Difference between revisions of "2020 AMC 10B Problems/Problem 14"
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+ | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #14]] and [[2020 AMC 12B Problems|2020 AMC 12B #11]]}} | ||
+ | |||
==Problem== | ==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? | 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? | ||
+ | |||
+ | <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); | ||
+ | </asy> | ||
+ | |||
+ | <math> \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | <asy> | ||
+ | real x=sqrt(3); | ||
+ | real y=2sqrt(3); | ||
+ | real z=3.5; | ||
+ | real a=x/2; | ||
+ | real b=0.5; | ||
+ | real c=3a; | ||
+ | pair A, B, C, D, E, F; | ||
+ | A = (1,0); | ||
+ | B = (3,0); | ||
+ | C = (4,x); | ||
+ | D = (3,y); | ||
+ | E = (1,y); | ||
+ | F = (0,x); | ||
+ | |||
+ | fill(A--B--C--D--E--F--A--cycle,grey); | ||
+ | fill(arc((2,0),1,0,180)--cycle,white); | ||
+ | fill(arc((2,y),1,180,360)--cycle,white); | ||
+ | fill(arc((z,a),1,60,240)--cycle,white); | ||
+ | fill(arc((b,a),1,300,480)--cycle,white); | ||
+ | fill(arc((b,c),1,240,420)--cycle,white); | ||
+ | fill(arc((z,c),1,120,300)--cycle,white); | ||
+ | draw(A--B--C--D--E--F--A); | ||
+ | draw(arc((z,c),1,120,300)); | ||
+ | draw(arc((b,c),1,240,420)); | ||
+ | draw(arc((b,a),1,300,480)); | ||
+ | draw(arc((z,a),1,60,240)); | ||
+ | draw(arc((2,y),1,180,360)); | ||
+ | draw(arc((2,0),1,0,180)); | ||
+ | label("2",(z,c),NE); | ||
+ | |||
+ | pair X,Y,Z; | ||
+ | X = (1,0); | ||
+ | Y = (2,0); | ||
+ | Z = (1.5,a); | ||
+ | pair d = 1.9*dir(7); | ||
+ | dot(X); | ||
+ | dot(Y); | ||
+ | dot(Z); | ||
+ | label("A",X,SW); | ||
+ | label("B",Y,SE); | ||
+ | label("C",Z,N); | ||
+ | draw(X--Y--Z--A); | ||
+ | label("1",(1.5,0),S); | ||
+ | label("1",(1.75,a/2),dir(30)); | ||
+ | draw(anglemark(Y,X,Z,8),blue); | ||
+ | label("$60^\circ$",anglemark(Y,X,Z),d); | ||
+ | </asy> | ||
+ | |||
+ | 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, <math>BC = 1</math>, since B is the center of the semicircle with radius 1 that C lies on, <math>AB = 1</math>, since B is the center of the semicircle with radius 1 that A lies on, and <math>\angle BAC = 60^\circ</math>, as a regular hexagon has angles of 120<math>^\circ</math>, and <math>\angle BAC</math> is half of any angle in this hexagon. Now, using the sine law, <math>\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}</math>, so <math>\angle ACB = 60^\circ</math>. Since the angles in a triangle sum to 180<math>^\circ</math>, <math>\angle ABC</math> is also 60<math>^\circ</math>. Therefore, <math>\triangle ABC</math> is an equilateral triangle with side lengths of 1. | ||
<asy> | <asy> | ||
Line 32: | Line 107: | ||
draw(arc((2,y),1,180,360)); | draw(arc((2,y),1,180,360)); | ||
draw(arc((2,0),1,0,180)); | draw(arc((2,0),1,0,180)); | ||
+ | pair G,H,I,J,K; | ||
+ | G = (2,0); | ||
+ | H = (2.5,a); | ||
+ | I = (1.5,a); | ||
+ | J = (1,0); | ||
+ | K = (3,0); | ||
+ | pair d = 2.7*dir(78); | ||
+ | dot(G); | ||
+ | dot(H); | ||
+ | dot(I); | ||
+ | dot(J); | ||
+ | dot(K); | ||
label("2",(z,c),NE); | label("2",(z,c),NE); | ||
+ | label("1",(1.5,0),S); | ||
+ | label("1",(2.5,0),S); | ||
+ | add(pathticks(G--J,1,0.5,0,3,red)); | ||
+ | add(pathticks(I--J,1,0.5,0,3,red)); | ||
+ | add(pathticks(G--I,1,0.5,0,3,red)); | ||
+ | add(pathticks(G--H,1,0.5,0,3,red)); | ||
+ | add(pathticks(G--K,1,0.5,0,3,red)); | ||
+ | add(pathticks(K--H,1,0.5,0,3,red)); | ||
+ | label("$60^\circ$",anglemark(H,G,I),d); | ||
+ | draw(anglemark(H,G,I,8),blue); | ||
+ | label("$60^\circ$",G,2*dir(146)); | ||
+ | draw(anglemark(I,G,J,8),blue); | ||
+ | label("$60^\circ$",G,2.8*dir(28)); | ||
+ | draw(anglemark(K,G,H,8),blue); | ||
+ | draw(G--J--I--G); | ||
+ | draw(G--H--K--G); | ||
+ | </asy> | ||
+ | |||
+ | 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 2== | ||
+ | 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> | </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== | ||
+ | https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10) | ||
+ | |||
+ | https://youtu.be/0xgTR3UEqbQ (for AMC 12) | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
− | + | https://youtu.be/oTqx8OqSMQI ~DSA_Catachu | |
==See Also== | ==See Also== | ||
+ | {{AMC12 box|year=2020|ab=B|num-b=10|num-a=12}} | ||
{{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:41, 14 September 2020
- The following problem is from both the 2020 AMC 10B #14 and 2020 AMC 12B #11, so both problems redirect to this page.
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 1
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 2
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
https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10)
https://youtu.be/0xgTR3UEqbQ (for AMC 12)
~IceMatrix
https://youtu.be/oTqx8OqSMQI ~DSA_Catachu
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.