|
|
| Line 1: |
Line 1: |
| − | <!-- #REDIRECT [[2020 AMC 10B Problems/Problem 14]] -->
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 16]] |
| − | ==Problem==
| |
| − | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length <math>2</math> 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\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</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
| |
| − | | |
| − | ==See Also==
| |
| − | | |
| − | {{AMC12 box|year=2020|ab=B|num-b=10|num-a=12}}
| |
| − | {{MAA Notice}}
| |
