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
Find the limit
Mathlover08092002   10
N 3 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 6 :-
Find the limit $\lim \limits_{n \to \infty} \sin{n!}$
[list=1]
[*] 1
[*] 0
[*] $\frac{\pi}{4}$
[*] None of the above
[/list]
10 replies
1 viewing
Mathlover08092002
Apr 9, 2020
quasar_lord
3 minutes ago
Find the number ways you can choose two distinct integers
Mathlover08092002   3
N 6 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 5 :-
What is the number of ways you can choose two distinct integers $a$ and $b$ (unordered i.e. choosing $(a, b)$ is same as choosing $(b, a)$ from the set $\{1, 2, \cdots , 100\}$ such that difference between them is atmost 10, i.e. $|a-b|\leq 10$
[list=1]
[*] ${{100}\choose{2}} -{{90}\choose{2}}$
[*] ${{100}\choose{2}} -90$
[*] ${{100}\choose{2}} -{{90}\choose{2}}-100$
[*] None of the above
[/list]
3 replies
Mathlover08092002
Apr 9, 2020
quasar_lord
6 minutes ago
Inequality
JK1603JK   2
N 9 minutes ago by lbh_qys
Source: unknown
Let a,b,c>=0 and ab+bc+ca>0 then prove \sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\ge 2\sqrt[4]{ab+bc+ca}+\sqrt{\frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{ab+bc+ca}}
2 replies
JK1603JK
Today at 2:58 AM
lbh_qys
9 minutes ago
Find the number of positive integral solutions
Mathlover08092002   1
N 9 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 3 :-
Find the number of positive integral solutions to the equation $\sum \limits_{i=1}^{2019} 10^{a_i}=\sum \limits_{i=1}^{2019} 10^{b_i}$, such that $a_1<a_2<\cdots <a_{2019}$ , $b_1<b_2<\cdots <b_{2019}$ and $a_{2019} < b_{2019}$
[list=1]
[*] 1
[*] 2
[*] 2019
[*] None of the above
[/list]
1 reply
Mathlover08092002
Apr 9, 2020
quasar_lord
9 minutes ago
Inequalities
sqing   2
N Today at 2:19 AM by sqing
Let $ a,b,c\geq 2  . $ Prove that
$$(a-1)(b^2-2)(c-1) -  \frac{1}{2}abc\geq -2$$$$(a^2-2)(b-1)(c^2-2) - \frac{3}{2}abc\geq -6$$
2 replies
sqing
Yesterday at 1:42 PM
sqing
Today at 2:19 AM
2019 Chile Classification / Qualifying NMO Juniors XXXI
parmenides51   5
N Today at 1:14 AM by liyufish
p1. Consider the sequence of positive integers $2, 3, 5, 6, 7, 8, 10, 11 ...$. which are not perfect squares. Calculate the $2019$-th term of the sequence.


p2. In a triangle $ABC$, let $D$ be the midpoint of side $BC$ and $E$ be the midpoint of segment $AD$. Lines $AC$ and $BE$ intersect at $F$. Show that $3AF = AC$.


p3. Find all positive integers $n$ such that $n! + 2019$ is a square perfect.


