Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Hard geometry
Lukariman   9
N 2 minutes ago by Captainscrubz
Given circle (O) and chord AB with different diameters. The tangents of circle (O) at A and B intersect at point P. On the small arc AB, take point C so that triangle CAB is not isosceles. The lines CA and BP intersect at D, BC and AP intersect at E. Prove that the centers of the circles circumscribing triangles ACE, BCD and OPC are collinear.
9 replies
+1 w
Lukariman
May 14, 2025
Captainscrubz
2 minutes ago
J
H
INMO 2007 Problem 6
Sathej   32
N 19 minutes ago by math_genie
Source: Inequality
If $ x$, $ y$, $ z$ are positive real numbers, prove that
\[ \left(x + y + z\right)^2 \left(yz + zx + xy\right)^2 \leq 3\left(y^2 + yz + z^2\right)\left(z^2 + zx + x^2\right)\left(x^2 + xy + y^2\right) .\]
32 replies
Sathej
Feb 4, 2007
math_genie
19 minutes ago
J
H
Prove that the triangle is isosceles.
TUAN2k8   3
N 22 minutes ago by TUAN2k8
Source: My book
Given acute triangle $ABC$ with two altitudes $CF$ and $BE$.Let $D$ be the point on the line $CF$ such that $DB \perp BC$.The lines $AD$ and $EF$ intersect at point $X$, and $Y$ is the point on segment $BX$ such that $CY \perp BY$.Suppose that $CF$ bisects $BE$.Prove that triangle $ACY$ is isosceles.
3 replies
TUAN2k8
2 hours ago
TUAN2k8
22 minutes ago
J
H
Simple but hard
Lukariman   3
N 41 minutes ago by Giant_PT
Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.
3 replies
Lukariman
Today at 2:47 AM
Giant_PT
41 minutes ago
J
H
Ah, easy one
irregular22104   2
N an hour ago by irregular22104
Source: Own
In the number series $1,9,9,9,8,...,$ every next number (from the fifth number) is the unit number of the sum of the four numbers preceding it. Is there any cases that we get the numbers $1234$ and $5678$ in this series?
2 replies
irregular22104
Wednesday at 4:01 PM
irregular22104
an hour ago
J
H
power of a point
BekzodMarupov   1
N an hour ago by nabodorbuco2
Source: lemmas in olympiad geometry
Epsilon 1.3. Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB at X and AC at Y. Let T be the intersection of EF with BC and let M be the midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
1 reply
BekzodMarupov
Today at 5:41 AM
nabodorbuco2
an hour ago
J
H
Functional Equation!
EthanWYX2009   5
N an hour ago by Miquel-point
Source: 2025 TST 24
Find all functions $f:\mathbb Z\to\mathbb Z$ such that $f$ is unbounded and
\[2f(m)f(n)-f(n-m)-1\]is a perfect square for all integer $m,n.$
5 replies
EthanWYX2009
Mar 29, 2025
Miquel-point
an hour ago
J
H
IMO Shortlist 2014 G3
hajimbrak   46
N an hour ago by Rayvhs
Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$.
(Here we always assume that an angle bisector is a ray.)

Proposed by Sergey Berlov, Russia
46 replies
hajimbrak
Jul 11, 2015
Rayvhs
an hour ago
J
H
Eight-point cicle
sandu2508   15
N an hour ago by Mamadi
Source: Balkan MO 2010, Problem 2
Let $ABC$ be an acute triangle with orthocentre $H$, and let $M$ be the midpoint of $AC$. The point $C_1$ on $AB$ is such that $CC_1$ is an altitude of the triangle $ABC$. Let $H_1$ be the reflection of $H$ in $AB$. The orthogonal projections of $C_1$ onto the lines $AH_1$, $AC$ and $BC$ are $P$, $Q$ and $R$, respectively. Let $M_1$ be the point such that the circumcentre of triangle $PQR$ is the midpoint of the segment $MM_1$.
Prove that $M_1$ lies on the segment $BH_1$.
