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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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jlacosta
Apr 2, 2025
0 replies
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Geometry
IstekOlympiadTeam   27
N a minute ago by SimplisticFormulas
Source: All Russian Grade 9 Day 2 P 3
An acute-angled $ABC \ (AB<AC)$ is inscribed into a circle $\omega$. Let $M$ be the centroid of $ABC$, and let $AH$ be an altitude of this triangle. A ray $MH$ meets $\omega$ at $A'$. Prove that the circumcircle of the triangle $A'HB$ is tangent to $AB$. (A.I. Golovanov , A.Yakubov)
27 replies
IstekOlympiadTeam
Dec 12, 2015
SimplisticFormulas
a minute ago
J
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Incenter and concurrency
jenishmalla   3
N 11 minutes ago by Captainscrubz
Source: 2025 Nepal ptst p3 of 4
Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
3 replies
jenishmalla
Mar 15, 2025
Captainscrubz
11 minutes ago
J
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Number Theory Chain!
JetFire008   1
N 17 minutes ago by whwlqkd
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
1 reply
JetFire008
23 minutes ago
whwlqkd
17 minutes ago
J
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Collinearity with orthocenter
math163   6
N 29 minutes ago by Nari_Tom
Source: Baltic Way 2017 Problem 11
Let $H$ and $I$ be the orthocenter and incenter, respectively, of an acute-angled triangle $ABC$. The circumcircle of the triangle $BCI$ intersects the segment $AB$ at the point $P$ different from $B$. Let $K$ be the projection of $H$ onto $AI$ and $Q$ the reflection of $P$ in $K$. Show that $B$, $H$ and $Q$ are collinear.

Proposed by Mads Christensen, Denmark
6 replies
math163
Nov 11, 2017
Nari_Tom
29 minutes ago
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No more topics!
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Convex quadrilateral
Leon   2
N Nov 10, 2006 by Ashegh
Source: 2002 Austrian-Polish, problem 6
The diagonals of a convex quadrilateral $ABCD$ intersect in the point $E$. Let $U$ be the circumcenter of the triangle $ABE$ and $H$ be its orthocenter. Similarly, let $V$ be the circumcenter of the triangle $CDE$ and $K$ be its orthocenter. Prove that $E$ lies on the line $UK$ if and only if it lies on the line $VH$.
2 replies
Leon
Sep 23, 2006
Ashegh
Nov 10, 2006
J
Convex quadrilateral
G H J
Reveal topic tags
geometry
circumcircle
geometry unsolved
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Source: 2002 Austrian-Polish, problem 6
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Leon
256 posts
Leon
#1 h  Sep 23, 2006, 8:18 PM • 2 Y
Y by Adventure10, Mango247
The diagonals of a convex quadrilateral $ABCD$ intersect in the point $E$. Let $U$ be the circumcenter of the triangle $ABE$ and $H$ be its orthocenter. Similarly, let $V$ be the circumcenter of the triangle $CDE$ and $K$ be its orthocenter. Prove that $E$ lies on the line $UK$ if and only if it lies on the line $VH$.
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Amir.S
786 posts
Amir.S
#2 h  Nov 7, 2006, 5:45 PM • 4 Y
Y by Adventure10, Mango247, Mango247, Mango247
easy problem.
I just Answered this one, to remove it from MATHLINKS Unsolved problems( those problem are VERY rare) , Even ashegh and lomos_lupin are able to solve it.

let $U,E,K$ be collinear hence $UK\perp CD\Longrightarrow \ \ 90^{\circ}-\angle A=\angle UEB=\angle DEK=90^{\circ}-\angle D\Longrightarrow \ \ \angle A=\angle D$ hence quadrilateral $A,B,C,D$ is cyclic.
and again with easy angle chasing you can prove that $H,E,V$ are collinear.
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Ashegh
858 posts
Ashegh
#3 h  Nov 10, 2006, 10:18 AM • 2 Y
Y by Adventure10, Mango247
i only give u the idea!

u know$U,G,H$,are colinear,and $\frac{UG}{GH}$, is constant.which $G$,

is the centroid.

u should know that projection keep it fix,and if u project the shape and

change triangle$EUH$,to an equalirateral triangle,u can firmly find the

answer.

sorry because i really dont have time to post the complete solution.

to amir.s:

i will kill u at school :lol:
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