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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:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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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
2025 Zhejiang Women's Mathematical Olympiad ,Q4
sqing   2
N 24 minutes ago by sqing
Source: China
Let $ a_1, a_2,\dots ,a_n\geq 0 $ and $ \sum _{i=1}^{n}a^3_i=n $ $(n\geq 3) .$ Prove that $$\sum_{1\le i<j<k\le n} \frac{1}{n-a_ia_ja_k}\leq \frac{n(n-2)}{6}$$
APMO 2012 #5
Inequalities Marathon
2 replies
1 viewing
sqing
Yesterday at 2:31 PM
sqing
24 minutes ago
Nice inequality
TUAN2k8   1
N 28 minutes ago by sqing
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
1 reply
TUAN2k8
32 minutes ago
sqing
28 minutes ago
Inequality
srnjbr   6
N 31 minutes ago by sqing
For real numbers a, b, c and d that a+d=b+c prove the following:
(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)>=0
6 replies
srnjbr
Oct 30, 2024
sqing
31 minutes ago
easy geo
ErTeeEs06   6
N 33 minutes ago by lksb
Source: BxMO 2025 P3
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\stackrel{\frown}{BC}, \stackrel{\frown}{CA}, \stackrel{\frown}{AB}$ of $\Omega$ not containing $A, B, C$ respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$ and the midpoint $M$ of $EF$ lie on a line.
6 replies
ErTeeEs06
Apr 26, 2025
lksb
33 minutes ago
trigonometric inequality
MATH1945   9
N 33 minutes ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
9 replies
MATH1945
May 26, 2016
sqing
33 minutes ago
Tangents inducing isogonals
nikolapavlovic   56
N an hour ago by Ilikeminecraft
Source: Serbian MO 2017 6
Let $k$ be the circumcircle of $\triangle ABC$ and let $k_a$ be A-excircle .Let the two common tangents of $k,k_a$ cut $BC$ in $P,Q$.Prove that $\measuredangle PAB=\measuredangle CAQ$.
56 replies
nikolapavlovic
Apr 2, 2017
Ilikeminecraft
an hour ago
Elementary Problems Compilation
Saucepan_man02   25
N an hour ago by trangbui
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
25 replies
Saucepan_man02
Monday at 1:44 PM
trangbui
an hour ago
partitioning 1 to p-1 into several a+b=c (mod p)
capoouo   5
N an hour ago by NerdyNashville
Source: own
Given a prime number $p$, a set is said to be $p$-good if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod{p}$.
Find all prime number $p$ such that $\{ 1, 2, \cdots, p-1 \}$ can be partitioned into several $p$-good sets.

Proposed by capoouo
5 replies
capoouo
Apr 21, 2024
NerdyNashville
an hour ago
Not homogenous, messy inequality
Kimchiks926   11
N an hour ago by Learning11
Source: Latvian TST for Baltic Way 2019 Problem 1
Prove that for all positive real numbers $a, b, c$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ the following inequality holds:
$$3(ab+bc+ca)+\frac{9}{a+b+c} \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2)+1$$
11 replies
Kimchiks926
May 29, 2020
Learning11
an hour ago
Problem 5
blug   2
N an hour ago by Jjesus
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
2 replies
+1 w
blug
May 19, 2025
Jjesus
an hour ago
D,E,F are collinear.
TUAN2k8   0
an hour ago
Source: Own
Help me with this:
0 replies
TUAN2k8
an hour ago
0 replies
NT Game in Iran TST
M11100111001Y1R   6
N 2 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 2
Suppose \( p \) is a prime number. We have a number of cards, each of which has a number written on it such that each of the numbers \(1, \dots, p-1 \) appears at most once and $0$ exactly once. To design a game, for each pair of cards \( x \) and \( y \), we want to determine which card wins over the other. The following conditions must be satisfied:

$a)$ If card \( x \) wins over card \( y \), and card \( y \) wins over card \( z \), then card \( x \) must also win over card \( z \).

$b)$ If card \( x \) does not win over card \( y \), and card \( y \) does not win over card \( z \), then for any card \( t \), card \( x + z \) must not win over card \( y + t \).

