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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
hard problem
Cobedangiu   5
N 7 minutes ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
+1 w
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
7 minutes ago
Nordic 2025 P3
anirbanbz   9
N 10 minutes ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
10 minutes ago
Aime type Geo
ehuseyinyigit   1
N 37 minutes ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
37 minutes ago
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N 40 minutes ago by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
40 minutes ago
Which numbers are almost prime?
AshAuktober   5
N an hour ago by Jupiterballs
Source: 2024 Swiss MO/1
If $a$ and $b$ are positive integers, we say that $a$ almost divides $b$ if $a$ divides at least one of $b - 1$ and $b + 1$. We call a positive integer $n$ almost prime if the following holds: for any positive integers $a, b$ such that $n$ almost divides $ab$, we have that $n$ almost divides at least one of $a$ and $b$. Determine all almost prime numbers.
original link
5 replies
AshAuktober
Dec 16, 2024
Jupiterballs
an hour ago
Inequality involving square root cube root and 8th root
bamboozled   1
N an hour ago by arqady
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the minimum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
1 reply
bamboozled
4 hours ago
arqady
an hour ago
If $b^n|a^n-1$ then $a^b >\frac {3^n}{n}$ (China TST 2009)
Fang-jh   16
N 2 hours ago by Aiden-1089
Source: Chinese TST 2009 6th P1
Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n-1,$ then $ a^b > \frac {3^n}{n}.$
16 replies
Fang-jh
Apr 4, 2009
Aiden-1089
2 hours ago
Confusing inequality
giangtruong13   1
N 2 hours ago by Natrium
Source: An user
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum: $$P= \sum_{cyc} \frac{a}{b} + \sum_{cyc} \frac{1}{a^3+b^3+abc}$$
1 reply
giangtruong13
Yesterday at 8:04 AM
Natrium
2 hours ago
Two equal angles
jayme   3
N 2 hours ago by jayme
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
3 replies
jayme
May 2, 2025
jayme
2 hours ago
50 points in plane
pohoatza   13
N 2 hours ago by cursed_tangent1434
Source: JBMO 2007, Bulgaria, problem 3
Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.
13 replies
pohoatza
Jun 28, 2007
cursed_tangent1434
2 hours ago
D1024 : Can you do that?
Dattier   5
N 2 hours ago by SimplisticFormulas
Source: les dattes à Dattier
Let $x_{n+1}=x_n^2+1$ and $x_0=1$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
5 replies
Dattier
Apr 29, 2025
SimplisticFormulas
2 hours ago
Hardest N7 in history
OronSH   25
N 2 hours ago by sansgankrsngupta
Source: ISL 2023 N7
Let $a,b,c,d$ be positive integers satisfying \[\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.\]Determine all possible values of $a+b+c+d$.
25 replies
OronSH
Jul 17, 2024
sansgankrsngupta
2 hours ago
Friends Status are changing
lminsl   64
N 3 hours ago by SteppenWolfMath
Source: IMO 2019 Problem 3
A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time:
[list]
[*] Three users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged.
[/list]
Initially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.

Proposed by Adrian Beker, Croatia
64 replies
lminsl
Jul 16, 2019
SteppenWolfMath
3 hours ago
well-known NT
Tuleuchina   9
N 4 hours ago by Blackbeam999
Source: Kazakhstan mo 2019, P6, grade 9
Find all integer triples $(a,b,c)$ and natural $k$ such that $a^2+b^2+c^2=3k(ab+bc+ac)$
9 replies
Tuleuchina
Mar 20, 2019
Blackbeam999
4 hours ago
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
9791 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
Z K Y
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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
Z K Y
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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.
Z K Y
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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
Z K Y
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VicKmath7
1389 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.
Z K Y
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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.
Z K Y
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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.
Z K Y
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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
Z K Y
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jayme
9791 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
Z K Y
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jayme
9791 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
993 posts
#14 • 1 Y
Y by RedFlame2112
nice

motivation
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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
8083 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
9791 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
9791 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
73 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
1523 posts
#27
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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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