15 replies
sandu2508
May 4, 2010
Mamadi
an hour ago
J
H
IMO Solution mistake
CHESSR1DER   1
N 2 hours ago by whwlqkd
Source: Mistake in IMO 1982/1 4th solution
I found a mistake in 4th solution at IMO 1982/1. It gives answer $660$ and $661$. But right answer is only $660$. Should it be reported somewhere in Aops?
1 reply
CHESSR1DER
5 hours ago
whwlqkd
2 hours ago
J
H
inequality
mathematical-forest   2
N 2 hours ago by mathematical-forest
For positive real intengers $x_{1} ,x_{2} ,\cdots,x_{n} $, such that $\prod_{i=1}^{n} x_{i} =1$
proof:
$$\sum_{i=1}^{n} \frac{1}{1+\sum _{j\ne i}x_{j} } \le 1$$
2 replies
mathematical-forest
Yesterday at 12:40 PM
mathematical-forest
2 hours ago
J
H
4-vars inequality
xytunghoanh   4
N 2 hours ago by JARP091
For $a,b,c,d \ge 0$ and $a\ge c$, $b \ge d$. Prove that
$$a+b+c+d+ac+bd+8 \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})$$.
4 replies
xytunghoanh
Yesterday at 2:10 PM
JARP091
2 hours ago
J
H
Sneaky one
Sunjee   2
N 2 hours ago by Sunjee
Find minimum and maximum value of following function.
$$f(x,y)=\frac{\sqrt{x^2+y^2}+\sqrt{(x-2)^2+(y-1)^2}}{\sqrt{x^2+(y-1)^2}+\sqrt{(x-2)^2+y^2}} $$
2 replies
Sunjee
3 hours ago
Sunjee
2 hours ago
J
H
J
H
2023 Hong Kong TST 3 (CHKMO) Problem 4
PikaNiko   3
N Apr 23, 2025 by lightsynth123
Source: 2023 Hong Kong TST 3 (CHKMO)
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
3 replies
PikaNiko
Dec 3, 2022
lightsynth123
Apr 23, 2025
J
2023 Hong Kong TST 3 (CHKMO) Problem 4
G H J
Reveal topic tags
geometry
L
G H BBookmark kLocked kLocked NReply
Source: 2023 Hong Kong TST 3 (CHKMO)
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PikaNiko
77 posts
PikaNiko
#1 h  Dec 3, 2022, 12:26 PM • 2 Y
Y by Mango247, Mango247
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LLL2019
834 posts
LLL2019
#2 h  Dec 4, 2022, 2:29 AM
Y by
I found this familiar during the test. Nevertheless, I couldn't find the synthetic solution, and I length chased it (compute $XE^2$ by LoC twice on $XEM$ and $DMC,$ where $X=AD\cap CN.$). In fact, this is equivalent to JMO 2011/5.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bxiao31415
2 posts
bxiao31415
#3 h  Nov 21, 2023, 2:15 PM
Y by
First, note that $(D,A;E,B)=C(D,A;M,\infty)=-1$, so $ABDE$ is harmonic. This implies that the tangents at $B$ and $E$ to $\Gamma$ and $AD$ are concurrent. Hence it suffices to show that $AD$, the tangent at $B$, and $CN$ are concurrent.

Let $P$ be the intersection of $AD$ and $CN$. $BN=NA$ and the fact that $CB$ and $AP$ are parallel tell us $\bigtriangleup CBN \equiv \bigtriangleup PAN$, so $CB = PA$. This then gives $\bigtriangleup ABP \equiv \bigtriangleup BAC$ as $\angle{CBA}=\angle{PAB}$. Hence $\angle{PBA}=\angle{CAB}=\angle{BCA}$ as $BA=BC$. Therefore, $PB$ is tangent to $\Gamma$, so $P$ is our desired concurrence point.
This post has been edited 1 time. Last edited by bxiao31415, Nov 21, 2023, 2:16 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lightsynth123
19 posts
lightsynth123
#4 h  Apr 23, 2025, 8:14 AM
Y by
Lemma
We just have to show that $CN$ intersects point $P$, where $P$ is the intersection of the tangent lines to $AD$.
Strategy
This post has been edited 5 times. Last edited by lightsynth123, Apr 23, 2025, 8:19 AM
Z K Y
N Quick Reply
G
H
=
a