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)
 
(19 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
The radius of circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>
+
== Problem==
 +
A square of side length <math>1</math> and a circle of radius <math>\dfrac{\sqrt{3}}{3}</math> share the same center. What is the area inside the circle, but outside the square?
  
−
Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}</math>. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle
+
<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>
  
−
Therefore the picture will look something like this:
+
==Solution 1==
 +
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:
  
−
[[Image:squarecircle.png]]
+
<center><asy>
 +
unitsize(5cm);
 +
defaultpen(linewidth(.8pt)+fontsize(10pt));
 +
dotfactor=3;
  
−
Then we proceed to find: 4 * (area of the sector marked off by the two radii drawn - area of the triangle with side on the square and the two radii).
+
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);
  
−
To find this, we do the following:
+
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></center>
  
−
First we realize that the radius perpendicular to the side of the square between the extra lines marking off the sector splits the chord in half. Let this half-length be <math>a</math>. Power of a point states that if two chords (AB and CD) intersect at X, then AX<math>\cdot</math>BX<math>=</math>CX<math>\cdot</math>DX. Applying power of a point to this situation, <math>a^2=(\frac{\sqrt{3}}{3}-\frac{1}{2})(\frac{\sqrt{3}}{3}+\frac{1}{2})</math>. (We know the center of a square to be halfway in each direction, if you know what I mean by direction).
+
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).
  
−
Solving, <math>a= \frac{\sqrt{3}}{6}</math>. The significance? This means the chord is equal to the radius of a circle, so the sector has angle <math>60^{\circ}</math>. Since this is a sixth of the circle, the sector has area <math>\frac{\pi}{6}\cdot \frac{\sqrt{3}}{3}^2=\frac{\pi}{18}</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>
  
−
Now we turn to the triangle.
+
<cmath>a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2</cmath>
  
−
Since it is equilateral, we know it has area equal to <math>\frac{\sqrt{3}}{4}</math> times the square of its sidelength. Therefore, our triangle has area <math>\frac{\frac{\sqrt{3}}{3}^2\sqrt{3}}{4}=\frac{\sqrt{3}}{12}</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>.
  
−
Putting it together, we get the answer to be <math>4( \frac{\pi}{18}-\frac{\sqrt{3}}{12} )=</math>
+
Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is
  
−
<cmath>\boxed{\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)}</cmath>
+
<cmath>\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}</cmath>
  
−
== See also ==
+
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>
−
[[Area of an equilateral triangle]]
+
 
 +
==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==
 +
{{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"