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You're at the center of a circular arena. A pair of lions are at the border, planning to catch you. One of them moves as fast as you, but the other moves slower than you. The three of you are confined to move within the arena (border included). A lion catches you if its position coincides with yours.

There are three possibilities:

  1. You can move wisely and never get caught.
  2. You will eventually get caught.
  3. It depends on how fast the slow lion can move.

Which one is correct, and why?

Hint

One lion doesn't catch you (see here or here). Two lions moving as fast as you will catch you very soon (see here).

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  • $\begingroup$ Where do the lions start on the border? Opposite sides? Both at the same point? Does it matter? $\endgroup$
    – will
    Commented Apr 7, 2022 at 15:35
  • $\begingroup$ @will It doesn't matter. $\endgroup$
    – Eric
    Commented Apr 7, 2022 at 15:38
  • $\begingroup$ Wouldn't your explanation to your other question answers this? $\endgroup$
    – justhalf
    Commented Apr 7, 2022 at 15:44
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    $\begingroup$ @justhalf No, there you have two fast lions. Here you have a slow one. $\endgroup$
    – Eric
    Commented Apr 7, 2022 at 15:49
  • $\begingroup$ Good point. My intuition is that a countably infinite strictly slower lions cannot catch you, so one fast slow and one slow lion equals one lion, which means escape for you. But I have no proof. $\endgroup$
    – justhalf
    Commented Apr 7, 2022 at 16:27

1 Answer 1

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I think that

the lions win

by the following strategy.

The fast lion moves directly to the centre of the arena, while the slow lion stays on the circumference. When the fast lion reaches the centre, you (the sheep) must be at another point.

After this, the fast lion starts to move outwards, always staying on the radius that connects the centre with the sheep. Thus, in polar coordinates, the fast lion's position changes argument exactly as fast as the sheep's position, and the fast lion has an extra speed component (since its modulus is smaller) which enables it to move outwards towards the sheep.

In this way, the fast lion pushes the sheep further and further to the boundary of the arena. In the one-lion problem, this wouldn't doom the sheep since it can always keep a finite distance away from the lion. But now, the slow lion is lurking on the boundary, waiting for the sheep to get close enough, with the fast lion on the inside, that its measly speed will be enough to reach the sheep. This must happen eventually, since the sheep's modulus approaches 1 (while its argument covers all possible values, it must keep moving to avoid the fast lion) and the slow lion's speed is positive.

This is inspired by

the third argument (one lion, two wolves) in this answer, as well as the fact that the relative circumferential starting points of the two lions don't matter.

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  • $\begingroup$ "This must happen eventually, since the sheep's modulus approaches 1", why does this guarantee the slow lion will eventually be at the same angle as the sheep? $\endgroup$
    – justhalf
    Commented Apr 7, 2022 at 20:34
  • $\begingroup$ @justhalf Because the sheep has to keep moving as fast as possible to avoid the fast lion, so it will cover all angles. $\endgroup$
    – Rand al'Thor
    Commented Apr 7, 2022 at 20:38
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    $\begingroup$ Is it not possible to go back and forth (but still approaching 1) in the strategy when avoiding the fast lion, therefore not covering all angles? Especially from the other answer, what matters is the perpendicularity, for which there are always two options. $\endgroup$
    – justhalf
    Commented Apr 7, 2022 at 20:48
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    $\begingroup$ I don't think the sheep has to approach the boundary of the arena. For instance, the sheep could pretend that it is confined in a smaller disk with half the radius of the real arena. The same strategy should enable it to avoid one lion without leaving the smaller disk. $\endgroup$
    – noedne
    Commented Apr 8, 2022 at 2:59
  • $\begingroup$ @noedne I don't think that's a problem - the slow lion could move inwards until it reaches that smaller disc. I'm more concerned with justhalf's idea, which I haven't managed to counter yet ... $\endgroup$
    – Rand al'Thor
    Commented Apr 8, 2022 at 5:55

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