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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Probability

Probability is a quantity that expresses the chance, or likelihood, of an event. It is most helpful to think of probability as a fraction.

The literal definition of probability is the chance of occurrence of an event. For example, if a person is standing at the intersection of two roads which direct towards North, South, East, and West. Thus, he has a total of $4$ choices (four different directions) to proceed. Now, if he wished to go towards a particular direction, then the probability of completing his wish is $\frac{1}{4}$ since he can only choose one out of the four directions.

Consider another example: A person has two different cars, a Toyota and Honda, which he uses randomly. It can then be said that the probability of using the Toyota is $\frac{1}{2}$ because out of his total of $2$ cars, he can randomly pick $1$ of them.

Hence, from the above examples, we can conclude that the probability of an event occurring is
$$ = \frac{\text{Number of desired or successful outcomes}}{\text{Total number of possible outcomes}}.$$

Important Definitions:
Elementary Events: An event containing only a single sample point. Also known as a simple event.
Ex. Tossing a coin and landing on heads.


Compound Events: Events that are not called elementary events are known as compound events, or in other words, events that contain more than one element.
Ex. Flipping heads on a coin and rolling a 5 on a dice.

Impossible Events: Events that occur with a probability of $0$.
Ex. Rolling a number greater than $6$ on a normal dice. 

Certain Events: Events that occur with a probability of $1$.
Ex. Flipping either heads or tails on a coin.

Complementary Event: The complement of an event is that specific event not happening. In other words, the complement is the exact opposite of a given event. 
Ex. The complement of landing heads on a coin is landing tails. 

Independent Events: Two events are independent if the result of the second event is not affected by the result of the first event. 
Ex. Flipping two coins.

Dependent Events: If two events are dependent, it means that the result of the first event affects the outcome of the second event.
Ex. Drawing marbles out of a box without replacement. 

Let's look at some introductory examples to use what we have learned.
Ex. 1: Find the probability of getting no heads after tossing a fair, two-sided coin three times.
Here, we see that we are looking at a compound and independent event. Thus, we can look at the probability of not flipping a head once, and repeat the process for the other two flips. We know that there is a $\frac{1}{2}$ probability of not flipping heads (flipping a tail). Thus, doing this a total of $3$ times would mean we need to multiply the three flips individually, giving us our answer of 
$$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \boxed{\frac{1}{8}}$$.

Ex. 2: What is the probability of rolling at least one $5$ from a roll of two die?
We will tackle this problem by looking at two different techniques. Firstly, we can think about this problem in different cases. 
Case 1. - Rolling one $5$ in total. 
We see that here, we have $2$ die to choose which of them will roll a $5$, and then a $\frac{1}{6}$ probability that it does indeed roll our desired outcome. Then, the other dice will have a $\frac{5}{6}$ probability to roll a value other than $5$. Multiplying these together, we get our probability of rolling one $5$ 
$$= 2 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{10}{36}$$

Case 2. - Rolling two $5$'s.
Here, we need both die to roll a $5$, and the probability of one dice rolling a $5$ is $\frac{1}{6}$, so two die rolling a $5$ would have a probability
$$= \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$

Now, because both cases are mutually exclusive, meaning that there is no intersection between the cases, we can simply add them up, getting our final answer of 
$$\frac{10}{36} + \frac{1}{36} = \boxed{\frac{11}{36}}$$

The other method I hinted at previously is known as complementary counting, and as the name suggests, it involves counting the complement, or the opposite, of the desired outcome. So, in this problem, we see that it asks how to roll at least one 5, so the complement is to roll no $5$'s. Each dice has a $\frac{5}{6}$ probability of rolling something other than a $5$, so the probability of two die rolling something other than a $5$ is $\frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36}$. 

Now, we need to go back to what the question is asking and find the probability of landing on at least one $5$, so we subtract what we found above from $1$, because $1$ is the total probability, and we got our same final answer 
$$1 - \frac{25}{36} = \boxed{\frac{11}{36}}$$

Leave a comment down below if you have any questions. 


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Equilateral and Equiangular Polygons

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