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Difference between revisions of "2016 AMC 10B Problems/Problem 21"
Line 29: Line 29:
 
With this shape we see that this shape can be cut into a right triangle and a semicircle.
 
With this shape we see that this shape can be cut into a right triangle and a semicircle.
 
The length of the hypotenuse of the triangle is <math>sqrt2</math> so using special right triangles, we see that
 
The length of the hypotenuse of the triangle is <math>sqrt2</math> so using special right triangles, we see that
−
the area of the triangle is<math>\frac{1}{2}</math> . The semicircle has the area of <math>\frac[1}{4}\pi</math>.
+
the area of the triangle is<math>\frac{1}{2}</math> . The semicircle has the area of <math>\frac[1}{4}/pi</math>.
  
 
But this is only <math>1</math> case. There are <math>4</math> cases in total so we have to multiply  
 
But this is only <math>1</math> case. There are <math>4</math> cases in total so we have to multiply  
−
<math>\frac{1}{2}+\frac[1}{4}\pi</math> by <math>4</math>.  
+
<math>\frac{1}{2}+\frac[1}{4}/pi</math> by <math>4</math>.  
  
−
After multiplying our answer is : <math>2+\pi</math>
+
After multiplying our answer is : <math>2+/pi</math>
  
−
<math>\textbf{(B)}2+\pi</math>
+
<math>\textbf{(B)}2+/pi</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2016|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:51, 21 February 2016

Problem

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$

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

Solution

WLOG note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore we can assume that $x, y \ge 0$ and multiply by $4$ at the end.

We can rearrange the equation to get $x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2$, which describes a circle with center $(\tfrac12, \tfrac12)$ and radius $\tfrac{\sqrt2}{2}.$ It's clear we now want to find the union of four circles with overlap.

[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted); for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy] There are several ways to find the area, but note that if you connect $(0, 1)$ to its other three respective points in the other three quadrants, you get a square of area $2$, along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2$ which is $\textbf{(B)}$.


Solution 2

Another way to solve this problem is using cases. Though this may seem tedious, we only have to do one case. The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible, 

we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive,

we get the equation $x^{2} +y^{2} -x-y$

Using the complete the square method, we get $(x-\frac{1}{2})^{2} + (y-\frac{1}{2})^{2}=\frac{1}{2}$ Therefore, the origin of this section of the shape is at $(\frac{1}{2}, \frac{1}{2})$. Using the equation we can also see that the radius has a length of $\frac{\sqrt{2}}{2}$ .

With this shape we see that this shape can be cut into a right triangle and a semicircle. The length of the hypotenuse of the triangle is $sqrt2$ so using special right triangles, we see that the area of the triangle is$\frac{1}{2}$ . The semicircle has the area of $\frac[1}{4}/pi$ (Error compiling LaTeX. Unknown error_msg).

But this is only $1$ case. There are $4$ cases in total so we have to multiply $\frac{1}{2}+\frac[1}{4}/pi$ (Error compiling LaTeX. Unknown error_msg) by $4$.

After multiplying our answer is : $2+/pi$

$\textbf{(B)}2+/pi$

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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