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XY is tangent to a fixed circle
a_507_bc   3
N 39 minutes ago by lksb
Source: Baltic Way 2022/15
Let $\Omega$ be a circle, and $B, C$ are two fixed points on $\Omega$. Given a third point $A$ on $\Omega$, let $X$ and $Y$ denote the feet of the altitudes from $B$ and $C$, respectively, in the triangle $ABC$. Prove that there exists a fixed circle $\Gamma$ such that $XY$ is tangent to $\Gamma$ regardless of the choice of the point $A$.
3 replies
a_507_bc
Nov 12, 2022
lksb
39 minutes ago
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One of the lines is tangent
Rijul saini   8
N an hour ago by ihategeo_1969
Source: LMAO 2025 Day 2 Problem 2
Let $ABC$ be a scalene triangle with incircle $\omega$. Denote by $N$ the midpoint of arc $BAC$ in the circumcircle of $ABC$, and by $D$ the point where the $A$-excircle touches $BC$. Suppose the circumcircle of $AND$ meets $BC$ again at $P \neq D$ and intersects $\omega$ at two points $X$, $Y$.

Prove that either $PX$ or $PY$ is tangent to $\omega$.

Proposed by Sanjana Philo Chacko
8 replies
Rijul saini
Wednesday at 7:02 PM
ihategeo_1969
an hour ago
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Cyclic quads jigsaw
Miquel-point   0
Apr 24, 2025
Source: Kürschák József Competition 2024/1
The quadrilateral $ABCD$ is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral $ABCD$. Prove that $ABCD$ is also a cyclic quadrilateral.
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Miquel-point
Apr 24, 2025
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Cyclic quads jigsaw
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Source: Kürschák József Competition 2024/1
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Miquel-point
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Miquel-point
#1 h  Apr 24, 2025, 11:55 PM
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The quadrilateral $ABCD$ is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral $ABCD$. Prove that $ABCD$ is also a cyclic quadrilateral.
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