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Difference between revisions of "2020 AMC 10B Problems/Problem 14"

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==Solution==
 
==Solution==
 +
<asy>
 +
real x=sqrt(3);
 +
real y=2sqrt(3);
 +
real z=3.5;
 +
real a=x/2;
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real b=0.5;
 +
real c=3a;
 +
pair A, B, C, D, E, F;
 +
A = (1,0);
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B = (3,0);
 +
C = (4,x);
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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));
 +
dot((2,0));
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dot((0.5,a));
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label("2",(z,c),NE);
 +
label("A",(2,0),S);
 +
label("B",(0.5,a),SW);
 +
label("C",(1.5,a),NE);
 +
label("1",(1.5,0),S);
 +
</asy>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:40, 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] 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); [/asy] $\textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi$

Solution

[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)); dot((2,0)); dot((0.5,a)); label("2",(z,c),NE); label("A",(2,0),S); label("B",(0.5,a),SW); label("C",(1.5,a),NE); label("1",(1.5,0),S); [/asy]

Video Solution

https://youtu.be/t6yjfKXpwDs

~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

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