Problem: ABCD is a square of side length 3β+1. Point P is on AC such that AP=2β. The square region bounded by ABCD is rotated 90β counterclockwise with center P, sweeping out a region whose area is c1β(aΟ+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 Aβ²,Bβ²,Cβ², and Dβ² be the images of A,B,C, and D, respectively, under the rotation. Because β³Aβ²PA and β³Cβ²PC are isosceles right triangles, points Aβ² and Cβ² are on lines AB and CD, respectively. Moreover, because AP=2β and PC=ACβAP=2β(3β+1)β2β=6β, it follows that AAβ²=2βAP=2 and CCβ²=2βCP=23β. By symmetry, points Bβ² and Dβ² are on lines CD and AB, respectively. Let Xξ =B and Yξ =Dβ² be the intersections of BC and Cβ²Dβ², respectively, with the circle centered at P with radius PB. Note that PDβ²=PD=PB, so this circle also contains Dβ². Therefore the required region consists of sectors APAβ²,BPX,CPCβ², and YPDβ², and triangles BPAβ²,CPX, YPCβ², and APDβ².
Sector APAβ² has area 41ββ (2β)2Ο=2Οβ, and sector CPCβ² has area 41ββ (6β)2Ο=23Οβ. Let H and I be the midpoints of AAβ² and BX, respectively. Then PH=AH=22ββAP=1, and PI=HB=ABβAH=3β. Thus β³BPH is a 30β60β90β triangle, implying that PB=2 and β³XPB is equilateral. Therefore congruent sectors BPX and YPDβ² each have area 61ββ 22Ο=32Οβ.
Congruent triangles BPAβ² and Dβ²PA each have altitude PH=1 and base Aβ²B=ABβAHβHAβ²=3ββ1, so each has area 21β(3ββ1). Congruent triangles CPX and Cβ²PY each have altitude PI=3β and base XC=BCβBX=3ββ1, so each has area 21β(3β3β).
