2021 AMC 10A Problem 21

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.