Fermat's Principle of Least Time - Analogy Confusion Question:
I was reading this analogy of Fermat's Principle of Least Time:

In Figure, our problem is to go from A to B in the shortest time. To illustrate that the best thing to do is not just to go in a straight line, let us imagine that a beautiful girl has fallen out of a boat, and she is screaming for help in the water at point B. The line marked x is the shoreline. We are at point A on land, and we see the accident, and we can run and can also swim. But we can run faster than we can swim. What do we do? Do we go in a straight line? (Yes, no doubt!) However, by using a little more intelligence we would realize that it would be advantageous to travel a little greater distance on land to decrease the distance in the water because we go so much slower in the water.

I am thinking that our speed on land is faster than in water so to reach in the least time, we must minimize our distance in the water. So we would take the path $AMB$ ($MB\perp x$).
Why would we take path $ACB$?
Is it because $AM$ is increased by a large factor?


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*The image is from The Feynman Lectures on Physics Volume 1
 A: For comparable distances, $AM \approx AC$ and $CB \approx MB$ you would think that both ACB and AMB will
take similar times since you are running and swimming for roughly equal distances. But you have said
that "AM is increased by a large factor" meaning AM >> AC (note that this condition would imply that the angle x will not necessarily be 90 degrees) and therefore MB > CB. It is then trivial to show
that since $t_{AC} < t_{AM}$ and $t_{CB} < t_{MB}$ and that since
$ t_{AMB} = t_{AM} + t_{MB}$
and
$ t_{ACB} = t_{AC} + t_{CB}$
then
$t_{ACB}$ will be smaller than $t_{AMB}$.
A: Yes, you're right. If you go to $M$ instead of $C$, the increased time on land is so much more than the decreased time in the water that the overall time is longer. You can show this mathematically by setting variables for the relevant lengths and speeds and then minimizing the total time required; you'll find that the location of $C$ is always between the foot of the perpendicular from $A$ to $x$ and the foot of the perpendicular from $B$ to $x$.
A: Which path you must take to save the girl before drowning (so in the shortest time) depends on the difference between your running velocity and your swimming velocity. If this difference is zero then you have to go to her in a straight line AB. When the difference is tiny, this straight path will be a bit different from the straight path. The bigger the difference the more the path will become one with one of both paths (swimming or running) perpendicular to the shore. There is no path that includes a point to the right of M (or to the left of the point beneath S).
