Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: -- Could someone please edit my solution to clarify it?)
(→‎Solution)
Line 72: Line 72:
 
J = (1,0);
 
J = (1,0);
 
K = (3,0);
 
K = (3,0);
 +
pair d = 2.7*dir(78);
 
dot(G);
 
dot(G);
 
dot(H);
 
dot(H);
Line 80: Line 81:
 
label("1",(1.5,0),S);
 
label("1",(1.5,0),S);
 
label("1",(2.5,0),S);
 
label("1",(2.5,0),S);
−
label("1",(1.25,0.5a),SE);
+
add(pathticks(G--J,1,0.5,0,3,red));
−
label("1",(2.75,0.5a),SW);
+
add(pathticks(I--J,1,0.5,0,3,red));
−
label("1",(2.75,0.5a),SW);
+
add(pathticks(G--I,1,0.5,0,3,red));
−
label("$60^\circ$",anglemark(H,G,I),2*N);
+
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);
 
draw(anglemark(H,G,I,8),blue);
 
draw(G--J--I--G);
 
draw(G--J--I--G);

Revision as of 19:43, 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)); 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("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); draw(G--J--I--G); draw(G--H--K--G); [/asy]

Since the area of a regular hexagon can be found with the formula $\frac{3\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\frac{\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}$. Since the area of a circle can be found with the formula $\pi r^2$, the area of a sixth of a circle with radius 1 is $\frac{\pi(1^2)}{6} = \frac{\pi}{6}$. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of $\frac{\sqrt{3}}{4}$, and one sixth of a circle with radius 1 colored white, with an area of $\frac{\pi}{6}$. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is $2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}$, which equals $\frac{\sqrt{3}}{2} + \frac{\pi}{6}$, and the total area colored white is $6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})$, which equals $3\sqrt{3} + \pi$. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is $6\sqrt{3} - (3\sqrt{3} + \pi)$, which equals $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_14&oldid=117053"