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

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

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

Be sure to mark your calendars for the following upcoming events:
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
An NT for a break
reni_wee   1
N 10 minutes ago by reni_wee
Source: ONTCP 2.4.1
Prove that there are no positive integers $x,k$ and $n \geq 2$ such that $x^2+1 = k(2^n -1)$.
1 reply
reni_wee
13 minutes ago
reni_wee
10 minutes ago
Inequality
lgx57   0
16 minutes ago
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
0 replies
lgx57
16 minutes ago
0 replies
p divides x^x-c
mistakesinsolutions   6
N 20 minutes ago by reni_wee
Show that for integer c and a prime p, $ p |x^x-c $ has a solution
6 replies
mistakesinsolutions
Jun 13, 2023
reni_wee
20 minutes ago
exponential diophantine in integers
skellyrah   1
N 22 minutes ago by skellyrah
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
1 reply
skellyrah
Yesterday at 7:04 PM
skellyrah
22 minutes ago
IMO 2017 Problem 4
Amir Hossein   117
N 27 minutes ago by ezpotd
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
117 replies
Amir Hossein
Jul 19, 2017
ezpotd
27 minutes ago
x^2+y^2+z^2+xy+yz+zx=6xyz diophantine
parmenides51   7
N an hour ago by Assassino9931
Source: Greece Junior Math Olympiad 2024 p4
Prove that there are infinite triples of positive integers $(x,y,z)$ such that
$$x^2+y^2+z^2+xy+yz+zx=6xyz.$$
7 replies
parmenides51
Mar 2, 2024
Assassino9931
an hour ago
Turkish JMO 2025?
bitrak   1
N 2 hours ago by blug
Let p and q be prime numbers. Prove that if pq(p+ 1)(q + 1)+ 1 is a perfect square, then pq + 1 is also a perfect square.
1 reply
bitrak
Yesterday at 2:04 PM
blug
2 hours ago
Combi Algorithm/PHP/..
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
5. [Czech and Slovak Republics 1997]
Each side and diagonal of a regular n-gon (n ≥ 3) is colored blue or green. A move consists of choosing a vertex and
switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the number of blue segments incident to each vertex even by following a sequence of moves. Also show that the final configuration obtained is uniquely determined by the initial coloring.
0 replies
CatalanThinker
2 hours ago
0 replies
Combi Proof Math Algorithm
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
3. [Russia 1961]
Real numbers are written in an $m \times n$ table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
0 replies
CatalanThinker
2 hours ago
0 replies
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   1
N 2 hours ago by Ritwin
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
1 reply
Primeniyazidayi
2 hours ago
Ritwin
2 hours ago
Easy Taiwanese Geometry
USJL   14
N 2 hours ago by Want-to-study-in-NTU-MATH
Source: 2024 Taiwan Mathematics Olympiad
Suppose $O$ is the circumcenter of $\Delta ABC$, and $E, F$ are points on segments $CA$ and $AB$ respectively with $E, F \neq A$. Let $P$ be a point such that $PB = PF$ and $PC = PE$.
Let $OP$ intersect $CA$ and $AB$ at points $Q$ and $R$ respectively. Let the line passing through $P$ and perpendicular to $EF$ intersect $CA$ and $AB$ at points $S$ and $T$ respectively. Prove that points $Q, R, S$, and $T$ are concyclic.