p4. In a party, there is a certain group of people, none of whom has more than $3$ friends in this. However, if two people are not friends at least they have a friend in this party. What is the largest possible number of people in the party?
5 replies
parmenides51
Oct 11, 2021
liyufish
Today at 1:14 AM
2^n = p+3^p
reeh_haan   8
N Today at 12:20 AM by MajesticCheese
Find all pairs $(p, n)$ of positive integers which satisfy the equation $$2^n = p+3^p$$
8 replies
reeh_haan
Dec 28, 2021
MajesticCheese
Today at 12:20 AM
2017 Mock ARML Team Round #7 Revenge of the incenters
parmenides51   1
N Yesterday at 10:16 PM by Giant_PT
Let $ABC$ be a triangle with side lengths $AB = 20$, $BC = 17$, $AC = 13$, incenter $I$ and circumcircle $\Gamma$, and let $\omega_A$ be the circle that is tangent to $AB$, $AC$, $\Gamma$ at $D$ ,$E$, $F$, respectively. If $I_F$ ,$ I_I$, $I_C$ denote the incenters of $CFI$, $FIE$, $ECF$, respectively, find the measure, in degrees, of the largest angle in $\vartriangle I_F I_II_C$.
1 reply
parmenides51
Jan 12, 2024
Giant_PT
Yesterday at 10:16 PM
Don Sirloin Bowel&#039;s Algorithm
BadAtCompetitionMath21420   1
N Yesterday at 10:07 PM by ohiorizzler1434
Can you sum
\begin{align*}
    \sum_{k=1}^n F^m_k?
\end{align*}
I coined a strategy Don Sirloin Bowel's algorithm because I found it and didn't see it anywhere else. Please tell me if this looks familiar because it can have absolutely scrumptious applications. (I hope that this stuff can help you sum that.) Starting with the "golden quadratic" or whatever the sigma it's called $x^2=x+1$. Multiplying by $x$ on both sides and resubstituting $x^2$, we find $x^3=x^2+x=(x+1)+x=2x+1$. Continuing this process, we find $x^4=3x+2$, $x^5=5x+3$, and so forth. We claim that $x^n=F_nx+F_{n-1}.$ We already have our base case. $x^n=F_nx+F_{n-1}$. Multiplying by $x$, $x^{n+1}=F_nx^2+F_{n-1}x=F_n(x+1)+F_{n-1}x=F_{n+1}x+F_{n}$. Hence, we have proved our claim by induction. Now, raising both sides to the $m$th power,
\begin{align*}
    x^{mn}&=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kx^m\\
    &=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^k(F_mx+F_{m-1})\\
    &=x\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kF_m+\sum_{k=0}^mF_{n}^{m-k}F_{n-1}^kF_{m-1}\\
    &=F_{mn}x+F_{mn-1}
\end{align*}So,
\begin{align*}
    F_{mn}=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kF_m
\end{align*}and
\begin{align*}
    F_{mn-1}=\sum_{k=0}^mF_{n}^{m-k}F_{n-1}^kF_{m-1}
\end{align*}Are these identities useful? And can you use them to compute the sum?
1 reply
BadAtCompetitionMath21420
Yesterday at 9:50 PM
ohiorizzler1434
Yesterday at 10:07 PM
Chessboard
Ecrin_eren   5
N Yesterday at 8:44 PM by TrendCrusher
On an 8×8 checkerboard, what is the minimum number of squares that must be marked (including the marked ones) so that every square has exactly one marked neighbor? (We define neighbors as squares that share a common edge, and a square is not considered a neighbor of itself.)

5 replies
Ecrin_eren
Tuesday at 8:55 PM
TrendCrusher
Yesterday at 8:44 PM
Why does the combined equation have two negative solutions?
Luking   1
N Yesterday at 7:30 PM by vanstraelen
It is known that the moving point $G(x,y)$ is on the curve $C_1: y^2 - x^2 = 1$. There is a parabola $C_2: x^2 = 4y$ with focus $F$. Two tangent lines to$C_2$ are drawn through a point $P$ on $C1$, and the tangent points are $A$ and $B$ respectively. The line $l$ parallel to the line $AB$ is tangent to $C_2$ at point $Q$. Question: When the line $l$ and $C_1$ have two intersection points, find the range of $|QF|$.\
This may make some of the information in the question useless, because I deleted the first two questions of this big question in order to avoid making the question too long and get straight to the point.\
According to the calculation, I get the analytical expression of the line $l$ as $y=\frac{x_0}{2} x - \frac{x_0^2}{4}$.\
At first, I thought it only needed to be not parallel to the parabola asymptotes.\
That is, the slope of the straight line $k \neq \frac{a}{b}$ , then $\frac{x_0}{2} \neq \frac{a}{b}$ , so $x_0 \neq \pm \frac{2a}{b} = \pm2$ ,and $x_0^2 \neq 4$.\
$|QF| = y_0 +1 \neq 5$.\
But when I checked the answer, it was wrong.\
It combines the straight line $l$ with the curve $C_1$ to get an equation and then uses Vieta's theorem.
\begin{cases}
y=\frac{x_0}{2} x - \frac{x_0^2}{4}
x^2=4y
\end{cases}
$$( 4 x_{0}^{2}-1 6 ) y^{2}-8 x_{0}^{2} y-x_{0}^{4}-4 x_{0}^{2}=0 $$The answer is that according to the question, the equation has two negative roots. I can't understand this, and this is exactly where my problem lies.\
Then it gets the following system of equations:
\begin{cases}
4x_0-16 \neq 0
\Delta > 0
x_1+x_2 <0
x_1\cdot x_2>0
\end{cases}
Solve, $2\sqrt{5}-2<x_0<4$.
So we get $|QF|=\frac {x_0^2}{4} + 1 \in (\frac {\sqrt{5}+1}{2},2)$.
I hope you can help me figure out why both roots of that equation are negative.
IMAGE
IMAGE
1 reply
Luking
Yesterday at 4:02 PM
vanstraelen
Yesterday at 7:30 PM
Rolling a die
Ecrin_eren   2
N Yesterday at 6:16 PM by tofubear
John has a standard die with values from 1 to 6. He plays a game where, after each roll, the value on the top face permanently increases by 1. After rolling the die three times, the probability that at least one face has a value of at least 7 is given as a/b , where a and b are coprime. What is the value of a/b ?
2 replies
Ecrin_eren
Yesterday at 5:05 PM
tofubear
Yesterday at 6:16 PM
Travel group
Ecrin_eren   0
Yesterday at 5:12 PM
A group of 5 students will form a travel group of 3 students. Each student is asked to select two other students they would like to travel with. Given that the probability of forming a 3-person group where each member has selected the other two is a/b , where a and b are coprime positive integers, what is the value of a/b ?

