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

Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
interesting geo config (2/3)
Royal_mhyasd   3
N 12 minutes ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
3 replies
Royal_mhyasd
Yesterday at 11:36 PM
Royal_mhyasd
12 minutes ago
J
H
interesting geo config (1\3)
Royal_mhyasd   2
N 14 minutes ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
2 replies
1 viewing
Royal_mhyasd
Yesterday at 11:18 PM
Royal_mhyasd
14 minutes ago
J
H
geometry problem with many circumcircles
Melid   0
36 minutes ago
Source: own
In scalene triangle $ABC$, which doesn't have right angle, let $O$ be its circumcenter. Circle $BOC$ intersects $AB$ and $AC$ at $A_{1}$ and $A_{2}$ for the second time, respectively. Similarly, circle $COA$ intersects $BC$ and $BA$ at $B_{1}$ and $B_{2}$, and circle $AOB$ intersects $CA$ and $CB$ at $C_{1}$ and $C_{2}$ for the second time, respectively. Let $O_{1}$ and $O_{2}$ be circumcenters of triangle $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2}$, respectively. Prove that $O, O_{1}, O_{2}$ are collinear.
0 replies
Melid
36 minutes ago
0 replies
J
H
Rootiful sets
InternetPerson10   38
N an hour ago by cursed_tangent1434
Source: IMO 2019 SL N3
We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_0, a_1, \cdots, a_n \in S$, all integer roots of the polynomial $a_0+a_1x+\cdots+a_nx^n$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^a - 2^b$ for positive integers $a$ and $b$.
38 replies
InternetPerson10
Sep 22, 2020
cursed_tangent1434
an hour ago
J
H
weird conditions in geo
Davdav1232   2
N an hour ago by teoira
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[ CL \cdot BD = BL \cdot CD. \]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
2 replies
1 viewing
Davdav1232
May 8, 2025
teoira
an hour ago
J
H
Long FE with f(0)=0
Fysty   4
N 2 hours ago by MathLuis
Source: Own
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f(0)=0$ and
$$f(f(x)+xf(y)+y)+xf(x+y)+f(y^2)=x+f(f(y))+(f(x)+y)(f(y)+x)$$for all $x,y\in\mathbb{R}$.
4 replies
Fysty
May 23, 2021
MathLuis
2 hours ago
J
H
Inspired by old results
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b> 0. $ Prove that
$$ \frac{a^3}{b^3+ab^2}+ \frac{4b^3}{a^3+b^3+2ab^2}\geq \frac{3}{2}$$$$\frac{a^3}{b^3+(a+b)^3}+ \frac{b^3}{a^3+(a+b)^3}+ \frac{(a+b)^2}{a^2+b^2+ab} \geq \frac{14}{9}$$
1 reply
sqing
3 hours ago
sqing
2 hours ago
J
H
Quadruple isogonal conjugate inside cyclic quad
Noob_at_math_69_level   8
N 3 hours ago by awesomeming327.
Source: DGO 2023 Team & Individual P3
Let $ABCD$ be a cyclic quadrilateral with $M_1,M_2,M_3,M_4$ being the midpoints of segments $AB,BC,CD,DA$ respectively. Suppose $E$ is the intersection of diagonals $AC,BD$ of quadrilateral $ABCD.$ Define $E_1$ to be the isogonal conjugate point of point $E$ in $\triangle{M_1CD}.$ Define $E_2,E_3,E_4$ similarly. Suppose $E_1E_3$ intersects $E_2E_4$ at a point $W.$ Prove that: The Newton-Gauss line of quadrilateral $ABCD$ bisects segment $EW.$

Proposed by 土偶 & Paramizo Dicrominique
8 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
3 hours ago
J
H
Interesting inequality
sqing   3
N 4 hours ago by sqing
Source: Own
Let $ a,b,c\geq 0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+ c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
3 replies
sqing
Yesterday at 2:54 AM
sqing
4 hours ago
J
H
2-var inequality
sqing   12
N 4 hours ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1} -\frac{b^4}{a+1} \leq 1 $$
12 replies
sqing
May 27, 2025
sqing
4 hours ago
J
H
Sum of whose elements is divisible by p
nntrkien   46
N 4 hours ago by Jackson0423
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
46 replies
nntrkien
Aug 8, 2004
Jackson0423
4 hours ago
J
H
Graph Theory
achen29   4
N 5 hours ago by ABCD1728
Are there any good handouts or even books in Graph Theory for a beginner in it? Preferable handouts which are extensive!
4 replies
achen29
Apr 24, 2018
ABCD1728
5 hours ago
J
H
J
H
cyclic quadrilateral starting with right triangle
parmenides51   4
N Apr 12, 2025 by Rounak_iitr
Source: Australian MO 2015
Let $ABC$ be a triangle with $ACB = 90^o$. The points $D$ and $Z$ lie on the side $AB$ such that $CD$ is perpendicular to $AB$ and $AC = AZ$. The line that bisects $BAC$ meets $CB$ and $CZ$ at $X$ and $Y$ , respectively. Prove that the quadrilateral $BXYD$ is cyclic.
4 replies
parmenides51
Sep 24, 2018
Rounak_iitr
Apr 12, 2025
J
cyclic quadrilateral starting with right triangle
G H J
Reveal topic tags
geometry
cyclic quadrilateral
perpendicular
equal segments
L
G H BBookmark kLocked kLocked NReply
Source: Australian MO 2015
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
parmenides51
30653 posts
parmenides51
#1 h  Sep 24, 2018, 3:11 PM • 2 Y
Y by Adventure10, Rounak_iitr
Let $ABC$ be a triangle with $ACB = 90^o$. The points $D$ and $Z$ lie on the side $AB$ such that $CD$ is perpendicular to $AB$ and $AC = AZ$. The line that bisects $BAC$ meets $CB$ and $CZ$ at $X$ and $Y$ , respectively. Prove that the quadrilateral $BXYD$ is cyclic.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9045 posts
Tintarn
#2 h  Sep 24, 2018, 3:29 PM • 2 Y
Y by Adventure10, Mango247
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anaskudsi
112 posts
Anaskudsi
#3 h  Sep 24, 2018, 4:11 PM • 2 Y
Y by Adventure10, Mango247
It is easy to prove that $Y$ is the midpoint of $CZ$. So $\angle{YDB} =\angle{CZA} =90-\frac{\angle{A}}{2}=180-(90+\frac{\angle{A}}{2})=180-\angle{AXB}$.
$\blacksquare$
This post has been edited 2 times. Last edited by Anaskudsi, Sep 24, 2018, 4:19 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sunken rock
4402 posts
sunken rock
#4 h  Sep 24, 2018, 5:03 PM • 2 Y
Y by Adventure10, Mango247
$AX\bot CZ\implies AC^2=AY\cdot AX=AD\cdot AB$, done.

