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
D1010 : How it is possible ?
Dattier   13
N 17 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
13 replies
Dattier
Mar 10, 2025
Dattier
17 minutes ago
Interesting inequality
sqing   1
N 17 minutes ago by sqing
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$ (a^2-1)(b^2-1) (1-ab)+\frac{27}{8}a^2b^2\leq 27$$$$ (a^2-1)(b^2-1)(1-a^2b^2 )+\frac{81}{4}a^2b^2     \leq 189$$$$ (a^2-1)(b^2-1)(1-a^2b^2 )+ 162  ab  \leq 513$$$$  (a^2-1)(b^2-1) (1-a^2b^2 )+21 a^2b^2\leq \frac{3219}{16}$$$$ (a^2-1)(b^2-1) (1-ab)+\frac{27}{8}a^2b^2\leq\frac{415+61\sqrt{61}}{18}$$
1 reply
1 viewing
sqing
an hour ago
sqing
17 minutes ago
Minimal Grouping in a Complete Graph
swynca   1
N 27 minutes ago by swynca
Source: 2025 Turkey TST P1
In a complete graph with $2025$ vertices, each edge has one of the colors $r_1$, $r_2$, or $r_3$. For each $i = 1,2,3$, if the $2025$ vertices can be divided into $a_i$ groups such that any two vertices connected by an edge of color $r_i$ are in different groups, find the minimum possible value of $a_1 + a_2 + a_3$.
1 reply
1 viewing
swynca
3 hours ago
swynca
27 minutes ago
Nice FE as the First Day Finale
swynca   1
N 35 minutes ago by swynca
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
1 reply
swynca
3 hours ago
swynca
35 minutes ago
Cn/lnn bound for S
EthanWYX2009   0
38 minutes ago
Source: 2025 March 谜之竞赛-2
Prove that there exists an constant $C,$ such that for all integer $n\ge 2$ and a subset $S$ of $[n],$ satisfy $a\mid\tbinom ab$ for all $a,b\in S,$ $a>b,$ then $|S|\le \frac{Cn}{\ln n}.$

Created by Yuxing Ye
0 replies
+1 w
EthanWYX2009
38 minutes ago
0 replies
Natural function and cubelike expression
sarjinius   2
N 43 minutes ago by Kaimiaku
Source: Philippine Mathematical Olympiad 2025 P8
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for all $m, n \in \mathbb{N}$, \[m^2f(m) + n^2f(n) + 3mn(m + n)\]is a perfect cube.
2 replies
sarjinius
Mar 9, 2025
Kaimiaku
43 minutes ago
hard problem
Noname23   3
N an hour ago by Noname23
problem
3 replies
Noname23
Sunday at 4:57 PM
Noname23
an hour ago
Roots, bounding and other delusions
anantmudgal09   28
N an hour ago by kes0716
Source: INMO 2021 Problem 6
Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients. Find all functions $f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ satisfying the following conditions:

[list]
[*] $f$ maps the zero polynomial to itself,
[*] for any non-zero polynomial $P \in \mathbb{R}[x]$, $\text{deg} \, f(P) \le 1+ \text{deg} \, P$, and
[*] for any two polynomials $P, Q \in \mathbb{R}[x]$, the polynomials $P-f(Q)$ and $Q-f(P)$ have the same set of real roots.
[/list]

Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha
28 replies
anantmudgal09
Mar 7, 2021
kes0716
an hour ago
square geometry bisect $\angle ESB$
GorgonMathDota   11
N an hour ago by miiirz30
Source: BMO SL 2019, G1
Let $ABCD$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to point $A$. Let $E$ be the intersection of $CM$ and $BD$, and let $S$ be the intersection of $MO$ and $AE$. Show that $SO$ is the angle bisector of $\angle ESB$.
11 replies
GorgonMathDota
Nov 8, 2020
miiirz30
an hour ago
Inspired by my own results
sqing   5
N an hour ago by sqing
Source: Own
Let $ a ,  b  $ be reals such that $ a+b+ab=1. $ Show that$$ 1-\frac{1 }{\sqrt2}\le \frac{1}{a^2+1}+\frac{1}{b^2+1}\le 1+\frac{1 }{\sqrt2} $$Let $ a ,  b\geq 0 $ and $ a+b+ab=1. $ Show that$$ \frac{3}{2}\le \frac{1}{a^2+1}+\frac{1}{b^2+1}\le 1+\frac{1 }{\sqrt2} $$
5 replies
sqing
Yesterday at 8:32 AM
sqing
an hour ago
Polygon formed by the edges of an infinite chessboard
AlperenINAN   1
N an hour ago by AlperenINAN
Source: Turkey TST 2025 P5
Let $P$ be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers $a_1,a_2,a_3$ represent the number of unit squares that have exactly $1,2\text{ or } 3$ edges on the boundary of $P$ respectively. Find the largest real number $k$ such that the inequality $a_1+a_2>ka_3$ holds for each polygon constructed with these conditions.
1 reply
AlperenINAN
3 hours ago
AlperenINAN
an hour ago
Interesting inequality
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$(a^2-1)(b^2-1) -6ab\geq-15$$$$(a^2-1)(b^2-1)  -7ab\geq  -\frac{58}{3}$$$$(a^3-1)(b^3-1)  -\frac{21}{4}a^2b^2\geq -35$$$$(a^3-1)(b^3-1)  -6a^2b^2\geq-\frac{2391}{49}$$
5 replies
1 viewing
sqing
5 hours ago
sqing
2 hours ago
Problem 2830
sqing   1
N 2 hours ago by invisibleman
Source: SXTB (2)2025
Let $ a,b>0 $ and $ \frac{1}{a^2+1}+ \frac{1}{b^2+1}=t $ $(1<t<2). $ Find the value range of $ a+b. $
h
1 reply
sqing
Yesterday at 8:15 AM
invisibleman
2 hours ago
Polynomials and powers
rmtf1111   26
N 2 hours ago by ihategeo_1969
Source: RMM 2018 Day 1 Problem 2
Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying
$$P(x)^{10}+P(x)^9 = Q(x)^{21}+Q(x)^{20}.$$
26 replies
rmtf1111
Feb 24, 2018
ihategeo_1969
2 hours ago
Converse of a classic orthocenter problem
spartacle   42
N Jan 27, 2025 by Saucepan_man02
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
42 replies
spartacle
Dec 14, 2020
Saucepan_man02
Jan 27, 2025
Converse of a classic orthocenter problem
G H J
G H BBookmark kLocked kLocked NReply
Source: USA TSTST 2020 Problem 6
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
spartacle
538 posts
#1 • 9 Y
Y by 606234, Math_olympics, justJen, samrocksnature, megarnie, CrazyInMath, Rounak_iitr, GeoKing, Funcshun840
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
This post has been edited 2 times. Last edited by v_Enhance, Mar 1, 2021, 5:21 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ABCDE
1963 posts
#2 • 16 Y
Y by Aryan-23, 606234, Math_olympics, centslordm, Mindstormer, nikitadas, samrocksnature, megarnie, PRMOisTheHardestExam, cosmicgenius, Vitriol, Mango247, sabkx, IAmTheHazard, OronSH, MS_asdfgzxcvb
We proceed using Cartesian coordinates. Let $\mathcal H$ be the rectangular circumhyperbola of $ABCD$, which is nondegenerate as lines $AP,BQ,CR$ are distinct. Let $\mathcal H$ be given by $xy=1$ and $A=(a,a^{-1}),B=(b,b^{-1}),C=(c,c^{-1}),D=(d,d^{-1})$ with $t=abcd$. Points $(e,e^{-1}),(f,f^{-1}),(g,g^{-1}),(h,h^{-1})$ on $\mathcal H$ form an orthocentric system iff $efgh=-1$ and are concyclic iff $efgh=1$. This fact is easy to check, so we leave the details to the reader. It remains to show that $t=1$.

We have that $P,Q,R\in\mathcal H$, so if $P=(p,p^{-1}),Q=(q,q^{-1}),R=(r,r^{-1})$ then $p=-at^{-1},q=-bt^{-1},r=-ct^{-1}$. The equations of lines $AP,BQ,CR$ are then given by $x+apy=a+p,x+bqy=b+q,x+cry=c+r$, so since they are concurrent we have that \[\begin{vmatrix}1&-a^2t^{-1}&a(1-t^{-1})\\1&-b^2t^{-1}&b(1-t^{-1})\\1&-c^2t^{-1}&c(1-t^{-1})\end{vmatrix}=\begin{vmatrix}1&ap&a+p\\1&bq&b+q\\1&cr&c+r\end{vmatrix}=0.\]Then, $a,b,c$ are distinct roots of $ut^{-1}x^2+v(1-t^{-1})x+w$ for some $u,v,w$ which are not all $0$, which means that $ut^{-1}=v(1-t^{-1})=w=0\implies t=1$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
spartacle
538 posts
#3 • 12 Y
Y by mira74, 606234, MarkBcc168, Kanep, Imayormaynotknowcalculus, Math_olympics, samrocksnature, Rg230403, icosahedraldiceqt, Lcz, megarnie, sabkx
What is this?