What is the maximum number of cards such that the game can be designed (i.e., one card does not defeat another unless the victory is symmetric or consistent)?
6 replies
M11100111001Y1R
Yesterday at 6:19 AM
sami1618
2 hours ago
Brilliant Problem
M11100111001Y1R   1
N 2 hours ago by aaravdodhia
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
1 reply
M11100111001Y1R
Yesterday at 7:28 AM
aaravdodhia
2 hours ago
Problem 3
blug   2
N 3 hours ago by LeYohan
Source: Polish Junior Math Olympiad Finals 2025
Find all primes $(p, q, r)$ such that
$$pq+4=r^4.$$
2 replies
blug
Mar 15, 2025
LeYohan
3 hours ago
AD is Euler line of triangle IKL
VicKmath7   16
N May 1, 2025 by ErTeeEs06
Source: IGO 2021 Advanced P5
Given a triangle $ABC$ with incenter $I$. The incircle of triangle $ABC$ is tangent to $BC$ at $D$. Let $P$ and $Q$ be points on the side BC such that $\angle PAB = \angle BCA$ and $\angle QAC = \angle ABC$, respectively. Let $K$ and $L$ be the incenter of triangles $ABP$ and $ACQ$, respectively. Prove that $AD$ is the Euler line of triangle $IKL$.

Proposed by Le Viet An, Vietnam
16 replies
VicKmath7
Dec 30, 2021
ErTeeEs06
May 1, 2025
AD is Euler line of triangle IKL
G H J
G H BBookmark kLocked kLocked NReply
Source: IGO 2021 Advanced P5
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VicKmath7
1391 posts
#1 • 1 Y
Y by amar_04
Given a triangle $ABC$ with incenter $I$. The incircle of triangle $ABC$ is tangent to $BC$ at $D$. Let $P$ and $Q$ be points on the side BC such that $\angle PAB = \angle BCA$ and $\angle QAC = \angle ABC$, respectively. Let $K$ and $L$ be the incenter of triangles $ABP$ and $ACQ$, respectively. Prove that $AD$ is the Euler line of triangle $IKL$.

Proposed by Le Viet An, Vietnam
This post has been edited 4 times. Last edited by VicKmath7, Oct 16, 2023, 5:55 PM
Z K Y
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827681
163 posts
#2
Y by
I was really really rude

Sorry @Vickmath
This post has been edited 1 time. Last edited by 827681, Oct 13, 2022, 2:06 PM
Z K Y
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827681
163 posts
#4
Y by
VicKmath7 wrote:
Yes, of course I know, but I believe that everyone who knows what IGO is will understand what A means in this case. Anyways, I will correct it.

Yes but even I got confused at first sight,it's reccomendable to write the full form cuz there are a lot of people who don't know abt IGO
Z K Y
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MP8148
888 posts
#5 • 2 Y
Y by leon.tyumen, MS_asdfgzxcvb
Not the best solution, but I guess it works. Ignore config issues (I don't think there are any but who knows).

diagram

Let $O,H,G$ denote the circumcenter, orthocenter, and centroid of $\triangle IKL$, respectively. We will show that
  • $A$, $O$, $H$ are collinear
  • $A$, $G$, $D$ are collinear
which imply the conclusion.

