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Geometry angle chasing olympiads
Foxellar   1
N 19 minutes ago by Ianis
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
1 reply
Foxellar
28 minutes ago
Ianis
19 minutes ago
Problem 4
codyj   87
N 38 minutes ago by ezpotd
Source: IMO 2015 #4
Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Greece
87 replies
codyj
Jul 11, 2015
ezpotd
38 minutes ago
IMO96/2 [the lines AP, BD, CE meet at a point]
Arne   47
N 2 hours ago by Bridgeon
Source: IMO 1996 problem 2, IMO Shortlist 1996, G2
Let $ P$ be a point inside a triangle $ ABC$ such that
\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]
Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point.
47 replies
Arne
Sep 30, 2003
Bridgeon
2 hours ago
A sharp one with 3 var (3)
mihaig   4
N 2 hours ago by aaravdodhia
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
4 replies
mihaig
Yesterday at 5:17 PM
aaravdodhia
2 hours ago
Cup of Combinatorics
M11100111001Y1R   1
N 3 hours ago by Davdav1232
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
1 reply
M11100111001Y1R
Yesterday at 7:24 AM
Davdav1232
3 hours ago
Bulgaria National Olympiad 1996
Jjesus   7
N 3 hours ago by reni_wee
Find all prime numbers $p,q$ for which $pq$ divides $(5^p-2^p)(5^q-2^q)$.
7 replies
Jjesus
Jun 10, 2020
reni_wee
3 hours ago
Can't be power of 2
shobber   31
N 3 hours ago by LeYohan
Source: APMO 1998
Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.
31 replies
shobber
Mar 17, 2006
LeYohan
3 hours ago
Brilliant Problem
M11100111001Y1R   4
N 3 hours ago by IAmTheHazard
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
4 replies
M11100111001Y1R
Yesterday at 7:28 AM
IAmTheHazard
3 hours ago
Own made functional equation
Primeniyazidayi   1
N 3 hours ago by Primeniyazidayi
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
1 reply
Primeniyazidayi
May 26, 2025
Primeniyazidayi
3 hours ago
not fun equation
DottedCaculator   13
N 4 hours ago by Adywastaken
Source: USA TST 2024/6
Find all functions $f\colon\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$,
\[f(xf(y))+f(y)=f(x+y)+f(xy).\]
Milan Haiman
13 replies
DottedCaculator
Jan 15, 2024
Adywastaken
4 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   12
N 4 hours ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
12 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
4 hours ago
Geometry with fix circle
falantrng   33
N 5 hours ago by zuat.e
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
33 replies
falantrng
Feb 25, 2018
zuat.e
5 hours ago
USAMO 2001 Problem 2
MithsApprentice   54
N 5 hours ago by lpieleanu
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
54 replies
MithsApprentice
Sep 30, 2005
lpieleanu
5 hours ago
Orthocenter config once again
Assassino9931   7
N Apr 10, 2025 by VicKmath7
Source: Bulgaria National Olympiad 2025, Day 2, Problem 4
Let \( ABC \) be an acute triangle with \( AB < AC \), midpoint $M$ of side $BC$, altitude \( AD \) (\( D \in BC \)), and orthocenter \( H \). A circle passes through points \( B \) and \( D \), is tangent to line \( AB \), and intersects the circumcircle of triangle \( ABC \) at a second point \( Q \). The circumcircle of triangle \( QDH \) intersects line \( BC \) at a second point \( P \). Prove that the lines \( MH \) and \( AP \) are perpendicular.
7 replies
Assassino9931
Apr 8, 2025
VicKmath7
Apr 10, 2025
Orthocenter config once again
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G H BBookmark kLocked kLocked NReply
Source: Bulgaria National Olympiad 2025, Day 2, Problem 4
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Assassino9931
1371 posts
#1 • 1 Y
Y by cubres
Let \( ABC \) be an acute triangle with \( AB < AC \), midpoint $M$ of side $BC$, altitude \( AD \) (\( D \in BC \)), and orthocenter \( H \). A circle passes through points \( B \) and \( D \), is tangent to line \( AB \), and intersects the circumcircle of triangle \( ABC \) at a second point \( Q \). The circumcircle of triangle \( QDH \) intersects line \( BC \) at a second point \( P \). Prove that the lines \( MH \) and \( AP \) are perpendicular.
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Primeniyazidayi
117 posts
#2 • 1 Y
Y by cubres
Try to show that P,E,F are collinear where E and F are the other altitudes.Then applying Brocard on (BEFC) shows that M is the orthocenter of the triangle PAH(which also means that H is the orthocenter of PAM).
As bonus:
K,the second intersection of AP and (ABC),is the A-queue point and A,K,E,H,F are concyclic.
P,K,A and M,H,K,A' are pairwise collinear where A' is the antipode of A wrt (ABC).
This post has been edited 4 times. Last edited by Primeniyazidayi, Apr 9, 2025, 1:33 PM
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aidenkim119
34 posts
#3 • 1 Y
Y by cubres
Let $K$ = $(AH) \cap (ABC)$, $X$ = $QD \cap (ABC)$, $H' = AH \cap (ABC), A' = KM \cap (ABC)$. A' is obviously the A antipode.

