# Pythagoras Puzzle

• Oct 13th 2010, 08:22 AM
Guffmeister
Pythagoras Puzzle
Hi!

So, I thought of something the other day that my friend once told me when I was in high school. I still remember it, but can't remember what he taught me about it, and can't find anything on Google. I thought I'd post it here to see if anyone had any thoughts.

Basically, imagine you had a square with sides equal to $\displaystyle x$, and you want to travel diagonally from the top left corner of the square, $\displaystyle A$, to the bottom right, $\displaystyle B$.

Now, the shortest route would be to go diagonally, and you can calculate that from Pythagoras' theorem as $\displaystyle \sqrt{2}x$.

Alternatively, you could go across from $\displaystyle A$, to the top right corner, then down to $\displaystyle B$, and will obviously have travelled $\displaystyle 2x$.

Now, instead, imagine going halfway between $\displaystyle A$ and the top right corner, then down to the centre, then across to the centre of the right hand side vertical line, then down to $\displaystyle B$. You'll again have travelled $\displaystyle 2x$. If you then repeated this, but going only a quarter of the way, then down $\displaystyle \frac{1}{4}x$, then across $\displaystyle \frac{1}{4}x$, and so on, you'll again travel $\displaystyle 2x$ towards $\displaystyle B$. If you repeated this to infinity, your path would eventually tend towards the hypotenuse, but the distance you travel would still be $\displaystyle 2x$ according to the pattern I described above, when it should tend towards $\displaystyle \sqrt{2}x$.

I wondered if anyone had any ideas about that. My friend was really really into maths, so it might be a famous problem that people have thought about, but I can't seem to find any information about it.

Thanks.
• Oct 13th 2010, 09:49 AM
It's a very nice example.

The horizontal distance x is divided into n equal parts.
The vertical distance x is also divided into n equal parts.

Hence the distance travelled, for any number of subdivisions n is

$\displaystyle \displaystyle\ n\left(\frac{x}{n}+\frac{x}{n}\right)$

You will always know the distance travelled,
however, if n is very large,
another observer may not spot the divisions for extremely large n
and conclude that the distance travelled is the straight diagonal.
• Oct 13th 2010, 11:55 AM
Traveller
That path is never same as the diagonal. No matter how fine it is made it will never be differentiable.
• Oct 14th 2010, 03:19 AM
Guffmeister
Sure it would! If the path you travel is equal to

$\displaystyle n\left (\frac{2x}{n} \right)$

and $\displaystyle n$ tends towards infinity, then it would become the diagonal. This is the nature of infinity.
• Oct 14th 2010, 03:28 AM
Guffmeister
Actually no, thinking about it. You can also zoom in infinitely and you'd get the same picture of steps. So you're right.

Interesting thing to think about though.