New Riddle Service .... Who Dis?

N.B.Given the new iteration of the Riddler, there are those who might surmise that I would change the name of my Riddler answer blog from on of the aliases of the Batman villain to some fiddling pun, like, I'm just spitballing here, I dunno... Stradi's Various Answers Shop, but I am lazy, so Ed Nigma lives on!

Suppose I start with a strip of paper that’s 10 inches long and 1 inch wide. I twist it once and attach the two 1-inch edges together, forming a Möbius strip. Finally, I draw a dot somewhere in the center of the strip—that is, the dot is half an inch away from both edges.

Naturally, the distance between any two points on my strip is the shortest path between them, as the ant crawls. By that, I mean that such paths can go over an edge to the other “side” of the strip (“side” in quotes because it’s all really the same side, after all), just as an ant would crawl around the edge of the paper. There is a unique point on the strip that is the farthest distance from my dot. What is this distance?

We can represent the strip of paper as a $10" \times 1"$ rectangle that has two (2) $10" \times 0.5"$ rectangles stacked immediately above and immediately below it. The two smaller rectangles represent the ``other side'' of the strip. Since this is a M\"{o}bius strip, the orientation of the left and right sides of the rectangles are reversed, so for instance lower left and upper right corners of the 10" \times 1" rectangle are the same point and same with the upper left and lower right. Meanwhile, the upper edge of the upper smaller rectangle and lower edge of the lower smaller rectangle are the same. Additionally, the midpoint of left side of the $10" \times 1"$ rectangle is simultaneous the upper right corner of the upper smaller rectangle and the lower right corner of the lower smaller rectangle, and vice versa with the midpoint of the right side of the larger rectangle. That was confusing enough, so let's set up a frame of reference and see if we can measure something, eh?

Assume the larger rectangle is given by $\{ (x,y) \mid 0 \leq x \leq 10, |y| \leq 0.5 \},$ with the upper and lower smaller rectangles given by $\{ (x,y) \mid 0 \leq x \leq 10, 0.5 \leq \pm y \leq 1 \},$ respectively. Finally, without loss of generality, assume that your mark is made at the origin. Then since the points $(0,0)$, $(10,1)$ and $(10,-1)$ are synonymous, the ``as the ant crawls'' metric is given by $$d(x,y) = \min \{ \sqrt{x^2+y^2}, \sqrt{ (x-10)^2 + (y-1)^2 }, \sqrt{ (x-10)^2 + (y+1)^2 } \}.$$ Maximizing this metric over the set $\{ (x,y) \mid 0 \leq x \leq 10, |y| \leq 1 \}$ gives the maximum possible distance away of $$d^* = \max_{ 0\leq x \leq 10, |y| \leq 1 } d(x,y) = 5.05,$$ which is obtained at the point $(x^*, y^*) = (5.05, 0)$ which is synonymous with the points $(4.95, \pm 1).$