Since he takes the same path both ways, you can assume he is traveling from 0 to 1 and that the total time is 1 unit. Say f(t) gives you his position when going up and g(t) his position when going down. Both of those must be continuous. So h(t) = f(t) – g(t) is also continuous. Since h(0) = -1 and h(1) = 1, by the IVT h(x) = 0 for some x, so at that point he is in the same position at the same time. ]]>

That said, I’m not sure what you mean about distracting detail or what special significance making the trip in one second has. Can you explain a bit more?

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