Problem:
Semicircle Ξ has diameter AB of length 14. Circle Ξ© lies tangent to AB at a point P and intersects Ξ at points Q and R. If QR=33β and β QPR=60β, then the area of β³PQR is cabββ, where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. What is a+b+c ?
Answer Choices:
A. 110
B. 114
C. 118
D. 122
E. 126
Solution:
Let O1β and O2β be the centers of Ξ and Ξ©, respectively, and let lines QR and AB intersect at point X. Without loss of generality, let Q lie between X and R. Then
[β³PQR]=[β³XPR]β[β³XPQ]=21ββ
XPβ
XRβ
sin(β PXQ)β21ββ
XPβ
XQβ
sin(β PXQ)=21ββ
XPβ
QRβ
sin(β PXQ),β
where the brackets indicate area. Let M be the midpoint of QRβ, which lies on line O1βO2β.
Triangle QO2βR is isosceles with base length 33β and vertex angle 120β, so O2βM=23β and O2βQ=O2βR=3. Also
O1βM=(O1βQ)2β(MQ)2β=72β(233ββ)2β=213β
so
O1βO2β=O1βMβO2βM=213ββ23β=5
Hence O1βP=(O1βO2β)2β(O2βP)2β=52β32β=4.
The similarity β³XMO1ββΌβ³O2βPO1β implies
sin(β PXQ)=sin(β PO2βO1β)=54β,
so
O1βX=sin(β PXQ)O1βMβ=213βΓ·54β=865β.
Therefore
XP=O1βXβO1βP=865ββ4=833β
so
[β³PQR]=21ββ
XPβ
QRβ
sin(β PXQ)=21ββ
833ββ
33ββ
54β=20993ββ.
The requested sum is 99+3+20=122β.
The problems on this page are the property of the MAA's American Mathematics Competitions