I'm sort of conflicted. I'm happy that I was able to use this sort of stuff on an actual test, but also, what?
This post has been edited 3 times. Last edited by spartacle, Dec 14, 2020, 11:14 PM
Reason: 3 -> $3$ because this is AoPS
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SnowPanda
186 posts
#4 • 4 Y
Y by Math_olympics, 606234, samrocksnature, swynca
Complex+bary bash
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
a1267ab
223 posts
#5 • 14 Y
Y by OlympusHero, spartacle, Gaussian_cyber, VipMath, Kanep, SnowPanda, Math_olympics, 606234, mijail, khina, PIartist, samrocksnature, Stuffybear, MS_asdfgzxcvb
My problem. I was very surprised to see it chosen as a problem 6. Here's my solution:

Let $T$ be the concurrency point. We have $AQ\parallel BP$ because both lines are perpendicular to $CD$. Since these lines are distinct there is a homothety $\Psi$ centered at $T$ with $\Psi(A)=P, \Psi(Q)=B$. By repeating this argument, we also have $\Psi(C)=R$, and then $\Psi(P)=A$. Hence $\Psi$ must be reflection about $T$ because $A\neq P$. Next, we have $PD\perp BC$ and $BC\parallel QR$, so $D$ is the orthocenter of $PQR$ by symmetry. Hence if $H$ is the orthocenter of $ABC$, then $T$ is the midpoint of $DH$. Now our setup is symmetric: we have four points $A, B, C, D$, forming four orthocenters $H, P, Q, R$ which are reflections of the previous four points about $T$.

Let $S$ be the center of mass of the four points $A, B, C, D$, and $O$ be the reflection of $T$ through $S$. We claim that $A, B, C, D$ are equidistant from $O$. Let $A', O', S', T', D'$ be the projections of the corresponding point onto line $BC$. Then $T'$ is the midpoint of $A'D'$, so $O'$ is the midpoint of $BC$ (because $S'=\frac{A'+D'+B+C}{4}$). Hence $OB=OC$. By the aforementioned symmetry, we easily show that $O$ is equidistant from $A, B, C, D$.
This post has been edited 1 time. Last edited by a1267ab, Dec 14, 2020, 6:06 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mira74
1010 posts
#6 • 32 Y
Y by JNEW, ProblemSolver2048, SnowPanda, samuel, The_Turtle, spartacle, ghu2024, nukelauncher, Math00954, 329020, amar_04, MP8148, Aryan-23, Ha_ha_ha, Bumblebee60, Kagebaka, franchester, 606234, Imayormaynotknowcalculus, Math_olympics, AforApple, mijail, parola, snakeaid, samrocksnature, IAmTheHazard, Mango247, Mango247, Mango247, sabkx, Funcshun840, MS_asdfgzxcvb
We have that if $T_A$ is the foot of the altitude from $D$ to $BC$, we have

$$\frac{\sin(BAP)}{\sin(CAP)}=\frac{BP}{CP}\cdot \frac{\sin(ABP)}{\sin(ACP)}=\frac{BT_A}{CT_A}\cdot\frac{CD}{BD}\cdot \frac{\sin(ABP)}{\sin(ACP)}=\frac{BT_A}{CT_A}\cdot\frac{CD}{BD}\cdot \frac{\cos(\angle(AB,CD))}{\cos(\angle(AC,BD))},$$so multiplying the similar terms, we get that $$\frac{BT_A}{CT_A}\cdot \frac{CT_B}{AT_B}\cdot \frac{AT_C}{BT_C}=1,$$where $T_B,T_C$ are defined similarly to $T_A$. Now, we get that $T_A,T_B,T_C$ are collinear by menalauas, and we are done by simson lines.
This post has been edited 3 times. Last edited by mira74, Dec 17, 2020, 2:39 AM
Reason: removed reference to title
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Monkey_king1
119 posts
#7 • 12 Y
Y by winnertakeover, IAmTheHazard, 606234, pandadude, OlympusHero, Kagebaka, Kanep, Math_olympics, Aryan-23, samrocksnature, kn07, gvole
I almost was going to use partagura's theorem, but then i realized it didnt work. Thus, I used lines that never met and fractions. Then I used fake numbers.

This was a tough problem. I also tried using Seward’s Theory of Triangular Motion, as well as Potomac Quadrangle Conjecture. The Circular Quadra-angle was pretty hard to find ngl.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Monkey_king1
119 posts
#9 • 12 Y
Y by winnertakeover, IAmTheHazard, 606234, ProblemSolver2048, mobro, Kagebaka, Kanep, Math_olympics, samrocksnature, jacoporizzo, kn07, gvole
Fun fact: Seward's Theory of Triangular Motion is named after William H. Seward, the Secretary of State during the Civil War who narrowly escaped assassination!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dantaxyz
410 posts
#11 • 8 Y
Y by mira74, RodwayWorker, mobro, SK_pi3145, ProblemSolver2048, Math_olympics, bobjoe123, samrocksnature
Another similar triangles/trigonometry solution(not my solution during the test):

Suppose the three lines meet at a point \(O\). Then since \(AQ || BP\) as both are perpendicular to \(CD\), \(\triangle AOQ \sim \triangle POB\). Thus, \(\frac{AO}{OP}=\frac{QO}{OB}\). We can set up a similar fomula on the other pairs \((B,C)\) and \((C,A)\) to get

\[\frac{AO}{OP}=\frac{QO}{OB}=\frac{CO}{OQ}=\frac{OP}{AO},\]
so \(AO=OP\). Note that \(\frac{AO}{OP}=\frac{AQ}{BP}\), so \(AQ=BP\). \(\dfrac{AQ}{2|\cos \angle CAD|}\) and \(\dfrac{CD}{2\sin \angle CAD}\) are both the diameter of \((ACD)\), so \(CD=2 AQ |\tan \angle CAD| =2BP |\tan \angle CBD|\). Thus \(\angle CAD\) and \(\angle CBD\) are either supplementary or equal. A quick case check on whether \(A\) and \(B\) are on the same side of \(CD\) finishes.
This post has been edited 1 time. Last edited by dantaxyz, Dec 14, 2020, 6:27 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aryan-23
558 posts
#13 • 6 Y
Y by naman12, ProblemSolver2048, SK_pi3145, A-Thought-Of-God, Math_olympics, samrocksnature
This is probably similar to others; but I decided to post it anyway. Really nice problem !

Let $S$ be the desired concurrency point.

Note that $BR \perp AD$ and $CQ \perp AD$, so $BR\parallel CQ$. This means $\triangle BSR \sim \triangle SQC$. So we have :

$$ \frac {BS}{QS} =  \frac {RS}{CS}  \stackrel {\triangle RSA \sim \triangle CSP}{=}  \frac {AS}{PS} \stackrel {\triangle BSP \sim \triangle QSA}{=}  \frac {QS}{BS} \implies QS=SB $$
Similarly, $AS=SP$ and $CS=SR$. Note that we now have :

$$ BR \stackrel {\parallel}{=} CQ \quad  BP \stackrel {\parallel}{=} AQ \quad  CP \stackrel {\parallel}{=} AR$$$$  BC \stackrel {\parallel}{=} RQ \quad  AB \stackrel {\parallel}{=} PQ \quad  CA \stackrel {\parallel}{=} PR$$
So $PD\perp QC$ , and so on, so $D$ is the orthocenter of $PQR$. (Note that this provides a shorter finish to the sol in #2)

Hence since $PQR \cong ABC$, we must have $H,S,D$ collinear with $HS=SD$. Finally note that the problem is now symmetric in $A,B,C,D$, so we can WLOG assume that $A,D$ lie on opposite sides of $BC$. Let $R_A$ and $R_D$ denote the circumradii of $\triangle ABC$ and $\triangle DBC$ respectively. Let $\measuredangle XYZ$ denote the acute angle $XYZ$. We have :

$$ BC = 2R_A \times \sin \measuredangle BAC = AH \times \tan \measuredangle BAC$$
Similarly $BC=PD \times \tan \measuredangle BDC$

Since $BC=PD$, we must have $\tan \measuredangle BDC = \tan \measuredangle BAC$.