Claim) $K H D M$ is cyclic.
pf) $\angle KQD = \angle KQX = \angle BQD - \angle BQK = \angle B - \angle BQK$
$\angle DHM = \angle H'HA' = \angle KCA + \angle H'AA' =\angle C - \angle BQK + \angle B - C = \angle B - \angle BQK = \angle KQD$

Ok.

Therefore $KHDQP$ is cyclic, so $\angle HKP = \angle HDM = 90,$ also $\angle AKH = 90. $

So $\angle AKP = 180$, so $A, K, P$ colinear. $MH \perp AK$, so problem solved.
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hukilau17
291 posts
#4 • 1 Y
Y by cubres
Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$$|a|=|b|=|c|=1$$$$m=\frac{b+c}2$$$$d=\frac{a^2+ab+ac-bc}{2a}$$$$h=a+b+c$$Now we find the coordinate of $Q$. Since $Q$ is on the unit circle we have $|q|=1$, and since line $AB$ is tangent to the circumcircle of $\triangle BDQ$, we have
$$\frac{(a-b)(d-q)}{(b-d)(b-q)} \in \mathbb{R}$$$$\frac{a^2+ab+ac-bc-2aq}{(a+c)(b-q)} = \frac{a^2q+2abc-abq-acq-bcq}{a(a+c)(b-q)}$$$$a^3+a^2b+a^2c-abc-2a^2q = a^2q+2abc-abq-acq-bcq$$$$q = \frac{a(a^2+ab+ac-3bc)}{3a^2-ab-ac-bc}$$Now we find the coordinate of $P$. Since $P$ is on line $BC$, we have $\overline{p} = \frac{b+c-p}{bc}$. Since $D,H,P,Q$ are concyclic, we have
$$\frac{(d-q)(h-p)}{(h-q)(d-p)} \in \mathbb{R}$$Now $\frac{(d-q)(a-b)}{(b-c)(q-b)} \in \mathbb{R}$, and $\frac{d-p}{b-c} \in \mathbb{R}$, so we write this as
$$\frac{(q-b)(h-p)}{(a-b)(h-q)} \in \mathbb{R}$$Now
$$q-b = \frac{a^3-2a^2b+a^2c+ab^2-2abc+b^2c}{3a^2-ab-ac-bc} = \frac{(a-b)^2(a+c)}{3a^2-ab-ac+bc}$$$$h-q = \frac{2a^3+a^2b+a^2c-ab^2-ac^2-b^2c-bc^2}{3a^2-ab-ac-bc} = \frac{(a+b)(a+c)(2a-b-c)}{3a^2-ab-ac-bc}$$So we write this as
$$\frac{(a-b)(h-p)}{(a+b)(2a-b-c)} \in \mathbb{R}$$And since $\frac{a-b}{a+b}$ is pure imaginary, we have
$$\frac{a+b+c-p}{2a-b-c} \in i\mathbb{R}$$$$\frac{a+b+c-p}{2a-b-c} = -\frac{\frac1a+\frac1b+\frac1c-\frac1b-\frac1c+\frac{p}{bc}}{\frac2a-\frac1b-\frac1c} = \frac{ap+bc}{ab+ac-2bc}$$$$(a+b+c-p)(ab+ac-2bc) = (ap+bc)(2a-b-c)$$$$p = \frac{(a+b+c)(ab+ac-2bc) - bc(2a-b-c)}{(ab+ac-2bc) + a(2a-b-c)} = \frac{a^2b+a^2c+ab^2-2abc+ac^2-b^2c-bc^2}{2(a^2-bc)}$$Then we have the vectors
$$h-m = \frac{2a+b+c}2$$$$a-p = \frac{2a^3-a^2b-a^2c-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} = \frac{(a-b)(a-c)(2a+b+c)}{2(a^2-bc)}$$and so
$$\frac{a-p}{h-m} = \frac{(a-b)(a-c)}{a^2-bc}$$which is pure imaginary. $\blacksquare$
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cj13609517288
1924 posts
#5 • 1 Y
Y by cubres
Rename $P$ to $X$, then rename $Q$ to $P$. Let $Q$ be the $A$-queue point. Then $X$ is the $A$-Ex point.

