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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 15"

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label("F",(3,0),E);
 
label("F",(3,0),E);
 
</asy>
 
</asy>
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We can see that this shape is made out of <math>12</math> of these shapes. <math>\angle{A} = 30^{\circ}</math> and <math>\angle{C} = 90^{\circ}</math>
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We can see that this shape is made out of <math>12</math> of these shapes. <math>\angle{A} = 30^{\circ}</math> and <math>\angle{C} = 90^{\circ}</math>. We also know that there is a square <math>AECF</math> and two triangles <math>AEB</math> and <math>ADF</math>. With these triangles and squares we find the area of one of these shapes to be <math>9-3\sqrt{3}</math>. <math>12(9-3\sqrt{3})=108-36\sqrt{3}</math>

Revision as of 17:04, 24 November 2021

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. $\angle{A} = 30^{\circ}$ and $\angle{C} = 90^{\circ}$. 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 $9-3\sqrt{3}$. $12(9-3\sqrt{3})=108-36\sqrt{3}$

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