Problem:
Point P is inside equilateral β³ABC. Points Q,R, and S are the feet of the perpendiculars from P to AB,BC, and CA, respectively. Given that PQ=1, PR=2, and PS=3, what is AB ?
Answer Choices:
A. 4
B. 33β
C. 6
D. 43β
E. 9 Solution:
Let the side length of β³ABC be s. Then the areas of β³APB, β³BPC, and β³CPA are, respectively, s/2,s, and 3s/2. The area of β³ABC is the sum of these, which is 3s. The area of β³ABC may also be expressed as (3β/4)s2, so 3s=(3β/4)s2. The unique positive solution for s is 43ββ.