# 2013 AIME II Problems/Problem 10

## Problem 10

Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

## Solution 1

Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\sqrt{13},0)$, and $A(4+\sqrt{13},0)$

The equation for Circle O is $x^2+y^2=13$, and let the slope of the line$AKL$ be $k$, then the equation for line$AKL$ is $y=k(x-4-\sqrt{13})$

Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$, according to Vieta's Formulas, we get

$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$, and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$

So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$

Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$

So the area $S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}$

Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$

So the answer is $104+26+13+3=\boxed{146}$.

## Solution 2

Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$.

$$\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}$$

Therefore, to maximize area of $\triangle BKL$, we need to maximize area of $\triangle OKL$.

$$\triangle OKL = \frac12 r^2 \sin{\angle KOL}$$

So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$.

Eventually, we get $$\triangle BKL= (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}$$

So the answer is $104+26+13+3=\boxed{146}$.