Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
P2 Geo that most of contestants died
AlephG_64   1
N 27 minutes ago by Funcshun840
Source: 2025 Finals Portuguese Mathematical Olympiad P2
Let $ABCD$ be a quadrilateral such that $\angle A$ and $\angle D$ are acute and $\overline{AB} = \overline{BC} = \overline{CD}$. Suppose that $\angle BDA = 30^\circ$, prove that $\angle DAC= 30^\circ$.
1 reply
AlephG_64
Yesterday at 1:23 PM
Funcshun840
27 minutes ago
Triangles with equal areas
socrates   11
N 43 minutes ago by Nari_Tom
Source: Baltic Way 2014, Problem 13
Let $ABCD$ be a square inscribed in a circle $\omega$ and let $P$ be a point on the shorter arc $AB$ of $\omega$. Let $CP\cap BD = R$ and $DP \cap  AC = S.$
Show that triangles $ARB$ and $DSR$ have equal areas.
11 replies
socrates
Nov 11, 2014
Nari_Tom
43 minutes ago
Alice in Wonderland
ilovemath04   26
N an hour ago by atdaotlohbh
Source: ISL 2019 C8
Alice has a map of Wonderland, a country consisting of $n \geq 2$ towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be “one way” only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns.

Alice wants to know whether there is at least one town in Wonderland with at most one outgoing road. Prove that she can always find out by asking at most $4n$ questions.
26 replies
ilovemath04
Sep 22, 2020
atdaotlohbh
an hour ago
Geometry
Captainscrubz   0
an hour ago
Source: Own
Let $D$ be any point on side $BC$ of $\triangle ABC$ .Let $E$ and $F$ be points on $AB$ and $AC$ such that $EB=ED$ and $FD=FC$ respectively. Prove that the locus of circumcenter of $(DEF)$ is a line.
Prove without using moving points :D
0 replies
Captainscrubz
an hour ago
0 replies
Pebble Game
oVlad   6
N 2 hours ago by The5within
Source: KöMaL A. 790
Andrew and Barry play the following game: there are two heaps with $a$ and $b$ pebbles, respectively. In the first round Barry chooses a positive integer $k,$ and Andrew takes away $k$ pebbles from one of the two heaps (if $k$ is bigger than the number of pebbles in the heap, he takes away the complete heap). In the second round, the roles are reversed: Andrew chooses a positive integer and Barry takes away the pebbles from one of the two heaps. This goes on, in each round the two players are reversing the roles. The player that takes the last pebble loses the game.

Which player has a winning strategy?

