Harvard-MIT Mathematics Tournament

Solution:
Answer: $\frac{1106}{5}$
Let $M$ and $N$ denote the perpendiculars from $X_{1}$ and $A$ to $BC$, respectively. Since triangle $ABC$ is isosceles, we have $M$ is the midpoint of $BC$. Moreover, since $AM$ is parallel to $X_{1}N$, we have $\frac{NC}{X_{1}C} = \frac{MC}{AC} \Leftrightarrow \frac{X_{1}N}{130} = \frac{70}{182} = \frac{5}{13}$, so $NC = 50$. Moreover, since $X_{1}N \perp BC$, we find $X_{1}C = 120$ by the Pythagorean Theorem. Also, $BN = BC - NC = 140 - 50 = 90$, so by the Pythagorean Theorem, $X_{1}B = 150$.

We want to compute $X_{2}X_{1} = X_{1}B \tan \left(\angle ABX_{1}\right)$. We have
$$\begin{aligned} \tan \left(\angle ABX_{1}\right) = \tan \left(\angle ABC - \angle X_{1}BC\right) & = \frac{\tan(\angle ABC) - \tan\left(\angle X_{1}BC\right)}{1 + \tan(\angle ABC)\tan\left(\angle X_{1}BC\right)} = \frac{\left(\frac{12}{5}\right) - \left(\frac{4}{3}\right)}{1 + \left(\frac{12}{5}\right)\left(\frac{4}{3}\right)} \\ & = \frac{\frac{16}{15}}{\frac{63}{15}} = \frac{16}{63} . \end{aligned}$$
Hence $X_{2}X_{1} = 150 \cdot \frac{16}{63}$, and by the Pythagorean Theorem again, $X_{2}B = 150 \cdot \frac{65}{63}$.

Next, notice that $\frac{AX_{n}}{AX_{n+2}}$ is constant for every nonnegative integer $n$ (where we let $B = X_{0}$). Indeed, since $X_{n}X_{n+1}$ is parallel to $X_{n+2}X_{n+3}$ for each $n$, the dilation taking $X_{n}$ to $X_{n+2}$ for some $n$ also takes $X_{k}$ to $X_{k+2}$ for all $k$.

Since $\triangle AX_{n+2}X_{n+3} \sim \triangle AX_{n}X_{n+1}$ with ratio $\frac{AX_{n}}{AX_{n+2}}$ for each even $n$, we can compute that $\frac{X_{n+2}X_{n+3}}{X_{n}X_{n+1}} = \frac{AX_{n+2}}{AX_{n}} = 1 - \frac{150 \cdot 65}{182}$ for every nonnegative integer $n$. Notice we use all three sides of the above similar triangles.

We now split our desired sum into two geometric series, one with the even terms and one with the odd terms, to obtain
$$\begin{gathered} BX_{1} + X_{1}X_{2} + \ldots = \left(BX_{1} + X_{2}X_{3} + \ldots\right) + \left(X_{1}X_{2} + X_{3}X_{4} + \ldots\right) = \frac{150}{\frac{150 \cdot \frac{65}{63}}{182}} + \frac{150 \cdot \frac{16}{63}}{\frac{150 \cdot \frac{65}{63}}{182}} \\ = \frac{\frac{79}{63} \cdot 150}{\frac{150 \cdot \frac{65}{63}}{182}} = \frac{1106}{5} . \end{gathered}$$