Tangent Circles

Circles $A$ and $B$ are tangent to each other and to line $L$ at three distinct points.
Circle $C$ is inside the space between $A$, $B$ and $L$, and tangent to all three.
Let $r_A$, $r_B$ and $r_C$ be the radii of $A$, $B$ and $C$ respectively.
Let $S(n) = \sum r_A + r_B + r_C$, for $0 \lt r_A \le r_B \le n$ where $r_A$, $r_B$ and $r_C$ are integers.
The only solution for $0 \lt r_A \le r_B \le 5$ is $r_A = 4$, $r_B = 4$ and $r_C = 1$, so $S(5) = 4 + 4 + 1 = 9$.
You are also given $S(100) = 3072$.
Find $S(10^9)$.

To solve this problem, we will use dynamic programming to compute the number of paths from $(0,0)$ to $(W,H)$.

First, let’s define a function `fibonacci(n)` that returns the n-th Fibonacci number. We can implement this function using the recursive definition of Fibonacci numbers:

“`python
def fibonacci(n):
if n <= 0: return 0 elif n == 1: return 1 else: return fibonacci(n-1) + fibonacci(n-2) ``` Next, we define a function `count_paths(W, H)` that computes and returns the number of paths from $(0,0)$ to $(W,H)$. We will use a 2D matrix to store the intermediate results. ```python def count_paths(W, H): # Create a (W+1)x(H+1) matrix to store the intermediate results dp = [[0] * (H+1) for _ in range(W+1)] # Initialize the base cases dp[0][0] = 1 # Compute the number of paths for each cell for i in range(W+1): for j in range(H+1): # Check if the distance is a Fibonacci number if fibonacci(i*i + j*j) != 0: # Count paths from the previous cells if i > 0:
dp[i][j] += dp[i-1][j]
if j > 0:
dp[i][j] += dp[i][j-1]

# Return the number of paths from (0,0) to (W,H)
return dp[W][H]
“`

Now, we can compute the value of `F(10000, 10000)` modulo `1_000_000_007` by calling the `count_paths` function.

“`python
result = count_paths(10_000, 10_000) % 1_000_000_007
print(result)
“`

This will give us the desired value of `F(10_000, 10_000) % 1_000_000_007`.

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