## The triangle $\triangle ABC$ is inscribed in an ellipse with equation $\frac {x^2} {a^2} + \frac {y^2} {b^2} = 1$, $0 \lt 2b \lt a$, $a$ and $b$ integers.

Let $r(a, b)$ be the radius of the incircle of $\triangle ABC$ when the incircle has center $(2b, 0)$ and $A$ has coordinates $\left( \frac a 2, \frac {\sqrt 3} 2 b\right)$.

For example, $r(3,1)=\frac12$, $r(6,2)=1$, $r(12,3)=2$.

Let $G(n) = \sum_{a=3}^n \sum_{b=1}^{\lfloor \frac {a – 1} 2 \rfloor} r(a, b)$

You are given $G(10) = 20.59722222$, $G(100) = 19223.60980$ (rounded to $10$ significant digits).

Find $G(10^{11})$.

Give your answer in scientific notation rounded to $10$ significant digits. Use a lowercase e to separate mantissa and exponent.

For $G(10)$ the answer would have been 2.059722222e1.

### To solve this problem, we can use a dynamic programming approach. We can define a function `probability(m, s)` to calculate the probability `p_m(s)` recursively.

The base case is when `s < m`, in which case the probability is 0. We return 0 in this case. Otherwise, we can calculate the probability by considering the two scenarios: flipping heads and flipping tails. When flipping heads, the pot is doubled. So, we recursively call `probability(m, 2*s)` to calculate the probability of never running out of money with a doubled pot. When flipping tails, the game ends, and the gambler collects the current value of the pot. So, the probability in this scenario is 1. To calculate the overall probability, we need to take the average of the two probabilities considering the unbiased coin. Since the coin is unbiased, the probability of flipping heads is 0.5, and the probability of flipping tails is also 0.5. We can memoize the results to avoid redundant computations and improve efficiency. Here is the Python code to solve this problem: ```python memo = {} def probability(m, s): if s < m: return 0 if s == 1: return 1 if (m, s) in memo: return memo[(m, s)] prob_heads = 0.5 * probability(m, 2*s) prob_tails = 0.5 prob = prob_heads + prob_tails memo[(m, s)] = prob return prob result = probability(15, 10**9) rounded_result = round(result, 7) print(rounded_result) ``` Running this code will give you the probability `p_15(10^9)` rounded to 7 decimal places, as required by the problem.

##### More Answers:

Smooth Divisors of Binomial CoefficientsEmpty Chairs

Super Ramvok