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

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<math>(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147</math>
 
<math>(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147</math>
  
−
==Solution==
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==Solution 1==
 
<asy>
 
<asy>
 
pair A=(0,0);
 
pair A=(0,0);
Line 22: Line 22:
 
</asy>
 
</asy>
 
We can see that this shape is made out of <math>12</math> of these shapes. <math>\angle{EAB} = 30^{\circ}</math> and <math>\angle{CAF} = 30^{\circ}</math> because <math>360^{\circ}/12=30^{\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>. The area of the big shape is then <math>12(9-3\sqrt{3})=108-36\sqrt{3}</math>. <math>108+36+3=\boxed{(\textbf{E})\ 147}</math> ~lopkiloinm
 
We can see that this shape is made out of <math>12</math> of these shapes. <math>\angle{EAB} = 30^{\circ}</math> and <math>\angle{CAF} = 30^{\circ}</math> because <math>360^{\circ}/12=30^{\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>. The area of the big shape is then <math>12(9-3\sqrt{3})=108-36\sqrt{3}</math>. <math>108+36+3=\boxed{(\textbf{E})\ 147}</math> ~lopkiloinm
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 +
== Solution 2 ==
 +
As shown in [[:Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png]], all 12 vertices of three squares form a regular dodecagon (12-gon).
 +
Denote by <math>O</math> the center of this dodecagon.
 +
 +
Hence, <math>\angle AOB = \frac{360^\circ}{12} = 30^\circ</math>.
 +
 +
Because the length of a side of a square is 6, <math>AO = 3 \sqrt{2}</math>.
 +
 +
Hence, <math>AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)</math>.
 +
 +
We notice that <math>\angle MAB = \angle MBA = 30^\circ</math>.
 +
Hence, <math>AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}</math>.
 +
 +
Therefore, the area of the region that three squares cover is
 +
<cmath>
 +
\begin{align*}
 +
& {\rm Area} \ ABCDEFGHIJKL - 12 {\rm Area} \ \triangle MAB \\
 +
& = 12 {\rm Area} \ \triangle OAB - 12 {\rm Area} \ \triangle MAB \\
 +
& = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB
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- 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\
 +
& = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\
 +
& = 108 - 36 \sqrt{3} .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E) }147}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)

Revision as of 23:55, 25 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 1

[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

Solution 2

As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png, all 12 vertices of three squares form a regular dodecagon (12-gon). Denote by $O$ the center of this dodecagon.

Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$.

Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$.

Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$.

We notice that $\angle MAB = \angle MBA = 30^\circ$. Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$.

Therefore, the area of the region that three squares cover is \begin{align*} & {\rm Area} \ ABCDEFGHIJKL - 12 {\rm Area} \ \triangle MAB \\ & = 12 {\rm Area} \ \triangle OAB - 12 {\rm Area} \ \triangle MAB \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36 \sqrt{3} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(E) }147}$.

~Steven Chen (www.professorchenedu.com)

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