China National Olympiad 2018 P4

For discussing Olympiad level Geometry Problems
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Atonu Roy Chowdhury
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China National Olympiad 2018 P4

Unread post by Atonu Roy Chowdhury » Mon Apr 09, 2018 9:31 am

$ABCD$ is a cyclic quadrilateral whose diagonals intersect at $P$. The circumcircle of $\triangle APD$ meets segment $AB$ at points $A$ and $E$. The circumcircle of $\triangle BPC$ meets segment $AB$ at points $B$ and $F$. Let $I$ and $J$ be the incenters of $\triangle ADE$ and $\triangle BCF$, respectively. Segments $IJ$ and $AC$ meet at $K$. Prove that the points $A,I,K,E$ are cyclic.
This was freedom. Losing all hope was freedom.

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Atonu Roy Chowdhury
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Location: Chittagong, Bangladesh

Re: China National Olympiad 2018 P4

Unread post by Atonu Roy Chowdhury » Mon Apr 09, 2018 9:33 am

China 2018 #4.png
China 2018 #4.png (56.56 KiB) Viewed 406 times
Let $X$ and $Y$ be the midpoints of shorter arcs $AD$ and $BC$ in the circles $\circ ADPE$ and $\circ BCPF$ respectively.
$Z$ is the intersection point of $XE$ and $YF$.
It is obvious that $X, P, Y$ are collinear.

Claim 1: $\triangle AXP$ and $\triangle BYP$ are similar.
Proof: $\angle AXP = \angle ADP = \angle PCB = \angle BYP$
and $\angle APX = \frac {\angle APD}{2} = \frac {\angle BPC}{2} = \angle BPY$

Claim 2: $XY||IJ$
Proof: $\frac{XI}{YJ} = \frac{XA}{YB} = \frac{PA}{PB} = \frac{\sin \angle PBA}{\sin \angle PAB} = \frac{\sin \angle PYZ}{\sin \angle PXZ} = \frac{XZ}{YZ}$

Now we get back to our problem.
$\angle IKA = \angle XPA = \frac {\angle DPA}{2} = \frac {\angle DEA}{2} = \angle IEA$
$Q.E.D.$
This was freedom. Losing all hope was freedom.

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Safwan
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Re: China National Olympiad 2018 P4

Unread post by Safwan » Fri Jan 18, 2019 9:18 pm

gettin' headaches seein' this one!

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