Submitted by András Imolay, Budapest
6 replies
oVlad
Mar 24, 2022
The5within
2 hours ago
Equality with Fermat Point
nsato   13
N 2 hours ago by Nari_Tom
Source: 2012 Baltic Way, Problem 11
Let $ABC$ be a triangle with $\angle A = 60^\circ$. The point $T$ lies inside the triangle in such a way that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Let $M$ be the midpoint of $BC$. Prove that $TA + TB + TC = 2AM$.
13 replies
nsato
Nov 22, 2012
Nari_Tom
2 hours ago
China 2017 TSTST1 Day 2 Geometry Problem
HuangZhen   46
N 3 hours ago by ihategeo_1969
Source: China 2017 TSTST1 Day 2 Problem 5
In the non-isosceles triangle $ABC$,$D$ is the midpoint of side $BC$,$E$ is the midpoint of side $CA$,$F$ is the midpoint of side $AB$.The line(different from line $BC$) that is tangent to the inscribed circle of triangle $ABC$ and passing through point $D$ intersect line $EF$ at $X$.Define $Y,Z$ similarly.Prove that $X,Y,Z$ are collinear.
46 replies
HuangZhen
Mar 7, 2017
ihategeo_1969
3 hours ago
Cool combinatorial problem (grid)
Anto0110   1
N 3 hours ago by Anto0110
Suppose you have an $m \cdot n$ grid with $m$ rows and $n$ columns, and each square of the grid is filled with a non-negative integer. Let $a$ be the average of all the numbers in the grid. Prove that if $m >(10a+10)^n$ the there exist two identical rows (meaning same numbers in the same order).
1 reply
Anto0110
Yesterday at 1:57 PM
Anto0110
3 hours ago
one nice!
teomihai   2
N 3 hours ago by teomihai
3 girls and 4 boys must be seated at a round table. In how many distinct ways can they be seated so that the 3 girls do not sit next to each other and there can be a maximum of 2 girls next to each other. (The table is round so the seats are not numbered.)
2 replies
teomihai
Yesterday at 7:32 PM
teomihai
3 hours ago
Find the constant
JK1603JK   0
3 hours ago
Source: unknown
Find all $k$ such that $$\left(a^{3}+b^{3}+c^{3}-3abc\right)^{2}-\left[a^{3}+b^{3}+c^{3}+3abc-ab(a+b)-bc(b+c)-ca(c+a)\right]^{2}\ge 2k\cdot(a-b)^{2}(b-c)^{2}(c-a)^{2}$$forall $a,b,c\ge 0.$
0 replies
1 viewing
JK1603JK
3 hours ago
0 replies
IMO ShortList 1999, number theory problem 1
orl   61
N 4 hours ago by cursed_tangent1434
Source: IMO ShortList 1999, number theory problem 1
Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p-1}$ is a divisor of $ (p-1)^{x}+1$.
61 replies
orl
Nov 13, 2004
cursed_tangent1434
4 hours ago
nice problem
hanzo.ei   2
N Mar 30, 2025 by Lil_flip38
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
2 replies
hanzo.ei
Mar 29, 2025
Lil_flip38
Mar 30, 2025
nice problem
G H J
G H BBookmark kLocked kLocked NReply
Source: I forgot
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
15 posts
#1 • 1 Y
Y by cubres
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
15 posts
#2
Y by
bump!!!!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lil_flip38
48 posts
#3 • 2 Y
Y by hanzo.ei, X.Luser
I dont understand why there are so many unnecessary points defined, but oh well
Let \(S\) be the midpoint of \(BC\), \(R\) be the midpoint of \(AH\) where \(H\) is the foot of the altitude, \(AI\cap BC=P,\), let \(V\) be the point on \(I\) such that \(SV=SD\). Let \(AV\cap BC=Q\). It is well known that \(\angle AVD=90^\circ\) Let \(U=AI\cap DV\), and \(AT\cap BC = H'\)
We first claim that \(V\) lies on \((DKL)\), and that the center is the midpoint of \(BC\). First, consider inversion about \(I\). \(K,L\) are inverses, because \(L\) is mapped to \(AI\cap (CDEI)\) which is \(K\) by Iran lemma. This implies that \((DKL)\) is orthogonal to \(I\), so the center of \(DKL\) lies on \(BC\) as \(ID\perp BC\). Also by Iran lemma, \(BL\perp AI, CK\perp AI\implies BL\parallel CK\). So if we now consider the perpendicular bisector of \(KL\) it passes through \(S\). Thus, \(S\) is the center of \((DKL)\). Now, by definition, \(V\) lies on this circle aswell. Again, it is well known that \(Q\) is the \(A\)-extouch point, so it also lies on \((DKL)\). Note that \((L,K;D,V)=-1\), so \(T\) lies on \(DV\), and \((T, O;D,V)=-1\). Now, as
\[-1=(T,O;D,V) \overset{A}{=}(H',P;D,Q)\]and as its well known that \(R,I,Q\) lie on a line, we also have:
\[-1=(H,A;\infty_{\perp BC},R)\overset{I}{=}(H,P;D,Q)\]It follows that \(H=H'\) as desired.
I believe that most of the well known lemmas i stated throughout the solution are in EGMO chapter 4
Z K Y
N Quick Reply
G
H
=
a