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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Counting

In this blog post, we will study the very Fundamental Principles of Counting
(i) Multiplication 
If one operation can be performed in $m$ ways and corresponding to each way of performing the first operation, a second operation can be performed in $n$ ways then the two operations can be performed in $m \cdot n$ ways.

In other words, if there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \cdot n$ ways of doing both. 

Here the different jobs/operations are mutually inclusive. It implies that all the jobs are being done in succession. In this case we use the 'and' operator to account for all scenarios, and remember 'and' refers to multiplication.

Example: 
A student has to select a letter from vowels and another letter from consonants, then in how many ways can he make this selection?
Solution:
Out of $5$ vowels he can select one vowel in $5$ ways and out of $21$ consonants he can select one consonant in $21$ ways. Thus a letter from the vowels and a letter from the consonants can be selected together in $5 \cdot 21 = 105$ ways. 

(ii) Addition
If there are two events such that one of them can happen in $m$ ways in the other in $n$ ways, then either of the two events can happen in $m+n$ ways. In this case, we use the operator 'or' between various jobs and 'or' indicates addition.

Example: 
A student has to select a letter either from vowels or from consonants, then in how many ways can he make this selection. 
Solution:
Out of $5$ vowels he select one letter in $5$ ways. Similarly, he can select one letter from $21$ consonants in $21$ ways. Thus he can select one letter in $5 + 21 = 26$ ways. 

Let us look at some additional examples.

Example: 
If a die is rolled and then a coin is tossed, find the number of all possible outcomes. 
Solution: 
A die can fall in the $6$ different ways showing $6$ different values $1,2,3,4,5,6$ and a coin can fall in $2$ different ways showing heads (H) or tails (T).
Thus, the number of total possible outcomes from a die and a coin $= 6 \cdot 2 = 12.$

Example:
How many four digit numbers can be formed using $5$ only once?
Solution:
Case 1: $5$ in Units place

8 9 9 1

$8 \cdot 9 \cdot 9 \cdot 1 = 648$
When $5$ is fixed at unit place, then there are only 8 digits available for thousands place (i.e. except $0$ and $5$)
          Again hundreds and tens places can be filled by any one of the digits from $0$ to $9$ except $5$.

Case 2: $5$ in Tens place

8 9 1 9

$8 \cdot 9 \cdot 1 \cdot 9 = 648$
When $5$ is fixed at tens place, then there are only $8$ digits available for thousands place and $9$ digits are available for each of the hundreds and units place.

Case 3: $5$ in Hundreds place

8 1 9 9

$8 \cdot 1 \cdot 9 \cdot 9 =648$
Here $5$ is fixed at hundreds place

Case 4: $5$ in Thousands place 

1 9 9 9

$1 \cdot 9 \cdot 9 \cdot 9  = 729$
Here $5$ is fixed at thousands place and each of the hundreds, tens and units place can be filled up in $9$ ways each.

Therefore the total required number of ways $ = 648 + 648 + 648 + 729 = 2673$ ways.

Example:
How many four digit numbers can be formed with the digits $0, 2, 3, 5, 8, 9$ if
     $(i)$ repetition of digits is allowed ?
     $(ii)$ repetition of digits s not allowed ?
Solution:
$(i)$

5 6 6 6

$5 \cdot 6 \cdot 6 \cdot 6 = 1080$, since zero cannot be placed in the thousands place. 

$(ii)$

5 5 4 3

$5 \cdot 5 \cdot 4 \cdot 3 = 300$ since zero cannot be place in the thousands place which leaves us with only $5$ available digits. For the hundreds place, we also have $5$ available digits (remember we can use zero but not the digit that we have placed in the thousands place). We then have $4$ digits for the tens place and $3$ digits for the unit place.

Leave a message if you have any questions.


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Equilateral and Equiangular Polygons

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