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 $m + n + p?$

$47$

$52$

$55$

$58$

$63$

Answer: C

Concepts:equiangular polygon equilateral triangle triangle area

Difficulty rating: 2150

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

$\begin{aligned} \frac{\sqrt3}{4}PQ^2 &=192\sqrt3, \\ \frac{\sqrt3}{4}YZ^2 &=324\sqrt3. \end{aligned}$

So $PQ=16\sqrt3$ and $YZ=36.$ To justify the perimeter relation, write the consecutive hexagon side lengths as $a,b,c,d,e,f.$ The two alternating-line triangles have side lengths $b+c+d$ and $c+d+e,$ while closure of the hexagon gives $a+f=c+d.$ Hence their side-length sum is $\begin{aligned} &(b+c+d)\\ &\quad+(c+d+e)\\ &=b+e+2(c+d)\\ &=(a+b+c)+(d+e+f), \end{aligned}$ the hexagon's perimeter. Therefore the perimeter 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

Answer: C

Difficulty rating: 2150

Solution:

Problem 21 in Other Years