If $\angle BAC + \angle BDC =180$, we are done. The other possibility is $\angle BAC = \angle BDC$. The second claim implies $P\in \odot (ABC)$, similarly for $Q,R$. But then $D$ lies inside $ABC$ so the angle condition can't hold. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MarkBcc168
1593 posts
#14 • 16 Y
Y by ProblemSolver2048, Aryan-23, 508669, Rg230403, amar_04, A-Thought-Of-God, tapir1729, Math_olympics, samrocksnature, Modesti, math31415926535, Geometry285, Mango247, puntre, ohiorizzler1434, MS_asdfgzxcvb
Solved this in under two minutes—disappointing :(.

Let $D_1$ be the isogonal conjugate of $D$ w.r.t. $\triangle ABC$. Assume for the contradiction that $D_1$ is not an infinity point.

Let $X=QR\cap BC$, $Y=PR\cap AC$, and $Z=PQ\cap AB$; they are colinear due to Desargues. However, from Property 1 here, we get that $\triangle XYZ$ is the pedal triangle of $D_1$ w.r.t. $\triangle ABC$.

But this means by Simson lines that $D_1\in\odot(ABC)$ or $D$ is an infinity point, contradiction.
This post has been edited 1 time. Last edited by MarkBcc168, Dec 15, 2020, 10:17 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rg230403
222 posts
#18 • 11 Y
Y by Aryan-23, A-Thought-Of-God, NJOY, p_square, jelena_ivanchic, Math_olympics, srijonrick, 606234, RudraRockstar, samrocksnature, MS_asdfgzxcvb
We have $AP,BQ,CR$ concurrent and $AR\parallel CP$, $AQ\parallel BP$ and $BR\parallel CQ$ so they concur at the midpoint of $AP,BQ,CR$(same idea as Evan's proof from Twitch stream till now). Now, we also get $D$ is the orthocenter of $PQR$ and thus, $PQRD$ is $ABCH$ reflected across some point. But, as they all lie on a rectangular hyperbola, and $PQAB$ is now a parallelogram, the intersection point is the center of the hyperbola and thus, is the Poncelet point of $D$ and thus lies on the NPC. Thus, $H$ reflected around it, must lie on $(ABC)$ so $D$ is on $(ABC)$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Orestis_Lignos
555 posts
#19 • 4 Y
Y by NJOY, Math_olympics, RudraRockstar, samrocksnature
USA TSTST 2020 P6 wrote:
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

W H A T

Let $O',O'',O'''$ be the circumcenters of triangles $ABD,ACD,BCD$ respectively. If two of them coincide, we are done. Suppose for the sake of contradiction that $O' \neq O'' \neq O''' \neq O'$. Also let $O$ be the concurrency point of $AP,BQ,CR$.

Since $CQ \parallel BR$ (both perpendicular to $AD$) and similar relations, we have $$\frac{AR}{CP}=\frac{AO}{PO}=\frac{AQ}{BP}=\frac{QO}{BO}=\frac{CQ}{BR}$$hence $$\frac{CQ}{BR}=\frac{AQ}{BP}=\frac{AR}{CP}$$
Let $X,Y,Z$ be the midpoints of $AD,BD,CD$. Using the $AH=2OM$ Lemma, we obtain $\frac{CQ}{BR}=\frac{O''X}{O'X}$ and similar relations. Cyclically we have $$\frac{O''X}{O'X}=\frac{O''Z}{O'''Z}=\frac{O'Y}{O'''Y}$$The first equality implies $XZ \parallel O'O'''$ and since $XZ \parallel AC$ and $O'O''' \perp BD$ we obtain $AC \perp BD$. Similarly the second implies $O'O'' \parallel YZ$ hence $AD \perp BC $. Therefore $D$ is the orthocenter of $ABC$, contradiction to the problem hypothesis.

Hence, done.
This post has been edited 2 times. Last edited by Orestis_Lignos, Dec 15, 2020, 10:22 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Supercali
1260 posts
#20 • 3 Y
Y by p_square, Math_olympics, samrocksnature
spartacle wrote:
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Let $X$ be the point of concurrence, and let $H$ be the orthocenter of $\triangle ABC$.

By Desargues on $\triangle AQR$ and $\triangle PBC$, we get $QR \parallel BC$. But we also know $BR \parallel CQ$ $\implies$ $\square BCQR$ is a parellelogram. Therefore $X$ is the common midpoint of $AP, BQ, CR$ $\implies$ $\triangle PQR$ is the reflection of $\triangle ABC$ in $X$. We have $PD \perp BC$ and so on $\implies$ $D$ is the orthocenter of $\triangle PQR$ $\implies$ $D$ is reflection of $H$ in $X$. Let $\mathcal{H}$ be the unique rectangular hyperbola passing through $A,B,C,D$. Then $P,Q,R,H$ also lie on $\mathcal{H}$ $\implies$ $X$ is the center of $\mathcal{H}$ $\implies$ $X$ lies on nine-point circle of $\triangle ABC$ $\implies$ $D$ lies on $\odot (ABC)$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mofumofu
179 posts
#22 • 10 Y
Y by Pluto1708, Aryan-23, timon92, Supercali, Imayormaynotknowcalculus, Greenleaf5002, Math_olympics, samrocksnature, Brian_Xu, khina
Is no one going to comment on how the title might potentially spoil the problem? To me, the title might do a wonderful job in skewing the thought process for other users trying this problem. Also, the censorship on this community is ridiculous; someone also posted on the TSTST P4 thread about a potentially spoiler title, but the post was deleted without any sort of address to it.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
spartacle
538 posts
#23 • 8 Y
Y by Rg230403, Imayormaynotknowcalculus, mira74, Idio-logy, Math_olympics, Aryan-23, samrocksnature, Mango247
mofumofu wrote:
Is no one going to comment on how the title might potentially spoil the problem? To me, the title might do a wonderful job in skewing the thought process for other users trying this problem. Also, the censorship on this community is ridiculous; someone also posted on the TSTST P4 thread about a potentially spoiler title, but the post was deleted without any sort of address to it.

This is a good point. I thought about this, but I thought that nearly everyone either would not know how to use that method or find it easy regardless of whether or not it was in the title.

I suppose this was probably bad judgement on my part. I've changed the title to "Converse of a classic orthocenter problem."

(Re TSTST P4 thread: I think that was a simple mistake, that thread has had some issues with posts being incorrectly deleted.)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
khina
993 posts
#24 • 3 Y
Y by SK_pi3145, Math_olympics, samrocksnature
solution

motivational remarks
This post has been edited 1 time. Last edited by khina, Dec 17, 2020, 6:44 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
v_Enhance
6858 posts
#25 • 10 Y
Y by Math_olympics, srijonrick, 606234, v4913, a_friendwr_a, PIartist, samrocksnature, Modesti, HamstPan38825, Rounak_iitr
Here is the current draft of the official solution; the solution here is the author's submitted approach.

