Tag Archives: square

Enclosing closed curves in squares

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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.

Childhood Maths – II

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I found two documents that I was very proud of as a child. Both were the result of trying to understand the kind of things Ramanujan did in free time, a result of the little AMTI books I read as a child. I will share the second document in this post and the other one was in the previous post.

Following document was created in MS Word on my old Windows XP desktop. The calculations were done using some Microsoft advanced calculator:

primes

I was not happy with the result though since the pattern didn’t continue which was supposed to continue according to Ramanujan.

Area of Rectangle

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When we learn to find area enclosed by a curve, we are told to divide area in rectangular elements.

integratuon

Taken from pp. 361 of Mathematics (Part – II), Class XII textbook, NCERT

But, how do we always know the value of area of rectangle? In this post, I will try to prove this well known fact in the spirit of Euclid and Cauchy.

Let’s define:

  • Boundary: A boundary is that which is an extremity of anything.
  • Figure: A figure is that which is contained by any boundary or boundaries.
  • Triangular region : A triangular region is a figure which is the union of a triangle and its interior. Also the sides of the triangle are called edges of the region and vertices of the triangle are called vertices of the region.
  • Polygonal region: A polygonal region is a plane figure which can be expressed as the
    union of a finite number of triangular regions, in such a way that if two of the
    triangular regions intersect, their intersection is an edge or a vertex of each of them.
  • Square region: It is the union of a square and its interior.

The structure in our geometry is

[\mathcal{S},\mathcal{L}, \mathcal{P}, d, m, \alpha]

where \mathcal{S} is the set of points, \mathcal{L} is the set of lines, \mathcal{P} is the set of planes, d is distance (a function satisfying first three properties of metric function), m is angular measure (a function defined for angles, with real numbers as values of the function, satisfying following postulates) and \alpha is the area function satisfying following postulates:

  1. \alpha is a function \mathcal{R} \rightarrow \mathbb{R}, where \mathcal{R} is the set of all polygonal regions and \mathbb{R}  is the set of all real numbers.
  2. For every polygonal region R, \alpha(R) > 0.
  3. If two triangular regions are congruent, then they have the same area.
  4. If two polygonal regions intersect only in edges and vertices (or do not intersect at all), then the area of their union is the sum of their areas.
  5. If a square region has edges of length 1, then its area is 1.

Proposition 1. If a square has edges of length 1/q (q a positive integer), then its area is 1/q^2.

Proof. A unit square region can be decomposed into q^2 square regions, all with the same edge 1/q as

sq1

Then all smaller square have the same area A (divide each square into triangles using diagonals and then use Postulate 3 to prove that all of them have same area). Therefore 1 = q^2A (from Postulate 4) and A = 1/q^2.

Proposition 2. If a square has edges of rational length p/q, then its area is p^2/q^2.

Proof. Such a square can be decomposed into p^2 squares, each of edge 1/q as:

sq2.png

If A is its area, then

A = p^2 \times \frac{1}{q^2} = \frac{p^2}{q^2}

Proposition 3.[Peter Lawes] If a square has edges of length a, then its area is a^2.

Proof. Given a square S_a with edges of length a. Given any rational number p/q,  let S_{p/q} be a square of edge p/q, with an angle in common with S_a, as:

sq3

Then, \frac{p}{q} < a hence S_{p/q} lies inside S_a. For some real number s (by using Postulate 4) we get:

\alpha(S_{p/q}) + s = \alpha(S_{a})

\Rightarrow \alpha(S_{p/q}) < \alpha(S_{a})

\Rightarrow \frac{p^2}{q^2}< \alpha(S_{a})

\Rightarrow\frac{p}{q} < \sqrt{\alpha(S_{a})}

But, selection of \frac{p}{q} being arbitrary, the upper-bound should be unique. Since there exists a unique supremum of the set consisting all possible side lengths of smaller square in \mathbb{R}, we can claim:

a = \sqrt{\alpha(S_{a})}

We can prove this claim by following the proof of statement: \sup\{x \in \mathbb{R} : 0 \leq x, x^2 < 2\} = \sqrt{2}. Hence:

a^2=\alpha(S_a)

Theorem: Area of rectangle is equal to the product of length of any two adjacent sides.

Proof. Given a rectangle of base b and altitude h, we construct a square of edge b + h, and decompose it into squares and rectangles as:

rect
Then from Postulate 4, we get:

(b+h)^2 = 2A + A_1 +A_2

b^2 + 2bh + h^2=2A +h^2+b^2

2bh = 2A

bh = A

 

REFERENCE:

Moise, Edwin (1990). Elementary Geometry from an Advanced Standpoint. Addison-Wesley Pub. Co.

Thanks to Dr. Shailesh Shirali for pointing out this beautiful book.