Let $AD$ be the altitude of an acute-angled triangle $ABC$ and $A'$ be the point on its circumcircle opposite to $A$. A point $P$ lies on the segment $AD$, and points $X$, $Y$ lie on the segments $AB$, $AC$ respectively in such a way that $\angle CBP = \angle ADY$, $\angle BCP = \angle ADX$. Let $PA'$ meet $BC$ at point $T$. Prove that $D$, $X$, $Y$, $T$ are concyclic.
Problem
Source: Sharygin Final 2024 8.3
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2024
math_comb01
02.08.2024 15:11
Cute! Let $E,F$ be the foot of altitude form $P$ to $AB,AC$ and let $(BPC) \cap AB,AC = U,V$ then by angle condition we know that $PDUX,PDVY$ are cyclic which implies that $UXVY$ is cyclic, now by angle chase we get $X$ belongs to the pedal circle of $P$, and so $XYDEF$ is cyclic, now let $(AEF) \cap (ABC) = K$ then by radical axes $AK,EF,BC$ concur at a point, say $R$, then $RD \cdot RT=RA \cdot RK = RE \cdot RF$ and so $EFTD$ is cyclic which implies $DXYT$ is cyclic as desired
sami1618
13.08.2024 20:12
Let $AD$ meet $(ABC)$ again at $H$ and let $Q$ be the isogonal conjugate of $P$ w.r.t. triangle $ABC$. Claim: $X$ is the foot of $Q$ onto $AB$ Triangle $AXD$ and $AQC$ are similar and as $\angle ADC=90^{\circ}$ we must have $\angle AXQ=90^{\circ}$ by spiral symmetry. Similarly $Y$ is also the foot of $Q$ onto $AC$. Now we claim the $T$ is the foot of $Q$ onto $BC$ which follows from $$\frac{A'Q}{QA}=\frac{BX}{XA}=\frac{BD\cdot \sin(\angle BDX)}{AD\cdot \sin(\angle ADX)}=\frac{HD}{CD}\cot(\angle PCD)=\frac{HD}{DP}=\frac{A'T}{TP}$$As the pedal triangle of two isogonal conjugates share a circumcircle, we are done.
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