Enclosing closed curves in squares

Standard

Let’s look at the following innocent looking question:

Is it possible to circumscribe a square about every closed curve?

The answer is YES! I found an unexpected and interesting proof in the book “Intuitive Combinatorial Topology ” by V.G. Boltyanskii and V.A. Efremovich . Let’s now look at the outline of proof for our claim:

1. Let any closed curve K be given. Draw any line l and the line l’ such that line l’ is parallel to l as shown in the fig 1.

capture11

2. Move the lines l and l’ closer to K till they just touch the curve K as shown in fig 2. Let the new lines be line m and line m’. Call these lines as the support lines of curve K with respect to line l.

capture21

3. Draw a line l* perpendicular to l and the line (l*)’ parallel to l* . Draw support lines with respect to line l* to the curve K as shown in the fig 3. Let the rectangle formed be ABCD .

capture31.png

4. The rectangle corresponding to a line will become square when AB and AD are equal . Let the length of line parallel to l (which is AB)  be h_1(\mathbf{l}) and line perpendicular to l (which is AD) be h_2(\mathbf{l}). For a given line n, define a real valued function f(\mathbf{n}) = h_1(\mathbf{n})-h_2(\mathbf{n}) on the set of lines lying outside the curve .  Now rotate the line l in an anti-clockwise direction till l coincides with l’. The rectangle corresponding to l* will also be ABCD (same as that with respect to l). When l coincides with l’, we can say that  AB = h_2(\mathbf{l^*}) and AD = h_1(\mathbf{l^*}).

capture41

5. We can see that when the line is lf(\mathbf{l}) = h_1(\mathbf{l})-h_2(\mathbf{l}). When we rotate l in an anti-clockwise direction the value of the function f changes continuously i.e. f is a continuous function (I do not know how to “prove” this is a continuous function but it’s intuitively clear to me; if you can have a proof please mention it in the comments). When l coincides with l’ the value of f(\mathbf{l^*}) = h_1(\mathbf{l^*})-h_2(\mathbf{l^*}). Since h_1(\mathbf{l^*}) = h_2(\mathbf{l}) and h_2(\mathbf{l^*}) = h_1(\mathbf{l}). Hence f(\mathbf{l^*}) = -(h_1(\mathbf{l}) - h_2(\mathbf{l})). So f is a continuous function which changes sign when line is moved from l to l’. Since f is a continuous function, using the generalization of intermediate value theorem we can show that there exists a line p between l and l* such that f(p) = 0 i.e. AB = AD.  So the rectangle corresponding to line p will be a square.

Hence every curve K can be circumscribed by a square.

Advertisements

6 responses

    • 1)We choose a line l
      2) Then we construct a rectangle ABCD like it was shown above
      3) We find the centre of ABCD ( point where diagonals meet )
      We rotate l about this centre which remains fixed as the line l rotates

      Like