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Lines concur on bisector of BAC
Invertibility   2
N an hour 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
an hour ago
NO_SQUARES
an hour ago
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angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N 2 hours 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
2 hours ago
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Problem 4 of Finals
GeorgeRP   2
N 3 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
3 hours ago
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Interesting functional equation with geometry
User21837561   3
N 3 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
3 hours ago
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greatest volume
hzbrl   1
N 3 hours ago by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
1 reply
hzbrl
Yesterday at 9:56 AM
hzbrl
3 hours ago
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(n+1)2^n, (n+3)2^{n+2} not perfect squares for the same n
parmenides51   3
N 3 hours ago by AylyGayypow009
Source: Greece JBMO TST 2015 p3
Prove that there is not a positive integer $n$ such that numbers $(n+1)2^n, (n+3)2^{n+2}$ are both perfect squares.
3 replies
parmenides51
Apr 29, 2019
AylyGayypow009
3 hours ago
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IMO 2010 Problem 3
canada   59
N 3 hours ago by pi271828
Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that \[\left(g(m)+n\right)\left(g(n)+m\right)\] is a perfect square for all $m,n\in\mathbb{N}.$

Proposed by Gabriel Carroll, USA
59 replies
canada
Jul 7, 2010
pi271828
3 hours ago
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Equilateral triangle formed by circle and Fermat point
Mimii08   2
N 3 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
3 hours ago
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two circumcenters and one orthocenter, vertices of parallelogram
parmenides51   4
N 4 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
4 hours ago
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m^4+3^m is a perfect square number
Havu   5
N 4 hours ago by MR.1
Find a positive integer m such that $m^4+3^m$ is a perfect square number.
5 replies
Havu
5 hours ago
MR.1
4 hours ago
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Determine all the 'good' numbers
April   4
N 4 hours ago by DottedCaculator
Source: CGMO 2004 P1
We say a positive integer $ n$ is good if there exists a permutation $ a_1, a_2, \ldots, a_n$ of $ 1, 2, \ldots, n$ such that $ k + a_k$ is perfect square for all $ 1\le k\le n$. Determine all the good numbers in the set $ \{11, 13, 15, 17, 19\}$.
4 replies
April
Dec 27, 2008
DottedCaculator
4 hours ago
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Classical factorial number theory
Orestis_Lignos   21
N 4 hours ago by MIC38
Source: JBMO 2023 Problem 1
Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$.

Nikola Velov, North Macedonia
21 replies
Orestis_Lignos
Jun 26, 2023
MIC38
4 hours ago
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Functional equation
Nima Ahmadi Pour   100
N 5 hours ago by jasperE3
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
100 replies
Nima Ahmadi Pour
Apr 24, 2006
jasperE3
5 hours ago
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geometry problem
Medjl   4
N Apr 10, 2025 by tigerBoss101
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
4 replies
Medjl
Feb 1, 2018
tigerBoss101
Apr 10, 2025
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geometry problem
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geometry
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Source: Netherlands TST for IMO 2017 day 3 problem 1
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Medjl
757 posts
Medjl
#1 h  Feb 1, 2018, 2:43 PM • 3 Y
Y by Muradjl, Adventure10, Mango247
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
This post has been edited 1 time. Last edited by Medjl, Feb 1, 2018, 3:04 PM
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sbealing
308 posts
sbealing
#2 h  Feb 1, 2018, 4:48 PM • 3 Y
Y by lolm2k, AlastorMoody, Adventure10
Let $C=PM \cap AK$ and $O$ be the midpoint of $AK$ then as $\angle OCP=\angle OLP=\angle OQP=90^{\circ}$ we have $OCLPQ$ is cyclic.
$$\angle LCM=\angle LCP=\angle LQP=\angle LAQ=\angle LAM$$So $LACM$ is cyclic and hence $\angle ALM=\angle ACM=90^{\circ}$ so $K,L,M$ are colinear as $AK$ is a diameter.
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Cycle
79 posts
Cycle
#3 h  May 12, 2020, 3:58 AM
Y by
Redefine $L$ to be the intersection of $KM$ with $\omega$ so that it suffices to show $PL$ is tangent to the circle. Letting $D=PM\cap AK$ and $O$ the center, we have
$$\angle PQL=\angle QKM=\angle LAQ=\angle PDL$$by alternate segment and cyclic quadrilaterals $LMDA$, and $ALQK$. Hence $PQDL$ is cyclic. But $\angle QDM=\angle QKM$ implying $\angle LDQ=2\angle QKL=\angle LOQ$ so $LODQP$ is cyclic. Then $\angle OLP=180^\circ-\angle OQP=90^\circ $, proving the claim.
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llplp
191 posts
llplp
#4 h  Aug 13, 2020, 10:43 PM
Y by
We use complex numbers with $a = 1, k = -1$. We redefine $L,Q = \ell, q$ to be arbitrary points on $\omega$ and $p = \frac{2 \ell q}{\ell + q}$, $M = KL \cap AS$ and we wish to show $PM \perp AK$. From the intersection formula we have $m = \frac{2\ell q + \ell - q}{\ell + q}$. Finally $m - p = \frac{\ell - q}{\ell + q} \in i \mathbb{R}$, so $p-m$ is perpendicular to the real axis, i.e $AK$.
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tigerBoss101
5 posts
tigerBoss101
#5 h  Apr 10, 2025, 10:53 AM
Y by
Let $T = PM \cap AK$ and $R = PM \cap KQ$. As $\angle MTK = 90^\circ = \angle MQK$, we have that $MQKT$ is cyclic. Power of a Point gives $RM \cdot RT = RQ \cdot RK = RL \cdot RA$ which means $ALMT$ is cyclic, and $R$ is the radical center of the three circles. Since $AM \perp KR$ and $RM \perp AK$, we get that $M$ is the orthocenter of $\triangle AKR$, which means $KM \perp RA$, so $K$, $L$, $M$ are collinear as $KL \perp RA$.
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