## An ant on the move

### Problem 460

Published on Saturday, 22nd February 2014, 01:00 pm; Solved by 210; Difficulty rating: 55%
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} ≥ 0 and `y`_{1} ≥ 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}≠`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)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 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 ≈ 0.6667. From P_{2}(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is sqrt(5) / 1.8205 ≈ 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) ≈ 2.960516287.

We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.

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