Problem:
The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD=3. Point E is chosen so that ED=5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of β³ABC?
Answer Choices:
A. 37120β
B. 39140β
C. 39145β
D. 37140β
E. 31120β
Solution:
Because β ACB is inscribed in a semicircle, it is a right angle. Therefore β³ABC is similar to β³AED, so their areas are related as AB2 is to AE2. Because AB2=42=16 and, by the Pythagorean Theorem,
AE2=(4+3)2+52=74
this ratio is 7416β=378β. The area of β³AED is 235β, so the area of β³ABC is 235ββ
378β=(D)37140ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions