Converging Perimeters, showing \(\pi\) = 4

The concept

So we know that \(\pi=3.14159...\), but here is an interesting example of how playing with infitite sums can give weird results. If we have a circle of diameter \(1\), its circumference is \(\pi\times 1=\pi\). Take a square of width \(1\) and repeatedly fold in the corners like in the picture. By changing the shape, you don't change the perimeter, as you always still use horizontal or vertical lines at right angles so save or gain no distance. Adding up all the little straight lengths, it looks like we get the same shape as the circle of the same size... right?

Converging folded square to circle
Converging folded square to circle

What is going on?

We see that by repeatedly folding the corners in of the square, it converges to the shape of the circle. So the perimeter of the original square (\(4\)), which remains the same through the 'folding', must be equal to the circumference of the circle... \[\pi=4\] Now we know that's not true, in fact the two shapes never exactly match, the deformed square is always going to be crinkly, not smooth like the circle, and that is where the extra length hides!

Other shapes

You can show lots of seemingly impossible equalities with this method though, by using a triangle instead of a circle, we can show that \[2+\sqrt{2}=4\]

Converging folded square to triangle
Converging folded square to triangle