2019 AMC 10B Problems/Problem 20

The following problem is from both the 2019 AMC 10B #20 and 2019 AMC 12B #15, so both problems redirect to this page.

Problem

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form $\frac{a}{b}\cdot\pi-\sqrt{c}+d,$ where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$ ?

$[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]$

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

Solution

Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ( $C$ ); and the four parts which are the corners of a circle inscribed in a square ( $D$ ). Then the area is $A + B - C + D$ (in $B-C$ , we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).

The area of $A$ is $\frac{1}{2} \pi \cdot 2^2 = 2\pi$ .

The area of $B$ is $\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}$ .

For the area of $C$ , the radius of $2$ , and the distance of $1$ (the smaller semicircles' radius) to $BC$ , creates two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, so $C$ 's area is $2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}$ .

The area of $D$ is $4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$ .

Hence, finding $A+B-C+D$ , the desired area is $\frac{7\pi}{3}-\sqrt{3}+4$ , so the answer is $7+3+3+4=\boxed{\textbf{(E) } 17}$ .

Note from ~milquetoast: I found this solution incredibly unspecific and difficult to understand, especially in defining C, because of the wording. I think what this solution is trying to say is the same as the first video solution down below.

Solution 2

First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$ .The middle semicircle has area $\frac12\pi$ and the other two have about half of their are inside the circle = $\frac14\pi\ + \frac14\pi\ + \frac12\pi\ = \pi$ . Then we subtract the part of the quartercircle that isn't in Circle $F$ . This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$ . The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: $\sqrt{2^2 - 1^2} = \sqrt{3}.$ This means that the length of the triangle is $2 - \sqrt{3}$ and so the area of the triangle is $\frac{2 - \sqrt{3}}{2}$ . For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$ : $2$ , while its height is the radius of the semicircles: $1$ , so the area is 1. The angle is $30^{\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\frac{1}{12}$ of the whole area, which is $4\pi$ so the area of the sector is $\frac{\pi}{3}$ and the area of the segment is $\frac{\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: $(\frac{2 - \sqrt{3}}{2})-(\frac{\pi}{3} - 1) = \frac{2 - \sqrt{3}}{2} + 1 - \frac{\pi}{3} = \frac{4 - \sqrt{3}}{2} - \frac{\pi}{3}.$

Since there are two, one for each side, we have to multiply it by 2, so we have ${4 - \sqrt{3}} - \frac{2\pi}{3}$ , which we subtract from $\pi$ which gets us $\frac{5\pi}{3} - 4 + \sqrt{3}$ which we subtract from $4\pi$ $=$ $\frac{12\pi}{3}$ , which is $\frac{7\pi}{3} + 4 - \sqrt{3}$ so we get $7+3+4+3=\boxed{\textbf{(E) } 17}$ .

Video Solution by Education, the Study of Everything

https://youtu.be/nhDYNevwH1g

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/t3EWtMnJu2Y?t=516

~pi_is_3.14

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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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