proof that A-O-H
proof that A-G-D

Additional fact: $\overline{KL}$, $\overline{OX}$, $\overline{BC}$ seem to concur, although I didn't use/prove it.
This post has been edited 1 time. Last edited by MP8148, Dec 31, 2021, 2:14 AM
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MatteD
90 posts
#6 • 4 Y
Y by Acrylic2005, SerdarBozdag, toilaDang, PHSH
[asy]
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import graph; size(12.cm); 
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 = 4.4, xmax = 9.2, ymin = 0.8, ymax = 4.;  /* image dimensions */
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draw((9.,1.)--(6.109430702956201,3.762535441352023), linewidth(1.2) + xdxdff); 
draw((6.109430702956201,3.762535441352023)--(6.793968020425234,0.9976472566588956), linewidth(1.2) + xdxdff); 
draw((6.109430702956201,3.762535441352023)--(5.430792468160217,0.9961934236079506), linewidth(1.2) + xdxdff); 
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draw((9.,1.)--(7.711717014885521,0.9986260402193841), linewidth(1.2) + xdxdff); 
draw((4.520861536062415,0.9952229780472427)--(5.827852937434829,1.7529086067332473), linewidth(1.2) + xdxdff); 
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draw((6.355540510643266,2.0588182632439107)--(6.6393827585556835,1.9451703298270977), linewidth(1.2) + green); 
draw((6.6393827585556835,1.9451703298270977)--(9.,1.), linewidth(1.2) + xdxdff); 
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dot((6.109430702956201,3.762535441352023),linewidth(3.pt) + dotstyle); 
label("$A$", (6.128176663628198,3.7891605035276705), NE * labelscalefactor); 
dot((4.520861536062415,0.9952229780472427),linewidth(3.pt) + dotstyle); 
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dot((9.,1.),linewidth(3.pt) + dotstyle); 
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 /* end of picture */
[/asy]
Let $M,N$ denote the reflections of $A$ with respect to $BI,CI$ respectively, and let $G,H$ be the centroid and orthocenter of $\triangle AMN$.
First, notice that by symmetry $IM=IA=IN$, so $I$ is the circumcenter of $\triangle AMN$ and $D$ is the midpoint of $MN$.
Since $\triangle BAC \sim \triangle BPA$, we have $\frac{BK}{BI}=\frac{BA}{BC}=\frac{BM}{BC}$, so $MK$ is parallel to $CI$. Since $CI \perp AN$, we also have $MK \perp AN$, and $NL \perp AM$ similarly, so $H=NL \cap MK$. $KILH$ is a parallelogram, and it is well-known that $\frac{IG}{IH}=\frac{1}{3}$, so $G$ is the centroid of $\triangle IKL$. The similarity of $\triangle BAC$ and $\triangle BPA$ also gives $\angle AKI = \angle AKB = \angle BIC$, and similarly we have $\angle BIC= \angle ILA$, which is enough to know that $A$ lies on the Euler line of $\triangle IKL$. Since $G \in AD$, we have that $AD$ is the Euler line of $\triangle IKL$. $\blacksquare$
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buratinogigle
2400 posts
#7 • 4 Y
Y by parmenides51, PHSH, hakN, MS_asdfgzxcvb
Here is my simple proof using barycentric coordinates.
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alinazarboland
172 posts
#8
Y by
Here's a solution me and my friends found at our 3 A.M :alien:
Claim 1.In $\triangle ABC$ ,let $P$ be a point such that $\angle PBA=\angle PCA=180-A$. Then $P$ is on the Euler line of $\triangle ABC$.
Proof.It's well-known that the locus of points $X$ such that $\angle XBA=\angle XCA$ is a hyperbola passing through $A,B,C,H,A'$ where $H$ is the orthocenter of $ABC$ and $A'$ is the antipod of $A$ in $(ABC)$ . So $P$ is on this hyperbola. Also, by the definition of $P$, we have $PC \cap AB$ lies on the perpendicular bisector of $AC$ . Now , applying Pascal on the hyperbola $AA'CPHBA$ we'll get that a line through $PC \cap AB$ and $HP \cap AA'$ is perpendicular to $AC$ and since $PC \cap AB$ lies on the perpendicular bisector of $AC$, $HP \cap AA'$ is in fact the circumcenter of $(ABC)$ . So $H,O,P$ are collinear and we're done.

Claim2 .In $\triangle ABC$ with incenter $I$ , If a circle with center $I$ intersects $AB,BC,CA$ at $X,{Y,Z},T$ , then $XY,ZT,AD$ are concurrent.
Proof. Let the lines through $D$parallel to $AB,AC$ intersects $XY,ZT$ at $E,F$ respectively. Since $CT=CZ$ and $DF||AC$ , we have $DZ=DF$, and similarly , $DE=DF$ . On the other hand , since $IY=IZ$ ,it follows that $DY=DZ$ . So $DE=DF$ and $\triangle DEF$ is an isosceles triangle. Also , with a little angle chasing , $AXY$ is an isosceles to , so since $DE||AB$ and $DF||AC$ , $\triangle AXY$ and $\triangle DEF$ are homothetic. the conclusion follows.


