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.

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**.

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 .

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 and line perpendicular to **l** (which is AD) be . For a given line **n**, define a real valued function on the set of lines lying outside the curve **K **. 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 and .

5. We can see that when the line is **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 . Since and . Hence . 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.

This was very interesting, and surprising!

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liked very much indeed as I know the intermediate value theorem…thanks…

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Rotate l with respect to which point?

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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

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So we rotate line m till it coincides with line m’? Also, why coincide with m’ and not the support lines of l* ?

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*support line

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