Difference between revisions of "2021 Fall AMC 10B Problems/Problem 18"
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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> | ||
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| + | ==Solution== | ||
| + | First note the useful fact that if <math>R</math> is the circumradius of a dodecagon (<math>12</math>-gon) the area of the figure is <math>3R^2.</math> If we connect the vertices of the <math>3</math> squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is <math>3\sqrt{2}.</math> Thus the area of the dodecagon is <math>3 \cdot (3\sqrt{2})^2 = 3 \cdot 18 = 54.</math> But, the problem asks for the area of figure of rotated squares. This area is the area of the dodecagon, which was found, subtracting the <math>12</math> isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this <math>x.</math> Then, we can use Law of Cosines, on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that <math>x = 3\sqrt{3} -3.</math> Realize that the <math>12</math> isosceles are congruent with an angle measure of <math>120^{\circ},</math> this means that we can create <math>4</math> congruent equilateral triangles with side length <math>3\sqrt3 - 3.</math> The area of the equilateral triangle is <math>\frac{\sqrt{3}}{4} \cdot (3\sqrt{3} -3)^2 = \frac{\sqrt{3}}{4} \cdot 36 - 18\sqrt{3} = \frac{36\sqrt{3} - 54}{4}.</math> Thus, the area of all the twelve small equilateral traingles are <math>4 \cdot \frac{36\sqrt{3} - 54}{4} = 36\sqrt{3} - 54</math>. Thus, the requested area is <math>54 - (36\sqrt{3} - 54) = 108 - 36\sqrt{3}.</math> Thus, <math>a+b+c = 108 + 36 + 3 = 147.</math> Thus, the answer is <math>\boxed{(\textbf{E}.)}.</math> | ||
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| + | ~NH14 | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=19|num-b=17}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=19|num-b=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:02, 23 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 
Solution
First note the useful fact that if 
is the circumradius of a dodecagon (
-gon) the area of the figure is 
If we connect the vertices of the 
squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is 
Thus the area of the dodecagon is 
But, the problem asks for the area of figure of rotated squares. This area is the area of the dodecagon, which was found, subtracting the isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this 
Then, we can use Law of Cosines, on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that 
Realize that the isosceles are congruent with an angle measure of 
this means that we can create 
congruent equilateral triangles with side length 
The area of the equilateral triangle is 
Thus, the area of all the twelve small equilateral traingles are 
. Thus, the requested area is 
Thus, 
Thus, the answer is 
~NH14
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
