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

Join our free webinar April 22 to learn about competitive programming!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Circumcircle excircle chaos
CyclicISLscelesTrapezoid   25
N 31 minutes ago by bin_sherlo
Source: ISL 2021 G8
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
25 replies
CyclicISLscelesTrapezoid
Jul 12, 2022
bin_sherlo
31 minutes ago
J
H
Inequalities
sqing   27
N 5 hours ago by Jackson0423
Let $ a,b $ be reals such that $ a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $ a,b $ be reals such that $ a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
27 replies
sqing
Apr 16, 2025
Jackson0423
5 hours ago
J
H
Inequalities
sqing   6
N 5 hours ago by sqing
Let $ a,b,c> 0 $ and $ ab+bc+ca\leq 3abc . $ Prove that
$$ a+ b^2+c\leq a^2+ b^3+c^2 $$$$ a+ b^{11}+c\leq a^2+ b^{12}+c^2 $$
6 replies
sqing
Today at 1:54 PM
sqing
5 hours ago
J
H
Problem of the Week--The Sleeping Beauty Problem
FiestyTiger82   1
N 5 hours ago by martianrunner
Put your answers here and discuss!
The Problem
1 reply
FiestyTiger82
6 hours ago
martianrunner
5 hours ago
J
H
Inequalities
sqing   4
N Today at 1:09 PM by sqing
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that$$ |a-b|+|b-2c|+|c-3a|\leq 5$$$$|a-2b|+|b-3c|+|c-4a|\leq \sqrt{42}$$$$ |a-b|+|b-\frac{11}{10}c|+|c-a|\leq \frac{29}{10}$$
4 replies
sqing
Today at 5:05 AM
sqing
Today at 1:09 PM
J
H
Inequalities
nhathhuyyp5c   2
N Today at 12:38 PM by pooh123
Let $a, b, c$ be non-negative real numbers such that $a^2 + b^2 + c^2 = 3$. Find the maximum and minimum values of the expression
\[ P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}. \]
2 replies
nhathhuyyp5c
Apr 20, 2025
pooh123
Today at 12:38 PM
J
H
Challenging Optimization Problem
Shiyul   5
N Today at 12:28 PM by exoticc
Let $xyz = 1$. Find the minimum and maximum values of $\frac{1}{1 + x + xy}$ + $\frac{1}{1 + y + yz}$ + $\frac{1}{1 + z + zx}$

Can anyone give me a hint? I got that either the minimum or maximum was 1, but I'm sure if I'm correct.
5 replies
Shiyul
Yesterday at 8:20 PM
exoticc
Today at 12:28 PM
J
H
Geometry Angle Chasing
Sid-darth-vater   1
N Today at 9:36 AM by vanstraelen
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)

For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
1 reply
Sid-darth-vater
Yesterday at 11:50 PM
vanstraelen
Today at 9:36 AM
J
H
Radical Axes and circles
mathprodigy2011   4
N Today at 7:53 AM by spiderman0
Can someone explain how to do this purely geometrically?
4 replies
mathprodigy2011
Today at 1:58 AM
spiderman0
Today at 7:53 AM
J
H
Combinatoric
spiderman0   0
Today at 7:46 AM
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
0 replies
spiderman0
Today at 7:46 AM
0 replies
J
H
BMT 2018 Algebra Round Problem 7
IsabeltheCat   5
N Today at 6:56 AM by P162008
Let $$h_n := \sum_{k=0}^n \binom{n}{k} \frac{2^{k+1}}{(k+1)}.$$Find $$\sum_{n=0}^\infty \frac{h_n}{n!}.$$
5 replies
IsabeltheCat
Dec 3, 2018
P162008
Today at 6:56 AM
J
H
J
H
ABCD cyclic --> incenters of BCD, etc. form rectangle
Arne   12
N Oct 23, 2023 by ismayilzadei1387
Source: Kedlaya; also: APMO 1996, Problem 3
If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.
12 replies
Arne
Jan 23, 2005
ismayilzadei1387
Oct 23, 2023
J
ABCD cyclic --> incenters of BCD, etc. form rectangle
G H J
Reveal topic tags
geometry
incenter
rectangle
cyclic quadrilateral
geometry proposed
Angle Chasing
L
G H BBookmark kLocked kLocked NReply
Source: Kedlaya; also: APMO 1996, Problem 3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Arne
3660 posts
Arne
#1 h  Jan 23, 2005, 5:15 PM • 2 Y
Y by Adventure10, Mango247
If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kueh
392 posts
kueh
#2 h  Jan 23, 2005, 5:58 PM • 6 Y
Y by Pluto1708, Adventure10, anonman, Mango247, Stuffybear, and 1 other user
By angle chasing.

