Problem:
Let ABCD be a parallelogram with ∠BAD<90∘. A circle tangent to sides DA,AB, and BC intersects diagonal AC at points P and Q with AP<AQ, as shown. Suppose that AP=3,PQ=9, and QC=16. Then the area of ABCD can be expressed in the form mn, where m and n are positive integers, and n is not divisible by the square of any prime. Find m+n.
Solution:
Answer (150):
Let X,Y, and Z denote the points where the circle is tangent to AD,BC, and AB, respectively. Let R be the foot of the perpendicular from C to line AD.
By Power of a Point, AX=AP⋅AQ=6 and CY=CQ⋅CP=20. Hence AR=AX+CY=26. By the Pythagorean Theorem
XY=CR=AC2−AR2=282−262=63.
Let BY=BZ=t. Then AB=6+t, and so again by the Pythagorean Theorem
63=(6+t)2−(6−t)2=26t.
Thus t=29, so BC=249, and the area of ABCD is 249⋅63=1473. The requested sum is 147+3=150.
