IGO 2016 Advanced/1

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IGO 2016 Advanced/1

Post Number:#1  Unread postby Thamim Zahin » Tue Jan 10, 2017 4:16 pm

1. Let the circles $w$ and $w'$ intersect in points $A$ and $B$. Tangent to circle $w$ at $A$ intersects $w'$ in $C$ and tangent to circle $w'$ at $A$ intersects $w$ in $D$. Suppose that the segment $CD$ intersects $w$ and $w'$ in $E$ and $F$, respectively (assume that $E$ is between $F$ and $C$). The perpendicular to $AC$ from $E$ intersects $w'$ in point $P$ and perpendicular to $AD$ from $F$ intersects $w$ in point $Q$ (The points $A, P$ and $Q$ lie on the same side of the line $CD$). Prove that the points $A, P$ and $Q$ are collinear.
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Re: IGO 2016 Advanced/1

Post Number:#2  Unread postby Thanic Nur Samin » Wed Jan 11, 2017 6:13 pm

It seems unbelievable that I missed this on the contest.

Using alternate segment theorem, it follows that $\triangle AEC ~ \triangle AFD$. From this, we see that $\angle AEF=\angle AFE$. Now, we get that $\angle APC=\angle AFD=\angle AEC$. However, $PE\perp AC$. So, $P$ is the reflection of $E$ over $AC$. Similarly, $Q$ is the reflection of $F$ over $AD$.

Now, the obvious part. $\angle PAQ=\angle PAC+\angle CAD+\angle DAQ=\angle ADC+\angle CAD+\angle ACD=180^{\circ}$. So, $P,A$ and $Q$ are collinear.

Also, I noticed that $AB$ is the symmedian of $\triangle ACD$. Is there any solution that lets us use that information?
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.
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Re: IGO 2016 Advanced/1

Post Number:#3  Unread postby ahmedittihad » Wed Jan 11, 2017 8:29 pm

This formulation has many interesting extra problems.
$1.$ $B$ is the miquel point of quadrilateral $PEFQ$.
$2.$ $AB$ intersect circle $EFBX$ = $K$. Then, $K$ is the orthocenter of $\triangle PQX$
$3.$ $AB$ is a symmedian of both $\triangle ACD$ and $\triangle BEF$.
Frankly, my dear, I don't give a damn.
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