BdMO National Secondary 2006/10
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
BdMO National Secondary 2006/10
Two circles with the centers $O$ and $C$,touch each other externally at $F$.The tangent $TT'$ touches two circles at $P$ and $Q$ respectively.Prove $\angle PFO+\angle QFC=1$ right angle
Last edited by samiul_samin on Sun Feb 25, 2018 7:00 am, edited 1 time in total.
Re: BdMO National Secondary 2006/10
Complicated. No $A$ found in the diagram.

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: BdMO National Secondary 2006/10
It suffices to prove that $\bigtriangleup PFQ$ is a right triangle.Now,let $OF$ and $CF$ extended meet the $2$ circles for the second time at $A$ and $B$ respectively.$\bigtriangleup APF$ and $\bigtriangleup BQF$ both are right triangles. So,by Alternate Segment Theorem,angle $FPQ$=angle $FAP$=$90AFP$.Let the radical axis of the $2$ circles meet $PQ$ at $M$.It is well known that $M$ is the midpoint of $PQ$.Since angle $AFM$=$90$°,angle $PFM$=$90AFP$=angle $FPM$.We therefore have $PM=QM=FM$,so $M$ is the circumcenter of $\bigtriangleup PFQ$,hence $\bigtriangleup PFQ$ is a right triangle.The rest is trivial.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Secondary 2006/10
Very nice solution by using radical axis.Here is the diagram:
Re: BdMO National Secondary 2006/10
If I just connect O and P , C and Q , they're both perpendiculer on TT'
Now solving the problem will be a whole lot easier
Now solving the problem will be a whole lot easier
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Secondary 2006/10
samiul_samin wrote: ↑Sat Feb 24, 2018 9:42 pmTwo circles with the centers $O$ and $C$,touch each other externally at $F$.The tangent $TT'$ touches two circles at $P$ and $Q$ respectively.Prove $\angle PFO+\angle QFC=1$ right angle Screenshot_201802232004061.png