Problem:
In the diagram below, ABCD is a rectangle with side lengths AB=3 and BC=11, and AECF is a rectangle with side lengths AF=7 and FC=9. The area of the shaded region common to the interiors of both rectangles is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let G be the intersection of AD and CF. Then △AGF∼△CGD, so
CGAG=DGFG=CDAF=37
It follows that there are constants x and y such that AG=7x,CG=3x,FG=7y, and DG=3y. Thus
7x+3y=11 and 7y+3x=9.
Adding the two equations and dividing by 10 gives x+y=2. Subtracting the second equation from the first and dividing by 4 gives x−y=21. Hence x=45 and y=43. Because AD∥BC and AE∥CF, the region interior to the two rectangles is a parallelogram, and thus the required area is AG⋅AB=7x⋅3=7⋅45⋅3=4105. The requested sum is 105+4=109.
Defining G as above, let t=DG so that AG=11−t and, by the Pythagorean Theorem, CG=32+t2. Because △AGF∼△CGD, it follows that
AFAG=CDCG and 711−t=39+t2
Solving for t gives t=49, from which the required area is 3⋅(11−49)=4105, as above.
