A triangle $ABC$ with orthocenter $H$ is given. $P$ is a variable point on line $BC$. The perpendicular to $BC$ through $P$ meets $BH$, $CH$ at $X$, $Y$ respectively. The line through $H$ parallel to $BC$ meets $AP$ at $Q$. Lines $QX$ and $QY$ meet $BC$ at $U$, $V$ respectively. Find the shape of the locus of the incenters of the triangles $QUV$.
Problem
Source: 2022 Israel TST 3 P1
Tags: geometry, incenter
RagvaloD
23.05.2022 01:30
$\angle XHY= 90^o-\angle HCA =\angle BAC, \angle HYX=\angle PYC= \angle ABC$ so $\triangle ABC \sim \triangle HYX$ Let $HQ$ intersect $XY$ in $E$ and $AH$ intersect $BC$ in $D$. Then $HQ \parallel DP \to \frac{HQ}{HE}=\frac{HQ}{DP}=\frac{AH}{AD}$ and so $Q$ is orthocenter of $ \triangle HXY$. So $QV \perp BH \to QV \parallel AC,QU \parallel AB$ So $\triangle ABC \sim \triangle QUV$ with $k=\frac{AD}{HD}$ So $r_{\triangle QUV}=\frac{HD}{AD}r_{\triangle ABC}$ ($r_{\triangle ABC}$ is inradius of $\triangle ABC$) and so locus on the incenters of triangles $QUV$ is line that parallel $BC$ and is on distance $\frac{HD}{AD}r_{\triangle ABC}$ from $BC$
atdaotlohbh
21.04.2024 23:10
Let the altitudes be $AD$, $BE$, $CF$. By simple angle chasing we get that $\angle FQP=\angle FYP$, so the quadraliteral $FQYP$ can be inscribed in a circle. Thus, $\angle QYF = \angle QPF$, so $QY \parallel AC$. Similarly we get that $QX \parallel BC$. Then triangle $UVQ$ can be achieved from $ABC$ by homothethy with center $P$ and constant koefficient, so the locus of incenters is a line