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.
Rounak_iitr
457 posts
Rounak_iitr
#5 h  Apr 12, 2025, 8:05 AM
Y by
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -16.02395521434761, xmax = 20.54733084835408, ymin = -11.469406754458138, ymax = 9.824057454373513; /* image dimensions */ pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen ffqqff = rgb(1,0,1); pen wwqqcc = rgb(0.4,0,0.8); draw((-5.244613601042284,6.437450095559159)--(-5.244613601042284,-6.562549904440841)--(10.755386398957713,-6.562549904440841)--cycle, linewidth(1) + red); /* draw figures */ draw((-5.244613601042284,6.437450095559159)--(-5.244613601042284,-6.562549904440841), linewidth(1) + red); draw((-5.244613601042284,-6.562549904440841)--(10.755386398957713,-6.562549904440841), linewidth(1) + red); draw((10.755386398957713,-6.562549904440841)--(-5.244613601042284,6.437450095559159), linewidth(1) + red); draw((-5.244613601042284,-6.562549904440841)--(1.115093289714018,1.2701882468196626), linewidth(1) + ffdxqq); draw((-5.244613601042284,-6.562549904440841)--(4.930712337333065,-1.8300022293708142), linewidth(1) + blue); draw(circle((-5.246810029613714,-0.06254990444084142), 6.500000371099872), linewidth(1) + qqccqq); draw((1.115093289714018,1.2701882468196626)--(-0.2772366139119786,-4.252221119494483), linewidth(1) + ffqqff); draw((-0.2772366139119786,-4.252221119494483)--(-5.244613601042284,6.437450095559159), linewidth(1) + ffdxqq); /* special point */ draw((-1.402545519265254,-1.830584459993069)--(4.930712337333065,-1.8300022293708142), linewidth(1) + ffqqff); draw(circle((1.765759738600801,-1.8305844599930685), 3.1683052578660558), linewidth(1) + blue); draw(circle((5.7763422580138135,-2.8417470888992677), 6.215726357362417), linewidth(1) + linetype("4 4") + red); draw((0.7972981170699148,-6.562549904440841)--(4.930712337333065,-1.8300022293708142), linewidth(1) + wwqqcc); draw((-0.2772366139119786,-4.252221119494483)--(10.755386398957713,-6.562549904440841), linewidth(1) + wwqqcc); draw((-0.2772366139119786,-4.252221119494483)--(0.7972981170699148,-6.562549904440841), linewidth(1)); /* dots and labels */ dot((-5.244613601042284,6.437450095559159),dotstyle); label("$A$", (-5.424966303225186,6.959465856772843), NE * labelscalefactor); dot((-5.244613601042284,-6.562549904440841),dotstyle); label("$B$", (-5.520452689811875,-7.4589785178172), NE * labelscalefactor); dot((10.755386398957713,-6.562549904440841),dotstyle); label("$C$", (11.285151349445298,-7.077032971470444), NE * labelscalefactor); dot((1.115093289714018,1.2701882468196626),dotstyle); label("$D$", (1.1635943712563193,2.0896601408517026), NE * labelscalefactor); dot((4.930712337333065,-1.8300022293708142),dotstyle); label("$Z$", (5.460481767657301,-1.5865657427358248), NE * labelscalefactor); dot((-0.2772366139119786,-4.252221119494483),dotstyle); label("$Y$", (-0.5790321839507456,-5.501507592790075), NE * labelscalefactor); label("$E$", (-2.1068143693377612,-1.8252817092025473), NE * labelscalefactor); dot((0.7972981170699148,-6.562549904440841),dotstyle); label("$X$", (0.39970327856281135,-7.363492131230511), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

$\color{red}\textbf{Claim:-}$ $B,X,Y,D$ are cyclic
$\color{blue}\textbf{proof:-}$ Let $AY \cap CD=E$ Now By Angle Chasing we get, $$\angle ACZ = \angle AZC = \angle AED = \angle CEY = \angle AXC = \theta$$Also we get, $$\angle CAY = \angle ZAY = \angle BCZ = \angle ECZ = 90-\theta \implies \boxed{AY \perp CZ}$$Now we have $A,D,Y,C$ and $D,E,Y,Z$ are cyclic also. Now let $\angle YDZ= \alpha$ By Angle Chasing we get, $$\angle YDZ = \angle YEZ = 90 - \angle EZY=90 - \angle EDY$$Since the points $A,D,Y,C$ are cyclic we get, $$\angle CAY = \angle CDY =90-\alpha=90-\theta \implies \boxed{\theta=\alpha}$$Therefore the points $B,X,Y,D$ are cyclic since $\angle CXY = \angle YDB.$
Z K Y
N Quick Reply
G
H
=
a