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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...

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...

Problems - 1

Let's look at some interesting applications of the basic arithmetic formulas in the problems below. Problem 1 - Evaluate $123456789 \cdot 999999999$.         Solution - $123456789 \cdot 999999999 = 123456789 \cdot (1000000000-1)$                         $=123456789000000000-123456789=\boxed{123456788876543211}$ Problem 2 - Find the value of $\frac{13579}{(-13579)^2+(-13578)(13580)}$.         Solution - Using $(a-b)(a+b) = a^2 - b^2$, we have $$\frac{13579}{(-13579)^2+(-13578)(13580)} = \frac{13579}{(13579)^2-(13579^2-1)} = \boxed{13579}$$ Problem 3 - Evaluate $\frac{83^3+17^3}{83 \cdot 66+17^2}$.         Solution - By the use of the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ $$\frac{83^3+17^3}{83 \cdot 66+17^2} = \frac{(83+17)(83^2-83 \cdot 17 + 17^2)}{83 \cdot 66 + 17^2} = \frac{100 \cdot (83 \cdot 66 + 17^2)}{83 \cdot 66 + 17^2} = \boxed{100}$$ Problem 4 ...

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