I am new to the forums, hopefully my first post won't disappoint.

This one was given to my class by my first-year University maths teacher.

Take a triangle with a hypotenuse going for a length of sqrt(2), and with the two other sides of length 1 above the hypotenuse.

Define the cut operation by travelling up from the base line of the triangle along all lines leaving it, half-way to the first vertex, then drawing a line coming down from that point making angle with the first line equal to the angle between the upper lines of the original triangle.

This process can be thought of as cutting new triangles into the old 1/2 way up the old edges, and the first step illustrated below.

The second step would contain another 4 smaller triangles embedded in the two new ones above, with side lengths 1/4 (and a hypotenuse of another length), etc.

Now the path along P

_{n} is given by taking the path along the cuts after n steps, and we use P

_{n} also to represent that path's length. P(1) is shown above.

Inspection shows that P

_{n} = 2 for all n, as you always double the number of lines you travel along, and halve the length of each one (and the original length was 1+1 = 2)

Hence lim

_{n->infinity} P

_{n}= 2

However this is clearly equal to the length of the hypotenuse, as the limiting Path is the hypotenuse.

Hence sqrt(2) = 2.