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I saw this interesting problem in a Mathematics book in Chinese(I translated it):

You and your friend is playing a game. There is a square swimming pool, and you are in the middle of it. Your friend is at one of the corners of the square pool. Your goal is to escape from your friend, and he cannot swim in the pool. His running speed is three times your swimming speed but you run faster than him. Can you escape from him?

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Here is a straightforward escape route.

Go to the centre of the pool. Divide the pool into octants, and swim directly across the middle of the octant opposite your friend. In this picture your friend is somewhere along AE, where E is the midpoint of side AB. You swim directly to the point F, which is the point one quarter along side CD. By rotating/reflecting, this method works wherever your friend is.

enter image description here

Let us set the side length of the pool equal to 4. The distance you swim is $\sqrt{5}$, so your friend can run a distance of $3\sqrt{5}$, or about $6.7$. However, they need to run at least $7$ to go along ADF or along EBCF.

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  • $\begingroup$ This is nice because it doesn't need a change in direction. It seems that here we can place F anywhere strictly between the midpoint and a distance 1/2 away from C (if I've worked that out correctly). $\endgroup$
    – hexomino
    Commented Jun 30, 2020 at 16:18
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    $\begingroup$ @hexomino I think it can be anywhere between the midpoint and a distance of $3-3/\sqrt{2}=0.8787$ from C. $\endgroup$
    – Jaap Scherphuis
    Commented Jun 30, 2020 at 20:14
  • $\begingroup$ Correct! You get the checkmark! $\endgroup$
    – Culver Kwan
    Commented Jul 1, 2020 at 3:15
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You can.

I started with two options, take his longest route or take your shortest

Your shortest would to go to the middle of one edge opposite of him, that would make your distance 1/2A, but add the slowness and it's 3/2A and he would have to go across one side and half of other, which would be 3/2A, which is same length.

His longest would be 2A, and for that it would take you about 2.1A (3/2 of the diagonal, but without the computer to get the exact number)

The way to run away would be to start his longest route and when he moves 1/2A you change direction to go to edge further from him. At that point your movement would be somewhat less than 3/2A and his would be more than 3/2A since he has to go around two corners and he's in the middle of one side

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