Trapezium | Geometry | PRMO-2018 | Problem 5

Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Can you now finish the problem ……….

Draw a perpendicular from C on AB at the point F

Can you finish the problem……..

Let the inner circle touches BC at E

Then CE=36 (as BE & BQ are tangents)

BE=49 (as CE & PC are tangents)

Let the radius of the inner circle be r

Therefore AQ=PD=r and AD=2r

Let draw a perpendicular from C on AB at the point F

So BF=(AB-AF)=(49-36)=13

BC=85

Now in the triangle CBF we have

\(CF^2=85^2-13^2\)

\(\Rightarrow CF=84\)

Therefore CF=PQ=84