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

m
(→‎Solution 1)
Line 14: Line 14:
 
draw(A--(0,3)--C--(3,0)--A, lightgray);
 
draw(A--(0,3)--C--(3,0)--A, lightgray);
 
draw(A--B--C--D--A);
 
draw(A--B--C--D--A);
−
label("A",A,W);
+
label("$A$",A,W);
−
label("B",B,N);
+
label("$B$",B,N);
−
label("C",C,N);
+
label("$C$",C,N);
−
label("D",D,E);
+
label("$D$",D,E);
−
label("E",(0,3),W);
+
label("$E$",(0,3),W);
−
label("F",(3,0),E);
+
label("$F$",(3,0),E);
 
</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

Revision as of 17:35, 6 June 2022

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 $a+b+c$?


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

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)

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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