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

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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}
 
{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}
==Problem==
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 +
== Problem ==
 
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  
 
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  
 
<cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath>
 
<cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath>
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label("$D$", (3,-1), S);
 
label("$D$", (3,-1), S);
 
</asy>
 
</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>
 
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17</math>
==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): <math>2\pi</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>.
  
Area of (B): <math>\frac{4\pi}{3}</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>.
  
Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is <math>2\sqrt{3}*1/2=\sqrt{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>.
  
Area of (D): <math>4*1-1/4*\pi*4=4-\pi</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>.
  
Total sum: <math>\frac{7\pi}{3}-\sqrt{3}+4</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>.
  
<math>7+3+3+4=\boxed{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.
  
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.  - iron
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~Ericshi1685
  
==See Also==
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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 ==
 
{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{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 (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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