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== | + | ==Solution 1== |
<asy> | <asy> | ||
pair A=(0,0); | pair A=(0,0); | ||
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</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 | ||
+ | |||
+ | == 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 | ||
+ | - 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 22:55, 25 November 2021
Problem
Three identical square sheets of paper each with side length are stacked on top of each other. The middle sheet is rotated clockwise about its center and the top sheet is rotated clockwise about its center, resulting in the -sided polygon shown in the figure below. The area of this polygon can be expressed in the form , where , , and are positive integers, and is not divisible by the square of any prime. What is ?
IMAGE
Solution 1
We can see that this shape is made out of of these shapes. and because . We also know that there is a square and two triangles and . With these triangles and squares we find the area of one of these shapes to be . The area of the big shape is then . ~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 the center of this dodecagon.
Hence, .
Because the length of a side of a square is 6, .
Hence, .
We notice that . Hence, .
Therefore, the area of the region that three squares cover is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)