Problem:
Let ABCDEF be an equiangular hexagon. The lines AB,CD, and EF determine a triangle with area 1923β, and the lines BC,DE, and FA determine a triangle with area 3243β. The perimeter of hexagon ABCDEF can be expressed as m+npβ, where m,n, and p are positive integers and p is not divisible by the square of any prime. What is m+n+p?
Answer Choices:
A. 47
B. 52
C. 55
D. 58
E. 63
Solution:
Because the given hexagon is equiangular, each of its interior angles measures 120β, so each angle adjacent to one of these measures 60β. This forces the three angles in every triangle in the figure shown below to be 60β.
Let x=BC+DE+FA and y=AB+CD+EF. Then the perimeter of the hexagon is x+y. The triangle determined by lines AB,CD, and EF has perimeter 2x+y, and the triangle determined by lines BC,DE, and FA has perimeter x+2y. Then
43ββ(32x+yβ)2=1923β and 43ββ(3x+2yβ)2=3243β
Thus
32x+yβ=768β=163β and 3x+2yβ=1296β=36
Adding these equations gives x+y=36+163β. The requested sum is 36+16+3=(C)55β.