2021 AMC 10A Problem 21

1. The angles of a hexagon add to $720^\circ$, so each angle of an equiangular hexagon is $720^\circ \div 6 = 120^\circ$. When you extend a side past a corner, the angle outside the hexagon is $180^\circ - 120^\circ = 60^\circ$.
2. Extend sides $AB$ and $CD$ until they meet at a point $X$. Triangle $XBC$ has $60^\circ$ angles at $B$ and at $C$, so its third angle is also $60^\circ$: it is equilateral, and each of its sides equals $BC$. So $BX = BC$.
3. The same thing happens where lines $CD$ and $EF$ meet (a small equilateral triangle on side $DE$) and where lines $EF$ and $AB$ meet (a small equilateral triangle on side $FA$).
4. So the big triangle made by lines $AB$, $CD$ and $EF$ has three $60^\circ$ angles and is equilateral. Its side along line $AB$ is made of three pieces: the small triangle's side next to $A$ (equal to $FA$), then $AB$, then $BX$ (equal to $BC$). So its side length is
$$s_1 = FA + AB + BC.$$
5. In the same way, the triangle made by lines $BC$, $DE$ and $FA$ is equilateral, and its side along line $DE$ is
$$s_2 = CD + DE + EF.$$
6. Adding: $s_1 + s_2 = AB + BC + CD + DE + EF + FA$, which is exactly the perimeter of the hexagon.
7. An equilateral triangle with side $s$ has area $\frac{\sqrt3}{4}s^2$.
- $\frac{\sqrt3}{4}s_1^2 = 192\sqrt3$ gives $s_1^2 = 768 = 256 \times 3$, so $s_1 = 16\sqrt3$.
- $\frac{\sqrt3}{4}s_2^2 = 324\sqrt3$ gives $s_2^2 = 1296$, so $s_2 = 36$.
8. The perimeter is $36 + 16\sqrt3$, so $m + n + p = 36 + 16 + 3 = 55$.

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

A hexagon with all angles $120^\circ$ is a big equilateral triangle with three small equilateral triangles cut off its corners, and this works in two ways, depending on which three sides you extend. Each big triangle's side uses three hexagon sides, and the two big triangles use different sides, so together they give the perimeter. You never need the six sides one at a time.

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 $\boxed{\textbf{(C)}\ 55}$.