Back to the problem.

By Claim 1 , Since $\angle AKI = \angle ALI = 90 - \angle A /2 = 180 - \angle KIL$ , $A$ lies on the Euler line of $\triangle KIL$.

Now , let the perpendicular bisector of $AI$ intersects $AC,AB$ at $M,N$ respectively. Since $MI=MA$ and $\angle AMI = 2\angle ALI$ , $M$ is the circumcenter of $\triangle ALI$. Similarly , $N$ is the circumcenter of $\triangle AKI$ . Now , let $M',N'$ be the reflection of $M,N$ to $IL,IK$ respectively . Since $MI=MK$ , $M'I = M'L$ and similar for $N$ . But $M',N'$ lie on $BC$ . So since $I$ is the center of $(MNM'N')$ , and since $MM',NN'$ are the perpendicular bisectors of $IL,IK$ respectively , by claim 2 , the circumcenter of $\triangle IKL$ lies on $AD$.

So $AD$ is the Euler line of $\triangle IKL$ and we're done.
This post has been edited 1 time. Last edited by alinazarboland, Jan 11, 2022, 8:25 PM
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Fedor Bakharev
181 posts
#9 • 1 Y
Y by amar_04
Proposed by Le Viet An -Vietnam
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LoloChen
479 posts
#11 • 1 Y
Y by buratinogigle
It seems that this one is not hard to bash.
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PHSH
60 posts
#12
Y by
motivation

We have that $APB$ is opositive similar to $CAB,$ hence ${KB}{/IB} = {AB}{/BC}.$ Hence, if we define $X$ as the reflection of $A$ over line $BI,$ we will have $KX \parallel IC.$ Similarly, $Y$ is the reflection of $A$ with respect to $CI,$ and $L$ is the intersection of $CI$ and the parallel of $BI$ through $Y.$ We are going to take this redefinition later on.

[asy]
import olympiad;
import graph;
import geometry;
size(250);
pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt;
A = dir(-240.54);
B = dir(-170.37);
C = dir(-10.43);
O = (0,0);
I = incenter(A,B,C);
D = foot(I,B,C);
X= 2*foot(A,B,I) - A;
Y = 2*foot(A,C,I) - A;
Bt= B*0.5 + I*0.5;
Mt = extension(I,Y,C,Bt);
L =extension(B,Mt,C,I);
Ct= C*0.5 + I*0.5;
Nt = extension(I,X,B,Ct);
K = extension(C,Nt,B,I);

dot("$K$",K,dir(120));
dot("$L$",L, dir(60));
dot("$Y$", Y, dir(-90));
dot("$X$", X, dir(X));
dot("$B$",B, dir(-90));
dot("$C$",C, dir(-90));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
draw(circle(I,distance(I,D)),dotted + green);
draw(A--B--C--cycle,gray);
draw(B--I,blue);
draw(C--I,blue);
draw(K--X, dotted+magenta);
draw(Y--L,dotted+magenta);
dot("$A$",A, dir(A)); 
[/asy]

Note now, that on triangle $\triangle AXY,$ we have that $BI$ and $CI$ are the perpendicular bissectors of $AX$ and $AY,$ respectively. Furthermore, $YL$ and $XK$ are altitudes of $\triangle AXY.$ So the new problem reads as follows

Let $\triangle AXY$ be an triangle with circumcenter $I$. Point $K$ is the intersection of the $X$-altitude with the perpendicular bissector of $AX;$ similarly point $L$ is the intersection of the $Y$-altitude with the perpendicular bissector of $AY.$ If $D$ is the midpoint of side $XY,$ show that $AD$ is the euler line of thriangle $IKL.$


So let $H$ be the orthocenter of $\triangle AXY.$ Also let $M$ and $N$ be the midpoints of sides $AX$ and $AY,$ respectively, and let $E$ and $F$ be the foots of $X$ and $Y$ respectively. Also let $T$ be the orthocenter of $IKL$

