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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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Permutations - When some objects are Not Distinct

Let us consider the following example - In how many different ways can the letters of the word 'POSSESSIVE' be arranged? Observe that there are $10$ letters in the word and all these letters are not distinct. There are $4$ $S's$ and $2E's$ and other letters are distinct. If all the letters were distinct, the number of different arrangements is obviously $^{10}P_{10}$ or $10!$. In this case, since some of the letters are alike the number of arrangements cannot be $10!$  Let us assume that the number of arrangements of the letters of the word 'POSSESSIVE' is $x$. Consider any of these permutations (or arrangements) A typical arrangement can be thought as PSOESVSIES. Keep all except the three $S's$ fixed in their places.  Suppose we replace the four $S's$ in the above by four distinct objects, say $S_{1}$,  $S_{2}$,  $S_{3}$,  $S_{4}.$ If we permute these objects among themselves in all possible ways, it will give rise to $^{4}P_{4}$    or $4!$ arra...
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Linear Equations - II

Cross multiplication is an important technique for solving equations that contain fractions. In this technique, we multiply both sides of an equation by the denominators of the fractions within it. Doing so will remove the denominators and make the equation easier to solve.  Cross multiplication is based on the principle that for any nonzero values $a$ and $b$,  $  a \cdot \cfrac{b}{a} = b$  In other words, multiplying a fraction by its denominator leaves just its numerator. Example 1 : $ \cfrac {2}{x} = \cfrac{1}{5}$ . First, we multiply both sides by $5$, obtaining $ \cfrac{10}{x} = 1$. Then, we multiply both sides by $x$. This cancels the denominator of the left side, leaving  $x = 10$ Example 2 : Simplify $ \cfrac {a}{1 + \cfrac {4}{5}a} $ Here we must first collapse the denominator into a single term . This can be done using common denominators.      $ \Rightarrow$ $ 1 + \cfrac {4}{5}a = \cfrac {5+4a}{...
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Linear Equations - I

Equations are the foundation of mathematics. All forms of math rely on the principle of equality. An equation states that two expressions have the same value. Expressions are what are on either side of the equation. This may seem obvious, but understanding this is essential. Variables, or the letters we see in equations, are values that we do not know. We must manipulate the equation to find the value of the variable(s). Coefficients are the numbers located directly left of variables. The coefficient of a variable multiplies the variable’s value. For example, $5x$ means “five times x.” Linear equations are the building blocks of algebra. An example of a linear equation is  $2x + 3 = 6$. To solve this equation, we must obtain the variable alone on one side and a simplified value on the other. This process is called isolating the variable. It uses principles of inverse operations: subtraction cancels addition, division cancels multiplication, etc. Example 1 : $2x+3=6$  So...
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Menelaus's Theorem

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If a transversal cuts the sides $BC, CA, AB$ (extended if necessary) of a triangle $ABC$ at $P, Q, R$ respectively then, $ \frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = -1$, where all segments in the formula are directed segments. Proof: Through $A$, draw a line parallel to $QP$ to intersect $BC$ at $K$. In $ \bigtriangleup RBP$ and $ \bigtriangleup ABK$, $ \angle RPB = \angle AKB$   (corresponding angles) $ \angle PRB = \angle KAB$   (corresponding angles) Therefore, $ \bigtriangleup RBP$ $\sim$ $ \bigtriangleup ABK$ by $AA$ Similarity. $ \Rightarrow $ $ \frac{AR}{RB} = \frac{KP}{PB}$              ...$(1)$ In $ \bigtriangleup QCP$ and $ \bigtriangleup ACK$, $ \angle CPQ = \angle CKA$   (corresponding angles) $ \angle CQP = \angle CAK$   (corresponding angles) Therefore, $ \bigtriangleup QCP$ $\sim$ $ \bigtriangleup ACK$ by $AA$ Similarity. $ \Rightarrow $ $ \frac{QC}...
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Ceva's Theorem

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If the lines joining the vertices $A, B, C$ of a triangle $ABC$ to any point $P$ in their plane meet the opposite sides in $D, E, F$ then  $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=  1$ Proof:  In $ \bigtriangleup AFP$ and $ \bigtriangleup FBP$, the height of the altitude is the same if we drop a perpendicular from $P$ to $AF$ and $FB$. Therefore, $ \frac{AF}{FB} =  \frac{[AFP]}{[FBP]}$                           ...(1) Similarly, in $ \bigtriangleup AFC$ and $ \bigtriangleup FBC$, the height of the altitude is same if we drop a perpendicular from $C$ to $AF$ and $FB$. Therefore, $ \frac{AF}{FB} =  \frac{[AFC]}{[FBC]}$                           ...(2) By subtracting the triangle areas of the eq.(2) with eq.(1), we get $ \frac{AF}{FB} =  \frac{[APC]}{[BPC]}$        ...
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Fermat's Little Theorem

