Difference between revisions of "2020 AMC 10B Problems/Problem 14"
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Since the area of a regular hexagon can be found with the formula <math>\frac{3\sqrt{3}s^2}{2}</math>, where <math>s</math> is the side length of the hexagon, the area of this hexagon is <math>\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}</math>. Since the area of an equilateral triangle can be found with the formula <math>\frac{\sqrt{3}}{4}s^2</math>, where <math>s</math> is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is <math>\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}</math>. Since the area of a circle can be found with the formula <math>\pi r^2</math>, the area of a sixth of a circle with radius 1 is <math>\frac{\pi(1^2)}{6} = \frac{\pi}{6}</math>. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of <math>\frac{\sqrt{3}}{4}</math>, and one sixth of a circle with radius 1 colored white, with an area of <math>\frac{\pi}{6}</math>. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is <math>2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}</math>, which equals <math>\frac{\sqrt{3}}{2} + \frac{\pi}{6}</math>, and the total area colored white is <math>6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})</math>, which equals <math>3\sqrt{3} + \pi</math>. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is <math>6\sqrt{3} - (3\sqrt{3} + \pi)</math>, which equals <math>\boxed{\textbf{(D) }3\sqrt{3} - \pi}</math>. | Since the area of a regular hexagon can be found with the formula <math>\frac{3\sqrt{3}s^2}{2}</math>, where <math>s</math> is the side length of the hexagon, the area of this hexagon is <math>\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}</math>. Since the area of an equilateral triangle can be found with the formula <math>\frac{\sqrt{3}}{4}s^2</math>, where <math>s</math> is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is <math>\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}</math>. Since the area of a circle can be found with the formula <math>\pi r^2</math>, the area of a sixth of a circle with radius 1 is <math>\frac{\pi(1^2)}{6} = \frac{\pi}{6}</math>. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of <math>\frac{\sqrt{3}}{4}</math>, and one sixth of a circle with radius 1 colored white, with an area of <math>\frac{\pi}{6}</math>. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is <math>2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}</math>, which equals <math>\frac{\sqrt{3}}{2} + \frac{\pi}{6}</math>, and the total area colored white is <math>6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})</math>, which equals <math>3\sqrt{3} + \pi</math>. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is <math>6\sqrt{3} - (3\sqrt{3} + \pi)</math>, which equals <math>\boxed{\textbf{(D) }3\sqrt{3} - \pi}</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | First, subdivide the hexagon into 24 equilateral triangles with side length 1: | ||
+ | <asy> size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); | ||
+ | draw((1,0)--(3,2sqrt(3))); | ||
+ | draw((3,0)--(1,2sqrt(3))); | ||
+ | draw((4,sqrt(3))--(0,sqrt(3))); | ||
+ | draw((2,0)--(3.5,3sqrt(3)/2)); | ||
+ | draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); | ||
+ | draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); | ||
+ | draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); | ||
+ | draw((2,0)--(0.5,3sqrt(3)/2)); | ||
+ | draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2)); | ||
+ | </asy> | ||
+ | Now note that the entire shaded region is just 6 times this part: | ||
+ | <asy> size(100); | ||
+ | fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); | ||
+ | fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white); | ||
+ | draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); | ||
+ | label("$1$",(2.25,7sqrt(3)/4),NE); | ||
+ | draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); | ||
+ | draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); | ||
+ | </asy> | ||
+ | The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: | ||
+ | <cmath> 2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}</cmath> | ||
+ | The arc that is not included has an area of: | ||
+ | <cmath> \frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}</cmath> | ||
+ | Hence, the area of the shaded region in that section is <cmath>\frac{\sqrt{3}}{2}-\frac{\pi}{6}</cmath> | ||
+ | For a final area of: | ||
+ | <cmath>6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}</cmath> | ||
+ | ~N828335 | ||
==Video Solution== | ==Video Solution== | ||
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{{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}} | ||
+ | {{AMC12 box|year=2020|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:53, 9 February 2020
Contents
[hide]Problem
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?
Solution
Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on, and , as a regular hexagon has angles of 120, and is half of any angle in this hexagon. Now, using the sine law, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.
Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagon is . Since the area of an equilateral triangle can be found with the formula , where is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixth of a circle with radius 1 is . In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of , and one sixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is , which equals , and the total area colored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is , which equals .
Solution
First, subdivide the hexagon into 24 equilateral triangles with side length 1: Now note that the entire shaded region is just 6 times this part: The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: The arc that is not included has an area of: Hence, the area of the shaded region in that section is For a final area of: ~N828335
Video Solution
~IceMatrix
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.