[asy]
import olympiad;
import graph;
import geometry;
size(250);
pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt;
A = dir(-240.54);
B = dir(-170.37);
C = dir(-10.43);
O = (0,0);
I = incenter(A,B,C);
D = foot(I,B,C);
X= 2*foot(A,B,I) - A;
Y = 2*foot(A,C,I) - A;
Bt= B*0.5 + I*0.5;
Mt = extension(I,Y,C,Bt);
L =extension(B,Mt,C,I);
Ct= C*0.5 + I*0.5;
Nt = extension(I,X,B,Ct);
K = extension(C,Nt,B,I);

pair M,N,E,F,H,T;
M = A*0.5 + X*0.5;
N = A*0.5 + Y*0.5;
E = foot(X,A,Y);
F = foot(Y,A,X);
H = extension(E,X,F,Y);
T = extension(A,D,K,A+K-Y);

draw(A--D, dotted);
draw(K--T--L,magenta);
draw(circumcircle(A,F,N), dotted + green);
draw(circumcircle(A,E,M),dotted + blue);
draw(A--X--Y--cycle,gray);
draw(M--I--K--cycle,orange);
draw(N--I--L--cycle,orange);
draw(X--E,red);
draw(F--Y,red);
dot("$A$", A, dir(A));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$D$",D, dir(-90));
dot("$M$",M, dir(30));
dot("$N$", N, dir(N));
dot("$E$",E, dir(E));
dot("$F$",F, dir(-10));
dot("$I$",I, dir(90));
dot("$K$",K,dir(260));
dot("$L$",L,-dir(L));
dot("$H$",H,dir(-90));
dot("$T$",T, dir(45));
[/asy]
First, note that the centroid of $\triangle IKL$ lies on $AD.$ Indeed, by definition $IKHL$ is an parallelogram, so the centroid of $\triangle IKL$ divides the segment $IH$ in the ratio $1{/3};$ hence the centroid of $IKL$ coincides with the centroid of $AXY$ which indeed lies on the $A$-median, $AD$.

Now it is sufficient to show that $T$ lies on $AD.$ On one hand, $AD$ is the $A$-median, so $T$ lies on $AD$ if and only if $[ATY]=[ATX].$ On the other hand, by construction $TK \parallel AY$ and $TL \parallel AX$ and hence $[ATY] = [AKY]$ and $[ATX] =[ALX]. $ Then it reduces to $$[AKY] = [ALX] \iff KE \cdot AY = LF \cdot AX.$$
So let $b =AX, c =AY,  p = XE, q = YF.$ Since $AEKM$ is cyclic by construction, it follows by power of a point that
$$ XH \cdot XE = XM \cdot XA = \frac{b^2}{2} \Rightarrow HE = \frac{2p^2 -b^2}{2p} \Rightarrow KE\cdot AY = \frac{(2p^2-b^2)c}{2p}$$Similarly, $LF \cdot AX =  \frac{(2q^2-c^2)b}{2q} $ and it is sufficient to show
$$(2p^2 -b^2)cq =(2q^2-c^2)bp $$which is obviously since $p = b\cdot \sin \angle XAY$ and $q = c \cdot \sin \angle XAY.$
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strong_boy
261 posts
#13 • 1 Y
Y by Mango247
official solution !
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FairyBlade
55 posts
#14
Y by
Consider point $R, S$ on side $AB, AC$ s.t. $RIA=SIA=90$, then i claim that $(RKIA)$ cyclic:
$$\angle AKI=\beta/ 2+\gamma /2=90-\alpha /2 = \angle ARI$$similarly the other one.now said $O_b, O_c$ the midpoint of $AR, AS$ we have that the axis of $KI, LI$ pass through this points.
Now consider point $X, Y$ as the second intersection of $(RKIA),(SLIA)$ with $AC, AB$. I claim that the height from $K$ and $L$ in triangle $\triangle KIL$ intersect side $AC, AB$ on $X,Y$:
$$\angle IKX=\angle AIX=\alpha/2=\angle KIL-90$$Now consider the point $E, F$ as the projection on the side $AB, AC$ of $I$, then we have that the bisector $BI\perp ED$ and so $ED$ is parallel to the axis of $KI$, and the height of $L$ wrt. $IK$. Now the problem can be finished by noting that $AX/AY=AO_b/AO_c=AE/AF=1$ and so by homothety in $A$ the point $O,H,D$ are aligned.
This post has been edited 2 times. Last edited by FairyBlade, Sep 26, 2022, 2:45 PM
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squarc_rs3v2m
46 posts
#15
Y by
Let $G, O, H$ be the centroid, circumcenter, orthocenter of $\triangle IKL$.
Claim 1: $A \in OH$.
proof
Claim 2: $G \in AD$.
proof
Combining these two claims clearly solves the problem.
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StefanSebez
53 posts
#16 • 1 Y
Y by GeoKing
I will share my proof of showing that centroid of $\Delta IKL$ lies on $AD$
This proof is very interesting and beautiful and as i have seen no one else posted it before
Also i found this without geogebra :gleam:

