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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Circular Permutations

In circular permutations, things are to be arranged in the form of a ring or a circle, e.g. arrangements of people around a circular table.

In circular permutation there are no end points, i.e. there are no beginning or ending positions. So, the number of circular permutations of n objects are $$\frac{n!}{n}=(n-1)!$$

Thus in a circular permutation, one thing is kept fixed and the remaining  $(n-1)$ things are arranged in $(n-1)!$ ways.

  • If the clockwise and counter clockwise orders are not distinguishable, then the number of ways = $\frac{1}{2}(n-1)!$


Let us look at a few examples. 

Ex: 
In how many ways can $6$ boys be seated at a circular table?
Sol: 
We keep one boy in a fixed position and to find the number of permutations to arrange the remaining 5 boys, we simply take $5!$, which gets us to our answer $\boxed {120} $. 

Ex: 
In how many ways can $6$ boys be arranged at a round table so that $2$ particular boys can be seated together?
Sol: 
Assume there are only $5$ boys since $2$ boys always remain together and these $2$ boys can be arranged mutually in $2$ ways. 
Thus, required number of permutations = $4! \cdot 2 = \boxed {48} $

Ex: 
In how many ways can $5$ men and $2$ women be arranged at a round table if the $2$ women can never be seated together?
Sol: 
Total number of arrangements $= 6! = 720$. Next, the number of arrangements in which the $2$ women are together $ = 2 \cdot 5! = 240 $
Thus, the number of arrangements in which the $2$ women are never together $ = 720 -240 = \boxed {480} $

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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

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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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