Let $T$ be the concurrency point, and let $H$ be the orthocenter of $\triangle ABC$.
[asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); pair D = dir(97); pair P = B+C+D; pair Q = C+A+D; pair R = A+B+D;
filldraw(R--B--D--cycle, invisible, palecyan); filldraw(Q--A--C--cycle, invisible, palered); filldraw(C--B--D--cycle, invisible, palegreen);
draw(R--foot(R,B,D), lightgrey+dashed); draw(B--foot(B,D,R), lightgrey+dashed); draw(D--foot(D,R,B), lightgrey+dashed);
draw(A--foot(A,C,Q), lightgrey+dashed); draw(C--foot(C,Q,A), lightgrey+dashed); draw(Q--foot(Q,A,C), lightgrey+dashed);
draw(C--foot(C,B,D), lightgrey+dashed); draw(B--foot(B,D,C), lightgrey+dashed); draw(D--foot(D,C,B), lightgrey+dashed);
draw(A--foot(A,B,C), lightgrey+dashed); draw(B--foot(B,C,A), lightgrey+dashed); draw(C--foot(C,A,B), lightgrey+dashed);
pair T = midpoint(A--P); pair S = (A+B+C+D)/4; pair O = 2*S-T; /* A' = foot A B C R270 T' = foot T B C R270 D' = foot D B C R300 O' = foot O B C R300 S' = foot S B C R270 T--Tp lightgrey dashed S--Sp lightgrey dashed O--Op lightgrey dashed */
draw(T--O, lightblue);
filldraw(A--B--C--cycle, invisible, lightgrey); draw(A--C, palered);
draw(A--Q, red+1); draw(B--P, red+1);
draw(A--P, blue); draw(B--Q, blue); draw(C--R, blue); pair H = A+B+C; draw(D--H, lightblue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(80)); dot("$P$", P, dir(285)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$T$", T, 1.8*dir(90)); dot("$S$", S, 1.8*dir(75)); dot("$O$", O, dir(315)); dot("$H$", H, dir(45));
/* TSQ Source:
A = dir 130 B = dir 210 C = dir 330 D = dir 97 R80 P = B+C+D R285 Q = C+A+D R = A+B+D
R--B--D--cycle 0.1 lightcyan / palecyan Q--A--C--cycle 0.1 lightred / palered C--B--D--cycle 0.1 lightgreen / palegreen
R--foot(R,B,D) lightgrey dashed B--foot(B,D,R) lightgrey dashed D--foot(D,R,B) lightgrey dashed
A--foot(A,C,Q) lightgrey dashed C--foot(C,Q,A) lightgrey dashed Q--foot(Q,A,C) lightgrey dashed
C--foot(C,B,D) lightgrey dashed B--foot(B,D,C) lightgrey dashed D--foot(D,C,B) lightgrey dashed
A--foot(A,B,C) lightgrey dashed B--foot(B,C,A) lightgrey dashed C--foot(C,A,B) lightgrey dashed
T = midpoint A--P 1.8R90 S = (A+B+C+D)/4 1.8R75 O = 2*S-T R315
T--O lightblue
A--B--C--cycle 0.1 yellow / lightgrey A--C palered
A--Q red+1 B--P red+1
A--P blue B--Q blue C--R blue H = A+B+C R45 D--H lightblue
*/ [/asy]

Claim: [Key claim] $T$ is the midpoint of $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$, $\overline{DH}$, and $D$ is the orthocenter of $\triangle PQR$.
Proof. Note that $\overline{AQ} \parallel \overline{BP}$, as both are perpendicular to $\overline{CD}$. Since lines $AP$ and $BQ$ are distinct, lines $AQ$ and $BP$ are distinct.
By symmetric reasoning, we get that $AQCPBR$ is a hexagon with opposite sides parallel and concurrent diagonals as $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$ meet at $T$. This implies that the hexagon is centrally symmetric about $T$; indeed \[ \frac{AT}{TP} = \frac{TQ}{BT} = \frac{CT}{TR} = \frac{TP}{AT} \]so all the ratios are equal to $+1$.
Next, $\overline{PD} \perp \overline{BC} \parallel \overline{QR}$, so by symmetry we get $D$ is the orthocenter of $\triangle PQR$. This means that $T$ is the midpoint of $\overline{DH}$ as well. $\blacksquare$


Corollary: The configuration is now symmetric: we have four points $A$, $B$, $C$, $D$, and their reflections in $T$ are four orthocenters $P$, $Q$, $R$, $H$.
Let $S$ be the centroid of $\{A, B, C, D\}$, and let $O$ be the reflection of $T$ in $S$. We are ready to conclude:
Claim: $A$, $B$, $C$, $D$ are equidistant from $O$.
Proof. Let $A'$, $O'$, $S'$, $T'$, $D'$ be the projections of $A$, $O$, $S$, $T$, $D$ onto line $BC$. Then $T'$ is the midpoint of $\overline{A'D'}$, so $S' = \tfrac14(A'+D'+B+C)$ gives that $O'$ is the midpoint of $\overline{BC}$.
Thus $OB = OC$ and we're done. $\blacksquare$
This post has been edited 2 times. Last edited by v_Enhance, Dec 17, 2020, 11:03 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rcorreaa
238 posts
#26 • 1 Y
Y by samrocksnature
Let $\mathcal{H}$ the rectangular circumhyperbola passing through $A,B,C,D,P,Q,R$ and let $X$ the concurrence point.

By Pascal’s Theorem on $APDRCB$, we have that $X,DP \cap BC, DR \cap AC$ are colinear. Similarly, by Pascal’s Theorem on $APDRCB$, $X, DP \cap BC, DR \cap AB$ are collinear. Thus, $DP \cap BC, DQ \cap AC, DR \cap AB$ are collinear. Hence, since these points are the orthogonal projections of $D$ WRT $BC,CA,AB$ (because $P,Q,R$ are the ortocenter of $BCD, CDA, DAB$, respectively), we are done by the converse of the Simson Line.

$\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TheUltimate123
1739 posts
#27 • 5 Y
Y by Pluto04, spartacle, samrocksnature, Tafi_ak, Rounak_iitr
Consider the rectangular circumhyperbola \(\mathcal H\) through the points \(A\), \(B\), \(C\), \(D\). Then \(\mathcal H\) also contains \(P\), \(Q\), \(R\). Let \(\overline{AP}\), \(\overline{BQ}\), \(\overline{CR}\) concur at \(O\).

Claim: \(O\) is the midpoint of segments \(AP\), \(BQ\), \(CR\).

Proof. First I contend \(BCQR\) is a parallelogram: indeed,
  • \(\overline{BR}\) and \(\overline{CQ}\) are both perpendicular to \(\overline{AD}\), so \(\overline{BR}\parallel\overline{CQ}\).
  • Similarly, \(\overline{BP}\parallel\overline{AQ}\) and \(\overline{CP}\parallel\overline{AR}\), so \(\overline{BC}\parallel\overline{QR}\) by Desargue's theorem on \(\triangle PBC\) and \(\triangle AQR\).
Then \(\overline{BQ}\cap\overline{CR}=O\) is the midpoint of \(\overline{BQ}\) and \(\overline{CR}\), so the claim follows symmetrically. \(\blacksquare\)

Claim: \(O\) is the center of \(\mathcal H\).

Proof. Let \(\mathcal H'\) be the image of \(\mathcal H\) under reflection in \(O\). Then \(\mathcal H\) and \(\mathcal H'\) both contain \(A\), \(B\), \(C\), \(D\), \(P\), \(Q\), \(R\), so they are the same hyperbola. \(\blacksquare\)

Finally \(\overline{PD}\perp\overline{BC}\parallel\overline{QR}\), so \(D\) is the orthocenter of \(\triangle PQR\). Therefore the reflection of \(D\) in \(O\) is the orthocenter \(H\) of \(\triangle ABC\). But it is well-known that the Poncelet point \(O\) lies on the nine-point circle of \(\triangle ABC\), so \(D\) lies on the circumcircle.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
606234
80 posts
#28 • 3 Y
Y by KaiDaMagical336, Lcz, samrocksnature
Basically like MarkBcc168's solution, but eh.

Solution: We will use the same notations as MarkBcc168 as Simson lines will come in handy, i.e. $X=QR\cap BC$, $Y=PR\cap AC$, $Z=PQ\cap AB$. Now, by Desargues on $\triangle{AQR}$ and $\triangle{PBC}$, we get that $X, Y, Z$ are collinear. Furthermore, from Delta 7.1, we have that points which have isogonal conjugates at infinity are the points on the circumcircle, but the Six Point Circle Theorem notes that this happens iff the pedal triangle is degenerate, which is the case here! That is, $\triangle{XYZ}$ is the degenerate pedal triangle of $D*$ with respect to $\triangle{ABC}$ where $D*$ is the isogonal conjugate of $D$ with respect to $\triangle{ABC}$, so we are done. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KevinCheng
4 posts
#29 • 3 Y
Y by qzc, SK_pi3145, samrocksnature
Let $\mathcal{H}$ be the rectangular circumhyperbola passing through $A,B,C,D,P,Q,R$ and let $T$ be the concurrency point. Let $X=BR\cap CQ,Y=CP\cap AR,Z=AQ\cap BP$ (they all lie on the line at infinity). By Brocard's Theorem we know that the polar of $T$ WRT $\mathcal{H}$ passes through $X,Y,Z$, so $T$ is the center of $\mathcal{H}$. Then we get $BR=CQ$. By Carnot's Theorem we get the (directed) distance from the circumcenter of $\triangle ABC$ and the circumcenter of $\triangle DBC$ to AD is equal. Therefore the circumcenter of $\triangle ABC$ and the circumcenter of $\triangle DBC$ coincide, so we are done. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jj_ca888
2726 posts
#30 • 1 Y
Y by samrocksnature
Idk if this sol has been posted before

Consider the two triangles $\triangle ABR$ and $\triangle PQC$. Note that $AR \parallel CP$ as both are perpendicular to $BD$. Similarly, $BR \parallel CQ$ are both are perpendicular to $AD$. Next, we perform Desargues. Note that by the problem, $AP, BQ, CR$ concur, hence $AB \cap PQ, AR \cap CP, BR \cap CQ$ are collinear. But by the previously derived parallellisms, the latter two are points at infinity, hence $AB \parallel PQ$.

