Difference between revisions of "Mock AIME II 2012 Problems/Problem 6"

Latest revision as of 02:06, 5 April 2012

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 $125+5+2=\boxed{132}$ .

@@ Line 3: / Line 3: @@
 A circle with radius <math>5</math> and center in the first quadrant is placed so that it is tangent to the <math>y</math>-axis.  If the line passing through the origin that is tangent to the circle has slope <math>\dfrac{1}{2}</math>, then the <math>y</math>-coordinate of the center of the circle can be written in the form <math>\dfrac{m+\sqrt{n}}{p}</math> where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, and <math> \text{gcd}(m,p)=1 </math>.  Find <math>m+n+p</math>.
+==Diagram used for both solutions==
 <asy>
 size(200);

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