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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Right Triangles

Let's take a look at right triangles and some of their special properties today. As a quick refresher, a right triangle is a triangle with a right angle. We can call the side opposite the right angle as the hypotenuse, and the two sides adjacent to the $90^{\circ}$ angle as the legs. In the image to the right, $AC$ is the hypotenuse, and the two legs are $AB$ and $BC$.


Pythagorean Theorem
One of the most famous and useful theorems in geometry is the Pythagorean Theorem. The theorem states that in right triangle $\triangle ABC$ with hypotenuse $c$ and legs $a$ and $b$, $$a^2+b^2=c^2$$

Lets apply this in a problem to see how it works. 

Problem 1: Let right triangle $\triangle ABC$ with right angle at $B$ have hypotenuse of $r+1$, and legs of length $7$ and $r$. Find $r$. 

Solution: We know that the sum of the squares of the legs is equal to the square of the hypotenuse. In other words, we have $$7^2 + r^2 = (r+1)^2.$$ Simplifying this and solving, we have $$49+r^2=r^2+2r+1 \Rightarrow 48 = 2r \Rightarrow \boxed{r=24}$$.

Special Right Triangles
There are some "special" right triangles whose properties are an absolute must-know in competition math.

1. $45^{\circ}-45^{\circ}-90^{\circ}$ Triangle, also known as an isosceles right triangle is a triangle where the two legs are congruent, and the non-right angles are both $45^{\circ}$. In such triangles, the ratio of the sides is $x - x - x \sqrt{2}$, where the hypotenuse is $x\sqrt{2}$, and both the legs are $x$.

2. $30^{\circ}-60^{\circ}-90^{\circ}$ Triangle is a right triangle where the hypotenuse is twice the length of the shortest side, which is opposite the $30^{\circ}$. In these triangles, the ratio of the side lengths is $x - x \sqrt{3} - 2x$, where the $x$ is opposite the $30^{\circ}$, the $x \sqrt{3}$ is opposite the $60^{\circ}$, and the hypotenuse is $2x$.


Let's see these concepts in action.

Problem 2: Given that $AB = 1$, $BD = BC$ and $EF = CF$ in the diagram below, find the length of $CF$.


Solution: Just looking at the diagram, we see many $30 - 60 - 90$ and $45 - 45 - 90$ triangles in this diagram. We can work our way around, starting with triangle $\triangle ABC$, and eventually get to $CF$. From $AB = 1$, we can say that the hypotenuse $BC = 2$ because we know the hypotenuse is twice the length of the side opposite the $30^{\circ}$. 
Next, we are given that $BD = BC$, so $\triangle BCD$ is an isosceles right triangle. Thus, we can says that $BC = BD = 2$, which leads us to $DC = 2 \sqrt{2}$ because of the side length ratios of a $45-45-90$ triangle. 
Now, in order to find $DE$, we just divide $DC$ by $\sqrt{3}$. Doing this, we get $$DE = \frac{CE}{2} = \frac{DC}{\sqrt{3}} = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{6}}{3} \Rightarrow \frac{CE}{2} = \frac{2 \sqrt{6}}{3} \Rightarrow CE = \frac{4 \sqrt{6}}{3}$$.
Now, we are almost done. We know that $\triangle CEF$ is a $45-45-90$ triangle, so in order to find $CF$, we just need to divide $CE$ by $\sqrt{2}$. Doing this, we get $$CF = \frac{CE}{\sqrt{2}} = \frac{\frac{4 \sqrt{6}}{3}}{\sqrt{2}} \Rightarrow CF = \boxed{\frac{4 \sqrt{3}}{3}}$$



There are some important results that are derived from Pythagorean Theorem. Feel free to reach out to me if you would like to see proofs behind each of these results. If not, I would highly recommend for you to memorize the following for easy application. 










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Equilateral and Equiangular Polygons

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