Points $P, Q$ are basically irrelevant, $K$ and $L$ are just points on $BI$ and $CI$ such that $\angle KAB=\frac{\angle C}{2}$ and $\angle LAC=\frac{\angle B}{2}$
As $\angle KAB=\angle ACL$ and $\angle KBA=\angle LAC$ we see that $\Delta KBA\sim \Delta LAC$
Because these triangles are oriented the same way we can now apply the gliding principle
Let $M, N, T$ be midpoints of $AC, AB, KL$
By the gliding principle $\Delta TNM\sim \Delta KBA\sim \Delta LAC$
Let $X=MT\cap AK$ and $Y=NT\cap AL$
$\angle YNM=\angle TNM=\angle LAC=\angle YAM$ so $ANYM$ cyclic
similarly $ANXM$ cyclic
Hence $ANXYM$ cyclic

Consider a homothety at $A$ with scaling ratio $2$
$N\mapsto B, M\mapsto C$ and hence $(ANXYM)\mapsto (ABC)$
Now let $AX$ and $AY$ intersect $(ABC)$ at $R$ and $S$
$X\mapsto R, Y\mapsto S$
Let $J=BS\cap CR$
$T\mapsto J$ so $A-T-J$ and $T$ is midpoint of $AJ$
$\angle JBC=\angle SBC=\angle SAC=\angle ABI=\angle IBC$
similarly $\angle JCB=\angle ICB$
This means that $J$ is the reflection of $I$ in $BC$, hence $I-D-J$ and $D$ is midpoint of $IJ$
Let $G=IT\cap AD$
Consider triangle $\Delta IJA$
$T$ is midpoint of $AJ$, $D$ is midpoint of $IJ$ hence $G$ is centroid of $\Delta IJA$
Now $\frac{GT}{GI}=\frac{1}{2}$
As $T$ is midpoint of $KL$ as well we see that $G$ is centroid $\Delta IKL$ as needed
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KST2003
173 posts
#17
Y by
Let $K' = \overline{CI} \cap \overline{AK}$, and $L' = \overline{BI} \cap \overline{AL}$. Note that $\triangle ABC \sim \triangle PBA \sim \triangle QAC$, so $\triangle BIC \sim \triangle BKA \sim \triangle ALC$. In particular, $\triangle IKK'$ and $\triangle ILL'$ are isosceles, and so $K$, $L$, $K'$ and $L'$ are concyclic.

By EGMO Theorem 10.5, the Euler line of $\triangle IKL$ must then be the radical axis $\ell$ of circles with diameters $KK'$ and $LL'$, call them $\omega_K$ and $\omega_L$. Obviously $A$ must lie on $\ell$ since $KK'L'L$ is cyclic, so we just need to show that $D$ has equal powers to both circles.

