2021 Fall AMC 12B Problems/Problem 15

Problem

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 $a+b+c$ ?

IMAGE

$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution

$[asy] pair A=(0,0); pair B=(1.732,3); pair C=(3,3); pair D=(3,1.732); draw(A--(0,3)--C--(3,0)--A, lightgray); draw(A--B--C--D--A); label("A",A,W); label("B",B,N); label("C",C,N); label("D",D,E); label("E",(0,3),W); label("F",(3,0),E); [/asy]$ We can see that this shape is made out of $12$ of these shapes. $\angle{EAB} = 30^{\circ}$ and $\angle{CAF} = 30^{\circ}$ because $360^{\circ}/12=30^{\circ}$ . We also know that there is a square $AECF$ and two triangles $AEB$ and $ADF$ . With these triangles and squares we find the area of one of these shapes to be $9-3\sqrt{3}$ . The area of the big shape is then $12(9-3\sqrt{3})=108-36\sqrt{3}$ . $108+36+3=\boxed{(\textbf{E})\ 147}$ ~lopkiloinm

Page

Toolbox

Search

2021 Fall AMC 12B Problems/Problem 15

Problem

Solution