Puzzle #1 – Calculating the Total Path Length of a Bouncing Line

Problem Statement

Consider the curve defined by the equation: xy=1

A straight line starts at the origin (0,0) and moves at a 45-degree angle towards the first intersection with the curve. When it reaches the curve, it turns clockwise by 90 degrees and heads towards the x-axis. After hitting the x-axis, it turns anticlockwise by 90 degrees and moves back toward the curve.

This bouncing pattern continues indefinitely.

Your Task

1. Find the total length of the first 100 segments of this path.
2. Find the exact total length of the first fifty billion segments.

Solution:

This puzzle involves a bouncing path along the hyperbola $$ xy = 1 $$ and the x-axis. Let’s break it down systematically.

Step 1: Understanding the Pattern

1. The line starts at the origin and follows the 45-degree path $$ y = x $$.
2. It meets the curve $$ xy = 1 $$, so at that point:
   $$ x = y $$
   $$ x^2 = 1 \Rightarrow x = 1, y = 1 $$
3. The line reflects at the curve and heads towards the x-axis. The reflection follows the property that the angle of incidence equals the angle of reflection.
4. When the line meets the x-axis, it turns anticlockwise by 90 degrees and starts the next segment towards the curve again.

Step 2: Finding the Sequence of Points

• The first intersection is at $$ (1,1) $$.
• The next x-axis intersection follows from symmetry, landing at $$ (2,0) $$.
• The process repeats with the line reaching $$ (2, 1/2) $$, then $$ (3,0) $$, then $$ (3, 1/3) $$, etc.
• The general pattern follows:
  $$ (n, \frac{1}{n}) \text{ on the curve, and } (n+1, 0) \text{ on the x-axis}. $$

Step 3: Calculating the Length of Each Segment

Each step consists of two parts:

1. A segment from $$ (n, 1/n) $$ to $$ (n+1,0) $$.
2. A segment from $$ (n+1,0) $$ to $$ (n+1, 1/(n+1)) $$.

Using the distance formula, the lengths are:

• First segment:
  $$ \sqrt{(n+1 – n)^2 + \left(0 – \frac{1}{n}\right)^2} = \sqrt{1 + \frac{1}{n^2}} $$
• Second segment:
  $$ \sqrt{(n+1 – n+1)^2 + \left(\frac{1}{n+1} – 0\right)^2} = \frac{1}{n+1} $$

Thus, the total length of the first 100 segments is:
$$ \sum_{n=1}^{50} \left( \sqrt{1 + \frac{1}{n^2}} + \frac{1}{n+1} \right) $$

Step 4: Large $$ n $$ Approximation

For large $$ n $$, the term $$ \frac{1}{n^2} $$ becomes small, so:
$$ \sqrt{1 + \frac{1}{n^2}} \approx 1 + \frac{1}{2n^2} $$
Thus, for 50 billion segments:
$$ \sum_{n=1}^{25 \text{ billion}} 1 + \frac{1}{2n^2} + \frac{1}{n+1} \approx 25 \text{ billion} + \sum_{n=1}^{25 \text{ billion}} \left(\frac{1}{2n^2} + \frac{1}{n+1}\right) $$

For large $$ n $$, this sum converges to a logarithmic growth with a small correction. Hence, the total length is approximately 50 billion.

Final Answers

1. First 100 segments: Approximately 100 units.
2. First 50 billion segments: Approximately 50 billion units.

This result is surprising yet elegant—the total path length grows linearly with the number of segments. 🚀