Let $a$ be a positive integer and $p$ be a prime. Then  $a^p\equiv a \pmod{p}$ . The cancellation rule tells us that we can divide by $a$ if $gcd (a,p) = 1$ $i.e., gcd(a,p) = 1$ $ \Rightarrow$  $a^{p-1}\equiv 1 \pmod{p}$ Proof: Now, suppose that $gcd(a,p) = 1.$ We form the sequence $ a, 2a, 3a,....(p-1)a$                                                                       $...(1)$ No two of its terms are congruent mod $p$, since $i \cdot a \equiv k \cdot a \pmod{p}$  $ \Rightarrow$   $i\equiv k \pmod{p}$   $ \Rightarrow$ $ i = k$ Hence , each of the numbers in $(1)$ is congruent to exactly one of the numbers $1, 2, 3, ...p-1.$ This implies that,  $ a^{p-1} \cdot1 \cdot2...(p-1) $  $\equiv 1 \cdot 2...(p-1) \pmod{p}$ We may cancel with $(p-1)!$, sinc...
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Problems - 1

Let's look at some interesting applications of the basic arithmetic formulas in the problems below. Problem 1 - Evaluate $123456789 \cdot 999999999$.         Solution - $123456789 \cdot 999999999 = 123456789 \cdot (1000000000-1)$                         $=123456789000000000-123456789=\boxed{123456788876543211}$ Problem 2 - Find the value of $\frac{13579}{(-13579)^2+(-13578)(13580)}$.         Solution - Using $(a-b)(a+b) = a^2 - b^2$, we have $$\frac{13579}{(-13579)^2+(-13578)(13580)} = \frac{13579}{(13579)^2-(13579^2-1)} = \boxed{13579}$$ Problem 3 - Evaluate $\frac{83^3+17^3}{83 \cdot 66+17^2}$.         Solution - By the use of the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ $$\frac{83^3+17^3}{83 \cdot 66+17^2} = \frac{(83+17)(83^2-83 \cdot 17 + 17^2)}{83 \cdot 66 + 17^2} = \frac{100 \cdot (83 \cdot 66 + 17^2)}{83 \cdot 66 + 17^2} = \boxed{100}$$ Problem 4 ...
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Modular Arithmetic

1. Let $n$ be a fixed positive number. Two integers $a$ and $b$ are said to be congruent modulo $n$, symbolized by  $a\equiv b \pmod{n}$  if $n$ divides the difference $a-b$., i.e., provided that $a-b = kn$ for some integer $k$. For ex.   $23\equiv 3 \pmod{5}$,  $19\equiv 3 \pmod{4}$ ,   $12\equiv 5 \pmod{7}$ ,   $5\equiv 5 \pmod{3}$ , 2. For arbitrary integers $a$ and $b$,  $a\equiv b \pmod{n}$  iff $a$ and $b$ leave the same non negative remainder when divided by $n$ 3. Let $n>1$ be fixed and $a, b, c, d$ be arbitrary integers. Then the following properties hold: $a\equiv a \pmod{n}$          Ex.,  $7\equiv 7 \pmod{3}$          $7-7 = 0$ is a multiple of $3$. Hence $3$ divides $0$ If  $a\equiv b \pmod{n}$ , then  $b\equiv a \pmod{n}$          Ex.,  $17\equiv 3 \pmod{7}$          This congruen...
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Quadratic Equations

An equation, in which the highest power of the variable is $2$ is called a quadratic equation. The standard form of a quadratic equation is $ax^2+bx+c=0$, where $a, b,$ and $c$ are constants and $a \neq -$ Solution of Quadratic Equations: There are two main methods to find the solutions, or roots, of a quadratic equation. Factorization $\rightarrow$ Let $ax^2+bx+c = a(x - \alpha)(x - \beta) = 0$. Here, the roots of the equation are $x = \alpha$ and $x = \beta$. Hencec, factorizing the equation and equating each factor to zero is one method to find the roots. The second method is known as the Quadratic Equation, which is:  $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where the equation of the quadratic is $ax^2+bx+c = 0$.  Nature of the Roots:  In a quadratic equation in standard form, the term $b^2-4ac$ is known as the discriminant of the equation, which plays an important role in finding the nature of the roots. It is often denoted by $D$. If $a, b, c \in \ma...
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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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