Problem:
Point P is inside equilateral â–³ABC. 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. 33​
C. 6
D. 43​
E. 9
Solution:
Let the side length of △ABC be s. Then the areas of △APB, △BPC, and △CPA are, respectively, s/2,s, and 3s/2. The area of △ABC is the sum of these, which is 3s. The area of △ABC may also be expressed as (3​/4)s2, so 3s=(3​/4)s2. The unique positive solution for s is (D)43​​.