2
$\begingroup$

Consider a square ABCD having length l and breadth. Now start folding the sides AB and AC so that the figure becomes something like this $$$$not to scale

All the vertical and horizontal folds/stairs are equal in length.

The perimeter of the figure is equal to the the perimeter of the square.

As we increase the number of divisions, the length of each fold/stair decreases. Let the number of stairs/folds be n. As $n\rightarrow\infty$ the figure becomes a right angled isosceles triangle BCD. The perimeter of triangle BCD should be equal to the perimeter of square ABCD since as we increase the number of folds/stairs the perimeter remains the same.

Can anyone please correct me where I have gone wrong?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ You are perfectly correct, except when you say that the perimeter of the limit is the limit of the perimeters. $\endgroup$ – AlainD Jun 7 '14 at 16:34
0
$\begingroup$

The question Is Pi equal to 4? Does the same thing, using a circle inside of a square and slowly removing rectangles from the sides of the square and then using $circumference=2\pi r$ to say that $\pi$ is equal to 4. The answers there will probably answer this quite well.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

The problem is the limit.

If a sequence of paths converges (even uniformly) to a path $P$, that doesn't mean that the sequence of their lengths converges to the length of $P$.

Not sure, but I think that if the curvature of the paths has constant sign the limit of the lengths is the length of the limit.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ No. You also need that the sequence of path derivatives converges to the derivatives of the path. Otherwise the limiting path will have every where a different direction than the target. This is the case in the example, the limiting path looks like the diagonal, but under a magnifing glass, the direction of any of the segments of the limit is either horizontal, either vertical, never at 45°. $\endgroup$ – AlainD Jun 7 '14 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.