2005 AMC 10A Problems/Problem 12

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

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2005 AMC 10A Problems/Problem 12

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