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Difference between revisions of "2020 AMC 12B Problems/Problem 11"

(→‎Solution)
(→‎Solution)
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draw((2,0)--(0.5,3sqrt(3)/2));
 
draw((2,0)--(0.5,3sqrt(3)/2));
 
draw((3.5,sqrt(3)/2)--(0.5,sqrt(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(200); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4));
 +
 +
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
 +
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>
  

Revision as of 20:53, 7 February 2020

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? [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] $\textbf{(A)}\ 6\sqrt3-3\pi \qquad\textbf{(B)}\ \frac{9\sqrt3}{2}-2\pi \qquad\textbf{(C)}\ \frac{3\sqrt3}{2}-\frac{\pi}{3} \qquad\textbf{(D)}\ 3\sqrt3-\pi \\ \qquad\textbf{(E)}\ \frac{9\sqrt3}{2}-\pi$

Solution

Work in progress

Subdivide the hexagon into 24 equilateral triangles with 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(200); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); 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]

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

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