Prove that the boy cannot escape the teacher I'm struggling with the following problem from Terence Tao's "Solving Mathematical Problems":

Suppose the teacher can run six times as fast as the
  boy can swim. Now show that the boy cannot escape. (Hint: Draw an
  imaginary square of sidelength 1/6 unit centred at $O$. Once the boy leaves
  that square, the teacher gains the upper hand.)

Here $O$ is the center of the swimming pool. This question is a follow up on the previous one, which is solved in the affirmative in the text

(Taylor 1989, p. 34, Q2). In the centre of a square swimming
  pool is a boy, while his teacher (who cannot swim) is at one corner
  of the pool. The teacher can run three times faster than the boy can swim,
  but the boy can run faster than the teacher can. Can the boy escape from
  the teacher? (Assume both persons are infinitely manoeuvrable.)


My attempt:
Since the boy can always swim back into the small square of sidelength 1/6 centered at $O$, I can't see how to apply the hint properly. Also, since the student's path need not even be smooth (it was taken as a polygonal chain in the previous question) I'm having difficulties writing data down clearly.
Any help would be appreciated. Thanks.

Consider a square centred at $(0,0)$ in the Cartesian plane. Without loss of generality let the square have a side length of 2 units.
Suppose that the teacher is at vertex  $(1,1)$.
Assume that there exists some point on the perimeter of the square with coordinates $(x,y)$ such that the student will arrive here before the teacher and so will escape. Due to the symmetry of the square, we can assume without loss of generality that $y=-1$. And so our point becomes $(x,-1)$.
Let the boy swim at $k$ $unit/s$ and let the teacher run at $6k$ $unit/s$
Then we have,
Distance from teacher= $2+(1-x)$  (distance travelled vertically down +distance travelled horizontally left towards point).
Distance from the student=$\sqrt{x^{2}+1}$
Time taken for teacher to reach this point $(t_{1})= \frac{3-x}{6k}$.
Time taken for student to reach this point $(t_{2})= \frac{\sqrt{x^{2}+1}}{k}$.
In order for the student to escape  $(t_{1})>(t_{2})$, 
hence, $\frac{3-x}{6k}>\frac{\sqrt{x^{2}+1}}{k}$.
Noting that $-1<x<0$ (i.e. the point we are looking for is in the third quadrant) reveals that this inequality does not hold in this interval.