Let $\omega_K$ intersect $\overline{BI}$ and $\overline{CI}$ at $W$ and $X$. Define $Y$ and $Z$ similarly for $\omega_L$. We claim that $D$ lies on $\overline{WX}$. Let $M_B$ be the midpoint of minor arc $AC$. By the incenter lemma, $M_BI = M_BC$, so $\triangle IWX \sim \triangle IK'K \sim \triangle IM_BC$, so $\overline{WX} \parallel \overline{M_BC}$. Let $I'$ be the reflection of $I$ over $\overline{BC}$. Then $\triangle BI'C$ and $\triangle BKA$ are spirally similar, so $\triangle BKI'$ and $\triangle BAC$ are spirally similar as well. In particular, $\angle BKI' = \angle BAC = \angle BM_BC$, so $\overline{DW} \parallel \overline{I'K} \parallel \overline{M_BC}$. This shows the collinearity.

Since $WXYZ$ is cyclic, it then follows that $D$ has equal power to both $\omega_K$ and $\omega_L$. Therefore, $D$ lies on $\ell$ as desired.
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bin_sherlo
733 posts
#18
Y by
Let $P,Q$ be the $A-$mixtilinear touch points to $AB,AC$ respectively. Let $E,F$ be the tangency points of the incircle with $AC,AB$. Let $O,H$ be the circumcenter and orthocenter of $\triangle IKL$.
Claim: $K=BI\cap (API)$ and $L=CI\cap (AQI)$.
Proof:
\[\measuredangle KAP=\measuredangle KAB=\frac{\measuredangle C}{2}=\measuredangle BIP=\measuredangle KIP\]\[\measuredangle QAL=\measuredangle CAL=\frac{\measuredangle B}{2}=\measuredangle QIC=\measuredangle QIL\]These yield the desired result.$\square$

Now we present a lemma.
Lemma: $ABC$ is a triangle whose circumcenter is $O$. Let $O'$ be the reflection of $O$ with respect to $BC$. The tangent to $(ABC)$ at $A$ intersect $BC$ at $D$. Points $E,F$ are taken on $O'C,O'B$ such that $EA=EC$ and $FA=FB$. Then, $D,E,F$ are collinear.
Proof: Let $X,Y$ be the intersection of the tangent to $(ABC)$ at $A$ with the tangent at $C,B$ respectively. By appyling Desargues theorem on $YOX$ and $BO'C,$ since $YB,OO',XC$ are concurrent, we get that these triangles are perpective. Hence $YO\cap BO'=F,OX\cap O'C=E,XY\cap BC=D$ are collinear.$\square$
Claim: $A,O,D$ are collinear.
Proof: Invert the diagram from $I$ with radius $ID$. Note that $K^*=IB\cap A^*P^*$ and $L^*=IC\cap A^*Q^*$. $O^*$ is the reflection of $I$ according to $K^*L^*$. Let $EF$ intersect $BC$ at $T$. $M,N,W$ are the midpoints of $IE,IF,IT$ respectively. $M,N,W$ are collinear. Let $K_1,L_1$ be the reflections of $I$ to $K^*,L^*$. By applying the lemma on $\triangle DEF,$ we get that $T,K_1,L_1$ are collinear. Take the homothety centered at $I$ with ratio $\frac{1}{2}$. Then, $W,K^*,L^*$ are collinear. Since $K^*L^*$ is the perpendicular bisector of $IO^*,$ we have $WT=WI=WO^*$. Hence $O^*$ is on $(TIDA^*)$ which proves the claim.$\square$
Claim: $A,H,D$ are collinear.
Proof: Let $(AIP)\cap AC=R$ and $(AIQ)\cap AB=S$.
\[\measuredangle HKI=90-(180-\measuredangle CIB)=\frac{\measuredangle A}{2}=\measuredangle RKI\]Thus, $H$ lies on $KR$. Similarily, $H$ lies on $LS$. So $H=KR\cap LS$.
Applying Desargues theorem on $RHS$ and $EDF,$ since $RH\cap ED,RS\cap EF,HS\cap DF$ is the line at infinity $A,H,D$ are collinear as desired.$\blacksquare$
This post has been edited 1 time. Last edited by bin_sherlo, Sep 15, 2024, 9:21 AM
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ErTeeEs06
69 posts
#19
Y by
Wow, what an incredibly cool geo problem! I've just skimmed through the other posts and seems like my solution is unique.

