2021 AMC 10A Problem 21

Every interior angle of an equiangular hexagon is $120^\circ$, so every exterior angle is $60^\circ$. Extending sides $AB$ and $CD$ until they meet cuts off a small triangle on side $BC$ whose two base angles are $60^\circ$: it is equilateral with side $BC$. The same happens at $DE$ and $FA$.

So the triangle formed by lines $AB, CD, EF$ is equilateral, and its side along line $AB$ runs from the corner cut off at $FA$ through $A$ and $B$ to the corner cut off at $BC$:
$$s_1 = FA + AB + BC.$$
Likewise the triangle formed by $BC, DE, FA$ is equilateral; its side along line $DE$ runs from the corner at $CD$ through $D$ and $E$ to the corner at $EF$:
$$s_2 = CD + DE + EF.$$

From the areas, $\frac{\sqrt3}{4}s_1^2 = 192\sqrt3$ gives $s_1^2 = 768 = 256 \cdot 3$, so $s_1 = 16\sqrt3$; and $\frac{\sqrt3}{4}s_2^2 = 324\sqrt3$ gives $s_2^2 = 1296$, so $s_2 = 36$.

The perimeter is $(FA + AB + BC) + (CD + DE + EF) = s_1 + s_2 = 36 + 16\sqrt3$. Hence $m + n + p = 36 + 16 + 3 = 55$.

The answer is $\boxed{\textbf{(C)}\ 55}$.

An equiangular hexagon is an equilateral triangle with three equilateral corners removed, in two different ways depending on which alternate sides you extend. Each big triangle's side is the sum of three consecutive hexagon sides, and the two triangles use complementary triples, so the perimeter is simply the sum of the two side lengths. You never need the individual sides.

• Trying to solve for the six individual side lengths, which the data does not determine, instead of grouping them into the two triangle sides.
• Simplifying $\sqrt{768}$ incorrectly (it is $16\sqrt3$, not $8\sqrt{12}$ or $24\sqrt3$), or adding areas instead of side lengths.