0 replies
Ecrin_eren
Yesterday at 5:12 PM
0 replies
Area problem
MTA_2024   1
N Yesterday at 4:46 PM by vanstraelen
Let $\omega$ be a circle inscribed inside a rhombus $ABCD$. Let $P$ and $Q$ be variable points on $AB$ and $AD$ respectively, such as $PQ$ is always the tangent line to $\omega$.
Prove that for any position of $P$ and $Q$ the area of triangle $\triangle CPQ$ is the same.
1 reply
MTA_2024
Mar 16, 2025
vanstraelen
Yesterday at 4:46 PM
Line Perpendicular to Euler Line
tastymath75025   54
N Mar 17, 2025 by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
54 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
Mar 17, 2025
Line Perpendicular to Euler Line
G H J
Source: USA TSTST 2017 Problem 1, by Ray Li
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
tastymath75025
3223 posts
#1 • 15 Y
Y by adamov1, Davi-8191, doxuanlong15052000, tenplusten, Mathuzb, nguyendangkhoa17112003, rashah76, mathematicsy, megarnie, samrocksnature, HWenslawski, son7, Adventure10, Mango247, duhGOAT
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
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
#2 • 23 Y
Y by tiwarianurag021999, huricane, Vrangr, jam10307, brokendiamond, brainiacmaniac31, abbosjon2002, green_leaf, amar_04, rashah76, brendan_cape, v4913, vvluo, Lamboreghini, megarnie, math31415926535, AllanTian, samrocksnature, son7, Kameawtamprooz, Adventure10, Mango247, cursed_tangent1434
First solution (power of a point) Let $\gamma$ denote the nine-point circle of $ABC$.

[asy] pair A = dir(125); pair B = dir(210); pair C = dir(330); pair M = midpoint(A--B); pair N = midpoint(A--C); pair O = origin; pair H = A+B+C;

draw(A--B--C--cycle, blue); pair E = foot(B, A, C); pair F = foot(C, A, B); draw(B--E, lightblue); draw(C--F, lightblue); pair R = extension(E, F, B, C); pair Q = -A+2*foot(O, A, R);

filldraw(unitcircle, invisible, blue); draw(A--R--E, heavygreen); pair P = extension(M, N, A, A+dir(90)*A); draw(A--P--N, red); filldraw(circumcircle(A, M, N), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavygreen);

filldraw(circumcircle(M, N, E), invisible, heavycyan);

dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(145)); dot("$N$", N, dir(20)); dot("$O$", O, dir(315)); dot("$H$", H, dir(H)); dot("$E$", E, dir(40)); dot("$F$", F, dir(F)); dot("$R$", R, dir(R)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P));

