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parmenides51   3
N Apr 19, 2025 by ihategeo_1969
Source: 2019 Geo Mock - Olympiad by Tovi Wen #5 https://artofproblemsolving.com/community/c594864h1787237p11805928
Let $ABC$ be an acute triangle with orthocenter $H$, and let $D$ denote the foot of the altitude from $B$ to $\overline{AC}$. Let the circle $\Omega$ with diameter $\overline{BC}$ intersect altitude $\overline{AH}$ at distinct points $X$ and $Y$. Suppose that the circle with center $C$ passing through point $X$ intersects segment $\overline{BD}$ at $L$. Line $\overline{CL}$ meets $\Omega$ at $P \neq C$. If $\overline{PX}$ intersects $\overline{AL}$ at $R$, and $\overline{PY}$ intersects $\overline{AL}$ at $S$, prove that $\Omega$ is tangent to the circumcircle of $\triangle DRS$.
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parmenides51
Nov 26, 2023
ihategeo_1969
Apr 19, 2025
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Source: 2019 Geo Mock - Olympiad by Tovi Wen #5 https://artofproblemsolving.com/community/c594864h1787237p11805928
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parmenides51
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parmenides51
#1 h  Nov 26, 2023, 10:42 PM
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Let $ABC$ be an acute triangle with orthocenter $H$, and let $D$ denote the foot of the altitude from $B$ to $\overline{AC}$. Let the circle $\Omega$ with diameter $\overline{BC}$ intersect altitude $\overline{AH}$ at distinct points $X$ and $Y$. Suppose that the circle with center $C$ passing through point $X$ intersects segment $\overline{BD}$ at $L$. Line $\overline{CL}$ meets $\Omega$ at $P \neq C$. If $\overline{PX}$ intersects $\overline{AL}$ at $R$, and $\overline{PY}$ intersects $\overline{AL}$ at $S$, prove that $\Omega$ is tangent to the circumcircle of $\triangle DRS$.
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aqwxderf
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aqwxderf
#2 h  Nov 27, 2023, 1:08 AM
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Very cute.

Let $\overline{AH} \cap \overline{BC} = E$.

Since $[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$ is the incenter of $\triangle PXY $.
$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$ and $S $ belong to the $P $ mixtilinear incircle of $\triangle PXY$.
Since $B $ is the midpoint of arc $XPY$ and $B - L - D \Rightarrow D$ belongs to the $P $ mixtilinear incircle of $\triangle PXY$.

This clearly implies that the circumcircle of $\triangle DRS$ is the $P $ mixtilinear incircle of $\triangle PXY$ thus tangent to $\Omega$.
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parmenides51
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parmenides51
#3 h  Nov 27, 2023, 8:13 PM
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older posted solutions from another thread in order to have them all in one place

by TheDarkPrince

by rocketscience

by math_pi_rate
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ihategeo_1969
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ihategeo_1969
#4 h  Apr 19, 2025, 12:16 AM
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Let $D$ be foot of $A$ to $\overline{BC}$.

Claim: $\measuredangle ALC=90^\circ$.
Proof: Say $L'$ is point on segment $\overline{BE}$ such that $\measuredangle AL'C=90^\circ$ so $\overline{CL'}$ is tangent to $(AEL')$ and also $\overline{CX}$ is tangent to $(BXD)$ and hence by PoP we have $CL'^2=CE \cdot CA=CD \cdot CB=CX^2$ and so $L \in (C,CX)$. $\square$

Claim: $L$ is incenter of $\triangle PXY$.
Proof: $\measuredangle XPL=\measuredangle XYC=\measuredangle CXY=\measuredangle LPY$ and see that $C$ is minor arc midpoint of $\widehat{XY}$ and so $L \in (C,CX)$ and $L$ obviously lies inside $\triangle PXY$ and so by IE lemma, the result follows. $\square$

Now $B$ is major arc midpoint of $\widehat{XY}$ and so $E=\overline{BC} \cap (PXY)$ and hence it is the $P$-Mixtilinear intouch point and by first claim $(RES)$ is $P$-Mixtilinear incircle and done.
This post has been edited 1 time. Last edited by ihategeo_1969, Apr 19, 2025, 12:16 AM
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