2020 AMC 10B Problems/Problem 14

The following problem is from both the 2020 AMC 10B #14 and 2020 AMC 12B #11, so both problems redirect to this page.

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?

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

Solution 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)); 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 Law of Sines, $\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}$ , 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 $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$ .

Solution 2

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: $2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$ The arc that is not included has an area of: $\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}$ Hence, the area of the shaded region in that section is $\frac{\sqrt{3}}{2}-\frac{\pi}{6}$ For a final area of: $6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}$ ~N828335

Solution 3

We can see there are six congruent "leaves" on the original figure and let's say the total area of them is $S$ . If we make a circle with radius one in the circle(the dashed one), we can get another six congruent "leaves", also they are same with the old six ones. So the area of the six new leaves can also be represented by $S$ . Then the area of the shaded region can be expressed in the following two ways:

1. $Area$ of hexagon - ( $Area$ of six semicircles $-$ $Area$ of six $old$ "leaves") $=$ $\frac{\sqrt{3}}{4}*4*6 - (3\pi - S);$

2. $Area$ of dashed circle $-$ $Area$ of six $new$ "leaves" = $\pi - S;$

Then we have $\frac{\sqrt{3}}{4}*4*6 - (3\pi - S) = \pi - S$

$S = 2\pi - 3\sqrt{3}$

Plug in $S$ back to either $1$ or $2$ , we can solve for the shaded area $= 3\sqrt{3} - \pi$ $\rightarrow\boxed{\mathrm{(D)}}$

--RyanZ@BRS

$[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(Arc((2,sqrt(3)), 1, 0, 360),dashed+linewidth(.5)); [/asy]$

Solution 4 (Variables)

We can set several variables.

Let $x$ be the area of the shaded region, and let $y$ be the area of one of the 'leaf-shaped' regions.

$[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)); label("$2$",(3.5,3sqrt(3)/2),NE); label("$x$",(2,1.6),N); label("$y$",(1.3,3.2),S); 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); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); draw((2.5,0.87)--(3,2sqrt(3)/2)); draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2)); draw((1,1.73)--(1.5,2.6)); draw((1.5,0.87)--(1,1.73)); draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2)); label("$1$", (1.2,2.2),W); [/asy]$

We then can connect the vertices of $x$ , and we end up with another hexagon, with side length $1$ . Since each of the regions that are outside of $x$ but inside of the smaller hexagon is half of $y$ , their total area is $\frac{6y}{2} = 3y$ . The smaller hexagon minus this region would be $x$ . So, $x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y$

Also, the area of the larger hexagon minus the semicircles then add back $6y$ would also be $x$ . (Note that by just subtracting the semicircles from the larger hexagon would result in an overlap of $6y$ .) So,

$x= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y$

Equating the equations, we have

$\begin{align*} \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y &= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y \\ \frac{3\sqrt{3}}{2} - 3y &= 6\sqrt{3} - 3\pi + 6y \\ \end{align*}$ Solving for $y$ , we have

$\begin{align*} 9y &= \frac{3\sqrt{3}}{2} - 6\sqrt{3} + 3\pi \\ &=-\frac{9\sqrt{3}}{2} + 3\pi \\ \end{align*}$

Then, we can plug this back into

$x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y$

Since $9y = -\frac{9\sqrt{3}}{2} + 3\pi$ , dividing by $3$ gives

$\begin{align*} 3y &= -\frac{9\sqrt{3}}{2} \times \frac{1}{3} + \pi \\ &=-\frac{3\sqrt{3}}{2} + \pi \\ \end{align*}$

Therefore, $x = \frac{3\sqrt{3}}{2} - (-\frac{3\sqrt{3}}{2} + \pi) \implies x= \boxed{\textbf{(D)} ~3\sqrt{3} - \pi}$ .

~MrThinker

Video Solution

https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10)

https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~IceMatrix

Video Solution

https://youtu.be/oTqx8OqSMQI

~DSA_Catachu

Video Solution

https://youtu.be/NsQbhYfGh1Q?t=2046

~ pi_is_3.14

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

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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2020 AMC 10B Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (Variables)

Video Solution

Video Solution

Video Solution

See Also