Question

Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a - b\sqrt{c},$ where $a,$ $b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a + b + c?$

$75$

$93$

$96$

$129$

$147$

Solution:

Because the three squares are rotated by $0^\circ,$ $30^\circ,$ and $60^\circ,$ the figure has $12$ -fold symmetry. Its $24$ vertices alternate every $15^\circ$ : outer vertices are the square corners at distance $3\sqrt2$ from the center, and inner vertices are edge crossings at distance $2\sqrt3.$

Connecting the center to all $24$ vertices splits the polygon into $24$ triangles, each with sides $3\sqrt2$ and $2\sqrt3$ and included angle $15^\circ.$ The total area is $24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ = 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}.$

This simplifies to $18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,$ so $a + b + c = 108 + 36 + 3 = 147.$

Thus, the correct answer is E.

2021 AMC 12B Fall Problem 15

Solution:

Problem 15 in Other Years