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### BdMO National Secondary 2006/10

Posted: 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

### Re: BdMO National Secondary 2006/10

Posted: Sat Feb 24, 2018 11:49 pm
Complicated. No \$A\$ found in the diagram.

### Re: BdMO National Secondary 2006/10

Posted: Sat Mar 03, 2018 5:11 pm
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.

### Re: BdMO National Secondary 2006/10

Posted: Mon Mar 05, 2018 2:03 pm
Very nice solution by using radical axis.Here is the diagram:
Screenshot_2018-03-05-13-58-40-1.png

### Re: BdMO National Secondary 2006/10

Posted: Tue Mar 06, 2018 3:08 am
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

### Re: BdMO National Secondary 2006/10

Posted: Mon Feb 18, 2019 11:42 pm
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
Screenshot_2019-02-18-23-33-59-1.png (13.3 KiB) Viewed 1638 times