I first envisioned this measuring the distance walking through a city. Because of the layouts of the streets there is no straight line for the shortest course. Without trimming any corners in streets, a meandering route through the city would be nearly the same distance traveled as a right triangle from point A to B. I'm just a Physics major and have had this idea in my head for quite some time and can't seem to get a real good answer to it anywhere on the net so I'd thought I'd try here. Next, I converted this to paper folds:
Take an isosceles right triangle and "fold" the right angle towards the hypotenuse until it just touches. This leaves 2 identical triangles that are scaled in size but also leaves an outer "edge" that is equal in length to the two sides. Repeating this again leaves 2 identical triangles for every original triangle prior to the fold but does not change the outer length from that of the original two sides. Repeating this to infinity begins to resemble the smaller and smaller pieces of the Fractal Staircase, but in the end one is left with two identical "lines" of seemingly equal length but one is actually root 2 longer than the other.
Is this an example of fractals and if not, is this a problem that is at least interesting?
Yeah, it is a kind of fractal, and it is a good example of how you gotta be careful evaluating limits. There's a similar construction showing pi = 4, starting with a circle inscribed in a square. The thing to notice is that the length of the "folded" sides is always 2 at every stage of construction. We see the area between the equal legs and the hypotenuses shrinking to zero and imagine that the length of the crinkled side must also go to zero. But look at the limits themselves, beginning with the formulas area= 1/2 base * height, and perimeter = sum of sides.
Say the original triangle is called stage 0 with leg length 1, and the first "fold" is stage 1 with leg length 1/2. At stage n, there are 2^n triangles, each of area =1/2 b*h= 1/2 product of two equal legs = 1/2 * (1/2)^n * (1/2)^n = (1/2)^(2n+1). Because there are 2^n of them, the total area at stage n is 2^n * (1/2)^(2n+1) = (1/2)^(n+1), which in the limit as n-> infinity approaches 0. But at stage n there are 2^n triangles each with total perimeter = hypotenuse + 2 length of leg = sqrt 2*(1/2)^n + 2 * (1/2)^n = (2+sqrt 2) * (1/2)^n. Since there are 2^n of them, the total perimeter is (2+sqrt 2) * (1/2)^n * 2^n = 2+sqrt 2.