Zig Zags of infinity on a triangle

Question

There's triangle below. The line is zig zags, and $P_1, P_2, P_3, ... P_n $ is where the zig zags is touching the hypotenuse.

I 2 questions:

1. Is the zig zags length infinite? To me, no because it's so much like the question if you cut a square in half each time, and then forever, but the length is not infinite.
2. If the zig zags length, is not infinite. Is the length of zig zags greater than the perimeter of the triangle? Wow, this is very tricky. Now it's using a zig zags series with infinity, like $\sum_{0}^{\infty}$. I can also find angles of each divded triangle. Finally I think the perimeter of the triangle is 3, because it's equilateral. How do I prove this, and how do I find if zig zags is greater than the perimeter?

Answer 1

For either adding the horizontal parts or the up-left ward parts of the zig-zag, each new part is half the length of the previous. So for the horizontal parts total length is $1+1/2+1/4+\cdots=2,$ and for the up-left ward parts total length is $\sqrt{2},$ making total ziz-zag length $2+\sqrt{2}$. Note that this is equal to the perimeter of the largest triangle.

Added: A good way to see the length of the up-left ward parts of the zig-zag should be $\sqrt{2}$ is to start at the lower right (the point forming the right angle of the largest triangle), and for each new piece sliding it over to line up with the end of the previous piece. That done one gets a line segment which is the diagonal of a 1 by 1 square.

A geometric argument can be given for the total length of the horizontal parts of the zig-zag: Leave the lowest one where it is, and rotate each other smaller part (hinged at its right end) downward until it hits the vertical side of the big triangle. Then we get total length is 1 from the horizontal leg and 1 more from the set of rotated parts which exactly fill out the vertical leg of the large triangle.

Answer 2

There's a cute non-explicit argument for finiteness here. First, look at the non-dotted lines. These have lengths which add up to a finite number - namely, the length of the hypoteneuse of the big triangle. Since corresponding sides in similar triangles scale identically, the sum of the dotted lengths must also be finite.

Of course this doesn't actually compute the sum of the dotted lengths, but I find it quite nice nonetheless. In particular, it works if we replace $\pi\over 4$ with whatever other angle we want.

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As to comparing the sum of the dotted lengths and the perimeter of the whole triangle, consider the following. Let $T_n$ be the right triangle in the diagram with right angle at the top right vertex and $P_n$ as the top left vertex. The top edge and (dotted) right edge have the same length, while the (dotted) hypoteneuse has the same length as the line from $P_n$ to $P_{n-1}$. This just leaves the dotted line touching $\mathcal{O}$. Putting this together, we get that the sum of the dotted lines' lengths is the same as the perimeter of the large triangle.

While more intricate than the previous part of this answer, this is again calculation-free - we just think about isosceles triangles, and draw the right ones.