Proposed by Li4 and usjl
14 replies
USJL
Jan 31, 2024
Want-to-study-in-NTU-MATH
2 hours ago
Problem 7
SlovEcience   6
N 3 hours ago by Li0nking
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
6 replies
SlovEcience
May 14, 2025
Li0nking
3 hours ago
Strange circles in an orthocenter config
VideoCake   1
N 3 hours ago by KrazyNumberMan
Source: 2025 German MO, Round 4, Grade 12, P3
Let \(\overline{AD}\) and \(\overline{BE}\) be altitudes in an acute triangle \(ABC\) which meet at \(H\). Suppose that \(DE\) meets the circumcircle of \(ABC\) at \(P\) and \(Q\) such that \(P\) lies on the shorter arc of \(BC\) and \(Q\) lies on the shorter arc of \(CA\). Let \(AQ\) and \(BE\) meet at \(S\). Show that the circumcircles of \(BPE\) and \(QHS\) and the line \(PH\) concur.
1 reply
VideoCake
Monday at 5:10 PM
KrazyNumberMan
3 hours ago
Inspired by Adhyayan Jana
sqing   0
3 hours ago
Source: Own
Let $a,b,c,d>0,a^2 + d^2+ad = b^2 + c^2  $ aand $ a^2 + b^2 = c^2 + d^2+cd$ Prove that $$ \frac{ab+cd}{ad+bc} =1$$
0 replies
sqing
3 hours ago
0 replies
Another perpendicular to the Euler line
darij grinberg   25
N Apr 30, 2025 by MathLuis
Source: German TST 2022, exam 2, problem 3
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$.

IMAGE
25 replies
darij grinberg
Mar 11, 2022
MathLuis
Apr 30, 2025
Another perpendicular to the Euler line
G H J
Source: German TST 2022, exam 2, problem 3
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darij grinberg
6555 posts
#1 • 11 Y
Y by megarnie, BVKRB-, buratinogigle, Sprites, TheCollatzConjecture, Hedra, HrishiP, HWenslawski, livetolove212, amar_04, RedFlame2112
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$.

[asy]
import olympiad;
unitsize(30);
pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2;
A = (-14.8, -6.6);
B = (-10.9, 0.3);
C = (-3.1, -7.1);
O = circumcenter(A,B,C);
H = orthocenter(A,B,C);
P = 1.2 * H - 0.2 * A;
Q = reflect(A, C) * P;
R = reflect(A, B) * P;
Y = foot(R, C, A);
Z = foot(Q, A, B);
P2 = foot(A, B, C);
Q2 = foot(P, C, A);
R2 = foot(P, A, B);
draw(B--(1.6*A-0.6*B));
draw(B--C--A);
draw(P--R, blue);
draw(R--Y, red);
draw(P--Q, blue);
draw(Q--Z, red);
draw(A--P2, blue);
draw(O--H, darkgreen+linewidth(1.2));
draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2));
draw(rightanglemark(R,Y,A,10), red);
draw(rightanglemark(Q,Z,B,10), red);
draw(rightanglemark(C,Q2,P,10), blue);
draw(rightanglemark(A,R2,P,10), blue);
draw(rightanglemark(B,P2,H,10), blue);
label("$\textcolor{blue}{H}$",H,NW);
label("$\textcolor{blue}{P}$",P,N);
label("$A$",A,W);
label("$B$",B,N);
label("$C$",C,S);
label("$O$",O,S);
label("$\textcolor{blue}{Q}$",Q,E);
label("$\textcolor{blue}{R}$",R,W);
label("$\textcolor{red}{Y}$",Y,S);
label("$\textcolor{red}{Z}$",Z,NW);
dot(A, filltype=FillDraw(black));
dot(B, filltype=FillDraw(black));
dot(C, filltype=FillDraw(black));
dot(H, filltype=FillDraw(blue));
dot(P, filltype=FillDraw(blue));
dot(Q, filltype=FillDraw(blue));
dot(R, filltype=FillDraw(blue));
dot(Y, filltype=FillDraw(red));
dot(Z, filltype=FillDraw(red));
dot(O, filltype=FillDraw(black));
[/asy]
Z K Y
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jayme
9801 posts
#2 • 3 Y
Y by buratinogigle, amar_04, RedFlame2112
Dear Darij,
very nice to hear You again on Mathlinks (AoPS)...