Now note that $P$ is the 11SLG4 point, because $PD\cap\Gamma$ has to be the isosceles trapezoid point. Since $Q$ lies on $(XHD)$, let's show that $XQDP$ is cyclic. Indeed,
\[\angle QXD=90^{\circ}-\angle QAD=\angle QPA'=\angle QPD,\]as desired. $\blacksquare$
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Assassino9931
1371 posts
#6 • 2 Y
Y by AlexCenteno2007, cubres
It is well known that if ray $MH^\to$ intersects the circumcircle of $ABC$ at $K$, then $\angle AKH = 90^{\circ}$. We have $\angle KQD = \angle BQD - \angle BQK = \angle ABC - \angle BAK = 90^{\circ} - \angle BAD - \angle BAK = 90^{\circ} - \angle KAH = \angle AHK = 180^{\circ} - \angle KHD$, so $KHDQ$ is cyclic. Hence $PKHDQ$ is cyclic, implying $\angle PKH = 180^{\circ} - \angle PDH = 90^{\circ}$, so $A$, $K$ and $P$ are collinear, implying $MH \perp AP$.
This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 10:01 PM
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wassupevery1
325 posts
#7 • 1 Y
Y by cubres
Diagram
Solution
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VicKmath7
1391 posts
#8 • 2 Y
Y by AlexCenteno2007, cubres
Trivial by angle-chasing, I am posting this just because the ratio lemma technique works nicely here.

Let $(AH) \cap (ABC)=T$, $AT \cap BC=P$, $R$ be such that $AR \parallel BC$ and redefine $Q=RD \cap (ABC)$. Clearly $PTHD$ is cyclic with diameter $PH$, $MH \perp AP$ and $(BDQ)$ touches $AB$ as $\angle BQD=\angle RCB=\angle ABC$. We are only left to show that $P \in (DHQ)$, so we will show that $PTDQ$ is cyclic. By Theorem 3.1 in mira74's ratio lemma handout., we have to show $f(T)f(Q)=f(P)f(D)$ (where $f(X)=\pm \frac{XB}{XC}$), and since $f(D)=f(Q)f(R)$ by Lemma 2.2, we need $f(T)=f(R)f(P)$. However, $f(R)f(A)=1$ and $f(P)=f(T)f(A)$ by Lemma 2.2, which implies that $f(T)=f(R)f(P)$.
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