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

## 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 byProblem 11 Followed byProblem 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

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

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