In the book, Solving Mathematical Problems: A personal perspective (written by Terry Tao), he discusses a problem named (on Analytic Geometry Chapter, page 79):
Problem 5.4 (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.)
A picture can be drawn as below, with the boy at O and the teacher at an arbitrary corner, say A. 
His solution as is (extremely paraphrased) starting from O, let the boy swim towards BD. Let the boy stop (at say X) when the teacher is at Y (A < Y < B) and AY = 3OX. Now tricking the teacher, make a straight dash towards CD.
I am having a much different solution (restricting the boys movement to a straight line).
Putting the square in Cartesian plane with O as the origin and and the teacher at D (Not A). Assuming the square is of lenght $2k$, the teachers coordinate is $( k, -k)$. Now, lets suppose the boy makes a dash to a point on AB (say P $(x, k)$ ) and reaches faster than the teacher. So to go to P, the teacher has to move in D->B->P, with a total distance of $3k-x$. The boy should go $\sqrt{x^2 + k^2}$. Now from the ratio of the speed of the two, it reduces to solving the inequality $\sqrt{x^2 + k^2} < \frac{3k -x}{3} = k - \frac{x}{3}$, which solves to $\frac{-3k}{4} < x < 0$.
So a straight dash to a bit left of the midpoint of AB is also probably a correct answer, isn't it ? [My main point of asking is not to check the answer numerically, which can be done by a sleeping baby with its left hand, but the method's correctenss]