Question

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 $a + b + c + d.$

Answer

Solution:

The nearest point $B$ lies on segment $OA$ with $OB = \sqrt{13}$ and $AB = 4.$ Triangles $OKL$ and $BKL$ share the base $KL,$ and their heights are the distances from $O$ and $B$ to the line through $A.$ For any point $P$ on line $OA,$ that distance is $PA\sin\varphi,$ where $\varphi$ is the angle between the two lines, so $\frac{[BKL]}{[OKL]} = \frac{AB}{AO} = \frac{4}{4 + \sqrt{13}}.$

Since $OK = OL = \sqrt{13},$ we have $[OKL] = \frac{13}{2}\sin(\angle KOL) \le \frac{13}{2},$ with equality when $\angle KOL = 90^\circ.$ Such a chord lies at distance $\sqrt{13/2}$ from $O,$ which is less than $OA,$ so a line through $A$ can achieve it.

The maximum area is $[BKL] = \frac{13}{2} \cdot \frac{4}{4 + \sqrt{13}} = \frac{26}{4 + \sqrt{13}} = \frac{26(4 - \sqrt{13})}{3} = \frac{104 - 26\sqrt{13}}{3},$ so $a + b + c + d = 104 + 26 + 13 + 3 = 146.$

2013 AIME II Problem 10

Solution:

Problem 10 in Other Years