Sequences in Geometry

This problem is a lot easier when it is placed on the coordinate plane. Let $C = (0, 0)$ and $A = (-5, 0)$ . We can easily get that $B = \left ( -\frac{16}{5}, \frac{12}{5} \right )$ . Notice that the distance from $C$ to $P_{n + 1}$ is $r_n$ , since $P_{n + 1}$ is on $G_n$ . Thus, the coordinates for $P_{n + 1}$ , where $n \geq 1$ , is $(-r_n, 0)$ .

Next, we find an equation for $l_{n+1}$ . It goes through $B$ and $P_{n + 1}$ , so we use the point-slope formula for the equation of a line: $y - y_1 = m(x - x_1)$ . We'll let the point be $P_{n + 1}$ , and the slope is

$m = \frac{\frac{12}{5} - 0}{-\frac{16}{5} + r_n} = \frac{12}{5r_n - 16}.$

Putting this all together, we have that the equation for $l_{n + 1}$ is

$\begin{aligned} y - y_1 &= m(x - x_1) \\ y &= \frac{12}{5r_n - 16} (x + r_n) \\ y &= \frac{12}{5r_n - 16}x + \frac{12r_n}{5r_n - 16}. \end{aligned}$

Multiplying $5r_n - 16$ by both sides and rearranging, we get the standard form equation:

$12x - (5r_n - 16)y + 12r_n = 0.$

To find $r_{n + 1}$ , we just find the distance from $C$ to $l_{n + 1}$ . To do this, we use a formula for the distance from a point to a line. The distance between the point $(p, q)$ and the line $Ax + By + C = 0$ is

$\frac{|Ap + Bq + C|}{\sqrt{A^2 + B^2}}.$

Since $C$ is the origin, the numerator of this is just $|C|$ . Plugging in using our equation for $l_{n + 1}$ , we get

$\begin{aligned} r_{n + 1} &= \frac{|C|}{\sqrt{A^2 + B^2}} \\ &= \frac{12r_n}{\sqrt{12^2 + (5r_n - 16)^2}} \\ &= \frac{12r_n}{\sqrt{144 + 25r_n^2 - 160r_n + 256}} \\ &= \frac{12r_n}{\sqrt{25r_n^2 - 160r_n + 400}}. \end{aligned}$

We cannot further simplify this beyond a trivial rationalization, and the recursion is already in the desired form, so we find that $a + b + c + d = 12 + 25 + 160 + 400 = \boxed{597}.$

Nice question. I used a lot of my mind. But then I finally solved. Nice question :)