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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Inradius of a Triangle

In $\triangle ABC$, let the name the incenter as point $I$ and the feet of radii as $D, E,$ and $F$ as shown below.

Since $I$ is the incenter, $ID = IE = IF = r$

Now, using the $A = \frac{bh}{2}$ formula on $\triangle ICB, \triangle IAC, \text{and } \triangle IBA$, we get  

$[ICB] = \frac{1}{2}ID \cdot BC = \frac{ra}{2}$
$[IAC] = \frac{1}{2}IE \cdot AC = \frac{rb}{2}$
$[IBA] =  \frac{1}{2}IE \cdot AB = \frac{rc}{2}$

Adding all three of our equations, we have 
$$\frac{ra}{2} + \frac{rb}{2} + \frac{rc}{2} = [ABC]$$

Letting $s = \text{semiperimeter} = \frac{a+b+c}{2}$, we get
$$[ABC] = r \cdot s$$

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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 ...
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Incenter/Excenter Lemma

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Given $\triangle ABC$ with incenter $I$, extend $AI$ to meet the circumcircle of $\triangle ABC$ at $L$. Then, reflect $I$ over $L$, and name this point $I_a$. Then:  $(i)$ $BICI_a$ is a cyclic quadrilateral with point $L$ as the center of the circumscribed circle.  $(ii)$ $BI_a$ and $CI_a$ bisect the exterior angles of $\angle B$ and $\angle C$,  respectively.  Proof $(i)$ Notice that the the question is equivalent to proving the distance from $L$ to points $B, I, C, I_a$ are equal. In other words, we need to show that  $$LB = LI = LC = LI_a$$ Firstly, it is obvious that $LI = LI_a$ because by definition, $I_a$ is the reflection of $I$ over $L$, and reflections preserve the length of the segment.  Now, we are left with proving that $LB=LI$, because similar calculations will provide $LI=LC$.  We see that most of our given information involves angles, so we focus on proving $\angle LBI = \angle LIB$. Firstly,  $$\angle LBC = \angl...
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