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2021 AMC Fall 10B Problem 18

2021 AMC Fall 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 4 of 5 · about 7 min · position P16-20

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 $a+b+c?$

A)
$75$
B)
$93$
C)
$96$
D)
$129$
E)
$147$

Answer

Key insight

The 24-gon has 12-fold symmetry: its 24 vertices alternate between square corners (every 30 degrees) and side intersections exactly halfway between, so it splits into 24 congruent triangles.

Solution

Put the common center at the origin with the bottom sheet axis-aligned, so its corners are at $(\pm3, \pm3)$ , at angles $45^\circ, 135^\circ, \dots$ from the center. The other sheets' corners sit at the same distance, every $30^\circ$ : the twelve corners are at $15^\circ, 45^\circ, 75^\circ, \dots$ .

Rotating the whole picture by $30^\circ$ sends each sheet to another sheet, so the polygon has $12$ -fold rotational symmetry; reflecting in the line at angle $30^\circ$ swaps the $0^\circ$ and $60^\circ$ sheets and fixes the $30^\circ$ sheet, so that line is a mirror as well. Hence the $24$ vertices are the $12$ corners together with $12$ intersection points, one on each bisecting ray at $0^\circ, 30^\circ, 60^\circ, \dots$ .

Consider the corner $C = (3, 3)$ of the bottom sheet and the intersection point $X$ on the ray at $30^\circ$ . $X$ lies on the bottom sheet's side $x = 3$ , so $X = (3, 3\tan 30^\circ) = (3, \sqrt3)$ .

Triangle $OCX$ has base $CX = 3 - \sqrt3$ along the line $x = 3$ and height $3$ (the distance from $O$ to that line), so its area is
$\frac12 \cdot 3 \cdot (3 - \sqrt3) = \frac{9 - 3\sqrt3}{2}.$
The polygon is made of $24$ such triangles (one for each corner-to-neighbor half-step), so its area is
$24 \cdot \frac{9 - 3\sqrt3}{2} = 108 - 36\sqrt3.$
Then $a + b + c = 108 + 36 + 3 = 147$ .

The answer is $\boxed{\textbf{(E)}\ 147}$ .

Why this works

Overlapping congruent shapes about a common center create rotational symmetry; find the smallest symmetric wedge and multiply. The key fact was that the boundary vertices alternate corner, intersection, corner, at equal angular spacing, which places each intersection on a known ray and makes its coordinates trivial.

Alternative approach

Same wedge via the law of sines: triangle $OCX$ has angles $15^\circ$ at $O$ , $45^\circ$ at $C$ and $120^\circ$ at $X$ with $OC = 3\sqrt2$ , so $CX = 3\sqrt2 \sin 15^\circ / \sin 120^\circ = 3 - \sqrt3$ , and the area follows as above.

The trap

Assuming the intersection points lie at the same distance from the center as the corners, or computing the union as the square area plus a guessed number of protruding triangles.

Common mistakes

Assuming the intersection points lie at the same distance from the center as the corners, or computing the union as the square area plus a guessed number of protruding triangles.
Using the regular dodecagon through the twelve corners ( $108$ ) as the answer, which drops the $12$ small triangles beyond it.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors) · Exploit symmetry to reduce work or pair up objects