49 lines
1.7 KiB
Markdown
49 lines
1.7 KiB
Markdown
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---
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id: 5900f5381000cf542c51004b
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title: 'Problem 460: An ant on the move'
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challengeType: 5
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forumTopicId: 302135
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dashedName: problem-460-an-ant-on-the-move
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---
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# --description--
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On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.
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In each step, the ant at point (x0, y0) chooses one of the lattice points (x1, y1) which satisfy x1 ≥ 0 and y1 ≥ 1 and goes straight to (x1, y1) at a constant velocity v. The value of v depends on y0 and y1 as follows: If y0 = y1, the value of v equals y0. If y0 ≠ y1, the value of v equals (y1 - y0) / (ln(y1) - ln(y0)).
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The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P1(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283. From P1(1, 3) to P2(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P2(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.
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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.
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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.
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Find F(10000). Give your answer rounded to nine decimal places.
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# --hints--
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`euler460()` should return 18.420738199.
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```js
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assert.strictEqual(euler460(), 18.420738199);
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```
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# --seed--
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## --seed-contents--
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```js
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function euler460() {
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return true;
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}
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euler460();
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```
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# --solutions--
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```js
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// solution required
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```
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