# Kissing Circles

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In mathematics there are number of named theorems, consider following:

Kissing Circle Theorem: If four circles are tangent to each other at six distinct points, and the circles have curvatures ki (for i = 1, …, 4) then:
$(k_1+ k_2 +k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2)$.
By solving this equation, one can construct a fourth circle tangent to three given, mutually tangent circles

The curvature (or bend) of a circle is defined as k = ±1/r, where r is its radius. The larger a circle, the smaller is the magnitude of its curvature, and vice versa.

This Theorem is more commonly known as : Descartes Theorem.

Author: Jiuguang Wang (http://bit.ly/1PZbRae)

For proof of this Theorem, refer: http://euler.genepeer.com/from-herons-formula-to-descartes-circle-theorem/

I want to share the poem explaining above Theorem: (published in Nature, June 1936)

The Kiss Precise

For pairs of lips to kiss maybe
Involves no trigonometry.
‘Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.

Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There’s now no need for rule of thumb.
Since zero bend’s a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

To spy out spherical affairs
An oscular surveyor
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.

Frederick Soddy

# NISER, Jatni

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It has been about a fortnight at our permanent campus. I just got inducted into “School of Mathematical Sciences”. Unfortunately no place on this earth is free from dirty politics. Working hard enough to survive.

Nevertheless, Jatni has beautiful hills to keep me alive. Hope to share some good mathematics next week.

Hill visible from NISER, Jatni

# A Curious Investigation

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As I always say:

Mathematicians are those weird beasts who enjoy being surrounded by problems.

My current field of interest is Diophantine Equations (DE). Those who ever studied Number theory know about the classic Pythagorean Triplets,  equivalent to finding possible integer solutions of $x^2+y^2 = z^2$.

Also, it is a standard exercise (involving Method of Infinite Descent) in DE to prove  that$x^2+y^2 = 3 z^2$ has no integer solutions. But in this blog post I intend  to discuss following sibling of such degree two DE:

Solve $x^2+y^2 = 2z^2$ for integers.

Clearly, $z\neq 0$, I can divide whole equation by $z^2$ and denote, $X=x/z$ and $Y=y/z$ to get:

Solve $X^2+Y^2 = 2$ for rational numbers.

Observe that $(1,1)$ is a solution of given equation, then any other solution $(a,b)$ will lie with $(1,1)$ on a line with rational slope (or infinite slope, a vertical line, trivial case). Furthermore, every line through $(1,1)$ with rational slope will intersect with the quadratic curve in exactly two points. Because every quadratic equation has either no solutions or two real solutions, and we already know that $(1,1)$ is one solution.

To find all solutions, we first look at the vertical line case by substituting $X=1$  and seeing what two solutions you get.  One will be $Y=1$, and the other gives a solution (which is the same solution in this case). Next, we take a line with rational slope $m$ through $(1,1)$, so that (using slope-intercept form):

$Y=m(X-1)+1$

Now solve this line  and given curve $X^2+Y^2 = 2$ (which is circle of radius $\sqrt{2}$). We will get:

$X = \frac{m^2-2m -1}{m^2+1}$

$Y=\frac{-m^2-2m+1}{m^2+1}$

Where, $m \in \mathbb{Q}$, thus like $x^2+y^2=z^2$, $x^2+y^2 = 2z^2$ has infinite integer solution.

Ending note:

$x^2+y^2 = 2$ has only 4 integer solutions.

# Archimedes Trisection of Angle

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The theorem that the general angle cannot be trisected with ruler and compass alone is true only when the ruler is regarded as an instrument for drawing a straight line through any two given points and nothing else.

By permitting other uses of the ruler the totality of possible constructions may be greatly extended. The following method for trisecting the angle, found in the works of Archimedes, is a good example.

METHOD:

Figure 1 : Archimedes’ trisection of an angle
[pp. 138, What is Mathematics?, © Oxford University Press Inc. ]

STEP – 1 Let an arbitrary angle x be given, as in Fig. 1.

STEP – 2 Extend the base of the angle to the left, and swing a semicircle with O as center and arbitrary radius r.

STEP – 3 Mark two points A and B on the edge of the ruler such that AB = r. (use compass to measure, thus using ruler to measure!!)

STEP – 4 Keeping the point B on the semicircle, slide the ruler into the position where A lies on the extended base of the angle x, while the edge of the ruler passes through the intersection of the terminal side of the angle x with the semicircle about O.

STEP – 5 With the ruler in this position draw a straight line, making an angle y with the extended base of the original angle x.

RESULT: This construction actually yields$y=x/3$

PROOF:

Figure 2: Labeling Archimedes’ trisection of an angle
[pp. 138, What is Mathematics?, © Oxford University Press Inc. ]

As in Fig 2,

$\Delta ABO$$\angle BAO = \angle BOA = y$ and $\angle ABO = \pi - z$

Similarly in , $\Delta BDO$$\angle BDO = \angle DBO = z$ and $\angle BOD = \pi - y - x$

From $\Delta ABO$,  we have, $2y-\pi - z = \pi \Rightarrow 2y = z \ldots (i)$

Similarly, from$\Delta BDO$, we have, $2z = y+x \ldots (ii)$

Now using $(ii)$ in$(i)$ , we can eliminate $z$ to get: $3y=x$