Question

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3},$ and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}.$ The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p},$ where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p?$

$47$

$52$

$55$

$58$

$63$

Solution:

Let the intersections of lines $AB,CD,EF$ form triangle $PQR$ , and let the intersections of lines $BC,DE,FA$ form triangle $XYZ$ . Because the hexagon is equiangular, all these outer triangles are equilateral.

For an equilateral triangle with side length $s$ , the area is $\frac{\sqrt3}{4}s^2$ . Hence

$\frac{\sqrt3}{4}PQ^2=192\sqrt3,\qquad \frac{\sqrt3}{4}YZ^2=324\sqrt3.$

So $PQ=16\sqrt3$ and $YZ=36$ . The perimeter of the hexagon is the sum of the side lengths cut out of these two equilateral triangles, which is

$PQ+YZ=16\sqrt3+36.$

Thus $m+n+p=36+16+3=55$ .

Thus, C is the correct answer.

2021 AMC 10A Spring Problem 21

Solution:

Problem 21 in Other Years