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

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

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.