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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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Combinations

Let $A, B, C$ be three letters , then we can combine any two of them in the following ways: $$ AB, BC, AC$$ Similarly, if $A, B, C, D$ are four letters, then we can combine any two of them in the following manner: $$ AB, AC, AD, BC, BD, CD$$ Similarly, we can combine any $3$ of $A, B, C, D$ as : $$ ABC, ABD, ACD, BCD$$ We can generalize by saying that the number of all combinations of $n$ distinct things taken $r$ at a time $(r \le n)$ is ${n}\choose{r}$  $= \frac{n!}{(n-r)!r!}$ Note In combinations, the order of the letters (or things) is not considered. Here, AB and BA are the same, so they are only counted once, unlike permutations. The term "combination" is generally used for selection of things and "permutations" are used for rearrangements. Combinations with Restrictions Number of combinations of $n$ things taken $r$ at a time in which $x$ particular things always occur is $${n-x}\choose{r-x}$$ Number of combinations of $n$ things t...
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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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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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Sequences and Series

 Let's take a look at some interesting sequences and series problems. 1. (USAMTS) Evaluate the value of  $$S = \sqrt{1+\cfrac{1}{1^2}+\cfrac{1}{2^2}} + \sqrt{1+\cfrac{1}{2^2}+\cfrac{1}{3^2}} + ... + \sqrt{1+\cfrac{1}{1999^2}+\cfrac{1}{2000^2}}$$ Solution Notice that $$1+\cfrac{1}{n^2}+\cfrac{1}{(n+1)^2} = \cfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2} = \cfrac{n^2+n+1)^2}{n^2(n+1)^2} = \Bigg(1+\cfrac{1}{n(n+1)} \Bigg)^2$$ Thus, $$S= \Bigg(1+\cfrac{1}{1 \cdot 2} \Bigg) + \Bigg(1+\cfrac{1}{2 \cdot 3} \Bigg) + \Bigg(1+\cfrac{1}{3 \cdot 4} \Bigg) + ... + \Bigg(1+\cfrac{1}{1999 \cdot 2000} \Bigg)$$ $$= 1999 + \Bigg(1 - \cfrac{1}{2} \Bigg) + \Bigg(\cfrac{1}{2} - \cfrac{1}{3} \Bigg) + ... + \Bigg(\cfrac{1}{1999} - \cfrac{1}{2000} \Bigg) = 1999 + 1 - \cfrac{1}{2000} = \boxed{\cfrac{3,999,999}{2000}}$$ 2. (HMMT) Find the value of  $$S = \cfrac{1}{3^2+1} + \cfrac{1}{4^2+2} + \cfrac{1}{5^2+3} + ...$$ Solution For this problem, we can use partial fraction decomposition and try to find an alt...
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