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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Clock Problems

When solving problems relating to clocks, keeping the following two points in mind will be helpful.
  • The Minute hand covers $360^{\circ}$ in 60 minutes, or $6^{\circ}$ per minute.
  • The Hour hand covers $360^{\circ}$ in 12 hours, or $30^{\circ}$ per hour.
Using these facts, the we can easily derive the results below:
  • In a period of 12 hours, the hour and minute hand coincide 11 times.
  • In a period of 12 hours, the hour and minute hand form a $180^{\circ}$ 11 times.
  • In a period of 12 hours, any other angle between the minute and hour hand is formed 22 times.
  • The time gap between any two coincidences is $\cfrac{12}{11}$ hours or $65\cfrac{5}{11}$ minutes.

One common problem involving clocks is to find the angle between the minute and hour hand at a given time. To do so, one can use the formula 
$$\theta = |5.5m - 30h |$$
Where $\theta =$ desired angle, and the time can be written as $h:m$.

Let's use what we have learned to solve a couple problems.
1. What is the angle between the minute and hour hands at the time 5:30?

    Solution: Here we use the formula $\theta = |5.5m - 30h |$ because we are asked to find the angle formed at 5:30 ($h=5, m=30$). 

Thus, we can simply plug in our values to get 
$$\theta = |5.5 \cdot 30 - 30 \cdot 5| = \boxed{15^{\circ}}$$


2. Find the minute value when the minute and hour hands form a $30^{\circ}$ angle given that the hour hand lies between the 4 and 5.

    Solution: We can again use the same formula, but here we are given the hour value ($h=4$) and the angle between the two hands ($\theta=30$), and we need to solve for the minute value, $m$.

Thus, we can simply plug in our formula to get 
$$ 30 = |5.5m - 30 \cdot 4| = 5.5m-120 \Rightarrow \boxed{m=27 \cfrac{3}{11}}$$

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Equilateral and Equiangular Polygons

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