1977 AHSME Problems/Problem 15

Revision as of 00:20, 18 May 2020 by Xwang2026 (talk | contribs) (Problem 15)

Solution

Solution by e_power_pi_times_i (Picture would be helpful.)


Draw perpendicular lines from the radii to the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Then because the triangle is equilateral, the lines divide the big triangle into a small triangle, three rectangles, and six small $30-60-90$ triangles. The length of one side is $2(3\sqrt{3})+6 = 6\sqrt{3}+6$. The perimeter is $\boxed{\textbf{(D) }18\sqrt{3}+18}$.