/* TSQ Source:

A = dir 125 B = dir 210 C = dir 330 M = midpoint A--B R145 N = midpoint A--C R20 O = origin R315 H = A+B+C

A--B--C--cycle blue E = foot B A C R40 F = foot C A B B--E lightblue C--F lightblue R = extension E F B C Q = -A+2*foot O A R

unitcircle 0.1 lightcyan / blue A--R--E heavygreen P = extension M N A A+dir(90)*A A--P--N red circumcircle A M N 0.1 lightred / red circumcircle A E F 0.1 lightgreen / heavygreen

circumcircle M N E 0.1 lightcyan / heavycyan

*/ [/asy]

Note that
  • $PA^2 = PM \cdot PN$, so $P$ lies on the radical axis of $\Gamma$ and $\gamma$.
  • $RA \cdot RQ = RE \cdot RF$, so $R$ lies on the radical axis of $\Gamma$ and $\gamma$.
Thus $\overline{PR}$ is the radical axis of $\Gamma$ and $\gamma$, which is evidently perpendicular to $\overline{OH}$.

Remark: In fact, by power of a point one may also observe that $R$ lies on $\overline{BC}$, since it is on the radical axis of $(AQFHE)$, $(BFEC)$, $(ABC)$. Ironically, this fact is not used in the solution.

Second solution (barycentric coordinates) Again note first $R \in \overline{BC}$ (although this can be avoided too). We compute the points in much the same way as before. Since $\overline{AP} \cap \overline{BC} = (0 : b^2 : -c^2)$ we have \[ P = \left( b^2-c^2 : b^2 : -c^2  \right) \](since $x=y+z$ is the equation of line $\overline{MN}$). Now in Conway notation we have \[ R = \overline{EF} \cap \overline{BC} = (0 : S_C : -S_B) 	=\left( 0 : a^2+b^2-c^2 : -a^2+b^2-c^2 \right). \]Hence \[ \overrightarrow{PR} 	= \frac{1}{2(b^2-c^2)} \left( b^2-c^2 , c^2-a^2 , a^2-b^2 \right).  \]On the other hand, we have $\overrightarrow{OH} = \vec A + \vec B + \vec C$. So it suffices to check that \[ \sum_{\text{cyc}} a^2\left( (a^2-b^2) + (c^2-a^2) \right) = 0 \]which is immediate.

Third solution (complex numbers) Let $ABC$ be the unit circle. We first compute $P$ as the midpoint of $A$ and $\overline{AA} \cap \overline{BC}$: \begin{align*} 	p &= \frac{1}{2} \left( a + \frac{a^2(b+c)-bc\cdot2a}{a^2-bc} \right) \\ 	&= \frac{a(a^2-bc)+a^2(b+c)-2abc}{2(a^2-bc)}. \end{align*}Using the remark above, $R$ is the inverse of $D$ with respect to the circle with diameter $\overline{BC}$, which has radius $\left\lvert \frac{1}{2}(b-c) \right\rvert$. Thus \begin{align*} 	r - \frac{b+c}{2} 	&= \frac{\frac14(b-c)\left( \frac1b-\frac1c \right)} 	{\overline{\frac{1}{2}\left( a-\frac{bc}{a} \right)}} \\ 	r &= \frac{b+c}{2} 	+ \frac{-\frac{1}{2} \frac{(b-c)^2}{bc}}{\frac1a-\frac{a}{bc}} \\ 	&= \frac{b+c}{2} + \frac{a(b-c)^2}{2(a^2-bc)} \\ 	&= \frac{a(b-c)^2+(b+c)(a^2-bc)}{2(a^2-bc)}. \end{align*}Expanding and subtracting gives \[ p-r = \frac{a^3-abc-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} 	= \frac{(a+b+c)(a-b)(a-c)}{2(a^2-bc)} \]which is visibly self-conjugate once the factor of $a+b+c$ is deleted.