Sincerely
Jean-Louis
Z K Y
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darij grinberg
6555 posts
#3 • 2 Y
Y by amar_04, RedFlame2112
Nice to see you again, Jean-Louis. I checked with your collection before posing this problem :)
This post has been edited 2 times. Last edited by darij grinberg, Mar 11, 2022, 1:55 PM
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799786
1052 posts
#5 • 1 Y
Y by RedFlame2112
Let $P$ varies on the plane. We only need to consider when segment $AP$ cuts segment $BC$.
Claim. (main claim) $YZ$ is perpendicular to a fixed line.
Proof
Then the rest is to move $P$ to a special point, then we are done! :)
(I don't think this is easy)
This post has been edited 2 times. Last edited by 799786, Mar 11, 2022, 2:58 PM
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darij grinberg
6555 posts
#6 • 1 Y
Y by RedFlame2112
wardtnt1234 wrote:
Then the rest is to move $P$ to a special point, then we are done! :)

This is easier said than done!

Checking that $YZ$ is parallel for all points $P$ satisfying $AP \perp BC$ is actually pretty easy without any trigonometry; just observe that if $P$ undergoes a homothety with center $A$, then $Q$, $R$, $Y$ and $Z$ undergo the same homothety. But I don't know of any choice of $P$ (other than $P = A$, which is too degenerate to be useful) that makes the problem significantly simpler.
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799786
1052 posts
#7 • 1 Y
Y by RedFlame2112
@above I have the same thought. I'm trying to find a good choice of position $P$.
(The best position so far I found is $P \equiv H$)
This post has been edited 1 time. Last edited by 799786, Mar 11, 2022, 3:01 PM
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VicKmath7
1391 posts
#8 • 1 Y
Y by RedFlame2112
By the way, it seems that this can be complex bashed (picking $o=0$, $h=a+b+c$), though I will try this later.
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Kvon
3 posts
#9 • 2 Y
Y by chystudent1-_-, RedFlame2112
First of all we define S= RY \cap QZ and X is the the projection of P onto AC (or the midpoint of PQ) and W is the projection of P onto AB. Now notice that PRSQ is a parallelogram and AXPW is a cyclic quad which also means that AXW is similar to ABC.

Let O' and H' be the cirumcenter and orthocenter of AXW. Now notice that O' is the midpoint of AP and H' is the midpoint of PS. To see the latter note that RS is parallel to WH' and PR is parallel to XH'.

Now we know that AS is parallel to O'H'. Now by an easy angle chase, we get:

90° - \angle AYZ=\angle ZYS = \angle ZAS=\angle BAS = \angle (O'H';AB)=\angle (OH;AC)

This means YZ \perp OH and we are finished.
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darij grinberg
6555 posts
#10 • 1 Y
Y by RedFlame2112
Nicely done, Kvon! This is the second of my proposed solutions.
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799786
1052 posts
#11 • 1 Y
Y by RedFlame2112
Continue from #7, let $P \equiv H$ then do like #9
Or try this: (also let $P \equiv H$)
Let $AH$ cuts $BC$ at $D$. $DM$ cuts $AC$ at $C'$, $DN$ cuts $AB$ at $B'$ then use Thales.
This post has been edited 1 time. Last edited by 799786, Mar 12, 2022, 4:08 AM
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jayme
9801 posts
#12 • 1 Y
Y by RedFlame2112
Dear Mathlinkers,

1. J, K the midpoints of AC, AB
2. V, W the feet of the perpendiculars to AC, AB though H.

If we prove that AY/AZ = KW/JV we are done...

Sincerely
Jean-Louis
This post has been edited 1 time. Last edited by jayme, Mar 12, 2022, 7:09 AM
Reason: typo
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jayme
9801 posts
#13 • 1 Y
Y by RedFlame2112
Dear Mathlinkers,

1. (QR) has a fix direction
2. we chioce for P the foot A' of the A-altitude. or H
3. J, K the midpoints of AC, AB
4. V, W the feet of the perpendiculars to AC, AB though A' or H;

one of these two points lead to a not so heavy calculation...

