Circles of Apollonius of a Triangle

Let the sides of the triangle be $a$ , $b$ , and $c$ such that $a < b < c$ , and let their opposite angles be $\angle A$ , $\angle B$ , and $\angle C$ , and let $\angle B$ be at the origin with side $c$ along the $x$ -axis, so that the coordinate of $A$ is $(c, 0)$ .

Then the circle of Apollonius through $C$ is the locus of points that are $ak$ away from $B$ and $bk$ away from $A$ , which by the distance equation can be expressed as $x^2 + y^2 = a^2k^2$ and $(x - c)^2 + y^2 = b^2k^2$ . Rearranging and combining gives $a^2b^2k^2 = b^2x^2 + b^2y^2 = a^2(x - c)^2 + a^2y^2$ , and further rearranging gives $(x + \frac{a^2c}{b^2 - a^2})^2 + y^2 = (\frac{abc}{b^2 - a^2})^2$ , a circle equation with a radius of $r_3 = \frac{abc}{b^2 - a^2}$ .

By a similar argument, the other two radii are $r_1 = \frac{abc}{c^2 - a^2}$ and $r_2 = \frac{abc}{c^2 - b^2}$ .

Since $\frac{1}{r_1} = \frac{c^2 - a^2}{abc}$ , $\frac{1}{r_2} = \frac{c^2 - b^2}{abc}$ , $\frac{1}{r_3} = \frac{b^2 - a^2}{abc}$ , and since $\frac{c^2 - a^2}{abc} = \frac{b^2 - a^2}{abc} + \frac{c^2 - b^2}{abc}$ , we have the relation:

$\frac{1}{r_1} = \frac{1}{r_2} + \frac{1}{r_3}$

where $r_1 < r_2$ and $r_1 < r_3$ . In this question, $r_1 = 15$ and $r_2 = 24$ , so $\frac{1}{15} = \frac{1}{24} + \frac{1}{r_3}$ , which solves to $r_3 = \boxed{40}$ .