(Actually, one can guess this factorization ahead of time by noting that if $A=B$, then $P=B=R$, so $a-b$ must be a factor; analogously $a-c$ must be as well.)
This post has been edited 3 times. Last edited by v_Enhance, Dec 23, 2019, 12:40 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WizardMath
2487 posts
#4 • 4 Y
Y by myang, samrocksnature, son7, Adventure10
My solution:

It is well known that $R$ is on the radical axis of the circumcircle and the npc of $ABC$. (Proof: radical axis on the these circles and the circle with $BC$ as a diameter.). Now doing the same trick with the circle with $AO$ as diameter and the npc and circumcircle of $ABC$ does the job as circle with diameter $AO$ is tangent to the circumcircle of $ABC$ by a simple homothety at $A$ or using collinearity of their centers.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BobsonJoe
811 posts
#6 • 3 Y
Y by samrocksnature, Adventure10, Mango247
see here
https://artofproblemsolving.com/community/c312162h1469210
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EulerMacaroni
851 posts
#8 • 6 Y
Y by anantmudgal09, Wizard_32, samrocksnature, son7, Adventure10, Mango247
What a beautiful problem! I add my sol

Let $X\equiv MN\cap EF$, $Y\equiv ME\cap NF$; then $AX$ is the polar of $Y$ with respect to the nine-point circle by Brokard's theorem. In addition, we know by Pappus that $Y\in OH$, so $AX$ is perpendicular to the Euler line. Then since $R\in BC$ we have $\overline{PX}$ bisects $\overline{AR}$ and from $AP\parallel RX$ we know quadrilateral $APRX$ is a parallelogram, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PepsiCola
85 posts
#9 • 4 Y
Y by samrocksnature, SerdarBozdag, Adventure10, Mango247
Draw the $9$-point circle, $(N)$. Radical Axis Theorem on $(ABC), (AMN), (N)$ implies $P$ lies on the radical axis of $(ABC)$ and $(N)$. Radical Axis Theorem on $(AEFH), (ABC), (N)$ implies $R$ lies on the radical axis of $(ABC)$ and $(N)$.
Then the radical axis of $(ABC), (N)$ is just $PR$, so $PR$ is perpendicular to the line connecting their centers, which is just the Euler Line! Done!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
droid347
2679 posts
#13 • 2 Y
Y by samrocksnature, Adventure10
Let $X$ be the midpoint of $BC$ and $Y=MN\cap AR$. $Q$ is the Miquel point of cyclic $BFEC$ so $R\in BC$ and $XQ\perp QR$. Let $L$ be the antipode of $A$; by the tangent $\angle XLO=\angle PAY$ and we can chase that $\angle XOL=\angle YPA=|B-C|$ so $\triangle XOL\sim \triangle YPA$. However, $HL=2XL$ and $RA=2YA$ so $\triangle HOL\sim \triangle RPA$ and thus $XQ\perp QR\implies PR\perp OH$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
enhanced
515 posts
#18 • 6 Y
Y by One2three, Vrangr, brokendiamond, amar_04, samrocksnature, Adventure10
We will prove a more general claim

In a $\Delta ABC$ with circumcenter $O$ and circumcircle $\Gamma$ and Let $K$ be a point on the line $AO$ and $K^*$ denote the isogonal conjugate of $K$ with respect to $\Delta ABC$ and let $O_1$ be the midpoint of $KK^*$ and $M,N$ are projection of $P$ in $AB,AC$ and $E,F$ are projection of $K^*$ in $AB,AC$ let tangent at $A$ to $\Gamma$ meet $MN$ at $P$ and $(AEF)\cap (ABC)=A,Q$ and $EF\cap AQ=R$ prove that $PR\perp OO_1$

