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

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The answer is 14. First, find the area of the semicircle by squaring the radius of the larger circle, multiplying it by pi, and dividing by 2. to find the grey parts in between the semi-circles, create a shape using two-quarter circles with radius's of 1 on the outside, and a rectangle with sides 2 and 1. Find the area, then subtract that area by two semicircles with a radius of 1. The difference would give you the grey area, then multiply the difference by two since there are two of them, the area is 4. Next, construct a 30, 60, 90 triangle using a line drawn from point F, intersecting the circle with a radius of 2 and a semicircle with radius 1 (it could be either the right or left one).  
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}
- I can't really word the rest without a picture, so if the rest of you editors can find a picture of the problem I would greatly appreciate it. -lialan2004
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== Problem ==
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As shown in the figure, line segment <math>\overline{AD}</math> is trisected by points <math>B</math> and <math>C</math> so that <math>AB=BC=CD=2.</math> Three semicircles of radius <math>1,</math> <math>\overarc{AEB},\overarc{BFC},</math> and <math>\overarc{CGD},</math> have their diameters on <math>\overline{AD},</math> and are tangent to line <math>EG</math> at <math>E,F,</math> and <math>G,</math> respectively. A circle of radius <math>2</math> has its center on <math>F. </math> The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form
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<cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath>
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where <math>a,b,c,</math> and <math>d</math> are positive integers and <math>a</math> and <math>b</math> are relatively prime. What is <math>a+b+c+d</math>?
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<asy>
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size(6cm);
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filldraw(circle((0,0),2), grey);
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filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));
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filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));
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filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));
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dot((-3,-1));
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label("$A$",(-3,-1),S);
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dot((-2,0));
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label("$E$",(-2,0),NW);
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dot((-1,-1));
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label("$B$",(-1,-1),S);
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dot((0,0));
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label("$F$",(0,0),N);
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dot((1,-1));
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label("$C$",(1,-1), S);
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dot((2,0));
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label("$G$", (2,0),NE);
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dot((3,-1));
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label("$D$", (3,-1), S);
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</asy>
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<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math>
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== Solutions ==
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=== Solution 1 ===
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Divide the circle into four parts: the top semicircle by connecting E, F, and G(<math>A</math>); the bottom sector (<math>B</math>), whose arc angle is <math>120^{\circ}</math> because the large circle's radius is <math>2</math> and the short length (the radius of the smaller semicircles) is <math>1</math>, giving a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle; the triangle formed by the radii of <math>A</math> and the chord (<math>C</math>); and the four parts which are the corners of a circle inscribed in a square (<math>D</math>). Then the area is <math>A + B - C + D</math> (in <math>B-C</math>, we find the area of the bottom shaded region, and in <math>D</math> we find the area of the shaded region above the semicircles but below the diameter).
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The area of <math>A</math> is <math>\frac{1}{2} \pi \cdot 2^2 = 2\pi</math>.
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The area of <math>B</math> is <math>\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}</math>.
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For the area of <math>C</math>, the radius of <math>2</math>, and the distance of <math>1</math> (the smaller semicircles' radius) to <math>BC</math>, creates two <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangles, so <math>C</math>'s area is <math>2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}</math>.
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The area of <math>D</math> is <math>4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi</math>.
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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>.
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=== Video Solution ===
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Video Solution from Youtube- https://www.youtube.com/watch?v=ZbWOZMfXtL8
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== See Also ==
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{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}
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{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 15:28, 3 January 2021

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]

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

Solutions

Solution 1

Divide the circle into four parts: the top semicircle by connecting E, F, and G($A$); 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 bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).

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 $7+3+3+4=\boxed{\textbf{(E) } 17}$.

Video Solution

Video Solution from Youtube- https://www.youtube.com/watch?v=ZbWOZMfXtL8

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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