$ ABC$ is an acute-angled triangle. $ M$ is the midpoint of $ BC$ and $ P$ is the point on $ AM$ such that $ MB = MP$. $ H$ is the foot of the perpendicular from $ P$ to $ BC$. The lines through $ H$ perpendicular to $ PB$, $ PC$ meet $ AB, AC$ respectively at $ Q, R$. Show that $ BC$ is tangent to the circle through $ Q, H, R$ at $ H$. Original Formulation: For an acute triangle $ ABC, M$ is the midpoint of the segment $ BC, P$ is a point on the segment $ AM$ such that $ PM = BM, H$ is the foot of the perpendicular line from $ P$ to $ BC, Q$ is the point of intersection of segment $ AB$ and the line passing through $ H$ that is perpendicular to $ PB,$ and finally, $ R$ is the point of intersection of the segment $ AC$ and the line passing through $ H$ that is perpendicular to $ PC.$ Show that the circumcircle of $ QHR$ is tangent to the side $ BC$ at point $ H.$
Problem
Source: IMO ShortList 1991 Problem 2 (JAP 5)
Tags: geometry, circumcircle, trigonometry, cyclic quadrilateral, IMO Shortlist
30.07.2005 06:32
Let $HQ,HR$ cut $BP,CP$ at $D,E$ respectively. Also let $CP, BP$ cut $AB,AC$ at $X,Y$ respectively. By Ceva, $XY//BC$, so $XP:PC=YP:PB$. As $MP=MB=MC$, so $\angle BPC=90^{\circ}$. Therefore $QD:DH=XP:PC=YP:PB=RE:EH$, and hence $QR//DE$. Since $PDHE$ is a cyclic quadrilateral, so $\angle HQR=\angle HDE = \angle HPE = 90^{\circ}-\angle PCH = \angle EHC$. Therefore $BC$ is tangent to the circle through $Q,H,R$ at $H$.
29.01.2009 16:42
This is a problem be impli from perpendicular circle
20.05.2014 10:38
Notations$HQ \cap BP=K,HR \cap PC=L$ From $CM=BM=PM$ it is easy to see that $\angle{BPC}=90^{\circ} \Rightarrow KPLH$ is a rectangle.Nice proof of $KL \parallel QR$ by yptsoi.I proved it using trigonometry letting $\angle{BHQ}=x$.After that it is just $\angle{BHK}=\angle{BPH}=\angle{KPH}=\angle{KLH}=\angle{QRH} \Rightarrow BC$ is tangent to $\odot{QRH}$ at $H$.
14.08.2024 01:24
unc im reviving the thread Let $HQ\cap BP=X$, $HR\cap CP=Y$. We claim $XY\parallel PQ$. We will show that $\frac{HX}{HQ}=\frac{HY}{HR}$. Extend $CP,BP$ to $AB,AC$ respectively at $D,E$. It is well-known, say by Ceva, that $DE\parallel BC$. Hence $\frac{CP}{CD}=\frac{BP}{BE}$. Yet as $BP\perp CP$ by Thales, we have $HQ\parallel CD$ and $HR\parallel BE$. This means that $\frac{HX}{HQ}=\frac{CP}{CD}=\frac{BP}{BE}=\frac{HY}{HR}$, hence we prove our lemma. Now, $\angle HPQ=\angle HXY=\angle BPH=\angle CHY$. By AST we are done.