If we prove that AY/AZ = KW/JV we are done...

Sincerely
Jean-Louis
This post has been edited 1 time. Last edited by jayme, Mar 12, 2022, 10:00 AM
Reason: typos
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khina
995 posts
#14 • 1 Y
Y by RedFlame2112
nice

motivation
Z K Y
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EulersTurban
386 posts
#15 • 1 Y
Y by RedFlame2112
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885;  /* image dimensions */

 /* draw figures */
draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); 
draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); 
draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); 
draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); 
draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); 
draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
 /* dots and labels */
dot((-4.356336480424067,10.601149161007438),dotstyle); 
label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); 
dot((-11.291822257619955,-6.524184426443613),dotstyle); 
label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); 
dot((9.757778495559513,-6.875988622852955),dotstyle); 
label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); 
dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); 
label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); 
dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); 
label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); 
dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); 
label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); 
dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); 
label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); 
dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); 
label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); 
dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); 
label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); 
dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); 
label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); 
dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); 
label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); 
dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); 
label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$.

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$.
From here we have that $p=-bc$.

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$, respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$, where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$.

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$, or we could say $q=\overline{r}$, which is going to help us in our calculations since from here we have that:
$$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$$
From here we see that:
$$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$$implying that $YZ \perp OH$, hence we are done :D
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jmiao
8082 posts
#16 • 1 Y
Y by RedFlame2112
jayme wrote:
Dear Mathlinkers,

Sorry if this is taken out of context, but what do you mean by Mathlinkers?
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i3435
1350 posts
#17 • 1 Y
Y by RedFlame2112
Let $O_A$ be the reflection of $O$ over $\overline{BC}$. Let $E,F$ be the feet from $P$ to $\overline{AC},\overline{AB}$ respectively. $EF=FY=EZ$. Let $P_{1}=\overline{CO}\cap\overline{BO_A}$ (a point at infinity), and let $P_2=\overline{BO}\cap\overline{CO_A}$. Then $(\overline{HO},\overline{HO_A});(\overline{HB},\overline{HC});(\overline{HP_1},\overline{HP_2})$ is an involution by DDIT. Let $m$ be the line at infinity. Then projecting from $H$ to $m$, $(\overline{HO}\cap m,\overline{HO_A}\cap m);(\overline{HB}\cap m,\overline{HC}\cap m);(P_1,P_2)$ is an involution. $(\overline{YZ}\cap m,\overline{FE}\cap m);(\overline{EY}\cap m,\overline{FZ}\cap m);(\overline{FY}\cap m,\overline{EZ}\cap m)$ is also an involution by DIT. However $\overline{FE}\perp\overline{HO_A},\overline{EY}\perp\overline{HB},\overline{FZ}\perp\overline{HC},\overline{FY}\perp\overline{CO},\overline{EZ}\perp\overline{BO}$ by angle chasing, so $\overline{YZ}\perp\overline{HO}$ as desired.
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livetolove212
859 posts
#18 • 2 Y
Y by amar_04, RedFlame2112
Nice problem!
I checked other special cases when $P$ moves on fixed line through $A$ and get 2 similar problems:
1) If $AP\perp OH$ then $YZ\perp BC.$
2) If $AP$ passes through nine-point circle's center then $YZ\parallel BC.$
Attachments:
This post has been edited 1 time. Last edited by livetolove212, Mar 14, 2022, 4:35 AM
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jayme
9801 posts
#19 • 1 Y
Y by RedFlame2112
Dear Darij,
has this nice problem an author?

Can you send me a scan of this problem if you have some time at <(jeanlouisayme@yahoo.fr>...

Very sincerely
Jean-Louis
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jayme
9801 posts
#20 • 2 Y
Y by darij grinberg, RedFlame2112
Dear Mathlinkers,

here

then

La perpendiculaire de Darij Grinberg

Sincerely
Jean-Louis
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armpist
527 posts
#21 • 1 Y
Y by RedFlame2112
Dear Darij, J-L, and MLs

Construct two perp lines at each vertex of ABC to the sides at that vertex.
We get two triangles with sides orthogonal to sides of ABC.
Thus their Euler Lines are perp to EL of ABC.
Now when a side of a triangle moves parallel to itself EL of this triangle also
moves parallel to itself.

