Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 10B Problems/Problem 16"
(→‎Solution 1)
 
(18 intermediate revisions by 11 users not shown)
Line 4: Line 4:
 
<math>\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}</math>
 
<math>\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}</math>
  
−
==Solution==
+
==Solution 1==
−
The radius of circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. 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:
+
The radius of the circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. 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:
  
 
<center><asy>
 
<center><asy>
Line 36: Line 36:
 
</asy></center>
 
</asy></center>
  
−
Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).
+
Then we proceed to find: 4 <math>\cdot</math> (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 <math>AB</math> in half. Let this half-length be <math>a</math>. Also note that <math>OX=\frac12</math> because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for <math>a.</math>
+
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits <math>AB</math> in half. Let this half-length be <math>a</math>. Also note that <math>OX=\frac12</math> because it is half the side length of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for <math>a.</math>
  
−
<cmath>a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{2} \right) ^2</cmath>
+
<cmath>a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2</cmath>
  
 
Solving, <math>a= \frac{\sqrt{3}}{6}</math> and <math>2a=\frac{\sqrt{3}}{3}</math>. Since <math>AB=AO=BO</math>, <math>\triangle AOB</math> is an equilateral triangle and the central angle is <math>60^{\circ}</math>. Therefore the sector has an area <math>\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}</math>.
 
Solving, <math>a= \frac{\sqrt{3}}{6}</math> and <math>2a=\frac{\sqrt{3}}{3}</math>. Since <math>AB=AO=BO</math>, <math>\triangle AOB</math> is an equilateral triangle and the central angle is <math>60^{\circ}</math>. Therefore the sector has an area <math>\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}</math>.
Line 48: Line 48:
 
<cmath>\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}</cmath>
 
<cmath>\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}</cmath>
  
−
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\cdot\left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}</math>
 +
 
 +
==Solution 2 (Answer Choices)==
 +
 
 +
At once, we can eliminate <math>(C)</math>, <math>(D)</math>, and <math>(E)</math> because the answer should be <math>a \times \pi + b</math>, where <math>a</math> and <math>b</math> are real numbers (not necessarily rational) not equal to <math>0</math>. Now, all that's left is <math>\frac{\pi}{3}-1</math> and <math>\frac{2 \pi}{9} - \frac{\sqrt{3}}{3}</math>. <math>\frac{\pi}{3}-1</math> is simply the area of the circle minus the square, which is definitely wrong, so the answer is <math>\boxed{\textbf{(B)} \frac{2 \pi}{9} - \frac{\sqrt{3}}{3}}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/FQO-0E2zUVI
 +
 
 +
~IceMatrix
  
 
==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}}

Latest revision as of 19:54, 24 March 2024

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 side length 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 $4\cdot\left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$

Solution 2 (Answer Choices)

At once, we can eliminate $(C)$, $(D)$, and $(E)$ because the answer should be $a \times \pi + b$, where $a$ and $b$ are real numbers (not necessarily rational) not equal to $0$. Now, all that's left is $\frac{\pi}{3}-1$ and $\frac{2 \pi}{9} - \frac{\sqrt{3}}{3}$. $\frac{\pi}{3}-1$ is simply the area of the circle minus the square, which is definitely wrong, so the answer is $\boxed{\textbf{(B)} \frac{2 \pi}{9} - \frac{\sqrt{3}}{3}}$

Video Solution

https://youtu.be/FQO-0E2zUVI

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&oldid=217399"