we have an isogonal triangle $ABC$ such that $BC=AB$. take a random $P$ on the altitude from $B$ to $AC$. The circle $(ABP)$ intersects $AC$ second time in $M$. Take $N$ such that it's on the segment $AC$ and $AM=NC$ and $M \neq N$.The second intersection of $NP$ and circle $(APB)$ is $X$ , ($X \neq P$) and the second intersection of $AB$ and circle $(APN)$ is $Y$ ,($Y \neq A$).The tangent from $A$ to the circle $(APN)$ intersects the altitude from $B$ at $Z$. Prove that $CZ$ is tangent to circle $(PXY)$.
Problem
Source: Iran 2nd round 2022 P6
Tags: geometry
Mahdi_Mashayekhi
10.05.2022 10:46
My Solution During Contest. Let circle $ABP$ meet $AZ$ at $T$ and $AB$ meet $CZ$ at $K$. we will show $K$ is tangency point. Claim $: C,B,T$ are collinear. Proof $:$ Note that $\angle TBZ = \angle TAP = \angle PNA = \angle PNM = \angle PMN = \angle PBA = \angle PBC$. Claim $: X,T,K$ are collinear. Proof $:$ Note that $\angle XTA = \angle XPA = \angle PNA + \angle PAN = \angle PAZ + \angle PAN = \angle CAZ = \angle ACZ$. Now Note that $AZC$ is isosceles so $TK || AC$ so $\angle ACZ = \angle TKZ$ and $\angle KTA = \angle TKC$ so $\angle XTA + \angle KTA = \angle TKZ + \angle TKC = \angle 180$. Now Note that $\angle KXP = \angle TXP = \angle TAP = \angle PNA = \angle PYB = \angle PYK$ so $PYXK$ is cyclic also Note that $\angle PKC = \angle PTA = \angle PBA = \angle PYB = \angle PYK$ so $CK$ is tangent to $PYXK$. we're Done (Funny fact this took me much less time than P1)
strong_boy
30.07.2022 11:15
It was a really easy question! Only the picture of the question was complicated and difficult
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