Problem:
Rectangle ABCD and a semicircle with diameter AB are coplanar and have nonoverlapping interiors. Let R denote the region enclosed by the semicircle and the rectangle. Line ℓ meets the semicircle, segment AB, and segment CD at distinct points N,U, and T, respectively. Line ℓ divides region R into two regions with areas in the ratio 1:2. Suppose that AU=84,AN=126, and UB=168. Then DA can be represented as mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Let O be the midpoint of segment AB. Then AO=(AU+UB)/2=126=AN and UO=42. Thus △AON is equilateral and ∠AON=60∘, implying that the ratio of the areas of sectors AON and NOB is 1:2. Let Q be the foot of the perpendicular from U to line DC. Because AU/UB=1/2, the ratio of the areas of rectangles AUQD and UQCB is also 1:2. Because line l divides region R into two parts with area ratio 1:2, it follows that triangles NUO and UQT have the same area. Let P be the foot of the perpendicular from N to line AB. Note that triangles NUP and UTQ are similar, with area ratio equal to
NP2UQ2=Area(NUP)Area(UTQ)=Area(NUP)Area(NOU)=UPUO or UQ=NP⋅UPUO
In △NOP,NO=AO=126,∠NPO=90∘, and ∠NOP=60∘. Hence NP=633,OP=63,UP=OP−UO=21, and UPUO=2142=2. Therefore
