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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Permutations

Each of the different arrangements that can be made out of a given number of things by taking some or all of them at a time is called a permutation.
Thus the permutation of the three letters $a, b, c$ taken two at a time are:
$$ab, ba, bc, cb, ac, ca$$
Therefore the number of permutations of three different things taken two a time is $_{3}P_{2}$ or $P(3, 2) = 6$.

We can generalize this to say $_{n}P_{r} =$ Total number of permutations of $n$ distinct things taken $r$ at a time.
$$_{n}P_{r} = \frac{n!}{(n-r)!}$$
Below are some extensions of the above concept.

  • Permutations of $n$ different things taken all at a time $$ = _{n}P_{n} = n!$$
Ex. Find the number of ways the letters of the word RAINBOW can be rearranged.
Sol. Every letter in the word RAINBOW is different, so to get the number of permutations, we would simply use $n!$. $$n = 7 \Rightarrow n! = \boxed{5040}$$
  • Permutations of $n$ different things taken $r$ at a time, when $k$ things never occurs $$ = _{n-k}P_{r}$$
Ex. How many $6$ digit telephone numbers can be constructed with the digits $0 - 9$ if the first two digits of each number are $3$ and $5$ respectively and no digit appears more than once?
Sol. Here, we have $10$ digits to choose $4$ digits from because the first two numbers in the telephone number are given. However, $2$ of the digits, namely $3$ and $5$, are not allowed to be picked again. Thus, using the formula above, we get $n = 10, r = 4,$ and $k=2$ and our answer is $$_{(10-2)}P_{4} = \boxed{1680}$$
  • Number of permutations of $n$ things, taken all at a time, of which $p$ are alike and of one kind, $q$ are alike and of a second kind, and the rest are different $$ = \frac{n!}{p!q!}$$
Ex. How many ways can the letters of the word BANANA be arranged in?
Sol. Here, we need to take a permutation of $6$ letters, but $3$ of them are A's and $2$ of them are N's. Thus, we get $n = 6, p = 3$, and $q=2$, which gives us our answer of $$\frac{6!}{3!2!} = \frac{6 \cdot 5 \cdot 4}{2} = \boxed{60}$$
  • Permutations of $n$ different things taken $r$ at a time, when each may be repeated any number of times in each arrangement $$=n^r$$
Ex. How many $5$-digit integers can be formed if all digits are from the set $\{0, 2, 3, 4, 5\}$ and if the digit can repeat?
Sol. Here, we approach this problem by looking at each of the 5 digits of our number. The left-most digit has $4$ options, because no integer can start with a $0$, and the other 4 digits can be any of the $5$ given numbers. Thus, we can use the formula explained above where $n = 5$ and $r = 4$ to get $$4 \cdot 5^4 = \boxed{2500}$$

Comments

  1. UnknownNovember 20, 2019 at 4:41 PM

    When you gave the Banana example, did you mean 2 N's when you say 2 B's?

    ReplyDelete
    Replies
    1. akaliaNovember 22, 2019 at 7:42 PM

      Yes, that is correct. Thanks for letting me know

      Delete

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Equilateral and Equiangular Polygons

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Incenter/Excenter Lemma

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