IGO 2016 Advanced/1

For discussing Olympiad level Geometry Problems
User avatar
Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

IGO 2016 Advanced/1

Unread post by 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.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

User avatar
Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

Re: IGO 2016 Advanced/1

Unread post by 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.

User avatar
ahmedittihad
Posts: 147
Joined: Mon Mar 28, 2016 6:21 pm

Re: IGO 2016 Advanced/1

Unread post by 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.

Post Reply