This tells us that $ABPQ$ is a parallelogram, since $AQ \parallel BP$ is clear as they are both perpendicular to $CD$. Therefore, by parallelogram diagonal bisection, it is clear that $T = AP \cap BQ \cap CR$ is actually the midpoint of each of the three segments.

Now, by orthocentric lengths,\[CD\cot{\angle DBC} = BP = AQ = CD\cot{\angle DAC}\]hence $\angle DBC = \angle DAC$ so $A, B, C, D$ indeed cyclic. $\blacksquare$

edit: oh, nvm, more or less, it has :/ I'm not liking this circumrectangular hyperbola stuff :/
This post has been edited 3 times. Last edited by jj_ca888, Mar 14, 2021, 1:25 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
i3435
1349 posts
#31 • 1 Y
Y by samrocksnature
Am I doing this right, idk. There's a lot of "by symmetry"s in this solution that I didn't add because I was lazy.

$\overline{AQ},\overline{PB}$ are both perpendicular to $\overline{CD}$, so they are both parallel. Desargues on $\triangle AQR$ and $\triangle PBC$ means that $\overline{QR},\overline{BC}$ meet at the line at infinity, so they are also parallel. Thus $\overline{PQ}||\overline{AB}$, but it was already said that $\overline{AQ}||\overline{PB}$, so $AQPB$ is a parallelogram. Let $S$ be the intersection of $\overline{AP},\overline{BQ},\overline{CD}$. Then $S$ is the midpoint of $AP,BQ,CR$.

$\overline{PD}\perp\overline{BC}$, so $\overline{PD}\perp\overline{QR}$. By symmetry $D$ is the orthocenter of $\triangle PQR$, so by a reflection around $S$, if $H$ is the orthocenter of $\triangle ABC$, then $HD$ has midpoint $S$. Plot this on the complex plane. Let $A,B,C,D$ be represented by $a,b,c,d$, where $a,b,c$ lie on the unit circle. $H$ is mapped to $a+b+c$, and $P$ is mapped to $b+c+d$ because $AHPD$ is a parallelogram. The centroid of $\triangle BCD$ is $\frac{b+c+d}{3}$, so the circumcenter of $\triangle BCD$ is $0$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Spacesam
597 posts
#32 • 3 Y
Y by samrocksnature, Mango247, Mango247
Let $Y$ be the concurrence point. We begin with the following key claim:

$\textbf{Claim:}$ $\triangle ABC$ and $\triangle PQR$ are symmetric about $Y$.

This is ratios. Observe that we have \begin{align*}
\frac{YA}{YP} = \frac{YQ}{YB} = \frac{AQ}{BP}, \frac{YQ}{YB} = \frac{YC}{YR} = \frac{QC}{BR}, \frac{YA}{YP} = \frac{YR}{YC} = \frac{AR}{CP}.
\end{align*}Now focus on \begin{align*}
\frac{YQ}{YB} = \frac{YA}{YP} = \frac{YC}{YR} = \frac{YR}{YC},
\end{align*}which yields $YR = YC$ and everything else follows. $\square$

Now we throw everything onto the complex plane. Let $a$, $b$, $c$ be on the unit circle with origin $0$, and let $d$ be the coordinate for $D$. Let $D'$ denote the reflection of $D$ over $Y$. Observe that by our parallel lines that $D$ is the orthocenter of $\triangle PQR$, and so $D'$ is the orthocenter of $\triangle ABC$. Thus we can calculate $d' = a + b + c$.

However, since $CDRD'$ is a parallelogram, we see that $r + c = d' + d$. Thus we have $r = a + b + d$. Since $A$ and $B$ lie on the unit circle, we know that this holds if and only if $D$ lies on the unit circle too. Thus 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.
tree_3
47 posts
#34
Y by
[asy]     size(8cm);     pen prim = blue;     pair A = dir(110);     pair B = dir(210);     pair C = dir(330);     pair D = dir(70) * 3/2;     pair E = foot(D, B, C);     pair F = foot(C, B, D);     pair G = foot(D, A, C);     pair H = foot(C, A, D);     pair P = extension(D, E, C, F);     pair Q = extension(D, G, C, H);     draw(A--B--C--cycle);     draw(CP( (C+D)/2, C));     draw(D--E); draw(C--F); draw(D--G);draw(C--H);     draw(A--D);     draw(D--B);     draw(G--E, prim);     draw(F--H, prim);     dot("$A$", A, dir(A));     dot("$B$", B, dir(B));     dot("$C$", C, dir(C));     dot("$D$", D, dir(D));     dot("$E$", E, S);     dot("$F$", F, dir(20));     dot("$G$", G, dir(220));     dot("$H$", H, dir(H-Q));     dot("$P$", P, dir(290)*2);     dot("$Q$", Q, dir(350));     draw(B--Q, prim); draw(A--P, prim);    [/asy]
Let $E, G$ be the feet from $D$ to $BC, AC,$ $H$ be the foot from $C$ to $AD,$ and $F$ be the foot from $C$ to $DB.$
Claim: $AP \cap BQ$ lies on $GE.$
Proof. Because of right angles, $DHGFEC$ is cyclic with diameter $DC.$ Now, we apply Pascals on $CEGDFH, CFHDEG$ to get $B, EG \cap FH, Q$ and $P, HF \cap GE, A$ are collinear. Therefore, $AP, GE, HF, BQ$ are concurrent. $\blacksquare$
Applying this claim symmetrically tells us that if the foot from $D$ to $AB$ is $I$ then $EG, GI, EI$ all pass through $AP \cap BQ \cap CR,$ which implies $EGI$ is collinear. By converse of Simson line $ABCD$ is cyclic as desired.
This post has been edited 1 time. Last edited by tree_3, Aug 2, 2021, 4:36 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8853 posts
#35
Y by
Hmm this was easier than expected ... 7200th post!
[asy]
size(250);
pair A = (-3.6488936772554, 4.7629376368042), B = (4.064712684455, 4.4134012725823), C = (4.709553074972, -3.7175408315205), D = (0.4691154449004, -5.9816327787115);
pair P = orthocenter(B, C, D), Q = orthocenter(A, C, D), R = orthocenter(A, B, D);
pair X = extension(B, Q, A, P);
draw(A--B--C--D--cycle, blue);
draw(circumcircle(A, B, C), lightblue);
draw(A--C, green); draw(B--D, green);
draw(A--P, orange); draw(B--Q, orange); draw(C--R, orange); draw(B--R--Q--C, red+dashed); draw(A--R, magenta+dashed); draw(C--P, magenta+dashed);
dot("$A$", A, NW);
dot("$B$", B, NE);
dot("$C$", C, SE);
dot("$D$", D, S);
dot("$Q$", Q, SE);
dot("$P$", P, SE);
dot("$R$", R, N);
dot("$X$", X, E);
[/asy]

Let the concurrency point be $X$. The crux of the problem lies in the following:

Claim. $BRQC$ is a parallelogram.

Proof. We will show that $RQ \parallel BC$, because $BR \parallel CQ$ is trivial (because they are both perpendicular to $\overline{AC}$ by definition.

First, observe that since $BP \parallel AQ$, we have $\triangle XAQ \sim \triangle XPB$. Similarly, $CP \parallel AR$ implies $\triangle ARX \sim \triangle PCX$. From these two relations, we have $$\frac{CP}{AR} = \frac{PH}{AH} = \frac{BP}{AQ}.$$However, because $AR \parallel PC$ and $AQ \parallel PB$, $\overline{CP} \cap \overline{AQ}, A, \overline{AR} \cap \overline{BP}, P$ are the vertices of a parallelogram.

