Problem:
In rectangle ABCD,AB=12 and BC=10. Points E and F lie inside rectangle ABCD so that BE=9,DF=8,BE∥DF,EF∥AB, and line BE intersects segment AD. The length EF can be expressed in the form mn−p, where m,n, and p are positive integers and n is not divisible by the square of any prime. Find m+n+p.
Solution:
Let line EF intersect AD at P1, and let line EF intersect BC at P2.
Let G be the foot of the perpendicular from E to AB, and let H be the foot of the perpendicular from F to CD. Note that △BEG is similar to △DFH, thus FHEG=DFBE. Therefore FH10−FH=89, and FH=1780. Note that
Extend AB and DC to X and Y respectively, so that BX=CY=EF. Then DYX is a right triangle and EFXB is a parallelogram. It follows that DX=17,XY=10, and DY=EF+12. By the Pythagorean Theorem, (EF+12)2=172−102, and thus EF=321−12, as before.
