Problem:
In triangle ABC,AB=1120​AC. The angle bisector of ∠A intersects BC at point D, and point M is the midpoint of AD. Let P be the point of intersection of AC and line BM. The ratio of CP to PA can be expressed in the form nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let [AMP]=x,[CMP]=y,[CMD]=z, and [BMD]=t.
Because M is the midpoint of AD, it follows that [AMB]=[BMD]=t.
The Angle Bisector Theorem yields
tz​=BDCD​=ABAC​=2011​
Also,
[CBM][CPM]​=MBPM​=[ABM][APM]​, or z+ty​=tx​.
Thus
PACP​=xy​=tz+t​=tz​+1=2011​+1=2031​
Hence m+n=51​.
OR
Through D draw a parallel to line BP intersecting line AC at Q. Then PQ=20k,QC=11k, and PA=20k, using the Angle Bisector Theorem and the fact that 3 or more parallel lines divide all transversals in the same proportions. Thus PACP​=20k20k+11k​=2031​ as in the previous solution.
