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
trolling geometry problem
iStud   4
N 31 minutes ago by GreenTea2593
Source: Monthly Contest KTOM April P3 Essay
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
4 replies
iStud
Yesterday at 9:28 PM
GreenTea2593
31 minutes ago
J
H
Interesting F.E
Jackson0423   13
N an hour ago by ja.
Show that there does not exist a function
\[ f : \mathbb{R}^+ \to \mathbb{R} \]satisfying the condition that for all \( x, y \in \mathbb{R}^+ \),
\[ f(x + y^2) \geq f(x) + y. \]

~Korea 2017 P7
13 replies
Jackson0423
Apr 18, 2025
ja.
an hour ago
J
H
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   51
N an hour ago by Blast_S1
Source: IMO 2021 P5
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.

Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.
51 replies
popcorn1
Jul 20, 2021
Blast_S1
an hour ago
J
H
basically INAMO 2010/6
iStud   4
N an hour ago by ja.
Source: Monthly Contest KTOM April P1 Essay
Call $n$ kawaii if it satisfies $d(n)+\varphi(n)=n+1$ ($d(n)$ is the number of positive factors of $n$, while $\varphi(n)$ is the number of integers not more than $n$ that are relatively prime with $n$). Find all $n$ that is kawaii.
4 replies
iStud
Yesterday at 9:31 PM
ja.
an hour ago
J
H
2021 EGMO P2: f(xf(x)+y) = f(y) + x^2 for rational x, y
anser   79
N 2 hours ago by NuMBeRaToRiC
Source: 2021 EGMO P2
Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation
\[f(xf(x)+y) = f(y) + x^2\]holds for all rational numbers $x$ and $y$.

Here, $\mathbb{Q}$ denotes the set of rational numbers.
79 replies
anser
Apr 13, 2021
NuMBeRaToRiC
2 hours ago
J
H
3D geometry theorem
KAME06   1
N 2 hours ago by mathuz
Let $M$ a point in the space and $G$ the centroid of a tetrahedron $ABCD$. Prove that:
$$\frac{1}{4}(AB^2+AC^2+AD^2+BC^2+BD^2+CD^2)+4MG^2=MA^2+MB^2+MC^2+MD^2$$
1 reply
KAME06
5 hours ago
mathuz
2 hours ago
J
H
IMO Shortlist 2012, Number Theory 6
mathmdmb   42
N 3 hours ago by ihategeo_1969
Source: IMO Shortlist 2012, Number Theory 6
Let $x$ and $y$ be positive integers. If ${x^{2^n}}-1$ is divisible by $2^ny+1$ for every positive integer $n$, prove that $x=1$.
42 replies
mathmdmb
Jul 26, 2013
ihategeo_1969
3 hours ago
J
H
GCD of a sequence
oVlad   7
N 4 hours ago by grupyorum
Source: Romania EGMO TST 2017 Day 1 P2
Determine all pairs $(a,b)$ of positive integers with the following property: all of the terms of the sequence $(a^n+b^n+1)_{n\geqslant 1}$ have a greatest common divisor $d>1.$
7 replies
oVlad
Yesterday at 1:35 PM
grupyorum
4 hours ago
J
H
Another System
worthawholebean   3
N 4 hours ago by P162008
Source: HMMT 2008 Guts Problem 33
Let $ a$, $ b$, $ c$ be nonzero real numbers such that $ a+b+c=0$ and $ a^3+b^3+c^3=a^5+b^5+c^5$. Find the value of
$ a^2+b^2+c^2$.
3 replies
worthawholebean
May 13, 2008
P162008
4 hours ago
J
H
Inequality with three conditions
oVlad   2
N 4 hours ago by Quantum-Phantom
Source: Romania EGMO TST 2019 Day 1 P3
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
2 replies
oVlad
Yesterday at 1:48 PM
Quantum-Phantom
4 hours ago
J
H
GCD Functional Equation
pinetree1   61
N 5 hours ago by ihategeo_1969
Source: USA TSTST 2019 Problem 7
Let $f: \mathbb Z\to \{1, 2, \dots, 10^{100}\}$ be a function satisfying
$$\gcd(f(x), f(y)) = \gcd(f(x), x-y)$$for all integers $x$ and $y$. Show that there exist positive integers $m$ and $n$ such that $f(x) = \gcd(m+x, n)$ for all integers $x$.

