Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Irrational Numbers

A number which cannot be written in the form $ \cfrac {p}{q}$  where $p$ and $q$ are integers and $ q $  $ \neq 0$.

 Example: $ \sqrt {2}$, $ \sqrt {3}$, $ \sqrt {6}$, $ \sqrt {7}$, $ \sqrt {8}$, $ \sqrt {10}$
Theorem: If $p$ divides $x^3$, then $p$ divides $x$, where $x$ is a positive integer and $p$ is a prime number. 

 Proof: 
Let $x = p_{1} p_{2}...p_{n}$ where $p_{1}, p_{2}, p_{3}.....p_{n}$ are primes, not necessarily distinct. 

 $ \rightarrow x^3 = p_{1}^3 p_{2}^3.....p_{n}^3$

 Given that $p$ divides $x^3$
By fundamental theorem, $p$ is one of the primes of $x^3$. 
By the uniqueness of fundamental theorem, the distinct primes of $x^3$ are same as the distinct primes of $x$. 

 $ \rightarrow p$ divides $x$

 Similarly if $p$ divides $x^2$, then $p$ divides $x$, where $p$ is a prime number and $x$ is a positive integer.
Example: Prove that $ \sqrt {2}$ is irrational. 

 Solution: Let us assume that $ \sqrt {2}$ is not irrational. 
So, $ \sqrt {2}$ is rational
$ \rightarrow$ $ \sqrt {2}$ = $\cfrac{x}{y} $ where $ x, y$ are integers and $ y \neq 0$

 Let $x$ and $y$ be co-primes.
Taking squares on both sides, 
$ \rightarrow$ $ 2 $ = $\cfrac{x^2}{y^2} $
$ \rightarrow$ $ 2y^2$ = $x^2 .....................(1)$
$ \rightarrow$ $ 2$ divides $x^2$
$ \rightarrow$ $ 2$ divides $x$
Thus for some integer $z$, 
$x=2z.................(2)$
from $(1)$ and $(2)$
$2y^2 = 4z^2$
$ \rightarrow$ $ y^2 = 2x^2$
$ \rightarrow 2 $ divides $y^2$
$ \rightarrow 2 $ divides $y$

 Thus $x$ and $y$ have at least $2$ as a common factor, which is a contradiction. 
Example: 
Show that $3 + \sqrt 2 $ is irrational

 Solution: 
Let us assume that $3 + \sqrt 2 $ is rational.
Thus, $3 + \sqrt 2 = \cfrac {p}{q}$ where $p$ and $q$ are integers. 
$ \rightarrow  \sqrt 2 = \cfrac {p}{q} - 3 $
$ \rightarrow  \sqrt 2 = \cfrac {p-3q}{q} $

 Since $p$ and $q$ are integers, $ \cfrac {p-3q}{q}$ is rational. 
But $\sqrt 2$ is rational

 It contradicts our assumption that $3 + \sqrt 2 $ is rational.
Thus our assumption is wrong. 

 Hence $3 + \sqrt 2 $ is irrational. 



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Equilateral and Equiangular Polygons

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