Difference between revisions of "2005 AMC 10A Problems/Problem 12"

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The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
 
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
  
[[Image:2005amc10a12.gif]]
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<asy>
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unitsize(1.5cm);
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defaultpen(linewidth(.8pt)+fontsize(12pt));
  
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math>
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pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);
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pair E=B+C;
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draw(D--E--B--O--C--B--A,linetype("4 4"));
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draw(Arc(O,1,0,60),linewidth(1.2pt));
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draw(Arc(O,1,120,180),linewidth(1.2pt));
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draw(Arc(C,1,0,60),linewidth(1.2pt));
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draw(Arc(B,1,120,180),linewidth(1.2pt));
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draw(A--D,linewidth(1.2pt));
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draw(O--dir(40),EndArrow(HookHead,4));
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draw(O--dir(140),EndArrow(HookHead,4));
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draw(C--C+dir(40),EndArrow(HookHead,4));
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draw(B--B+dir(140),EndArrow(HookHead,4));
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label("2",O,S);
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draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);
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draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);
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</asy>
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<math> \textbf{(A) }\frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \textbf{(B) } \frac{2}{3}\pi\qquad \textbf{(C) } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \textbf{(D) } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \textbf{(E) } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math>
  
 
==Solution==
 
==Solution==
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This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
 
This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
  
So the answer is:
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So the answer is <math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \boxed{\textbf{(B) }\frac{2}{3}\pi} </math>
 
 
<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math>
 
 
 
==See Also==
 
*[[2005 AMC 10A Problems]]
 
 
 
*[[2005 AMC 10A Problems/Problem 11|Previous Problem]]
 
  
*[[2005 AMC 10A Problems/Problem 13|Next Problem]]
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=11|num-a=13}}
  
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:43, 13 December 2021

Problem

The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $2$?

[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(12pt));  pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180); pair E=B+C;  draw(D--E--B--O--C--B--A,linetype("4 4")); draw(Arc(O,1,0,60),linewidth(1.2pt)); draw(Arc(O,1,120,180),linewidth(1.2pt)); draw(Arc(C,1,0,60),linewidth(1.2pt)); draw(Arc(B,1,120,180),linewidth(1.2pt)); draw(A--D,linewidth(1.2pt)); draw(O--dir(40),EndArrow(HookHead,4)); draw(O--dir(140),EndArrow(HookHead,4)); draw(C--C+dir(40),EndArrow(HookHead,4)); draw(B--B+dir(140),EndArrow(HookHead,4));  label("2",O,S); draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar); draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar); [/asy]

$\textbf{(A) }\frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \textbf{(B) } \frac{2}{3}\pi\qquad \textbf{(C) } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \textbf{(D) } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \textbf{(E) } \frac{2}{3}\pi+\frac{\sqrt{3}}{2}$

Solution

The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four $60^\circ$ sectors with a radius of $\frac{2}{2}=1$ minus the area of a small equilateral triangle.

This is equivalent to the area of four $60^\circ$ sectors with a radius of $1$.

So the answer is $4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \boxed{\textbf{(B) }\frac{2}{3}\pi}$

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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