You can show that the incentres of ABC (K), BCD (L), B and C are cyclic because angle CLB = 180 - DCB/2 - DBC/2 = 90 + BDC/2, and similarly, angle CKB = 90 + BAC/2. But BAC = BDC. Thus CLB = CKB and so B,C,K,L are concyclic. Also if we call the incentre of ADC (M), D,C,L,M is cyclic. angle KLC = 180 - angle LBC, angle MLC = 180 - MDC, thus angle KLM = angle KBC + angle MDC = (ADC+ABC)/2 = 90, which is what we want.

PS: Can someone give me a crashcourse on latex? or point to me where?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
darij grinberg
#3 h  Jan 23, 2005, 6:01 PM • 2 Y
Y by Adventure10, Mango247
It can be proved by complex numbers... See also

Paul Yiu, Notes on Euclidean Geometry, paragraph 10.2.6 (page 154);
Barry Wolk, Hyacinthos message #8154,
Alexey A.Zaslavsky, Hyacinthos message #8178.

darij
This post has been edited 1 time. Last edited by darij grinberg, Jun 15, 2006, 3:54 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pavel kozlov
615 posts
pavel kozlov
#4 h  Mar 1, 2006, 8:06 PM • 1 Y
Y by Adventure10
There is the converse of this fact (from Russian Summer TST 2005):
Let $ABCD$ be a quadrilateral. The incentres of the triangles $ABC, BCD, CDA, DAB$ form a rectangle. Proove, that $ABCD$ is cyclic.
It's very interesting and rather difficult problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9782 posts
jayme
#5 h  Mar 20, 2018, 7:51 AM • 3 Y
Y by govind7701, Adventure10, Mango247
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20rectangle%20de%20Ryokan%20Maruyama.pdf

Sincerely
Jean-Louis
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
purplish
299 posts
purplish
#6 h  May 10, 2019, 7:08 AM • 3 Y
Y by Pluto1708, Adventure10, Mango247
We use complex numbers. Plot the points on the unit circle. Then we can choose $a,b,c,d$ such that $A=a^2,B=b^2,C=c^2,D=d^2$, and \[I_{ABC}=-bc-ca-ab,I_{BCD}=-cd-db-bc,I_{CDA}=da+ac-cd,I_{DAB}=-ab+bd+da.\]Then we can compute \[I_{ABC}+I_{CDA}=-ab-bc-cd+da=I_{BCD}+I_{DAB}\]and
\begin{align*} \left|I_{ABC}-I_{CDA}\right|^2&=\left|-bc-2ca-ab-da+cd\right|^2=\frac{1}{abcd}\left(-bc-2ca-ab-da+cd\right)\left(-ad-2bd-cd-bc+ab\right)\\ &=\frac{1}{abcd}\left(-cd-2db-bc-da+ab\right)\left(-ab-2ac-ad-bc+cd\right)=\left|-cd-2db-bc-da+ab\right|^2=\left|I_{BCD}-I_{DAB}\right|^2 \end{align*}so the incenters form a rectangle.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zuss77
520 posts
zuss77
#7 h  Oct 13, 2020, 4:28 PM • 1 Y
Y by yellowhanoi
Labels are on the diagram.
By Fact 5 $AMNB$ is cyclic. And so is $AMQD$.
$\angle NMQ=\angle B/2 +\angle D/2=90^\circ$. And similarly the others.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
denery
180 posts
denery
#8 h  Dec 15, 2020, 6:19 AM
Y by
i only could find those right angles on the rectangle by drawing external tangents
and since it is a quadrilateral
it must be a square or rectangle
and by length chasing interestingly we could say that that square/rectangle is definitely composed of rectangles
This post has been edited 1 time. Last edited by denery, Dec 15, 2020, 6:21 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
primesarespecial
364 posts
primesarespecial
#9 h  Jul 12, 2021, 1:19 PM
Y by
We use complex numbers on this one.
We just easily prove this a parallelogram and then just show perpendicularity once or prove diagonals equal in length. :blush:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sunken rock
4384 posts
sunken rock
#10 h  Jul 13, 2021, 4:50 PM
Y by
My solution at https://artofproblemsolving.com/community/c6t48f6h24483_abcd_cyclic_gt_incenters_of_bcd_etc_form_rectangle
I hope you will enjoy it!

Best regards,
sunken rock
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
archimedes26
612 posts
archimedes26
#11 h  Jul 13, 2021, 9:42 PM
Y by
https://groups.io/g/euclid/topic/81006484#1476
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
archimedes26
612 posts
archimedes26
#12 h  Jul 13, 2021, 9:45 PM
Y by
Another property.
https://geometry-diary.blogspot.com/2020/11/1521.html
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ismayilzadei1387
219 posts
ismayilzadei1387
#13 h  Oct 23, 2023, 7:20 AM • 1 Y
Y by FriIzi
Also China 1986 TsT
Z K Y
N Quick Reply
G
H
=
a