Hare and Snell's law

Question

A hare from point 0 needs to cross a field with a length Δ and with two parallel straight edges as quickly as possible. The hare runs in a straight line perpendicular to the starting edge of the field and comes across the edge of a round section (point 1) where the hare can run faster (this round section of radius r is in the middle of the field). The hare changes direction of running, crosses a round section along a chord, runs out of it with a change of direction (point 2) and finishes along the shortest path (point 3). Determine the hare's running trajectory. 
Why is the Snell's law fulfilled at point 2, but not at point 1? Where is my mistake or misunderstanding of physics (Fermat's principle)? See please the picture and the Prime 6 sheet in attach.

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LouP · Accepted Answer

I don't think there is any mistake in Valery's reasoning. Rather, as he describes the problem, "The hare runs in a straight line perpendicular to the starting edge of the field and comes across the edge of a round section (point 1)." As given, point 1 is not a point to be determined by optimization, but a fixed point determined by the starting point and a perpendicular line. As such, any approach to the circle ending up at point 1 will have the same remaining optimal path that goes through point 2. There is no reason that Snell's law should apply at point 1 since point 1 is fixed and the incident angle theta 1 depends only on the defined approach.

Werner solved a different problem where the path leaving the starting line is not constrained to be a perpendicular. Now, both points 1 and 2 are part of the optimization, and Snell's law will be satisfied at both points.

I think both are right - just that each addressed distinct problems.

Lou

Werner_E · Answer

The mistake of reasoning consists in the fact that the fastest way does not run from point 0 vertically down to Point 1!
If you take that into account, you no longer come into conflict with Snell and Fermat 😉

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