# Difference between revisions of "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: (A), the bottom sector with arc length 120 degrees: (B), 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). The area is just (A) + (B) - (C) + (D).

Area of (A): $2\pi$

Area of (B): $\frac{4\pi}{3}$

Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is $2\sqrt{3}*1/2=\sqrt{3}$

Area of (D): $4*1-1/4*\pi*4=4-\pi$

Total sum: $\frac{7\pi}{3}-\sqrt{3}+4$ $7+3+3+4=\boxed{17}$

For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 