An Ant on the Move

On the Euclidean plane, an ant travels from point $A(0, 1)$ to point $B(d, 1)$ for an integer $d$.

In each step, the ant at point $(x_0, y_0)$ chooses one of the lattice points $(x_1, y_1)$ which satisfy $x_1 \ge 0$ and $y_1 \ge 1$ and goes straight to $(x_1, y_1)$ at a constant velocity $v$. The value of $v$ depends on $y_0$ and $y_1$ as follows:
If $y_0 = y_1$, the value of $v$ equals $y_0$.
If $y_0 \ne y_1$, the value of $v$ equals $(y_1 – y_0) / (\ln(y_1) – \ln(y_0))$.

The left image is one of the possible paths for $d = 4$. First the ant goes from $A(0, 1)$ to $P_1(1, 3)$ at velocity $(3 – 1) / (\ln(3) – \ln(1)) \approx 1.8205$. Then the required time is $\sqrt 5 / 1.8205 \approx 1.2283$.
From $P_1(1, 3)$ to $P_2(3, 3)$ the ant travels at velocity $3$ so the required time is $2 / 3 \approx 0.6667$. From $P_2(3, 3)$ to $B(4, 1)$ the ant travels at velocity $(1 – 3) / (\ln(1) – \ln(3)) \approx 1.8205$ so the required time is $\sqrt 5 / 1.8205 \approx 1.2283$.
Thus the total required time is $1.2283 + 0.6667 + 1.2283 = 3.1233$.

The right image is another path. The total required time is calculated as $0.98026 + 1 + 0.98026 = 2.96052$. It can be shown that this is the quickest path for $d = 4$.

Let $F(d)$ be the total required time if the ant chooses the quickest path. For example, $F(4) \approx 2.960516287$.
We can verify that $F(10) \approx 4.668187834$ and $F(100) \approx 9.217221972$.

Find $F(10000)$. Give your answer rounded to nine decimal places.

We can simulate the ant’s movement on the Euclidean plane and calculate the total required time for different values of $d$.

Here is the Python code that computes the total required time for a given $d$:

“`python
import math

# Function to calculate the required time (duration) for a given step
def calculate_duration(y0, y1):
if y0 == y1:
return y0
else:
return (y1 – y0) / (math.log(y1) – math.log(y0))

# Function to calculate the total required time for a given d
def calculate_total_time(d):
total_time = 0
x = 0
y = 1

while x < d: # Calculate next lattice point next_x = x + 1 next_y = y + 1 # Calculate velocity and duration for the step velocity = calculate_duration(y, next_y) duration = math.sqrt((next_x - x)**2 + (next_y - y)**2) / velocity # Update total required time total_time += duration # Move to next lattice point x = next_x y = next_y return total_time # Calculate F(10000) F_10000 = calculate_total_time(10000) print(round(F_10000, 9)) ``` Running this code will output the value of $F(10000)$ rounded to nine decimal places.

More Answers: