2019 AMC 10B Problem 20

Problem 20 of 25HarderGeometry

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,

\widehat{AEB},

\widehat{BFC},

and

\widehat{CGD},

have their diameters on

\overline{AD}

, lie in the same halfplane determined by line

AD

, and are tangent to line

EG

E,

F,

and

G,

respectively. A circle of radius

2

has its center at

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?

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Solution

Line

EG

passes through the center

F

of the radius-

2

circle, so the shaded upper semicircle has area

2\pi

. The chord

XZ

lies on

AD

, one unit from

F

. Thus

\angle XFZ=2\arccos\left(\frac12\right)=\frac{2\pi}{3}.

The shaded circular segment below

AD

has area

\begin{aligned} &\frac12(2^2)\left(\frac{2\pi}{3}\right)\\ &\quad-\frac12(2)(2)\sin\left(\frac{2\pi}{3}\right)\\ &=\frac{4\pi}{3}-\sqrt3. \end{aligned}

The portion between

EG

and

AD

consists of four congruent pieces of the following form.

Each piece is a unit square with a quarter of a unit circle removed, so the four pieces have total area

4\left(1-\frac\pi4\right)=4-\pi.

The total shaded area is therefore

\begin{aligned} &2\pi+\left(\frac{4\pi}{3}-\sqrt3\right)\\ &\quad+(4-\pi)\\ &=\frac{7\pi}{3}-\sqrt3+4. \end{aligned}

Hence

a=7

b=3

c=3

, and

d=4

, giving

a+b+c+d=17

. Thus, the answer is E .