# 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 (1), the bottom sector with arc length 120 degrees (2), the triangle formed by the radii of (2) and the chord (3), and the four parts which are the corners of a circle inscribed in a square. The area is just (1) + (2) - (3) + (4).

Area of (1): $2\pi$

Area of (2): $\frac{4\pi}{3}$

Area of (3): 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 (4): $4*1-1/4*\pi*4=4-\pi$

Total sum: $\frac{7\pi}{3}+\sqrt{3}+4$

$7+3+3+4=\boxed{17}$

$[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("A",(-3,-1),S); dot((-2,0)); label("E",(-2,0),NW); dot((-1,-1)); label("B",(-1,-1),S); dot((0,0)); label("F",(0,0),N); dot((1,-1)); label("C",(1,-1), S); dot((2,0)); label("G", (2,0),NE); dot((3,-1)); label("D", (3,-1), S); [/asy]$ -Arpitr20

## Solution 2

I don't have enough time to post it yet, but here is the outline: Divide it into the top semicircle, the bottom sector, and two rectangles. Using these rectangles we can find the area of those weird shapes that are shaded. Adding everything together, and we can finish. Sorry for not the full solution.

## See Also

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 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 byProblem 14 Followed byProblem 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

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

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