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Contests & Programs AMC and other contests, summer programs, etc.
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Circumcircle of MUV tangent to two circles at once
MathMystic33   1
N 35 minutes ago by ariopro1387
Source: Macedonian Mathematical Olympiad 2025 Problem 1
Given is an acute triangle \( \triangle ABC \) with \( AB < AC \). Let \( M \) be the midpoint of side \( BC \), and let \( X \) and \( Y \) be points on segments \( BM \) and \( CM \), respectively, such that \( BX = CY \). Let \( \omega_1 \) be the circumcircle of \( \triangle ABX \), and \( \omega_2 \) the circumcircle of \( \triangle ACY \). The common tangent \( t \) to \( \omega_1 \) and \( \omega_2 \), which lies closer to point \( A \), touches \( \omega_1 \) and \( \omega_2 \) at points \( P \) and \( Q \), respectively. Let the line \( MP \) intersect \( \omega_1 \) again at \( U \), and the line \( MQ \) intersect \( \omega_2 \) again at \( V \). Prove that the circumcircle of triangle \( \triangle MUV \) is tangent to both \( \omega_1 \) and \( \omega_2 \).
1 reply
MathMystic33
3 hours ago
ariopro1387
35 minutes ago
Concurrency of tangent touchpoint lines on thales circles
MathMystic33   0
an hour ago
Source: 2024 Macedonian Team Selection Test P4
Let $\triangle ABC$ be an acute scalene triangle. Denote by $k_A$ the circle with diameter $BC$, and let $B_A,C_A$ be the contact points of the tangents from $A$ to $k_A$, chosen so that $B$ and $B_A$ lie on opposite sides of $AC$ and $C$ and $C_A$ lie on opposite sides of $AB$. Similarly, let $k_B$ be the circle with diameter $CA$, with tangents from $B$ touching at $C_B,A_B$, and $k_C$ the circle with diameter $AB$, with tangents from $C$ touching at $A_C,B_C$.
Prove that the lines $B_AC_A, C_BA_B, A_CB_C$ are concurrent.
0 replies
MathMystic33
an hour ago
0 replies
Equal areas of the triangles on the parabola
NO_SQUARES   0
an hour ago
Source: Regional Stage of ARO 2025 10.10; also Kvant 2025 no. 3 M2837
On the graphic of the function $y=x^2$ were selected $1000$ pairwise distinct points, abscissas of which are integer numbers from the segment $[0; 100000]$. Prove that it is possible to choose six different selected points $A$, $B$, $C$, $A'$, $B'$, $C'$ such that areas of triangles $ABC$ and $A'B'C'$ are equals.
A. Tereshin
0 replies
NO_SQUARES
an hour ago
0 replies
Concurrency from symmetric points on the sides of a triangle
MathMystic33   0
an hour ago
Source: 2024 Macedonian Team Selection Test P3
Let $\triangle ABC$ be a triangle. On side $AB$ take points $K$ and $L$ such that $AK \;=\; LB \;<\;\tfrac12\,AB,$
on side $BC$ take points $M$ and $N$ such that $BM \;=\; NC \;<\;\tfrac12\,BC,$ and on side $CA$ take points $P$ and $Q$ such that $CP \;=\; QA \;<\;\tfrac12\,CA.$ Let $R \;=\; KN\;\cap\;MQ,
\quad
T \;=\; KN \cap LP, $ and $ D \;=\; NP \cap LM, \quad
E \;=\; NP \cap KQ.$
Prove that the lines $DR, BE, CT$ are concurrent.
0 replies
MathMystic33
an hour ago
0 replies
Grouping angles in a pentagon with bisectors
Assassino9931   2
N an hour ago by Assassino9931
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
2 replies
Assassino9931
May 9, 2025
Assassino9931
an hour ago
Geometric inequality with 2 orthocenters and midpoint of the side
NO_SQUARES   0
an hour ago
Source: Regional Stage of ARO 2025 10.5; also Kvant 2025 no. 3 M2836
The heights $BD$ and $CE$ of the acute-angled triangle $ABC$ intersect at point $H$, the heights of the triangle $ADE$ intersect at point $F$, point $M$ is the midpoint of side $BC$. Prove that $BH + CH \geqslant 2 FM$.
A. Kuznetsov
0 replies
NO_SQUARES
an hour ago
0 replies
Taking antipode on isosceles triangle's circumcenter
Nuran2010   1
N an hour ago by Sadigly
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.
1 reply
Nuran2010
May 11, 2025
Sadigly
an hour ago
Proving ZA=ZB
nAalniaOMliO   7
N 2 hours ago by nAalniaOMliO
Source: Belarusian National Olympiad 2025
Point $H$ is the foot of the altitude from $A$ of triangle $ABC$. On the lines $AB$ and $AC$ points $X$ and $Y$ are marked such that the circumcircles of triangles $BXH$ and $CYH$ are tangent, call this circles $w_B$ and $w_C$ respectively. Tangent lines to circles $w_B$ and $w_C$ at $X$ and $Y$ intersect at $Z$.
Prove that $ZA=ZH$.
Vadzim Kamianetski
7 replies
nAalniaOMliO
Mar 28, 2025
nAalniaOMliO
2 hours ago
Tangents involving a centroid with an isosceles triangle result
pithon_with_an_i   2
N 2 hours ago by Funcshun840
Source: Revenge JOM 2025 Problem 5, Revenge JOMSL 2025 G5, Own
A triangle $ABC$ has centroid $G$. A line parallel to $BC$ passing through $G$ intersects the circumcircle of $ABC$ at a point $D$. Let lines $AD$ and $BC$ intersect at $E$. Suppose a point $P$ is chosen on $BC$ such that the tangent of the circumcircle of $DEP$ at $D$, the tangent of the circumcircle of $ABC$ at $A$ and $BC$ concur. Prove that $GP = PD$.

Remark 1
Remark 2
2 replies
pithon_with_an_i
5 hours ago
Funcshun840
2 hours ago
orthocenter on sus circle
DVDTSB   1
N 2 hours ago by Double07
Source: Romania TST 2025 Day 2 P1
Let \( ABC \) be an acute triangle with \( AB < AC \), and let \( O \) be the center of its circumcircle. Let \( A' \) be the reflection of \( A \) with respect to \( BC \). The line through \( O \) parallel to \( BC \) intersects \( AC \) at \( F \), and the tangent at \( F \) to the circle \( \odot(BFC) \) intersects the line through \( A' \) parallel to \( BC \) at point \( M \). Let \( K \) be a point on the ray \( AB \), starting at \( A \), such that \( AK = 4AB \).
Show that the orthocenter of triangle \( ABC \) lies on the circle with diameter \( KM \).

Proposed by Radu Lecoiu

1 reply
DVDTSB
Today at 12:18 PM
Double07
2 hours ago
a