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.