From this, we devise that $\angle QAR = \angle CPB$, so we have $\triangle ARQ \sim \triangle PCB$. It follows that $\angle ARQ = \angle BCR$. Combined with $\angle ARX = \angle XCR$ which is already known, we have $\angle QRX = \angle HCX$, implying $\overline{QR} \parallel \overline{BC}$, as desired. $\blacksquare$

To finish, observe that $$\frac{CQ}{\cos \angle ACD} = \frac{CQ}{\sin \angle QDC} = \frac{CD}{\sin \angle DQC} = \frac{CD}{\sin \angle DAC},$$so $$CQ = \cos \angle ACD \left(\frac{CD}{\sin \angle DAC}\right) = \cos \angle ACD \left(\frac{AD}{\sin \angle ACD}\right) = \frac{AD}{\tan \angle ACD}.$$Similarly, $$\frac{BR}{\cos \angle ABD} = \frac{BR}{\sin \angle RAB} = \frac{AB}{\sin \angle ARB} = \frac{AB}{\sin \angle ADB},$$so $$BR = \cos \angle ABD\left(\frac{AB}{\sin \angle ADB}\right) = \cos \angle ABD\left(\frac{AD}{\sin \angle ABD}\right) = \frac{AD}{\tan \angle ABD}.$$But because $BQ=CR$ due to parallelogram $BQRC$, we must have $\tan \angle ABD = \tan \angle ACD$. This implies that $\angle ABD = \angle ACD$, or $ABCD$ is cyclic, as desired. $\square$

Remark. When $\angle ABD$ or $\angle ACD$ is obtuse, it actually follows that $\sin \angle RAB = -\cos \angle ABD$, in which case we obtain $\tan \angle ABD = -\tan \angle ACD$, implying $\angle ABD + \angle ACD = 180^{\circ}$. This resolves any potential configuration issues.
This post has been edited 2 times. Last edited by HamstPan38825, Aug 2, 2021, 8:04 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GeronimoStilton
1521 posts
#36 • 1 Y
Y by centslordm
Let $\mathcal H$ be the rectangular circumhyperbola of $ABC$ including $D$. Clearly $\mathcal H$ includes the orthocenters of $\triangle ABC, \triangle BCD, \triangle CDA, \triangle DAB$. Let the orthocenter of $\triangle ABC$ be $H$. Let the common intersection of $AP, BQ, CR$ be $T$. Note that $BR\perp AD\perp CQ$ so $\triangle TRB\sim \triangle TCQ$. In particular, we have $RT/TB = CT/CQ$, so by analogous results, we have
\[\frac{RT}{TC} = \frac{RT}{TB}\cdot \frac{TB}{TP} \cdot \frac{TP}{TC} = \frac{CT}{CQ}\cdot \frac{TQ}{TA}\cdot \frac{TA}{TR} = \frac{CT}{TR}.\]Thus $CT=TR$, meaning $T$ is the midpoint of $CR$. But $T$ is similarly the midpoint of $AP$ and $BQ$, meaning that reflection through $T$ must be an involution on $\mathcal H$. Thus $T$ is the Poncelet point of $\mathcal H$, meaning that, as the reflection of $R$ over $T$, $C$ must lie on $(ABD)$, 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.
Eyed
1065 posts
#37
Y by
Define $S$ as the orthocenter of $ABC$, and $M$ as the concurrency point of $AP, BQ, CR$. First, observe that $AQ\perp CD$, $BP\perp CD$, so $AQ || BP$. Similarly, $AR || CP, BR || CQ$, so
\[\frac{AM}{MQ} = \frac{MP}{BM}, \frac{AM}{MR} = \frac{MP}{CM}\Rightarrow \frac{MR}{MQ} = \frac{CM}{BM}\]However $\frac{CM}{MQ} = \frac{MR}{BM}$ so $MB = MQ$. Therefore, $M$ is the midpoint of $CR, BQ, AP$. Now, if we denote $S'$ as the reflection of $D$ over $M$, then $AS' || DP, BS' || DQ, CS' || DR$, but we also have $AS || DP, BS || DQ, CS || DR$, so $S = S'$. Therefore, $M$ is also the midpoint of $DS$.

However, consider the circum-rectangular hyperbola going through $ABCDPQRS$. Since $M$ is the midpoint of $4$ different line segments on the hyperbola, this means $M$ is the poncelet point. This means $M$ lies on the nine point circle of $ABC$, so the reflection of $S$ over $M$ lies on the circumcircle of $(ABC)$. Therefore, $A,B,C,D$ are cyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathscrazy
113 posts
#38 • 1 Y
Y by rama1728
Solved with rama1728
Let the lines $AP$, $BQ$, $CR$ concur at $X$
We begin by proving some claims:
Claim 1: X is midpoint of $AP$, $BQ$, $CR$
Proof:
Note, $AQ||BP$ as both are perpendicular to $CD$. Also, $AP \cap BQ = X$. $ \Rightarrow \frac{PX}{XA} =\frac{BX}{XQ} $
Similarly, $BR||CQ$ as both are perpendicular to $AD$. Also $ BQ \cap CR = X$.$\Rightarrow \frac{BX}{XQ} = \frac{RX}{XC} $
Hence, $\frac{PX}{XA} =  \frac{RX}{CR} \Rightarrow AC || PR $
Also, $ AR||CP$ as both are perpendicular to $BD$.
Hence, $ \square ARCP $ is a parallelogram!
Hence, $X$ which is intersection of it's diagonals is indeed the midpoint of the diagonal.
Hence, $X$ is midpoint of $AP$ and $CR$.
We can similarly prove that $X$ is midpoint of $BQ$
Hence proved claim 1.$\blacksquare$
Claim 2: $D$ is orthocenter of $PQR$.
Proof:
We proved that $ \square ARCP $ is a parallelogram. We can similarly prove that $\square AQBP $ and $ \square BRQC $ is parallelogram.
$PD$ is perpendicular to $BC$, as $P$ is orthocenter of $BCD$.
Hence, $PD$ is perpendicular to $ QR $, as $ BC||QR$ using the parallelogram $\square BRQC $
We can similarly prove that $QD$ is perpendicular to $PR$, and $RD$ is perpendicular to $PQ$.
hence, $D$ is orthocenter of $PQR$.
hence Proved claim 2.$\blacksquare$
Claim 3: $X$ is midpoint of $DS$ where $S$ is orthocenter of $\triangle ABC$
Proof:
From the three parallelograms, $ \square ARCP, \square AQBP, \square BRQC $, we get that,$PQ=AB, QR=BC, PR=AC$
Hence, $ \triangle ABC \sim \triangle PQR $
Further, $AP, BQ, CR$ intersect at $X$
Hence, there is a homothety centered at $X$ mapping $ \triangle ABC $ to $ \triangle PQR $ with scale factor $1$.
Hence, under this homothety, the orthocenter of $ \triangle ABC$ gets mapped to orthocenter of $\triangle PQR $ .
Hence, using claim 2, $S$ gets mapped to $D$ under this homothety.
Hence, $SD$ has midpoint X due to the homothety at $X$ with scale factor $1$.
Hence proved claim 3.$\blacksquare$
Claim 4: $\square PQRS $ is cyclic quadrilateral.
Proof:
We present a proof using complex numbers. We let $a$ be the complex number representing $A$ and similar notation throughout the proof.
We can wlog assume that $a,b,c$ lie on the unit circle.
It's equivalent to prove that $d$ also lies on the unit circle.
As $a,b,c$ lie on the unit circle, orthocenter of $\triangle ABC $ which is $S$ will be $s=a+b+c$.
$X$ is midpoint of $SD$ using claim 3.
Hence, $x = \frac{s+d}{2} = \frac{a+b+c+d}{2}$
Also, $X$ is midpoint of $AP$, $BQ$, $CR$ using claim 1.
Hence, we get that $ p=b+c+d, q=a+c+d , r=a+b+d $ (Note $a,b,c$ lie on the unit circle )
Hence, obviously $d$ lies on the unit circle.
Hence, $A,B,C,D$ lie on the unit circle and hence are cyclic. $\blacksquare$
Note that, a Homothety centered at $X$ with scale factor $1$ sends $\square ABCD $ to $\square PQRS $.
As $\square PQRS $ is cyclic from claim 3, hence so is $\square ABCD $!
Hence proved ! $\blacksquare$
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blackbluecar
302 posts
#39
Y by
First, let's rename the points to make things more clear. Replace $D$ with $P$ and let $H_A$, $H_B$, and $H_C$ be the orthocenters of $\triangle PBC$, $\triangle PCA$, and $\triangle PAB$. The problem statement tells us that $AH_A$, $BH_B$, and $CH_C$ all concur at a point $X$ and we would like to show $P$ lies on the circumcircle $\Omega$ of $\triangle ABC$.

