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 $7+3+3+4=\boxed{\textbf{(E) } 17}$ .

Solution 2 (Solution 1 but quicker)

Do everything in solution $1$ up to the part where you find the number where it is neither a fraction, radical, or number with $\pi$ . With the numbers we have so far, we can deduce that $a + b + c = 16$ . Using a bit of logic, and noticing that 16 is the second-largest answer, we can conclude that the answer is $\boxed{\textbf{(A) }17}$ because the dimensions of a geometric figure cannot be $0$ or below.

Video Solution

Video Solution from Youtube- https://www.youtube.com/watch?v=ZbWOZMfXtL8

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

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 20

Contents

Problem

Solution

Solution 2 (Solution 1 but quicker)

Video Solution

See Also