In this problem we move sides of the two triangles orthogonal to ABC.

Friendly,
M.T.


Darij, is there a connect to your paper with Nikos' solution and JPE's
note about directrix of hyperbola? BTW, where are Nikos and JPE
these days?
This post has been edited 1 time. Last edited by armpist, Apr 5, 2022, 9:19 PM
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darij grinberg
6555 posts
#22 • 1 Y
Y by RedFlame2112
@armpist: What is the exact connection between the two orthogonal triangles and the line $YZ$? How do you make $YZ$ the Euler line of any triangle? (Incidentally, when I was thinking up problems for this exam, one of the things I looked are were these two orthogonal triangles... but I didn't see any relation.)

PS. I just recalled that you asked me a while ago about Thebault's isogonal triangles theorem, but when I wanted to respond, your email wasn't active any more. Here is a recent reference with a very nice proof (using transformations, but it shouldn't be hard to rewrite it in purely Euclidean terms): Waldemar Pompe, Three reflections, 2021-12-19. This was, of course, not the proof I've had in mind, but I doubt mine was any simpler.
This post has been edited 1 time. Last edited by darij grinberg, Apr 5, 2022, 9:20 PM
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Imposter-xDDDDD
3 posts
#23 • 1 Y
Y by Mango247
EulersTurban wrote:
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885;  /* image dimensions */

 /* draw figures */
draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); 
draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); 
draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); 
draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); 
draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); 
draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
 /* dots and labels */
dot((-4.356336480424067,10.601149161007438),dotstyle); 
label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); 
dot((-11.291822257619955,-6.524184426443613),dotstyle); 
label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); 
dot((9.757778495559513,-6.875988622852955),dotstyle); 
label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); 
dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); 
label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); 
dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); 
label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); 
dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); 
label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); 
dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); 
label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); 
dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); 
label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); 
dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); 
label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); 
dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); 
label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); 
dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); 
label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); 
dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); 
label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$.

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$.
From here we have that $p=-bc$.

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$, respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$, where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$.

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$, or we could say $q=\overline{r}$, which is going to help us in our calculations since from here we have that:
$$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$$
From here we see that:
$$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$$implying that $YZ \perp OH$, hence we are done :D
why did u define P? If u define P,your solution is only the case of P to be on $(ABC)$.Can u explain clearly,pls?
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mathleticguyyy
3217 posts
#24
Y by
Imposter-xDDDDD wrote:
EulersTurban wrote:
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -36.44294096692108, xmax = 34.68535963021886, ymin = -20.993035080393877, ymax = 21.535374415023885;  /* image dimensions */

