BdMO Online Forum

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

Complicated. No $A$ found in the diagram.

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 $FAP$=$90-AFP$.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 $PFM$=$90-AFP$=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.

Very nice solution by using radical axis.Here is the diagram:

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

samiul_samin wrote: ↑

Sat Feb 24, 2018 9:42 pm

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 Screenshot_2018-02-23-20-04-06-1.png