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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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Combinations

Let $A, B, C$ be three letters , then we can combine any two of them in the following ways: $$ AB, BC, AC$$ Similarly, if $A, B, C, D$ are four letters, then we can combine any two of them in the following manner: $$ AB, AC, AD, BC, BD, CD$$ Similarly, we can combine any $3$ of $A, B, C, D$ as : $$ ABC, ABD, ACD, BCD$$ We can generalize by saying that the number of all combinations of $n$ distinct things taken $r$ at a time $(r \le n)$ is ${n}\choose{r}$  $= \frac{n!}{(n-r)!r!}$ Note In combinations, the order of the letters (or things) is not considered. Here, AB and BA are the same, so they are only counted once, unlike permutations. The term "combination" is generally used for selection of things and "permutations" are used for rearrangements. Combinations with Restrictions Number of combinations of $n$ things taken $r$ at a time in which $x$ particular things always occur is $${n-x}\choose{r-x}$$ Number of combinations of $n$ things t...
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Circular Permutations

In circular permutations, things are to be arranged in the form of a ring or a circle, e.g. arrangements of people around a circular table. In circular permutation there are no end points, i.e. there are no beginning or ending positions. So, the number of circular permutations of n objects are $$\frac{n!}{n}=(n-1)!$$ Thus in a circular permutation, one thing is kept fixed and the remaining  $(n-1)$ things are arranged in $(n-1)!$ ways. If the clockwise and counter clockwise orders are not distinguishable, then the number of ways = $\frac{1}{2}(n-1)!$ Let us look at a few examples.  Ex:   In how many ways can $6$ boys be seated at a circular table? Sol:   We keep one boy in a fixed position and to find the number of permutations to arrange the remaining 5 boys, we simply take $5!$, which gets us to our answer $\boxed {120} $.  Ex:  In how many ways can $6$ boys be arranged at a round table so that $2$ particular boys can be seat...
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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_{...
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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 i...
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Popular posts from this blog

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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Irrational Numbers

A number which cannot be written in the form $ \cfrac {p}{q}$  where $p$ and $q$ are integers and $ q $  $ \neq 0$. Example: $ \sqrt {2}$, $ \sqrt {3}$, $ \sqrt {6}$, $ \sqrt {7}$, $ \sqrt {8}$, $ \sqrt {10}$ Theorem: If $p$ divides $x^3$, then $p$ divides $x$, where $x$ is a positive integer and $p$ is a prime number.  Proof :  Let $x = p_{1} p_{2}...p_{n}$ where $p_{1}, p_{2}, p_{3}.....p_{n}$  are primes, not necessarily distinct.  $ \rightarrow x^3 = p_{1}^3 p_{2}^3.....p_{n}^3$ Given that $p$ divides $x^3$ By fundamental theorem, $p$ is one of the primes of $x^3$.  By the uniqueness of fundamental theorem, the distinct primes of $x^3$ are same as the distinct primes of $x$.  $ \rightarrow p$ divides $x$ Similarly if $p$ divides $x^2$, then $p$ divides $x$, where $p$ is a prime number and $x$ is a positive integer. Example: Prove that  $ \sqrt {2}$ is irrational.  Solution : Let us assume that  $ \sqrt ...
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Incenter/Excenter Lemma

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Given $\triangle ABC$ with incenter $I$, extend $AI$ to meet the circumcircle of $\triangle ABC$ at $L$. Then, reflect $I$ over $L$, and name this point $I_a$. Then:  $(i)$ $BICI_a$ is a cyclic quadrilateral with point $L$ as the center of the circumscribed circle.  $(ii)$ $BI_a$ and $CI_a$ bisect the exterior angles of $\angle B$ and $\angle C$,  respectively.  Proof $(i)$ Notice that the the question is equivalent to proving the distance from $L$ to points $B, I, C, I_a$ are equal. In other words, we need to show that  $$LB = LI = LC = LI_a$$ Firstly, it is obvious that $LI = LI_a$ because by definition, $I_a$ is the reflection of $I$ over $L$, and reflections preserve the length of the segment.  Now, we are left with proving that $LB=LI$, because similar calculations will provide $LI=LC$.  We see that most of our given information involves angles, so we focus on proving $\angle LBI = \angle LIB$. Firstly,  $$\angle LBC = \angl...
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