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Difference between revisions of "2010 AMC 10B Problems/Problem 16"

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Putting it together, we get the answer to be <math>4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}</math>
 
Putting it together, we get the answer to be <math>4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}</math>
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==Solution 2 (Intense Sketch)==
 
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After finding the equilateral triangle, you know there is going to be a factor of <math>\sqrt{3}</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:12, 13 August 2019

Problem

A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}$

Solution 1

The radius of the circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$. Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}$. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:

[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; real r=sqrt(1/3); pair O=(0,0); pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5); pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5); pair V=(0,0.5); path outer=Circle(O,r); draw(outer); draw(W--X--Y--Z--cycle); draw(O--A); draw(O--B); draw(V--O); pair[] ps={A,B,V,O}; dot(ps); label("$O$",O,SW); label("$\frac{\sqrt{3}}{3}$",O--B,SE); label("$A$",A,NW); label("$B$",B,NE); label("$X$",V,NW); label("$a$",B--V,S); label("$\frac12$",O--V,W); [/asy]

Then we proceed to find: 4 $\cdot$ (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).

First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$. Also note that $OX=\frac12$ because it is half the sidelength of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for $a.$

\[a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2\]

Solving, $a= \frac{\sqrt{3}}{6}$ and $2a=\frac{\sqrt{3}}{3}$. Since $AB=AO=BO$, $\triangle AOB$ is an equilateral triangle and the central angle is $60^{\circ}$. Therefore the sector has an area $\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}$.

Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is

\[\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}\]

Putting it together, we get the answer to be $4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$

See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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