Tricoloured Coin Fountains

An arrangement of coins in one or more rows with the bottom row being a block without gaps and every coin in a higher row touching exactly two coins in the row below is called a fountain of coins. Let $f(n)$ be the number of possible fountains with $n$ coins. For $4$ coins there are three possible arrangements:

Therefore $f(4) = 3$ while $f(10) = 78$.
Let $T(n)$ be the number of all possible colourings with three colours for all $f(n)$ different fountains with $n$ coins, given the condition that no two touching coins have the same colour. Below you see the possible colourings for one of the three valid fountains for $4$ coins:

You are given that $T(4) = 48$ and $T(10) = 17760$.
Find the last $9$ digits of $T(20000)$.

To solve this problem, we need to find the expected value of the distance between the centers of the circumcircle and the incircle, given certain constraints on the radii of the circles.

In order to find $\mathbb{E}(d)$, we can simulate the process of choosing random integers for the radii and compute the distance $d$ for each set of radii. By averaging these distances over a large number of simulations, we can approximate the expected value $\mathbb{E}(d)$.

Here’s a step-by-step Python code to solve this problem:

“`python
import math
import random

def calculate_distance(ra, rb, rc):
# Given the radii of the three circles, calculate the distance between their centers

# Calculate the semi-perimeter of the triangle formed by the radii
s = (ra + rb + rc) / 2

# Calculate the inradius of the triangle
inradius = math.sqrt((s – ra) * (s – rb) * (s – rc) / s)

# Calculate the distance between the centers of the circumcircle and the incircle
distance = math.sqrt(ra**2 + inradius**2) – inradius

return distance

def simulate_expected_value(num_simulations):
total_distance = 0

for _ in range(num_simulations):
# Choose random radii: ra < rb < rc ra = random.randint(1, 100) rb = random.randint(ra + 1, 100) rc = random.randint(rb + 1, 100) # Check if gcd(ra, rb, rc) = 1 if math.gcd(ra, math.gcd(rb, rc)) == 1: distance = calculate_distance(ra, rb, rc) total_distance += distance # Calculate the expected value expected_value = total_distance / num_simulations return expected_value # Set the number of simulations to perform num_simulations = 100000 # Simulate the expected value expected_value = simulate_expected_value(num_simulations) # Round the result to eight decimal places and print it print("{:.8f}".format(expected_value)) ``` Running this code will perform the simulations and output the expected value of $d$ rounded to eight decimal places. The more simulations you run, the more accurate the result will be.

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