 /* draw figures */
draw(circle((-0.6693999912874199,-0.8590481719792875), 12.038672002589127), linewidth(0.4) + red); 
draw((-4.356336480424067,10.601149161007438)--(-11.291822257619955,-6.524184426443613), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(9.757778495559513,-6.875988622852955), linewidth(0.4)); 
draw((9.757778495559513,-6.875988622852955)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
draw((-0.6693999912874218,-0.8590481719792886)--(-4.551580259909669,-1.0809275443305555), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-4.737241326966146,-12.18963908568131)--(18.00387568922102,6.175538566300643), linewidth(0.4)); 
draw((-4.356336480424067,10.601149161007438)--(-4.737241326966146,-12.18963908568131), linewidth(0.4)); 
draw((-11.291822257619955,-6.524184426443613)--(-12.33939587824527,-9.110888269108811), linewidth(0.4)); 
draw((-19.941550429524398,-6.032137452536313)--(-2.3781638546770263,8.151629966975745), linewidth(0.4)); 
draw((18.00387568922102,6.175538566300643)--(-2.7454920553265385,14.5787001247749), linewidth(0.4)); 
draw((-2.7454920553265385,14.5787001247749)--(-4.356336480424067,10.601149161007438), linewidth(0.4)); 
 /* dots and labels */
dot((-4.356336480424067,10.601149161007438),dotstyle); 
label("$A$", (-4.17520668819245,11.042557225106401), NE * labelscalefactor); 
dot((-11.291822257619955,-6.524184426443613),dotstyle); 
label("$B$", (-11.093037490394702,-6.043092004493747), NE * labelscalefactor); 
dot((9.757778495559513,-6.875988622852955),dotstyle); 
label("$C$", (9.939025284085972,-6.414519161658968), NE * labelscalefactor); 
dot((-4.551580259909669,-1.0809275443305555),linewidth(4pt) + dotstyle); 
label("$H$", (-4.360920266775061,-0.7038266202437011), NE * labelscalefactor); 
dot((-0.6693999912874218,-0.8590481719792886),linewidth(4pt) + dotstyle); 
label("$O$", (-0.4609351165402335,-0.4716846470154382), NE * labelscalefactor); 
dot((-4.737241326966146,-12.18963908568131),linewidth(4pt) + dotstyle); 
label("$P$", (-4.546633845357671,-11.800212940554667), NE * labelscalefactor); 
dot((-19.941550429524398,-6.032137452536313),linewidth(4pt) + dotstyle); 
label("$R$", (-19.77514728913176,-5.671664847328527), NE * labelscalefactor); 
dot((18.00387568922102,6.175538566300643),linewidth(4pt) + dotstyle); 
label("$Q$", (18.203279531012154,6.539002944478101), NE * labelscalefactor); 
dot((-2.3781638546770263,8.151629966975745),linewidth(4pt) + dotstyle); 
label("$Y$", (-2.1787857184293835,8.535423914241163), NE * labelscalefactor); 
dot((-2.7454920553265385,14.5787001247749),linewidth(4pt) + dotstyle); 
label("$Z$", (-2.550212875594605,14.942542375341217), NE * labelscalefactor); 
dot((-12.33939587824527,-9.110888269108811),linewidth(4pt) + dotstyle); 
label("$K$", (-12.160890567244715,-8.735938893941597), NE * labelscalefactor); 
dot((6.6333171811274365,-3.0070502596903337),linewidth(4pt) + dotstyle); 
label("$L$", (6.828322842827241,-2.6538191953611094), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Here is a quick solution using complex numbers.

First of all notice that, by homothethy, if we choose any point $P$ and fix it in place and scale $BC$ up and down $OH$ will scale as well, but $YZ$ stays fixed in place.
Thus now choose $P$ to be on $(ABC)$.

Now let's throw the configuration onto the complex plane, where $(ABC)$ will be the unit circle and $a=1$.
From here we have that $p=-bc$.

Let $K$ and $L$ be the feet from $P$ onto $AB$ and $AC$, respectively. then by definition we must have that $p+r=2k$ and $p+q=2l$, where $k=\frac{1}{2}(1+b+p-\overline{p}b)$ and $l=\frac{1}{2}(1+c+p-\overline{p}c)$.

From here we have that $r=1+b+\frac{1}{c}$ and $q=1+c+\frac{1}{b}$, or we could say $q=\overline{r}$, which is going to help us in our calculations since from here we have that:
$$y=\frac{1}{2}(1+c+r-qc) \; \; \text{and} \; \; z=\frac{1}{2}(1+b+q-rb)$$
From here we see that:
$$\frac{y-z}{\overline{y-z}}=-\frac{\frac{(b-c)(b+c+1)(bc+1)}{bc}}{\frac{(b-c)(bc+b+c)(bc+1)}{bc}} = - \frac{b+c+1}{bc+b+c} = -\frac{h}{\overline{h}}$$implying that $YZ \perp OH$, hence we are done :D
why did u define P? If u define P,your solution is only the case of P to be on $(ABC)$.Can u explain clearly,pls?

