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

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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==
 
==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).
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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 (1): <math>2\pi</math>
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Area of (B): <math>2\pi</math>
  
 
Area of (2): <math>\frac{4\pi}{3}</math>
 
Area of (2): <math>\frac{4\pi}{3}</math>
  
Area of (3): 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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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>
  
Area of (4): <math>4*1-1/4*\pi*4=4-\pi</math>
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Area of (D): <math>4*1-1/4*\pi*4=4-\pi</math>
  
 
Total sum: <math>\frac{7\pi}{3}-\sqrt{3}+4</math>
 
Total sum: <math>\frac{7\pi}{3}-\sqrt{3}+4</math>

Revision as of 21:13, 14 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 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 (B): $2\pi$

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

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

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

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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