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Lines concur on bisector of BAC
Invertibility   2
N 10 minutes ago by NO_SQUARES
Source: Slovenia 2025 TST 3 P2
Let $\Omega$ be the circumcircle of a scalene triangle $ABC$. Let $\omega$ be a circle internally tangent to $\Omega$ in $A$. Tangents from $B$ touch $\omega$ in $P$ and $Q$, such that $P$ lies in the interior of $\triangle{}ABC$. Similarly, tangents from $C$ touch $\omega$ in $R$ and $S$, such that $R$ lies in the interior of $\triangle{}ABC$.

Prove that $PS$ and $QR$ concur on the bisector of $\angle{}BAC$.
2 replies
Invertibility
24 minutes ago
NO_SQUARES
10 minutes ago
angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N an hour ago by breloje17fr
Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
5 replies
parmenides51
Dec 11, 2018
breloje17fr
an hour ago
Problem 4 of Finals
GeorgeRP   2
N 2 hours ago by Assassino9931
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]
2 replies
GeorgeRP
Sep 10, 2024
Assassino9931
2 hours ago
Interesting functional equation with geometry
User21837561   3
N 2 hours ago by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Today at 8:14 AM
Double07
2 hours ago
Equilateral triangle formed by circle and Fermat point
Mimii08   2
N 2 hours ago by Mimii08
Source: Heard from a friend
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.

Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).

Prove that triangle A2B2C2 is equilateral.

2 replies
Mimii08
Yesterday at 10:36 PM
Mimii08
2 hours ago
two circumcenters and one orthocenter, vertices of parallelogram
parmenides51   4
N 3 hours ago by AylyGayypow009
Source: Greece JBMO TST 2015 p2
Let $ABC$ be an acute triangle inscribed in a circle of center $O$. If the altitudes $BD,CE$ intersect at $H$ and the circumcenter of $\triangle BHC$ is $O_1$, prove that $AHO_1O$ is a parallelogram.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
3 hours ago
Find the angle alpha [Iran Second Round 1994]
Amir Hossein   4
N Today at 1:36 PM by Mysteriouxxx
In the following diagram, $O$ is the center of the circle. If three angles $\alpha, \beta$ and $\gamma$ be equal, find $\alpha.$
IMAGE
4 replies
Amir Hossein
Nov 26, 2010
Mysteriouxxx
Today at 1:36 PM
official solution of IGO
ABCD1728   7
N Today at 12:32 PM by ABCD1728
Source: IGO official website
Where can I get the official solution of IGO for 2023 and 2024, there are some inhttps://imogeometry.blogspot.com/p/iranian-geometry-olympiad.html, but where can I find them on the official website, thanks :)
7 replies
ABCD1728
May 4, 2025
ABCD1728
Today at 12:32 PM
Combo geo with circles
a_507_bc   10
N Today at 12:30 PM by EthanWYX2009
Source: 239 MO 2024 S8
There are $2n$ points on the plane. No three of them lie on the same straight line and no four lie on the same circle. Prove that it is possible to split these points into $n$ pairs and cover each pair of points with a circle containing no other points.
10 replies
a_507_bc
May 22, 2024
EthanWYX2009
Today at 12:30 PM
Really fun geometry problem
Sadigly   6
N Today at 11:45 AM by farhad.fritl
Source: Azerbaijan Senior MO 2025 P6
In an acute triangle $ABC$ with $AB<AC$, the foot of altitudes from $A,B,C$ to the sides $BC,CA,AB$ are $D,E,F$, respectively. $H$ is the orthocenter. $M$ is the midpoint of segment $BC$. Lines $MH$ and $EF$ intersect at $K$. Let the tangents drawn to circumcircle $(ABC)$ from $B$ and $C$ intersect at $T$. Prove that $T;D;K$ are colinear
6 replies
Sadigly
Yesterday at 4:29 PM
farhad.fritl
Today at 11:45 AM
find angle
TBazar   5
N Today at 9:56 AM by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
5 replies
TBazar
Yesterday at 6:57 AM
TBazar
Today at 9:56 AM
Grouping angles in a pentagon with bisectors
Assassino9931   0
Today at 9:28 AM
Source: Al-Khwarizmi International Junior Olympiad 2025 P2
Let $ABCD$ be a convex quadrilateral with \[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]The line through \(C\), parallel to \(AD\), intersects the external angle bisector of \(\angle ABC\) at point \(T\). Prove that the angles $\angle ATB$, $\angle TBC$, $\angle BCD$, $\angle CDA$, $\angle DAT$ can be divided into two groups, so that the angles in each group have a sum of $270^{\circ}$.

Miroslav Marinov, Bulgaria
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Assassino9931
Today at 9:28 AM
0 replies
Tangent circles
Sadigly   1
N Today at 9:19 AM by lbh_qys
Source: Azerbaijan Junior MO 2025 P6
Let $T$ be a point outside circle $\omega$ centered at $O$. Tangents from $T$ to $\omega$ touch $\omega$ at $A;B$. Line $TO$ intersects bigger $AB$ arc at $C$.The line drawn from $T$ parallel to $AC$ intersects $CB$ at $E$. Ray $TE$ intersects small $BC$ arc at $F$. Prove that the circumcircle of $OEF$ is tangent to $\omega$.
1 reply
Sadigly
Today at 8:06 AM
lbh_qys
Today at 9:19 AM
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
G H J
G H BBookmark kLocked kLocked NReply
Source: Bulgaria National Olympiad 2025, Day 2, Problem 4
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Assassino9931
1330 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
98 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
33 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
288 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
1915 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
1330 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
323 posts
#7 • 1 Y
Y by cubres
Diagram
Solution
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VicKmath7
1389 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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