# Mock AIME II 2012 Problems/Problem 6

## Problem

A circle with radius $5$ and center in the first quadrant is placed so that it is tangent to the $y$-axis. If the line passing through the origin that is tangent to the circle has slope $\dfrac{1}{2}$, then the $y$-coordinate of the center of the circle can be written in the form $\dfrac{m+\sqrt{n}}{p}$ where $m$, $n$, and $p$ are positive integers, and $\text{gcd}(m,p)=1$. Find $m+n+p$.

## Diagram used for both solutions $[asy] size(200); real y = (5+sqrt(125))/2; defaultpen(linewidth(0.8)); draw(origin--(0,15),EndArrow); draw(origin--(15,0),EndArrow); draw(circle((5,y),5)); draw(origin--(15,15/2)); label("\beta",(3/2,3/8),dir(aTan(1/2)/2)); label("y=\frac{1}{2}x",(11,5),S); dot((5,y)); draw(origin--(5,y)--(0,y),linetype("4 4")); pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4")); [/asy]$

## Solution 1

Since the circle has a radius of $5$, is tangent to the y-axis, and has its center in the first quadrant, its center has coordinates $(5, a)$ for some positive $a$. Also, since the circle is tangent to the line $y=\dfrac{1}{2}x$, the distance from the center to that line must be $5$. The equation for the line can be rewritten as $x-2y+0=0$. Using the point-to-line formula, we must have $\frac{\left|5-2a+0\right|}{\sqrt{1^2+(-2)^2}}=5$. Simplifying this, we have $\left|5-2a\right|=5\sqrt{5}=\sqrt{125}$. This gives two solutions for $a$. We have $a=\frac{5-\sqrt{125}}{2}$ or $a=\frac{5+\sqrt{125}}{2}$. We reject the negative value, and so we have $a=\frac{5+\sqrt{125}}{2}$, and $5+125+2=\boxed{132}$.

## Solution 2

Let $\beta$ be the angle between the line $y=\dfrac{1}{2}x$ and the $x$-axis. Then we have $\tan\beta=\dfrac{1}{2}$, so $\cos\beta=\dfrac{2\sqrt{5}}{5}$ and $\sin\beta=\dfrac{\sqrt{5}}{5}$. Drawing in the line from the center of the circle to the origin, we see that this line bisects the angle between the $y$-axis and the line $y=\dfrac{1}{2}x$, so its tangent is equal to $\tan\left(\dfrac{90-\beta}{2}\right)=\dfrac{\sin(90-\beta)}{1+\cos(90-\beta)}=\dfrac{\cos\beta}{1+\sin\beta}=\dfrac{2\sqrt{5}/5}{1+\sqrt{5}/5}=\dfrac{2\sqrt{5}}{5+\sqrt{5}}$. Now using the upper right triangle, we see that $\tan\left(\dfrac{90-\beta}{2}\right)$ also equals $\dfrac{5}{y}$, where $y$ is the $y$-coordinate of the center of the circle. Thus, we have $\dfrac{2\sqrt{5}}{5+\sqrt{5}}=\dfrac{5}{y}$, so $\dfrac{y}{5}=\dfrac{5+\sqrt{5}}{2\sqrt{5}}\left(\dfrac{\sqrt{5}}{\sqrt{5}}\right)=\dfrac{\sqrt{125}+5}{10}$ and $y=\dfrac{\sqrt{125}+5}{2}$. Our answer is thus $125+5+2=\boxed{132}$.