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 $1,$ $AEB,$ $BFC,$ and $CGD,$ 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

$\dfrac{a}{b}\cdot\pi-\sqrt{c}+d,$

where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d?$

$13$

$14$

$15$

$16$

$17$

Solution:

Put $A=(0,0),\ B=(2,0),\ C=(4,0),\ D=(6,0),$ so the semicircles are centered at $(1,0),(3,0),(5,0)$ and their tops are $E=(1,1),\ F=(3,1),\ G=(5,1).$ The circle has center $F=(3,1)$ and radius $2,$ so it passes through $E$ and $G,$ and has area $4\pi.$

The middle semicircle $BFC$ lies entirely inside the circle, removing area $\dfrac{\pi}{2}.$ By symmetry the outer semicircles each contribute the same overlap $R=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}$ inside the circle.

The shaded area is $4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4.$ Hence $a=7,\ b=3,\ c=3,\ d=4,$ so $a+b+c+d=17.$

Thus, E is the correct answer.

2019 AMC 12B Problem 15

Solution:

Problem 15 in Other Years