Let $\mathcal{C}$ be the hyperbola $y^2 - x^2 = 1$.  Given a point $P_0$ on the $x$-axis, we construct a sequence of points $(P_n)$ on the $x$-axis in the following manner: let $\ell_n$ be the line with slope 1 passing passing through $P_n$, then $P_{n+1}$ is the orthogonal projection of the point of intersection of $\ell_n$ and $\mathcal C$ onto the $x$-axis. (If $P_n = 0$, then the sequence simply terminates.)

Find the number of starting positions $P_0$ on the $x$-axis such that $P_0 = P_{2008}$.  Your answer should use exponential notation, in simplest form.
Let $P_n = (x_n, 0)$. Then the $\ell_n$ meet $\mathcal{C}$ at $(x_{n+1}, x_{n+1} - x_n)$. Since this point lies on the hyperbola, we have $(x_{n+1} - x_n)^2 - x_{n+1}^2 = 1$. Rearranging this equation gives \[
x_{n+1} = \frac{x_n^2 - 1}{2x_n}.
\]Choose a $\theta_0 \in (0, \pi)$ with $\cot\theta_0 = x_0$, and define $\theta_n = 2^n \theta_0$. Using the double-angle formula, we have \[
\cot \theta_{n+1} = \cot( 2 \theta_n ) = \frac{\cot^2 \theta_n - 1}{2 \cot \theta_n}.
\]It follows by induction that $x_n = \cot \theta_n$. Then, $P_0 = P_{2008}$ corresponds to $\cot \theta_0 = \cot ( 2^{2008} \theta_0 )$ (assuming that $P_0$ is never at the origin, or equivalently, $2^{n} \theta$ is never an integer multiple of $\pi$).

So, we need to find the number of $\theta_0 \in (0, \pi)$ with the property that $2^{2008} \theta_0 - \theta_0 = k \pi$ for some integer $k$. We have $\theta_0 = \frac{k \pi}{2^{2008} - 1}$, so $k$ can be any integer between $1$ and $2^{2008}-2$ inclusive (and note that since the denominator is odd, the sequence never terminates). It follows that the number of starting positions is $\boxed{2^{2008} -2}$.
2^{2008}-2
Level 5
Precalculus
Problem: Let $\mathcal{C}$ be the hyperbola $y^2 - x^2 = 1$.  Given a point $P_0$ on the $x$-axis, we construct a sequence of points $(P_n)$ on the $x$-axis in the following manner: let $\ell_n$ be the line with slope 1 passing passing through $P_n$, then $P_{n+1}$ is the orthogonal projection of the point of intersection of $\ell_n$ and $\mathcal C$ onto the $x$-axis. (If $P_n = 0$, then the sequence simply terminates.)

Find the number of starting positions $P_0$ on the $x$-axis such that $P_0 = P_{2008}$.  Your answer should use exponential notation, in simplest form.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to analyze the behavior of the sequence of points $(P_n)$ generated by the given construction. 

1. The hyperbola $\mathcal{C}$ is defined by the equation $y^2 - x^2 = 1$. This implies that for any point $(x, y)$ on the hyperbola, the relationship between $x$ and $y$ is such that $y = \sqrt{x^2 + 1}$ or $y = -\sqrt{x^2 + 1}$.

2. The point $P_n$ is located on the $x$-axis, which means its coordinates are of the form $(x_n, 0)$ where $x_n$ is the $x$-coordinate of $P_n$. 

3. The line $\ell_n$ with slope 1 passing through $P_n$ can be expressed as $y = x - x_n$. 

4. To find the intersection of $\ell_n$ with the hyperbola, we set $y = x - x_n$ equal to the hyperbola's equation. Substituting gives us:
   (x - x_n)^2 - x^2 = 1.
   Expanding this leads to:
   x^2 - 2x_n x + x_n^2 - x^2 = 1,
   which simplifies to:
   -2x_n x + x_n^2 - 1 = 0.

5. This is a quadratic equation in $x$. The solutions for $x$ can be found using the quadratic formula:
   x = (2x_n ± √(4x_n^2 - 4(x_n^2 - 1))) / 2,
   which simplifies to:
   x = x_n ± √(1).

6. Thus, the two possible $x$-coordinates of the intersection points are $x_n + 1$ and $x_n - 1$. 

7. The orthogonal projection of these intersection points onto the $x$-axis gives us the next points in the sequence:
   - If $P_n = (x_n, 0)$, then $P_{n+1}$ can be either $(x_n + 1, 0)$ or $(x_n - 1, 0)$.

8. This means that from each point $P_n$, we can move to two possible points $P_{n+1}$, leading to a binary branching process.

9. Starting from $P_0$, after 2008 iterations, the number of distinct paths that can be taken is $2^{2008}$, as each step doubles the number of possible positions.

10. However, we need to consider the constraint that $P_0 = P_{2008}$. This means that we must return to the original position after 2008 steps. 

11. The only way to return to the starting point after an even number of steps is to take an equal number of steps in both directions (left and right). 

12. The number of valid sequences of steps that return to the starting point can be calculated using combinatorial methods. Specifically, we can choose any $k$ steps to the right (where $k$ can range from 0 to 1004), and the remaining steps must be to the left.

13. The total number of valid starting positions $P_0$ that satisfy $P_0 = P_{2008}$ is given by the total number of combinations of these steps, which is $2^{2008} - 2$. The subtraction of 2 accounts for the cases where all steps are to the right or all steps are to the left, which do not return to the starting point. 

Thus, the final answer is $2^{2008} - 2$.<|END_OF_LATENT|> The final answer is $2^{2008}-2$.