Lemma: If hexagon $ABCDEF$ obeys $AB \parallel DE$, $BC \parallel EF$, and $CD \parallel FA$, and $AD$, $BE$, and $CF$ concur at $X$, then hexagon $ABCDEF$ is rotationally symmetric around $X$. Indeed, by similar triangles, we have \[ \frac{AX}{XD} = \frac{BX}{XE} = \frac{CX}{XF} = \frac{DX} {XA} \implies AX = DX\]So, $X$ is the midpoint of $AD$, $BE$, and $CF$ which clearly implies the result.

Notice that $AH_B \parallel BH_A$, $BH_C \parallel CH_B$, and $CH_A \parallel AH_C$ and $AH_A$, $BH_B$, and $CH_C$ concur. Thus, hexagon $AH_BCH_ABH_C$ is rotationally symmetric around $X$. Thus, $\triangle ABC$ and $\triangle H_AH_BH_C$ are reflections of each other over $X$. Note that $BC \parallel H_BH_C$. Now, assume for the asked of contradiction that $P$ does not lie on $\Omega$. Let $K_B$ and $K_C$ be reflections of $H_B$ and $H_C$ over line $AP$. We now have that $BAPK_B$, $CAPK_P$, and $BCK_BK_C$ are all cyclic. Since $P$ does not lie on $\Omega$, it must follow that $BAPK_B$ and $CAPK_C$ have distinct circumcircles. Since $BCK_BK_C$ is cyclic, it follows that $BK_B$ and $CK_C$ intersect on the radical axis of $BAPK_B$ and $CAPK_C$ which is $AP$. But, this is clearly a contradiction since $BK_B \parallel CK_C$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NoctNight
108 posts
#40
Y by
Step 1: $AR\parallel PC, QA \parallel BP, CQ\parallel RB$.
Proof: We have $AR\perp BD$ and $PC\perp BD$ due to orthocentres, so $AR\parallel BD$ and the rest follow by symmetry.

Step 2: $ARPC$ is a parallelogram
Proof: Let $AP, BQ, CR$ concur at $X$. Due to the parallel lines:
$$\frac{AX}{PX}=\frac{QX}{BX}=\frac{CX}{RX}=\frac{PX}{AX}$$so $AX=PX$ giving $RX=CX$ by symmetry so $ARPC$ is a parallelogram.

Step 3: $ABCD$ cyclic.
Proof: We have
$$\cot \angle BAD=\frac{AR}{BD}=\frac{CP}{BD}=\cot \angle BCD$$so $\angle BAD=\angle BCD$ so $ABCD$ is cyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bigtaitus
72 posts
#41 • 1 Y
Y by Vahe_Arsenyan
Notice how distance $AQ=DC\cot \angle DAC,$ and similar for all the other $Vertice-Orthocenter$ kinda distances. We'll show that $AQ=BP$ and $AR=PC,$ which will then solve the problem.
Denote the interesection of $AP,BQ,CR$ by $S.$ Notice that $$CP\parallel AR\implies \frac{SC}{SR}=\frac{SP}{SA},$$and $$AQ\parallel BP\implies \frac{SP}{SA}=\frac{SB}{SQ}.$$Thus $\frac{SC}{SR}=\frac{SB}{SQ},$ so we get that $\triangle SRQ\sim \triangle SCB, $ which implies that $RQ\parallel BC \implies RBCQ$ parallelogram, which implies that $S$ is the midpoint of $BQ.$ So now, $AQ\parallel BP\implies BP=AQ,$ and $AR\parallel BC\implies RA=PC,$ which ends the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5000 posts
#43
Y by
Let $T$ be the concurrency point.

Claim: $ARBPCQ$ has opposite sides parallel.
Proof: Note that both $\overline{AR}$ and $\overline{CP}$ are both perpendicular to $\overline{BD}$, so $\overline{AR} \parallel \overline{CP}$. The other parallelisms follow similarly. $\blacksquare$

Claim: $T$ is the midpoint of $\overline{AP},\overline{BQ},\overline{CR}$.
Proof: From the parallelism, we have $\triangle TAR \sim \triangle TPC$, so $\frac{TA}{TP}=\frac{TR}{TC}$. Continuing this cyclically, we have
$$\frac{TA}{TP}=\frac{TR}{TC}=\frac{TB}{TQ}=\frac{TP}{TA}=\frac{TC}{TR}=\frac{TQ}{TB},$$so the ratios are all $1$. $\blacksquare$

We now use complex numbers, denoting points with their lowercase versions and setting $(ABC)$ as the unit circle. We can calculate the circumcenter of $(BCD)$ as
$$\frac{\begin{vmatrix}b&1&1\\c&1&1\\d&d\overline{d}&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\c&\frac{1}{c}&1\\d&\overline{d}&1\end{vmatrix}}=\frac{\begin{vmatrix}b&0&1\\c&0&1\\d&d\overline{d}-1&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\c&\frac{1}{c}&1\\d&\overline{d}&1\end{vmatrix}}=\frac{(b-c)(d\overline{d}-1)}{\frac{(b-c)(b+c)}{bd}-\frac{d(b-c)}{bc}-\overline{d}(b-c)}=\frac{bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}.$$Since $H=3G-2O$, it follows that $P=b+c+d-\tfrac{2bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}$, and $T$, calculated as the midpoint of $\overline{AP}$, is thus
$$\frac{a+b+c+d}{2}-\frac{bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}.$$We obtain cyclic variations for $T$ from $\overline{BQ}$ and $\overline{CQ}$. Therefore, if $d\overline{d}-1 \neq 0$, we require
$$\frac{ab}{a+b-d-ab\overline{d}}=\frac{ac}{a+c-d-ac\overline{d}} \iff ab+bc-bd-abc\overline{d}=ac+bc-cd-abc\overline{d} \iff (a-d)(b-c)=0,$$absurd. Thus $d\overline{d}-1=0$, so $ABCD$ cyclic. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_comb01
659 posts
#44
Y by
Let $\mathcal{H}$ denote the circumrectangular hyperbola through $A,B.C,D,P,Q,R$
Firstly, we claim $T$ is midpoint of $AP,BQ,CR$, which follow from length chasing with parallel lines.
Next, we claim that $T$ is center of $\mathcal{H}$, which follows from the fact that if we reflect everything about $T$ then we get same hyperbola.
So $D$ must be the orthocenter of $PQR$, now since it is well known poncelet points lie on nine point circle, $D \in (ABC)$. Done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1720 posts
#45 • 1 Y
Y by dolphinday
Call $X$ the concurrency point. Since $AQ\perp CD\perp BP,$ we get $\frac{AX}{PX}=\frac{QX}{BX}.$ Similarly, $\frac{QX}{BX}=\frac{CX}{RX}=\frac{PX}{AX},$ thus all ratios must be $-1$ in directed lengths as clearly they cannot be $1.$ Then $ARBPCQ$ is centrally symmetric about $X.$

Now consider the rectangular hyperbola through $A,B,C,D.$ It must then pass through $P,Q,R.$ Now if its center is some point $Y\ne X,$ the reflection of hexagon $ARBPCQ$ over $Y$ is a different hexagon that still has all vertices on the hyperbola. However, this reflected hexagon is also a translation of $ARBPCQ.$ Since there is at most one conic through six points, this implies that the hyperbola is invariant under some translation, which is impossible.

Thus $X$ is the center of the hyperbola. Now we know that $X$ lies on the nine-point circle of $BCD,$ so the reflection of $P$ over $X$ lies on $(BCD).$ But this point is $A$ 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.
awesomehuman
496 posts
#46 • 1 Y
Y by OronSH
$\triangle ARB$ and $\triangle PCQ$ are perspective and $BR\parallel CQ$ and $AR\parallel CP$, so there is a homothety from $\triangle ARB$ to $\triangle PCQ$.
Therefore, $AB\parallel QP$. Thus, $ABPQ$ is a parallelogram. So, using signed lengths, $BP=AQ$. Note that using directed angles, $AQ=CD\cot \measuredangle DAC$ and $BP = CD \cot \measuredangle DBC$. Thus, $\measuredangle DAC = \measuredangle DBC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YaoAOPS
1486 posts
#47
Y by
I don't think its been noted here but the reason the center of the circle is the reflection of the concurrence over the centroid follows by complex bash to get that $s = \frac{1}{2}(a + b + c + d)$ :) which is also called the anticenter. This is also a motivation for considering such a point, by knowing a priori the converse solution.

Anyways, ye old hyperbola nuke.

Let $\mathcal{H}$ be the the circumrectangular hyperbola through $A, B, C, D$. Then $P, Q, R$ must lie on the hyperbola.
Then note that $AQ \parallel BP$, $BR \parallel CQ$, $CP \parallel AR$.