Ankan Bhattacharya
61 replies
pinetree1
Jun 25, 2019
ihategeo_1969
5 hours ago
J
H
J
H
Geometry
youochange   8
N Apr 7, 2025 by RANDOM__USER
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
8 replies
youochange
Apr 6, 2025
RANDOM__USER
Apr 7, 2025
J
Geometry
G H J
Reveal topic tags
geometry
L
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
youochange
160 posts
youochange
#1 h  Apr 6, 2025, 11:27 AM • 2 Y
Y by PikaPika999, Rounak_iitr
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
This post has been edited 1 time. Last edited by youochange, Apr 6, 2025, 11:28 AM
Reason: Y
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
youochange
160 posts
youochange
#2 h  Apr 6, 2025, 12:40 PM • 1 Y
Y by PikaPika999
Bump :first:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
youochange
160 posts
youochange
#3 h  Apr 6, 2025, 2:10 PM • 1 Y
Y by PikaPika999
Helpmmmmmmme
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Double07
76 posts
Double07
#4 h  Apr 6, 2025, 5:24 PM • 3 Y
Y by PikaPika999, RANDOM__USER, youochange
Try complex bashing:

Take $(ABC)$ to be the unit circle and WLOG suppose $m=1$.

Denote by $S=AP\cap(ABC), S\neq A$ and $R=AP'\cap (ABC), R\neq A$.

Since $P'$ is the reflection of $P$ over $AM\implies \widehat{SAM}=\widehat{MAR}\implies M$ is the midpoint of arc $\widehat{RS}$, so $r\cdot s=m^2=1\implies r=\frac{1}{s}=\overline{s}$.

Compute $p=\frac{2bc}{b+c}$ and $s=\frac{ab+ac-2bc}{2a-b-c}$, so $r=\frac{b+c-2a}{2bc-ab-ac}$.

Since $M, N, P$ are collinear and $|m|=|n|=1\implies -mn=\frac{p-m}{\overline{p}-\overline{m}}\implies n=\frac{2bc-b-c}{b+c-2}$.

$Q=AR\cap MN\implies q=\frac{ar(m+n)-mn(a+r)}{ar-mn}=\frac{a(b+c-2a)(2bc-2)-(2bc-b-c)(2abc-a^2b-a^2c+b+c-2a)}{a(b+c-2a)(b+c-2)-(2bc-b-c)(2bc-ab-ac)}=$
$=\frac{(2a+2bc-ab-ac-b-c)(ab+ac+2a-2abc-b-c)}{2(a-bc)(2a+2bc-ab-ac-b-c)}=\frac{ab+ac+2a-2abc-b-c}{2(a-bc)}$.

Compute $K=AM\cap BC\implies k=\frac{am(b+c)-bc(a+m)}{an-bc}=\frac{ab+ac-bc-abc}{a-bc}$.

Now, to prove that $A, N, Q, K$ are concyclic, we need to prove that $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}\in\mathbb{R}$.

$a-n=\frac{ab+ac+b+c-2a-2bc}{b+c-2}$

$a-q=\frac{2a^2-ab-ac-2a+b+c}{2(a-bc)}=\frac{(a-1)(2a-b-c)}{2(a-bc)}$

$k-q=\frac{ab+ac+b+c-2a-2bc}{2(a-bc)}$

$k-n=\frac{(ab+ac-bc-abc)(b+c-2)-(2bc-b-c)(a-bc)}{(a-bc)(b+c-2)}=\frac{(b-1)(c-1)(2bc-ab-ac)}{(a-bc)(b+c-2)}$

So $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}=\frac{(a-bc)(ab+ac+b+c-2a-2bc)^2}{(a-1)(b-1)(c-1)(2a-b-c)(2bc-ab-ac)}$, which is real by conjugating, so we're done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RANDOM__USER
7 posts
RANDOM__USER
#5 h  Apr 6, 2025, 9:08 PM • 2 Y
Y by PikaPika999, youochange
Hmm, very interesting problem, sadly I only have minor results that might be useful. :(

