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orthocenter on sus circle
DVDTSB   1
N an hour 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
+1 w
DVDTSB
Today at 12:18 PM
Double07
an hour ago
Incircle triangles inequality
MathMystic33   0
an hour ago
Source: 2025 Macedonian Team Selection Test P5
Let $\triangle ABC$ be a triangle with side‐lengths $a,b,c$, incenter $I$, and circumradius $R$. Denote by $P$ the area of $\triangle ABC$, and let $P_1,\;P_2,\;P_3$ be the areas of triangles $\triangle ABI$, $\triangle BCI$, and $\triangle CAI$, respectively. Prove that
\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]
0 replies
MathMystic33
an hour ago
0 replies
Collinearity of intersection points in a triangle
MathMystic33   0
an hour ago
Source: 2025 Macedonian Team Selection Test P1
On the sides of the triangle \(\triangle ABC\) lie the following points: \(K\) and \(L\) on \(AB\), \(M\) on \(BC\), and \(N\) on \(CA\). Let
\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]and let the line \(CS\) meet \(AB\) at \(Q\). Prove that the points \(P\), \(Q\), and \(R\) are collinear.
0 replies
MathMystic33
an hour ago
0 replies
Brazilian Locus
kraDracsO   15
N an hour ago by Ilikeminecraft
Source: IberoAmerican, Day 2, P4
Let $B$ and $C$ be two fixed points in the plane. For each point $A$ of the plane, outside of the line $BC$, let $G$ be the barycenter of the triangle $ABC$. Determine the locus of points $A$ such that $\angle BAC + \angle BGC = 180^{\circ}$.

Note: The locus is the set of all points of the plane that satisfies the property.
15 replies
kraDracsO
Sep 9, 2023
Ilikeminecraft
an hour ago
Circumcircle of MUV tangent to two circles at once
MathMystic33   0
2 hours ago
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 \( \omega_2 \).
0 replies
MathMystic33
2 hours ago
0 replies
Roots of unity
Henryfamz   0
3 hours ago
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
0 replies
Henryfamz
3 hours ago
0 replies
Aime 2005a #15
4everwise   22
N 3 hours ago by Ilikeminecraft
Source: Aime 2005a #15
Triangle $ABC$ has $BC=20$. The incircle of the triangle evenly trisects the median $AD$. If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n$.
22 replies
4everwise
Nov 10, 2005
Ilikeminecraft
3 hours ago
Nice geometry...
Sadigly   1
N 4 hours ago by aaravdodhia
Source: Azerbaijan Senior NMO 2020
Let $ABC$ be a scalene triangle, and let $I$ be its incenter. A point $D$ is chosen on line $BC$, such that the circumcircle of triangle $BID$ intersects $AB$ at $E\neq B$, and the circumcircle of triangle $CID$ intersects $AC$ at $F\neq C$. Circumcircle of triangle $EDF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. Lines $FD$ and $IC$ intersect at $Q$, and lines $ED$ and $BI$ intersect at $P$. Prove that $EN\parallel MF\parallel PQ$.
1 reply
Sadigly
Sunday at 10:17 PM
aaravdodhia
4 hours ago
AM=CN in Russia
mathuz   25
N 4 hours ago by Ilikeminecraft
Source: AllRussian-2014, Grade 11, day1, P4
Given a triangle $ABC$ with $AB>BC$, $ \Omega $ is circumcircle. Let $M$, $N$ are lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. $K(.)=MN\cap AC$ and $P$ is incenter of the triangle $AMK$, $Q$ is K-excenter of the triangle $CNK$ (opposite to $K$ and tangents to $CN$). If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.

M. Kungodjin
25 replies
mathuz
Apr 29, 2014
Ilikeminecraft
4 hours ago
Simson lines on OH circle
DVDTSB   2
N 4 hours ago by SomeonesPenguin
Source: Romania TST 2025 Day 2 P4
Let \( ABC \) and \( DEF \) be two triangles inscribed in the same circle, centered at \( O \), and sharing the same orthocenter \( H \ne O \). The Simson lines of the points \( D, E, F \) with respect to triangle \( ABC \) form a non-degenerate triangle \( \Delta \).
Prove that the orthocenter of \( \Delta \) lies on the circle with diameter \( OH \).

Note. Assume that the points \( A, F, B, D, C, E \) lie in this order on the circle and form a convex, non-degenerate hexagon.

Proposed by Andrei Chiriță
2 replies
DVDTSB
Today at 12:10 PM
SomeonesPenguin
4 hours ago
a