## Mathematical Training of Lifeguards ### Optimal path

For simplicity, we will start from a particular case: the lifeguard $$A$$ is exactly on the edge of the coast and starts to run along that line until reaching point $$C$$, and then he swims towards the distressed swimmer $$B$$. How should we choose $$C$$, so that $$ACB$$ is the fastest path? Let $$v_{1}$$ be the running velocity and $$v_{2}$$($$< v_{1}$$) the swimming velocity. We want $$C$$ such that (see next picture), for any points in the coast line, $$D$$ on the right of $$C$$ and $$E$$ on the left of $$C$$, the paths $$ADB$$ and $$AEB$$ would take longer than the path $$ACB$$.

$$F$$ is the point on the coast $$AD$$ and $$FB$$ is perpendicular to $$AD$$. In the dotted line, we have represented the perpendiculars of $$CB$$ through $$C$$ and $$D$$, and its parallel $$CB$$ through $$E$$, with $$G$$, $$H$$, $$I$$ points of intersection clearly identified in the picture.

The time spend in the path $$ACB$$ is $$\frac{\overline{AC}}{v_{1}} + \frac{\overline{CB}}{v_{2}}$$. And the time in $$ADB$$ is $$\frac{\overline{AD}}{v_{1}} + \frac{\overline{DB}}{v_{2}}$$ These two paths have a common subpath: $$AC$$. We want to find a condition on $$C$$ so that we can assure that $$CB$$ is faster than $$CDB$$ for every $$D$$ on the right of $$AC$$. Obviously, $$HB$$ is faster than $$DB$$: they are in the same environment, hence, the velocity is the same, and $$DB$$ is the hypotenuse of a triangle from which $$HB$$ is a cathetus, and consequently longer than $$HB$$. If the time spent on $$CH$$ ($$= \frac{\overline{CH}}{v_{2}}$$) is less than or equal to the time spent on $$CD$$ ($$= \frac{\overline{CD}}{v_{1}}$$), we can conclude that the path $$CB$$ is faster. But $$\frac{\overline{CH}}{v_{2}} \leq \frac{\overline{CD}}{v_{1}}$$ is equivalent to $$\frac{\overline{CH}}{\overline{CD}} \leq \frac{v_{2}}{v_{1}}$$. And the triangle $$CHD$$ is similar to the triangle $$CFB$$ (they are both rectangular and they have a common angle). This means that one can say that if the "slope" of $$CB$$, taking the coast line as the reference axis, given by $$\frac{\overline{CF}}{\overline{CB}}$$, is less than or equal to $$\frac{v_{2}}{v_{1}}$$, then the path $$ADB$$ will be slower than the path $$ACB$$.

A similar approach, now applied to $$AEB$$, given that the rectangular triangle $$EIC$$ is also similar to $$CFB$$, allows us to conclude that, if $$\frac{\overline{CF}}{\overline{CB}}$$ is great than or equal to $$\frac{v_{2}}{v_{1}}$$, the path $$ACB$$ is faster than the path $$AEB$$. We may now say that, if $$\frac{\overline{CF}}{\overline{CB}}$$ is exactly $$\frac{v_{2}}{v_{1}}$$1, then the path through $$C$$ is faster than the ones on the left and on the right.

1 Which is equivalent to $$\frac{v_{2}}{\small\mbox{slope of }CB}$$ of being equal to $$v_{1}$$. We should note that this condition seems to be asymmetric with respect to $$v_{1}$$ and $$v_{2}$$, but $$v_{1}$$ can also be written as $$\frac{v_{1}}{\small\mbox{slope of }AC}$$, since, in this particular case, in which $$A$$ is the line of the water, the slope of $$AC$$ is $$1$$.