Claim 1: If \(D\) is the midpoint of \(BC\), then \(ADNK\) is cyclic.
Proof: We intersect \(AN\) with \(BC\) at a point \(F\). Then because \(BCMN\) is harmonic (the tangents from \(B\) and \(C\) intersect on \(NM\)) it must be that if we project this harmonic quad from \(A\) onto \(BC\) that \((B,C;F,K)=-1\). Now using a very well known property of harmonic sets, we know that \(BF \cdot FC = FD \cdot FK\). However, due to PoP we know that \(BF \cdot FC = AF \cdot FN\), thus \(AF \cdot FN = FD \cdot FK\), meaning that, indeed \(ADNK\) is cyclic. \(\square\)

Now for another cool observation,

Claim 2: The problem is equivelent to showing that \(PDQP'\) is cyclic.
Proof: Assume \(PDQP'\) is cyclic, then \(\angle{DQA} = \angle{P'PD}\). If \(X = PP' \cup AK\), then \(\angle{PXK} = \frac{\pi}{2}\) and \(\angle{PDK} = \frac{\pi}{2}\). Thus \(\angle{DPP'} = \angle{DKA}\) and thus \(\angle{DKA} = \angle{DQA}\) which means that \(AQKD\) is cyclic. Taking into account the result that \(ADNK\) is cyclic, we obtain that \(ANKQ\) is cyclic. \(\square\)

And finally the last observation I think is note worthy is the following,

Claim 3: If \(E\) is the intersection of \(AP\) and \((ABC)\), then \(E, N, K\) and \(E,F,M\) are colinear.
Proof: Quite trivial through harmonics and projective ideas. \(\square\)
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lolsamo
11 posts
lolsamo
#6 h  Apr 6, 2025, 10:54 PM • 2 Y
Y by RANDOM__USER, youochange
Person above is just done, $\angle QAK=\angle PAK=\angle KNM=\angle KNQ$, as desired
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Captainscrubz
57 posts
Captainscrubz
#7 h  Apr 7, 2025, 4:45 AM • 2 Y
Y by RANDOM__USER, youochange
ig the simplest solution take $\sqrt bc$ inversion
Suppose $X$ is a point in the plane let after inversion it be $X'$
forgive me cuz I used $P'$ as a point after inversion of $P$ while there was $P'$ in the problem :P

So $P \rightarrow P'$ where $P'$ will be the $A-$Humpty point
$M'$ will be a random point on $\overrightarrow{CB}$
$N'=(M'AP')\cap BC$ ,$P''=$ reflection of $P'$ in $AM'$ , $Q'=AP''\cap (M'AP')$ and $K'=AM'\cap (ABC)$
We need to prove that $Q'-K'-N'$
Let $E$ be the reflection of $P'$ in $BC$ see that $E$ will lie on $(ABC)$

We will use phantom points here
Let $K^*=AM'\cap EN'$
and Let $D$ be the midpoint of $BC$
So-
$$\angle EN'D=\angle DN'P'=\angle M'AP'$$$$\implies (AK^*N'D)$$$$\implies \angle AK^*E =\angle AK^*N'=\angle ADC=\angle ABC+ \angle BAD=\angle EBC$$$$\therefore K^*\equiv K'$$$$\therefore \angle M'N'K'=\angle K'AP'=\angle M'AQ'$$$$\blacksquare$$
Attachments:
This post has been edited 2 times. Last edited by Captainscrubz, Apr 7, 2025, 4:47 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SimplisticFormulas
95 posts
SimplisticFormulas
#8 h  Apr 7, 2025, 7:00 AM • 2 Y
Y by RANDOM__USER, youochange
unless im seriously mistaken, the simplest solution is using projective
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RANDOM__USER
7 posts
RANDOM__USER
#9 h  Apr 7, 2025, 7:38 AM • 1 Y
Y by youochange
Yea, that seems to be the correct solution! That is essentially my solution in addition to the comment that I somehow didn't notice to finish of my solution :)
This post has been edited 1 time. Last edited by RANDOM__USER, Apr 7, 2025, 7:39 AM
Z K Y
N Quick Reply
G
H
=
a