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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Distance and Work Problems: Basics

Distance and Work problems are another type of problems that you can almost be sure will appear on all math competitions at every level. It is imperative for young olympians to understand them well and get comfortable as early as you can. This post will touch upon the very basics of Speed/Distance/Time and Work problems. There will be future posts where I will discuss different variations and complexity levels within this category.

 Distance Problems
The key in solving Speed/Distance/Time problems is to understand the relationship between each of these concepts. The basic formula to understand here is

Distance = Rate $\cdot$ Time
$D = R \cdot T$

It is highly recommended to make use of various tools like diagrams and charts to organize the given information leading up to the solution. The formula can be applied to different flavors of such problems. For example, if you know the time and rate a person is traveling on a bus, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled, you could quickly figure out the speed at which she traveled simply by reconfiguring the formula.

Work
Work can be found by multiplying the amount of people working by the rate at which they work by the amount of time they work for. Simplifying this, 
Work = People $\cdot$ Rate $\cdot$ Time
$W = P \cdot R \cdot T$

Let's look at a few examples to help us better understand these formulas.

Example 1: If a car drives at a speed of $60$ mph, and covers a distance of $90$ miles, how long was the car driving for? Express your answer in hours. (mph = miles per hour, m = miles, and h = hours).

Solution: 
We know that this is a distance problem, because the problem talks about distance covered by a moving object. We can start plugging in what we know into our formula. To begin with, the distance ($D$) is given to be $90$ miles, and the speed ($S$) is set at $60$ mph. We are asked to find the time, so we can simply plug our values into our formula. $$90 \text{m} = 60\text{mph} \cdot T \Rightarrow T = \frac{90\text{m}}{60\text{mph}} = \boxed{\frac{3}{2}\text{h}}$$


Example 2: Tim can eat $6$ cookies every $3$ minutes. How long would it take him to eat $114$ cookies, assuming he continues to eat cookies at a constant rate? Express you answer in minutes.

Solution: 
This problem will use the work formula, so we can start by filling out what we know. We know that we need to find the time he takes to eat $114$ cookies, which means we will need to first find the rate at which he eats. We know $W = P \cdot R \cdot T$, where $P = 1$ person, $W = 6$ cookies, and $T=3$ minutes. Plugging these values in, 
$$W = P \cdot R \cdot T \Rightarrow 6 = 1 \cdot R \cdot 3 \Rightarrow R = 2$$
In other words, Tim eats $2$ cookies every minute. We can take the rate we found, and use it to find our desired answer. We need to find how long it takes for him to eat $114$ cookies, so $W = 114$. Additionally, $P = 1$, because he is working alone to eat all those cookies. Lastly, $R = 2$, as we know from our work above. We now have everything we need to solve the question. Plugging in our values, we get 
$$W = P \cdot R \cdot T \Rightarrow 114 = 1 \cdot 2 \cdot T \Rightarrow \boxed{T = 57\text{ mins}}$$

Example 3: Andrew went on a $4$ day trip with his friends. On the first day, he drove for $3$ hours and traveled a total of $215$ miles. The next day, he drove for $9$ hours, and covered a total of $729$ miles. The third day, he only drove for $3$ hours, and traveled $171$ miles. On the last day, he drove $305$ miles in $5$ hours. What was his average speed over the past $4$ days? Express your answer in miles per hour.

Solution: 
In order to find average, the total distance and the total time must first be calculated. 

$D = 215 + 729 + 171 + 305 = 1420$ miles
$T = 3 + 9 + 3 + 5 = 20$ hours

Using our formula for distance, $D = R \cdot T$, and plugging in $D = 1420$ and $T = 20$, we get $$D = R \cdot T \Rightarrow 1420 = R \cdot 20 \Rightarrow R = \boxed{71 \text{ mph}}$$ 

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Equilateral and Equiangular Polygons

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