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)
(→‎Solution: Explain why two equilateral triangles exist in each sixth of the hexagon)
Line 37: Line 37:
  
 
==Solution==
 
==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));
 +
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>
 
real x=sqrt(3);
 
real x=sqrt(3);
Line 89: Line 140:
 
label("$60^\circ$",anglemark(H,G,I),d);
 
label("$60^\circ$",anglemark(H,G,I),d);
 
draw(anglemark(H,G,I,8),blue);
 
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--J--I--G);
 
draw(G--H--K--G);
 
draw(G--H--K--G);

Revision as of 08:51, 8 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)); 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, $BC = 1$, since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$, since B is the center of the semicircle with radius 1 that A lies on, and $\angle BAC = 60^\circ$, as a regular hexagon has angles of 120$^\circ$, and $\angle BAC$ is half of any angle in this hexagon. Now, using the sine law, $\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}$, so $\angle ACB = 60^\circ$. Since the angles in a triangle sum to 180$^\circ$, $\angle ABC$ is also 60$^\circ$. Therefore, $\triangle ABC$ is an equilateral triangle with side lengths of 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)); 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); 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 $\frac{3\sqrt{3}s^2}{2}$, 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=117428"