Problem:
Let ABCDEF be an equiangular hexagon. The lines AB,CD, and EF determine a triangle with area 1923​, and the lines BC,DE, and FA determine a triangle with area 3243​. The perimeter of hexagon ABCDEF can be expressed as m+np​, 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 120∘, so each angle adjacent to one of these measures 60∘. This forces the three angles in every triangle in the figure shown below to be 60∘.
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
43​​(32x+y​)2=1923​ and 43​​(3x+2y​)2=3243​
Thus
32x+y​=768​=163​ and 3x+2y​=1296​=36
Adding these equations gives x+y=36+163​. The requested sum is 36+16+3=(C)55​.
