Difference between revisions of "2019 AMC 10B Problems/Problem 20"

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Hence, finding <math>A+B-C+D</math>, the desired area is <math>\frac{7\pi}{3}-\sqrt{3}+4</math>, so the answer is <math>7+3+3+4=\boxed{\textbf{(E) } 17}</math>.
 
Hence, finding <math>A+B-C+D</math>, the desired area is <math>\frac{7\pi}{3}-\sqrt{3}+4</math>, so the answer is <math>7+3+3+4=\boxed{\textbf{(E) } 17}</math>.
 
==Video Solution==
 
For those who want a video solution: https://youtu.be/VP7ltu-XEq8
 
  
 
==See Also==
 
==See Also==

Revision as of 23:14, 24 February 2019

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 ($B$), whose arc angle is $120^{\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$, giving a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle; 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 ($D$). Then the area is $A + B - C + D$ (in $B-C$, we find the area of the shaded region above the semicircles but below the diameter, and in $D$ we find the area of the bottom shaded region).

The area of $A$ is $\frac{1}{2} \pi \cdot 2^2 = 2\pi$.

The area of $B$ is $\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}$.

For the area of $C$, the radius of $2$, and the distance of $1$ (the smaller semicircles' radius) to $BC$, creates two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, so $C$'s area is $2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}$.

The area of $D$ is $4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$.

Hence, finding $A+B-C+D$, the desired area is $\frac{7\pi}{3}-\sqrt{3}+4$, so the answer is $7+3+3+4=\boxed{\textbf{(E) } 17}$.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 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

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