Prove that the boy cannot escape the teacher
Solution 1:
Definitions and assumptions (without loss of generality)
Without loss of generality, we can consider the teacher is on the bottom half corner (we know the teacher is in a corner and the problem is symmetrical).
Lets call the inner square the square of 1/6 unit of side, centered at the origin.
It takes at least $6\times \frac{1}{12} = \frac{1}{2}$ unit of time for the kid to reach the side of the inner square.
In that time, the teacher goes to the center of the bottom side of the pool, no matter where the boy goes.
When the kid leaves the inner square, teacher always go as close as possible to the kid. If the kid goes back inside the inner square, the teacher goes back to his original position.
Solving the problem
We will now show that in this situation, the boy cannot escape.
Case 1 : The boy exits the inner square at its bottom square
Obviously, the boy can't escape by the bottom side of the pool, since the teacher is here.
If he tries to escape by one side, it will take at least $6\times 5/12 = 2.5$ units of time. The teacher can be anywhere on the side in 1.5 unit of time.
If he tries to escape by the top, it will take at least $6\times 7/12 = 3.5$ units of time. The teacher can be anywhere on the top in 2.5 units of time.
Case 2 : The boy exits the inner square by one side (for example, the right side).
Again, the boy can't escape by the bottom because the teacher is here.
If the boy tries to escape on the right side, it will take at least $6\times 5/12 = 2.5$ units of time. The teacher can be anywhere on the side in 1.5 unit of time.
If the boy tries to escape on the left side, it will take at least $6\times 7/12 = 3.5$ units of time. The teacher can be anywhere on the side in 3.5 unit of time, even if he start by going through the right.
If the boy tries to escape on the top side , it will take at least $6\times 5/12 = 2.5$ units of time. The teacher can be anywhere on the side in 2.5 unit of time.
Case 3 : The boy exits on the top side
If the boy exits the inner square on the right half of the top, then, the teacher goes right. If the boy exits on the left half, the teacher goes left.
For the sake of the argument, let's say the boy exits the inner square on the right side (the situation is symmetric if he exits on the left).
The boy can't escape by the bottom (obvious).
If the boy tries to escape on the right side, it will take at least $6\times 5/12 = 2.5$ units of time. The teacher can be anywhere on the side in 1.5 unit of time.
If the boy tries to escape on the top side, it will take at least $6\times 5/12 = 2.5$ units of time. The teacher can be anywhere on the side in 2.5 unit of time.
If the boy tries to escape on the top half of the left side, it will take him at least $6\times 1/2 = 3$ units of time. The teacher can be on the top half of the left side in 3 units of time, even if he starts going on the right.
If the boy tries to escape on the bottom half of the left side of the pool. (this one is a bit more tricky)
Let $x$ be the distance between the escape point and the middle of the left side. The minimum time the kid takes to go there is :
$$T_{kid} = 6\times \sqrt{0.5^2 + (\frac{1}{12}+x)^2}$$
The time the teacher takes to get there, if he starts by going right is :
$$T_{teacher} = 3+x$$
We can show that $T_{kid}\geq T_{teacher}$, or
$$f(x) = T_{kid}- T_{teacher} = 6\times \sqrt{0.5^2 + (\frac{1}{12}+x)^2} - 3- x \geq 0$$
It could be done analytically, but I used wolfram alpha here to show it.
Conclusion
Since wherever the kid exits the inner square, he loses, the proof is complete.
Solution 2:
The boy has no incentive to change direction since he loses time.
Which direction should he pick?
- In the 6×speed case, no matter which direction the boy chooses, the teacher can get there faster.
- In the 3×speed case, there are numerous directions where the boy can escape... just not the opposite diagonal corner.
