29. [16] Let $A B C$ be a triangle such that $A B=A C=182$ and $B C=140$. Let $X_{1}$ lie on $A C$ such that $C X_{1}=130$. Let the line through $X_{1}$ perpendicular to $B X_{1}$ at $X_{1}$ meet $A B$ at $X_{2}$. Define $X_{2}, X_{3}, \ldots$, as follows: for $n$ odd and $n \geq 1$, let $X_{n+1}$ be the intersection of $A B$ with the perpendicular to $X_{n-1} X_{n}$ through $X_{n}$; for $n$ even and $n \geq 2$, let $X_{n+1}$ be the intersection of $A C$ with the perpendicular to $X_{n-1} X_{n}$ through $X_{n}$. Find $B X_{1}+X_{1} X_{2}+X_{2} X_{3}+\ldots$
Answer: $\frac{1106}{5}$
Let $M$ and $N$ denote the perpendiculars from $X_{1}$ and $A$ to $B C$, respectively. Since triangle $A B C$ is isosceles, we have $M$ is the midpoint of $B C$. Moreover, since $A M$ is parallel to $X_{1} N$, we have $\frac{N C}{X_{1} C}=\frac{M C}{A C} \Leftrightarrow \frac{X_{1} N}{130}=\frac{70}{182}=\frac{5}{13}$, so $N C=50$. Moreover, since $X_{1} N \perp B C$, we find $X_{1} C=120$ by the Pythagorean Theorem. Also, $B N=B C-N C=140-50=90$, so by the Pythagorean Theorem, $X_{1} B=150$.
We want to compute $X_{2} X_{1}=X_{1} B \tan \left(\angle A B X_{1}\right)$. We have
$$
\begin{aligned}
\tan \left(\angle A B X_{1}\right)=\tan \left(\angle A B C-\angle X_{1} B C\right) & =\frac{1+\tan (\angle A B C) \tan \left(\angle X_{1} B C\right)}{\tan (\angle A B C)-\tan \left(\angle X_{1} B C\right)}=\frac{\left(\frac{12}{5}\right)-\left(\frac{4}{3}\right)}{1+\left(\frac{12}{5}\right)\left(\frac{4}{3}\right)} \\
= & \frac{16}{\frac{15}{15}}=\frac{16}{63} .
\end{aligned}
$$

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 $\frac{A X_{n}}{A X_{n+2}}$ 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 $\triangle A X_{n+2} X_{n+3} \sim \triangle A X_{n} X_{n+1}$ with ratio $\frac{A X_{n}}{A X_{n+2}}$ for each even $n$, we can compute that $\frac{X_{n+2} X_{n+3}}{X_{n} X_{n+1}}=$ $\frac{A X_{n+2}}{A X_{n}}=1-\frac{150 \cdot \frac{65}{63}}{182}$ 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
$$
\begin{array}{c}
B X_{1}+X_{1} X_{2}+\ldots=\left(B X_{1}+X_{2} X_{3}+\ldots\right)+\left(X_{1} X_{2}+X_{3} X_{4}+\ldots\right)=\frac{150}{\frac{150 \cdot \frac{65}{63}}{182}}+\frac{150 \cdot \frac{16}{63}}{\frac{150 \cdot \frac{65}{63}}{182}} \\
=\frac{\frac{79}{63} \cdot 150}{\frac{150 \cdot \frac{65}{63}}{182}}=\frac{1106}{5}
\end{array}
$$