PROOF
By six point circle theorem we see that $(EFMN)$ is cyclic with center $O_1$ and denote it's circumcircle by $\gamma$
Since $K \in AO$ it is easy to see that $MN\parallel BC$ so $(ABC),(AMNK)$ are tangent to each other,on the other hand
By power of a point
$(RQ)(RA)=(RE)(RF)$
$PA^2=(PM)(PN)$
so $PR$ is the radical axis of $\Gamma,\gamma$,so $PR\perp OO_1$
and we are done $\blacksquare$
When,$K\equiv O$,we get the problem ;)
This post has been edited 9 times. Last edited by enhanced, Dec 14, 2017, 3:05 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PROF65
2016 posts
#19 • 3 Y
Y by samrocksnature, Adventure10, Mango247
enhanced wrote:
...
$(PM)(PN)=PA^2$
this result doesn't occur unless $NM\parallel BC$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Magikarp1
587 posts
#20 • 5 Y
Y by wu2481632, amar_04, benk25, samrocksnature, Adventure10
$M$ and $N$ are the midpoints of $AB$ and $AC$ so $NM\parallel BC$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
One2three
17 posts
#21 • 2 Y
Y by samrocksnature, Adventure10
@PROF65 I don't see any flaw in enhanced's proof,
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
tworigami
844 posts
#22 • 2 Y
Y by samrocksnature, Adventure10
Since there was some controversy, here is a little bit more explanation: (Basically v_enhance's solution)

We have $AMON$ cyclic and internally tangent to $\Gamma$ as they share a diameter. Hence, we have from power of a point,

$PA^2 = PM \cdot PN$.

Additionally, let $\gamma$ denote the nine-point circle of $\triangle{ABC}$. One key lemma which was left out as it is very well known is that the nine-point circle of a triangle is the triangle passing through the three midpoints, the altitudes, and the midpoints of $AH, BH, CH$. It is easy to see from a homothety centered at $H$ with dilation factor $2$ that the center $N$ of the nine point circle is the midpoint of $OH$, hence $O, N, H$ are collinear. Also, it is ovious that $AEHF$ is cyclic, so from power of a point, we also have

$RA \cdot RQ = RE \cdot RF$.

The key observation is that $Pow_{\gamma}P = Pow_{\Gamma}P$, and similarly, $Pow_{\gamma}R = Pow_{\Gamma}R$. Thus, both points lie on the radical axis of circles $\Gamma, \gamma$, and as two points uniquely determine a line, $PR$ is the radical axis of $\Gamma$ and $\gamma$. The definition of the radical axis is that it is perpendicular to the line connecting the centers of the circles, so we see that $\overline{OH} \perp \overline{PR}$ and from the previous collinearity with $H$, the desired result follows. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_pi_rate
1218 posts
#23 • 2 Y
Y by samrocksnature, Adventure10
Nice and easy. Here's my solution: Let $AP \cap BC=T$, and let $X$ be the point such that $ATRX$ is a parallelogram. Also let $K$ be the midpoint of $RX$. From here, we get that $X$ lies on $EF$, and so $K$ is the point where $EF$ and $MN$ meet. But, as $P$ is the midpoint of $AT$, we get that $APRK$ is also a parallelogram, or equivalently that $AK \parallel PR$. Restating the problem in terms of $\triangle DEF$, where $D$ is the foot of $A$-altitude, we are left with the following problem:-
New problem wrote:
Let $O$ be the circumcenter, $\omega$ be the circumcircle, $I$ be the incenter, $I_A$ be the $A$-excenter of $\triangle ABC$. Let $BI_A \cap \omega=M$ and $CI_A \cap \omega=N$. Finally let $BC \cap MN=K$. Then show that $I_AK$ is perpendicular to line $OI$.
Let $BN \cap CM=L$. Note that, by Brokard's Theorem, $I_AK$ is the polar of $L$ wrt $\omega$, and so $I_AK$ is perpendicular to $OL$. So it suffices to show that $O,I,L$ are collinear. Let $BI \cap \omega=U$ and $CI \cap \omega=V$. Then by Pascal's Theorem on $UMCVNB$, we get the desired result. $\blacksquare$
This post has been edited 1 time. Last edited by math_pi_rate, Nov 9, 2018, 5:39 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RC.
439 posts
#24 • 5 Y
Y by AlastorMoody, MarkBcc168, samrocksnature, Adventure10, Mango247
Let \(A'\) denote the antipode of \(A\) in \(\Gamma\) and \(AP \cap BC= X\). Then, \(\Delta AHA' \sim \Delta ARX \Rightarrow \angle AHO= \angle PRX \therefore PR\perp OH.\)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9767 posts
#25 • 2 Y
Y by samrocksnature, Adventure10
Dear RC.
very nice and elegant proof...

Sincerely
Jean-Louis
Z K Y
G
H
=
a