Let $ ABC$ be an acute triangle with $ \angle BAC=60^{\circ}$. Let $ P$ be a point in triangle $ ABC$ with $ \angle APB=\angle BPC=\angle CPA=120^{\circ}$. The foots of perpendicular from $ P$ to $ BC,CA,AB$ are $ X,Y,Z$, respectively. Let $ M$ be the midpoint of $ YZ$. a) Prove that $ \angle YXZ=60^{\circ}$ b) Prove that $ X,P,M$ are collinear.
Problem
Source: Indonesia TST 2009 First Stage Test 3 Problem 3
Tags: geometry, power of a point, radical axis, geometry proposed
plane geometry
28.12.2008 09:21
Construct equilateral triangle ABC’ , ACB’ Then we have A,P,B,C’ A,P,C,B’ are concyclic ∠YXZ=∠AC’P+∠AB’P=180-120=60 Denote N=XP∩YZ ZN/NY=XZsin∠ZXP/XY sin∠YXP =XZ sin∠PAC/XY sin∠PAB= sin∠PAB/ sin∠120 * sin∠120/ sin∠PAC * sin∠PAC/ sin∠PAB=1 N=N Done!
Attachments:
Ronald Widjojo
31.05.2010 15:18
For the second part, can be solved by noticing that $M$ lies on radical axis of 2 circles that intersect in $X$ and $P$. ($MY$ and $MZ$ is tangent to $2$ circles)
arshakus
31.05.2010 16:58
hi, can you write more clear on which radical axis(please, write the radius and the center of that circle)
nawaites
14.07.2014 17:39
Anyone can give proof for B????