Claim: If the diagonals of equiangular hexagon $ABCDEF$ concur at some point $O$, then hexagon is symmetric about $O$.
Proof. Let $X = AB \cap CD$ and $Y = AF \cap DE$. Note that $AXDY$ is a paralellogram.
We claim that $G = BE \cap CF$ lies on $XY$.
Let $G_B = BE \cap XY$ and $G_C = CF \cap XY$. Then it follows that \[ \frac{XG_B}{G_BY} = \frac{XB}{YE} = \frac{XC}{YF} = \frac{XG_C}{YG_C} \]so $G_B = G_C = G$.
Anyways, if $G$ lies on $AD$, it follows that $G = XY \cap AD$, which is the midpoint of $AD$. By symmetry, it follows that $G$ is the midpoints of $BE, CF$ which suffices. $\blacksquare$
As such, since $AQCPBR$ is equiangular, it follows that their concurrence point $K$ is their symmetry point.
This implies that $(AQCPBR)$ is fixed under reflection about $K$, which implies that $K$ is the Poncelet point. Then $A$ is the reflection of $P$ across $K$, and thus $ABCD$ must be cyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1979 posts
#48
Y by
Pure moving points? :o
spartacle wrote:
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu

Fix a line $\ell$ passing through the orthocentre $H$ of $\triangle ABC$. Suppose it meets $(ABC)$ at $X$ and $Y$.

Animate point $D$ on line $\ell$. We will show that $D$ must equal one of $X, Y, \ell_{\infty}$ for the concurrence to hold. This combined with varying $\ell$ implies the conclusion.
Each of these four cases work is obvious: $X$ and $Y$ work for analogous reasons as the midpoints of the segments $AP, BQ, CR$ in that case coincides with the point with vector $(A+B+C+D)/2$, $\ell_{\infty}$ works trivially as each of $P,  Q, R$ coincide with the point at infinity in the direction perpendicular to $\ell$.

Finally, observe that as $H \in \ell$, the locii of each of $P, Q, R$ are hyperbolas $\mathcal{H}_a, \mathcal{H}_b, \mathcal{H}_c$ passing through $A, B, C$ respectively with the maps $D \mapsto P, D \mapsto Q, D \mapsto R$ all projective, hence lines $AP, BQ, CR$ are each degree $1$ moving lines. The concurrence is a cubic condition which can have at most $3$ solutions, else it is an identity.

However, the concurrence is never a concurrence: this can be seen by letting $D = \ell \cap BC$ --- points $Q, R$ lie on the $A$-altitude and $AP$ becomes the $A$-altitude, so concurrence requires $Q=R=D=AH \cap BC$. Thus, for every choice of $\ell$ other than the $A$ altitude, the concurrence is not always true, similarly taking cases for $B$ and $C$ tells that the concurrence is never identically true for any line $\ell$. This completes the proof.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
StefanSebez
53 posts
#49 • 1 Y
Y by GeoKing
Let's first prove the converse of the problem. Assume $ABCD$ is cyclic and define $H_A$ orthocenter of $\triangle BCD$ and similar. Then in complex numbers midpoint of $AH_A$ is $\frac{a+h_A}{2}=\frac{a+b+c+d}{2}$ which is the same for other points. Therefore $AH_A,BH_B,CH_C,DH_D$ are concurrent.
Let $\mathcal{H}$ be a conic which passes through $A,B,C,D,Q$. This is a rectangular hyperbola so orthocenters $P,R\in \mathcal{H}$. Since $AP,CR,DQ$ are concurrent there is an involution $\Phi:\mathcal{H}\rightarrow \mathcal{H}$ which swaps $(A,P),(C,R),(B,Q)$. Therefore $(DB,DQ;DA,DC)\stackrel{\Phi}{=}(DQ,DB;DP,DR)=(DB,DQ;DR,DP)$. Now let $D'=(ABC)\cap BD$ and suppose $D'\neq D$ and define $P',Q',R'$ accordingly. Since converse of the problem is true we have $BQ',AP',CR'$ concurrent so analogously $(D'B,D'Q';D'A,D'C)=(D'B,D'Q';D'R',D'P')=(DB,DQ;DR,DP)=(DB,DQ;DA,DC)$. This means that $A,B,C,D,D',DQ_{\infty}$ are on a conic. $B,D,D'$ are collinear so this conic must be union of 2 lines which is impossible unless $BD_{\infty}=DQ_{\infty}$ (since certainly $DQ_{\infty}\neq AC_{\infty}$) which would mean $BD\perp AC$. Then we could just do the same thing for $C$ and $A$. If both fail then we have $AB\perp CD,CB\perp AD$ which is a contradiction since $ABCD$ is not an orthocentric system.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Batsuh
152 posts
#50
Y by
I'm sure this solution was found already, but posting for storage.

Let $T$ be the concurrency point of $AP$, $BQ$, $CR$. Note that $AR \cap PC$, $BP\cap AQ$, $BR \cap CQ$ are all points at infinity, so they lie on the line at infinity. Hence, by the converse of Pascals theorem, the points $A$,$B$,$C$,$D$,$P$,$Q$,$R$ lie on a conic $\mathcal{H}$, which incidentally is a rectangular hyperbola.

Next, note that $$\frac{AT}{TP} = \frac{RT}{TC} = \frac{BT}{TQ} = \frac{TP}{TA}$$This implies that $T$ is the midpoint of $AP$, $BQ$ and $CR$. We now show that $T$ is the center of $\mathcal{H}$. Consider the triangles $\triangle BRQ$ and $\triangle BCQ$. It's easy to see that the nine point circle of these two are tangent to each other at $T$. On the other hand, we know that the center of $\mathcal{H}$ lies on these nine point circles. This means that $T$ is the center of $\mathcal{H}$.

Let $S$ be the orthocenter of $\triangle ABC$. Then, the two triangles $\triangle PQR$ and $\triangle ABC$ are reflections of each other at $T$. This means that their orthocenters, which are $D$ and $S$ respectively, are also reflections of each other through $T$.

To finish, since $T$ lies on the nine point circle of $\triangle ABC$, the reflection of $S$, which is $D$, must lie on $(ABC)$. We're done.
This post has been edited 1 time. Last edited by Batsuh, Oct 3, 2024, 8:00 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AlperenINAN
82 posts
#51 • 2 Y
Y by bin_sherlo, egxa
Let $QR \cap BC=\{X\}$, $PR \cap AC=\{Y\}$, $PQ \cap AB=\{Z\}$. Now we'll introduce the following lemma.

Lemma: Let $l_a$ be the line through $X$ and perpendicular to $BC$. Define $l_b$ and $l_c$ similarly. Then the lines $l_a, l_b$, and $l_c$ are concurrent on the isogonal conjugate of $D$.

Proof: Let $\mathcal{H}$ be the circumrectangular hyperbola passing through $A, B, C, D$. Then from the property of the rectangular hyperbola, $P, Q, R \in \mathcal{H}$. Now consider Pascal's theorem on the points $(PQRABC)$. From this, we obtain that $XZ, RA$, and $PC$ are concurrent. (possibly on the point at infinity) Since both of the lines $RA$ and $PC$ are perpendicular to $BD$, we obtain that $XZ$ is perpendicular to $BD$. Similarly, $XY$ is perpendicular to $CD$ and $YZ$ is perpendicular to $AD$. Thus $XYZ$ is the pedal triangle of the isogonal conjugate of $D$ wrt. triangle $ABC$.

From Desargues' theorem; $AP$, $BQ$ and $CR$ are concurrent if and only if $X$, $Y$ and $Z$ are collinear. Thus this concurrency is equivalent to the six-point circle of $D$ being a line, which is equivalent to $A, B, C, D$ being concyclic as desired.
This post has been edited 3 times. Last edited by AlperenINAN, Nov 17, 2024, 7:04 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Saucepan_man02
1298 posts
#52
Y by
Well, thats a nice problem...

Let the point of concurrence be $X$
Consider the rectangular hyperbola $\mathcal{H}$, which passes though $A, B, C, D, P, Q, R$. By Pascal's theorem on $APDRCB$, we have: $X, BC \cap PC, AB \cap DR$ to be collinear. Similarly, by Pascal on $PDQACB$, we have $X, BC \cap PD, CA \cap DQ$ to be collinear. Therefore, $BC \cap PC, CA \cap DQ, AB \cap DR$ are collinear. Since the pedal triangle of $D$ wrt $\triangle ABC$ is degenerate, we have $D \in (ABC)$ (due to converse of Simson).
Z K Y
N Quick Reply
G
H
=
a