma shown above, we have $ PQ= \frac{CD \cdot AB}{2R}$ and $ QR= \frac{AD \cdot BC}{2R}$. Hence, $ PQ =QR \Leftrightarrow \frac{AB}{BC} = \frac{AD}{DC}$. Let $ S$ denote the intersection of the angle bisectors of $ \angle ABC$ and $ \angle CDA$. If $ S \in AC$, then $ \frac{AB}{BC} = \frac{AS}{SC} = \frac{AD}{DC}$. If $ PQ=QR$, then $ \frac{AS_1}{S_1C} = \frac{AB}{BC} = \frac{AD}{DC} = \frac{AS_2}{S_2C}$, hence $ S_2=S_1=S$, where $ S_1$ and $ S_2$ denote the intersections of the angle bisectors of angles $ \angle ABC$ and $ \angle CDA$ with the side $ AC$. Thus, the problem is proved.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
serialk11r
1449 posts
#10 • 1 Y
Y by Adventure10
Hmmm I came up with this, which uses a bit more of Simson line properties.

Let the angle bisector of $ \angle{ABC}$ intersect $ AC$ at $ N$, let the midpoint of $ AC$ be $ M$, let $ E$ and $ F$ be points diametrically opposite on the circumcircle of $ ABCD$ such that $ EF$ is the perpendicular bisector of $ AC$, with $ E$ on the same side of $ AC$ as $ B$. Let $ B'$ be the point diametrically opposite to $ B$.

$ FDQN$ is cyclic, $ FDMN$ is a rectangle (cyclic), so $ \angle{FND} = \angle{FQD} = \angle{MDQ}$.
$ \angle{FND} = \angle{FBD} + \angle{BFE}$
$ \angle{B'BF} = \angle{BFE}$, since both are diameters.
Thus $ \angle{B'BD} = \angle{BNE} = \angle{MDQ}$.

$ P,Q,R$ are collinear and are the Simson line of $ D$. $ AC$ is the Simson line of $ B'$. By a well known property of Simson lines, $ \angle{PQA} = \frac12\overarc{B'D}$, but $ \frac12\overarc{B'D} = \angle{B'BD} = \angle{BNE} = \angle{MDQ}$ and $ \angle{PQA} = \angle{PDA} = \angle{CDR}$, since $ PDR$ is directly similar to $ ADC$. Thus we have a spiral similarity with center $ D$ mapping $ A,M,C$ to $ P,Q,R$ respectively, and since $ M$ is the midpoint of $ AC$, $ Q$ is the midpoint of $ PR$ so we are done.

The argument should work in reverse as well, with some tweaking here and there.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hatchguy
555 posts
#11 • 4 Y
Y by AllanTian, Adventure10, Mango247, LeYohan
The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$.

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$. (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$. Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MariusBocanu
429 posts
#12 • 1 Y
Y by Adventure10
Use that a quadrilateral is harmonic if and only if the diagonals are symmedians(easy to prove with polarity and crossed ratio).
Now, the problem is equivalent to this:
$\triangle{ABC}$,$D$ is on the circumcircle of $\triangle{ABC}$(on the arc$BC$). Denote $X,Y,Z$ the projections of $D$ onto $AB,BC,CA$. $XY=YZ$if and only if $D$ is on the symmedian with respect to $A$.
Proof: See that $BXDY$ and $DYZC$ are inscribed in circles with diameter$BD$,respectively$CD$. We have from Simson's line that $X-Y-Z$are collinear. In $\triangle{XBY}$(aplying law of sines) $XY=BDsinB$, and in $\triangle{YZC}$ we have$YZ=DCsinC$. So, $XY=YZ$ if and only if $\frac{BD}{CD}=\frac{sinC}{sinB}$, but it happens only for symmedian(note that $\frac{sinBAM}{sinCAM}=\frac{sinB}{sinC}$, where $M$ is the midpoint of $BC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arshakus
769 posts
#13 • 1 Y
Y by Adventure10
hey guys)
I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic.
Can it be so?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
nickthegreek
35 posts
#14 • 2 Y
Y by Adventure10, Mango247
A different approach:


To begin with, it is clear that triangles $\bigtriangleup  DRP$ and $\bigtriangleup  DAC$ similar. (This follows by a very simple angle chasing, as in the above posts). This also gives us the fact that triangles $\bigtriangleup  DAR$ and $\bigtriangleup  DPC$ are similar, therefore $\frac{DA}{DC}=\frac{RA}{PC} (1)$

Also, points $P,Q,R$ are collinear (Simson's line)

It remains to prove the equivalence $PQ=QR \Leftrightarrow \frac{BA}{BC}=\frac{DA}{DC}$

Let's focus on triangle $\bigtriangleup RBP$ . Points $A,Q,C $ are collinear and lie on the lines $BR,RP,PB$ respectively. Applying Menelaus' Theorem, we have:
$\frac{AR}{AB}\cdot \frac{CB}{CP}\cdot \frac{QP}{QR}=1$

Therefore, $QP=QR\Leftrightarrow \frac{QP}{QR}=1\Leftrightarrow \frac{AB}{BC}=\frac{RA}{PC}\Leftrightarrow \frac{AB}{BC}= \frac{DA}{DC}$,

since $\frac{DA}{DC}=\frac{RA}{PC}$ by relation (1). The conclusion follows immediately.

Nick
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
StefanS
149 posts
#15 • 1 Y
Y by Adventure10
nickthegreek wrote:
A different approach:
Your solution is actually identical to hatchguy's. Take a look:
hatchguy wrote:
The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$

Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$.

Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$. (1)

By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain:

$\frac{AB*RQ*PC}{AR*PQ*BC}=1$. Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.
Z K Y
G
H
=
a