We start with the following lemma.
Lemma: Let $\triangle ABC$ be a triangle with $\angle BAC>90^\circ$ and $D$ be a point such that $\angle DBA=\angle DCA=180^\circ-\angle BAC$. Then $D$ is on the Euler line of triangle $ABC$.
Proof: Let $DB$ and $DC$ intersect $(ABC)$ again in $E, F$ and let $H, O$ be the orthocenter and circumcenter of $\triangle ABC$ respectively. Then its obvious by the angle condition that $AB\parallel CE$ and $AC\parallel BF$. Let $HB, HC$ intersect $(ABC)$ again in $X, Y$. Then $XB\perp AC$ so $XB\perp BF$. Pascal on $EBXFCY$ now gives $D, H, O$ collinear so $D$ is on the Euler line.

CLaim: $A$ is on the Euler line of $\triangle IKL$.
Proof: Angle chase $$\angle AKI=180^\circ-\angle AKB=180^\circ-(90^\circ+\frac{\angle APB}{2})=90^\circ-\frac{\angle BAC}{2}=180^\circ-\angle KIL$$Analogously $\angle ALI=180^\circ-\angle KIL$. Applying the above lemma with $IKL$ as reference triangle we see that $A$ is on the Euler line of $\triangle IKL$ which finishes the proof of this claim.

Now denote the circumcircles of triangles $DKP$ and $DLQ$ as $\omega_1$ and $\omega_2$ respectively. Let $\omega_1$ and $\omega_2$ intersect again in $S$ and let $BK$ intersect $\omega_1$ again in $X$ and $CL$ intersect $\omega_2$ again in $Y$. We calculate $$\angle XDC=\angle XDP=180^\circ-\angle BKP=90^\circ-\frac{\angle ACB}{2}$$This implies $DX\perp CI$. So $X$ is on the $C$-intouch chord of $\triangle ABC$, since it is also on line $BI$ we see that $X$ is the Iran lemma point and $\angle AXB=90^\circ$. We calculate $$\angle IAX=\angle AIB-90^\circ=\frac{\angle ACB}{2}$$so $\angle CAX=\frac{\angle BAC-\angle BCA}{2}=\frac{\angle CAP}{2}$ so $AX$ is the angle bisector of $\angle PAC$. If we let $AX, CI$ intersect each other at $E$ then $E$ is the incenter of $\triangle ACP$. Now we see $\angle EPK=90^\circ=\angle EXP$ so $E$ is on $\omega_1$ and it is the antipode of $K$. If we define $F$ similarly as the incenter of $\triangle ABQ$ then we now have $$\angle EXF=\angle EXK=90^\circ=\angle LYF=\angle EYF$$, so $EFYX$ is cyclic. This implies that $AE\cdot AX=AF\cdot AY$ and therefore $A$ is on the radax of $\omega_1$ and $\omega_2$.

Let $O$ be the circumcenter of $\triangle IKL$ and $KO, LO$ intersect $\omega_1, \omega_2$ in $M, N$ respectively. Angle chase gives $$\angle NDY=\angle NLY=\angle OLI=90^\circ-\angle LKI$$and similarly $\angle MDX=90^\circ-\angle KLI$. $AXDY$ is a parallelogram so $$\angle XDY=\angle XAY=180^\circ-\angle XIY=180^\circ-\angle KIL$$Adding these angles gives $$\angle MDN=\angle MDX+\angle XDY+\angle YDN=90^\circ-\angle LKI+90^\circ-\angle LKI+180^\circ-\angle KIL=180^\circ$$so $M, D, N$ are collinear. Now $$\angle NMO=\angle DMK=\angle DPK=\frac{\angle BAC}{2}$$and by symmetry $\angle MNO=\frac{\angle BAC}{2}$, so $|OM|=|ON|$. Also clearly $|OK|=|OL|$ so $O$ is on the radax of $\omega_1$ and $\omega_2$. But we already know $A$ and $D$ are on that radax so $A, O, D$ are collinear. Also we already had $A$ on the Euler line so this implies $D$ is on the Euler line also and that is exactly what we wanted to prove.
This post has been edited 1 time. Last edited by ErTeeEs06, May 1, 2025, 9:03 AM
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