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Difference between revisions of "2012 AMC 12A Problems/Problem 12"

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<math> \textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5} </math>
 
<math> \textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5} </math>
  
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== Solution ==
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== Solution 1 ==
  
 
<center><asy>
 
<center><asy>
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<cmath>s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}</cmath>
 
<cmath>s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}</cmath>
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== Solution 2 ==
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Using the diagram above, we look at the top-right vertex of the square. Let us call this point <math>(x,y)</math>. Then, we that since the square is symmetrical over the y-axis, that the y value is equal to <math>2x+1</math>, since we can multiply the x value(which is half of <math>s</math>) by two to get <math>s</math>, and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle, <math>x^2 + y^2 = 4</math>, and the line <math>y=2x+1</math>. Substituting the second equation into the first, we get:
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<math>5x^2 +4x -3 = 0</math>
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Using the quadratic formula, we arrive with <math>x=\frac{-4 \pm 2\sqrt{19}}{10}</math>. However, recall that the x value is only one half of the side length. Multiplying this value by <math>2</math>, then, and using only the positive root(since the top right vertex of the square has a positive x value), we get:
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<math>\frac{-4 + 2\sqrt{19}}{5} \Rightarrow \boxed{\textbf{(D)}}</math>
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== Solution 3 ==
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Like the previous solutions, we can say that the square is symmetrical with respect to the y-axis.
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Since CD passes through point (0,1), let C be (-x,1) and D be (x,1)
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Point B is <math>\left(-x, \sqrt{4-x^2}\right)</math> and Point A is <math>\left(x,\sqrt{4-x^2}\right)</math>. Now, we see that AB = 2x and BC = <math>\sqrt{4-x^2}</math>-1
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Setting those two equal, we get the same equation as the previous solutions: <math>5x^2 +4x -3 = 0</math> and then solve the same way.
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-Conantwiz2023
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:42, 3 January 2020

Problem

A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square?

$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$

Solution 1

[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]

The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem.

\begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*}

Use the quadratic formula.

\[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]

Solution 2

Using the diagram above, we look at the top-right vertex of the square. Let us call this point $(x,y)$. Then, we that since the square is symmetrical over the y-axis, that the y value is equal to $2x+1$, since we can multiply the x value(which is half of $s$) by two to get $s$, and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle, $x^2 + y^2 = 4$, and the line $y=2x+1$. Substituting the second equation into the first, we get:

$5x^2 +4x -3 = 0$

Using the quadratic formula, we arrive with $x=\frac{-4 \pm 2\sqrt{19}}{10}$. However, recall that the x value is only one half of the side length. Multiplying this value by $2$, then, and using only the positive root(since the top right vertex of the square has a positive x value), we get:

$\frac{-4 + 2\sqrt{19}}{5} \Rightarrow \boxed{\textbf{(D)}}$

Solution 3

Like the previous solutions, we can say that the square is symmetrical with respect to the y-axis. 

Since CD passes through point (0,1), let C be (-x,1) and D be (x,1)

Point B is $\left(-x, \sqrt{4-x^2}\right)$ and Point A is $\left(x,\sqrt{4-x^2}\right)$. Now, we see that AB = 2x and BC = $\sqrt{4-x^2}$-1

Setting those two equal, we get the same equation as the previous solutions: $5x^2 +4x -3 = 0$ and then solve the same way.


-Conantwiz2023

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 12 Problems and Solutions

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