triangle with A=60

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triangle with A=60

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Source: Indonesia TST 2009 First Stage Test 3 Problem 3

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#1 h Dec 28, 2008, 12:04 AM • 2 Y

Y by Adventure10, Mango247

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.

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#2 h Dec 28, 2008, 7:21 AM • 2 Y

Y by Adventure10, Mango247

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!

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#3 h May 31, 2010, 9:37 AM • 2 Y

Y by Adventure10, Mango247

the first problem a)

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#4 h May 31, 2010, 12:18 PM • 1 Y

Y by Adventure10

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)

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#5 h May 31, 2010, 1:58 PM • 2 Y

Y by Adventure10, Mango247

hi, can you write more clear on which radical axis(please, write the radius and the center of that circle)

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#6 h Jul 14, 2014, 2:39 PM • 2 Y

Y by Adventure10, Mango247

Anyone can give proof for B????

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