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:

The boundary can be split into $24$ congruent triangles. Each has angles $15^\circ$ , $45^\circ$ , and $120^\circ$ .

For one such triangle, draw the altitude from the center-side direction. The altitude is $3$ , half the side length of a square. The adjacent right triangle has a $30^\circ$ angle, so the part cut off from a length $3$ base is $3\tan30^\circ=\sqrt3$ .

Thus each small triangle has base $3-\sqrt3$ and height $3$ , giving area $\frac{3(3-\sqrt3)}{2}=\frac{9-3\sqrt3}{2}$ .

The total area is $24\cdot\frac{9-3\sqrt3}{2}=108-36\sqrt3.$ Hence $a+b+c=108+36+3=147$ .

Thus, the answer is E .

2021 AMC 10B Fall Problem 18

Solution:

Problem 18 in Other Years