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# Difference between revisions of "2021 AMC 10A Problems/Problem 21"

## Problem

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 $m + n + p$?

$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

## Diagram

$[asy] /* Made by MRENTHUSIASM */ size(250); path P1, P2; P1 = scale(16sqrt(3))*polygon(3); P2 = shift(3,3)*scale(36)*rotate(180)*polygon(3); draw(P1, dashed+black); draw(P2, dashed+black); pair A, B, C, D, E, F; E = intersectionpoints(P1,P2)[0]; F = intersectionpoints(P1,P2)[1]; A = intersectionpoints(P1,P2)[2]; B = intersectionpoints(P1,P2)[3]; C = intersectionpoints(P1,P2)[4]; D = intersectionpoints(P1,P2)[5]; filldraw(A--B--C--D--E--F--cycle,yellow); dot("E",E,1.5*dir(0),linewidth(4)); dot("F",F,1.5*dir(60),linewidth(4)); dot("A",A,1.5*dir(120),linewidth(4)); dot("B",B,1.5*dir(180),linewidth(4)); dot("C",C,1.5*dir(-120),linewidth(4)); dot("D",D,1.5*dir(-60),linewidth(4)); dot(16sqrt(3)*dir(90)^^16sqrt(3)*dir(210)^^16sqrt(3)*dir(330),linewidth(4)); dot((3,3)+36*dir(30)^^(3,3)+36*dir(150)^^(3,3)+36*dir(270),linewidth(4)); [/asy]$ ~MRENTHUSIASM

## Solution

Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.

The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles.

We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively.

By the equilateral triangles, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{\textbf{(C)} ~55}.$

~sugar_rush (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

~ pi_is_3.14

~IceMatrix

~MRENTHUSIASM