Question

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 $EG$ at $E,F,$ 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 $a+b+c+d?$

$13$

$14$

$15$

$16$

$17$

Answer: E

Concepts:sector area decomposition tangent circles

Difficulty rating: 2380

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 .

2019 AMC 10B Problem 20

Answer: E

Difficulty rating: 2380

Solution:

Problem 20 in Other Years