Given an acute triangle $ABC, H$ is the foot of the altitude drawn from the point $A$ on the line $BC, P$ and $K \ne H$ are arbitrary points on the segments $AH$ and$ BC$ respectively. Segments $AC$ and $BP$ intersect at point $B_1$, lines $AB$ and $CP$ at point $C_1$. Let $X$ and $Y$ be the projections of point $H$ on the lines $KB_1$ and $KC_1$, respectively. Prove that points $A, P, X$ and $Y$ lie on one circle.
Problem
Source: Ukraine TST 2015 p6
Tags: geometry, Concyclic, perpendicular, altitude
math_pi_rate
01.05.2020 21:41
Nice problem! Here's my solution: It's well known that $HA$ must bisect $\angle B_1HC_1$. Let $U,V=BC,AH \cap B_1C_1$, and suppose the perpendicular bisector of $HV$ meets lines $KB_1,KC_1$ at $M,N$. Since $BC$ (i.e. line $KU$) is parallel to $MN$, so we have $$-1=(B_1,C_1;U,V) \stackrel{K}{=} (M,N;\infty_{MN},KV \cap MN)$$which means that $KV$ bisects $MN$. Also, the reflection of $U$ in $MN$ (i.e. $H$) lies on $KH$, which combined with $KH \parallel MN$ gives that $MN$ bisects $KV$. In particular, we have that $KMVN$ is a parallelogram, which easily translates to the fact that $KHMN$ is an isosceles trapezoid. Since $XY$ is the Simson line of $H$ wrt $\triangle KMN$, so $XY$ must pass through the midpoint $T$ of segment $HV$. Then we get $$-1=(H,V;P,A) \Rightarrow TP \cdot TA=TH^2=TX \cdot TY \Rightarrow APXY \text{ is cyclic.} \quad \blacksquare$$
hectorraul
02.05.2020 17:29
Great Solution!! math_pi_rate
Sugiyem
02.05.2020 18:01
Define $D=AH\cap B_{1}C_{1}$, $E=B_{1}C_{1}\cap BC$, $F=XY\cap AH$, $G=KY\cap HX $, $I=KX\cap HY$, $J=XY\cap GI$ and $L=XY\cap BC$. $\textbf{Claim 1.}$ $D,J,K$ are collinear. $\textbf{Proof:}$ Notice that: $$-1=(C_{1},B_{1},D,E)\stackrel{K}{=}(Y,X;KD\cap XY,L)$$But it's easy to check that $(Y,X,J,L)=-1$, so $J=KD\cap XY$ and the claim is proven. Now note that $GYIX$ and $KYXH$ are both cyclic because $\angle GYI=\angle GXI=\angle KYH=\angle KXH=90^{\circ}$. Name the circumcircle of $GYIX$ and $KYXH$ as $\Omega$ and $\Gamma$ respectively. By brokard at quadrilateral $GXIY$, we have $KJ$ is the polar of $H$ wrt $\Omega$. As $K,J,D$ are collinear, we have that $D$ lies on the polar of $H$ wrt $\Omega$ Define $F'$ as the midpoint of $HD$, then we'll have that $F'H^{2}=\text{Pow}_{\Omega}(F')$. Now, consider the radical center of $\Gamma$, $\Omega$ and degenerate circle $H$. It's not hard to see that $F$ is the radical center. ($XY$ is the radical axis of $\Gamma$ and $\Omega$, $AH$ is the radical axis of $H$ and $\Gamma$) So, $FH^{2}=\text{Pow}_{\Omega}(F)$, and we'll have $F=F'$. Therefore, $F$ is the midpoint of $HD$. As $(H,D;P,A)=-1$, we'll have $FH^{2}=FP\times PA$. But simple angle chasing shows that $FH^{2}=FX\times FY$. Hence, $FX\times FY=FP\times PA$. So, $A,P,X,Y$ are concyclic and the problem is now done.