2021 AMC 12B Fall Problem 15

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

By symmetry the polygon’s vertices alternate between outer square-corners and inner edge-crossings, every

15^\circ

The outer vertices are at distance

3\sqrt2

from the center and the inner ones at distance

2\sqrt3

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

\begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned}

This simplifies to

18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,

so

a + b + c = 108 + 36 + 3 = 147.

Thus, the correct answer is E.

Problem 15 in other years