Problem: ABCD is a square of side length 3+1. Point P is on AC such that AP=2. The square region bounded by ABCD is rotated 90∘ counterclockwise with center P, sweeping out a region whose area is c1(aπ+b), where a,b, and c are positive integers and gcd(a,b,c)=1. What is a+b+c ?
Answer Choices:
A. 15
B. 17
C. 19
D. 21
E. 23 Solution:
Assume that the vertices of ABCD are labeled in counterclockwise order. Let A′,B′,C′, and D′ be the images of A,B,C, and D, respectively, under the rotation. Because △A′PA and △C′PC are isosceles right triangles, points A′ and C′ are on lines AB and CD, respectively. Moreover, because AP=2 and PC=AC−AP=2(3+1)−2=6, it follows that AA′=2AP=2 and CC′=2CP=23. By symmetry, points B′ and D′ are on lines CD and AB, respectively. Let X=B and Y=D′ be the intersections of BC and C′D′, respectively, with the circle centered at P with radius PB. Note that PD′=PD=PB, so this circle also contains D′. Therefore the required region consists of sectors APA′,BPX,CPC′, and YPD′, and triangles BPA′,CPX, YPC′, and APD′.
Sector APA′ has area 41⋅(2)2π=2π, and sector CPC′ has area 41⋅(6)2π=23π. Let H and I be the midpoints of AA′ and BX, respectively. Then PH=AH=22AP=1, and PI=HB=AB−AH=3. Thus △BPH is a 30−60−90∘ triangle, implying that PB=2 and △XPB is equilateral. Therefore congruent sectors BPX and YPD′ each have area 61⋅22π=32π.
Congruent triangles BPA′ and D′PA each have altitude PH=1 and base A′B=AB−AH−HA′=3−1, so each has area 21(3−1). Congruent triangles CPX and C′PY each have altitude PI=3 and base XC=BC−BX=3−1, so each has area 21(3−3).
