Question

Let ABC be a triangle such that AB= 5, AC= 8, and ∠BAC= 60◦. Let P be a point inside the triangle such that ∠APB = ∠BPC = ∠CPA. Lines BP and AC intersect at E, and lines CP and AB intersect at F. The circumcircles of triangles BPF and CPE intersect at points P and Q not equal P. Then QE + QF = m/n, where m and n are positive integers with gcd(m,n) = 1. Compute 100m+n.

Answer 1

The length from the circumcentre P of the triangle to points E, D and F =>

Let P be the center of the circumcenrer and PA be a perpendicular to FAE.

∡FAP = 0.5 × ∡FAE = 30°,

since, ∠APB = ∠BPC = ∠CPA.

we obtain:

PE = 8cos30° = 6.93...

$QE + QF = 6.93×2 = 13.9 = \dfrac{139}{10}$ ,

m = 139, n = 10

since, $gcd(m,n) = 1,$

100m + n = 100.139 + 1.10 = 13910