Intuitively, Y and Z move linearly at the same velocity as a function of P, and moreover they coincide at A when P=A. This means that, for any choice of P, the line YZ will always have the same slope.
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mikestro
75 posts
#25
Y by
My student's solution:
https://youtu.be/d4B2KZXWV2w
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jkmmm3
56 posts
#26
Y by
It is clear that we just need to solve this problem when $P = H$. Then by homothety, $\angle AZY$ will be constant so $ZY \perp OH$ no matter our choice of $P$.
Since $Q$ and $R$ are the reflections of $H$ across lines $AB$ and $AC$ respectively, they're both on the circumcircle of $\triangle ABC$. We now define points $D$ and $E$ to be the foot of the perpendicular from $C$ to $AB$ and $B$ to $AC$ respectively. Now, let $D'$ the foot of the perpendicular from $D$ to $AC$ and $E'$ be the foot of the perpendicular from $E$ to $AB$. Because $HD = DR$, It follows that $ED' = D'Y$ because $RY \parallel D'D \parallel EH$.

We utilize cartesian coordinates where $A = (0, 0), B = (b,d), C = (c,0)$. The orthocenter is just $H = (b, \frac{bc - b^2}{d})$ by power of point. $OH$ is just the Euler line, so we will use the more convenient centroid $G = (\frac{b+c}{3}, \frac{d}{3})$.

We first calculate $Y$. The line $CD$ has the equation $y = -\frac{b}{d} (x-c)$, while the line $AB$ has the equation $y = \frac{c-b}{d}x$. Solving this system of equations, we get that $D = (\frac{cb^2}{b^2+d^2}, \frac{cbd}{b^2+d^2})$. Now, $AC$ is just the x-axis, so $D' = (\frac{cb^2}{b^2+d^2})$. Reflecting $E = (b,0)$ across $D'$, we get that $Y = (\frac{2cb^2 - b^3 - bd^2}{b^2+d^2}, 0)$.

We now calculate $Z$. Because $EE' \parallel CD$, $\frac{AE'}{AD} = \frac{AE}{AC} = \frac{b}{c}$. So, $\vec{E}' = \frac{b}{c} \vec{D} = (\frac{b^3}{b^2+d^2}, \frac{b^2d}{b^2+d^2})$. Reflecting $D$ across $E'$, we can get that $Z = (\frac{(2b-c)b^2}{b^2+d^2}, \frac{(2b-c)bd}{b^2+d^2})$.

The slope of the line $XY$ can then be calculated to be $\frac{(2b-c)d}{3b^2 - 3bc + d^2}$. We now calculate the slope of the Euler line $GH$, which is just $\frac{\frac{3bc - 3b^2 - d^2}{3d}}{\frac{2b - c}{3}} = \frac{3bc - 3b^2 - d^2}{(2b-c)d}$. Multiplying these slopes results in $-1$, which finishes.
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MathLuis
1556 posts
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Absolutely beautiful problem.
Let $PQ,PR$ meet $AC,AB$ at $V,W$ respectively then let $RY \cap QZ=S$, notice $SRPQ$ is a parallelogram, but also that $BWVC$ is cyxlic from antiparallels or some $\sqrt{bc}$+homothety, let $M,N$ midpoints of $AP,SP$ then $M,N$ are in fact circumcenter, orthocenter of $\triangle AVW$ respectively so now notice internal angle bisector of the angle $\angle (MN,OH)$ is parallel to the external angle bisector of $\angle BAC$, this is due to anti-parallels transform and so if you consider $\infty_{\perp YZ}$ then $A\infty_{\perp YZ}$ is symetric to $AS$ on the external angle bisector of $\angle BAC$ but as a result it is parallel to $OH$ thus $YZ \perp OH$ as desired thus we are done :cool:.
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