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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. 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). Unfinished
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}
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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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=== Solution 2 (Video Solution but quicker and if you don't know how to find d) ===
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Do everything in the video solution up to the part where you find the number where it is neither a fraction, radical, or number with <math>\pi</math>. With the numbers we have so far, we can deduce that <math>a + b + c = 16</math>. Using a bit of logic, and noticing that 16 is the second-largest answer, we can conclude that the answer is <math>\boxed{\textbf{(E) }17}</math> because the dimensions of a geometric figure cannot be <math>0</math> or below.
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~Ericshi1685
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=== Video Solution ===
 +
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 01:16, 19 October 2020

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

Solution 2 (Video Solution but quicker and if you don't know how to find d)

Do everything in the video solution up to the part where you find the number where it is neither a fraction, radical, or number with $\pi$. With the numbers we have so far, we can deduce that $a + b + c = 16$. Using a bit of logic, and noticing that 16 is the second-largest answer, we can conclude that the answer is $\boxed{\textbf{(E) }17}$ because the dimensions of a geometric figure cannot be $0$ or below.

~Ericshi1685

Video Solution

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

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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

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