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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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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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Percentages

Percent is another way of saying 'for every hundred'.     Any division where the divisor is $100$ is a percentage.         For example: $ \cfrac {20}{100} = 20 \%$  or                             $ \cfrac {45}{100} = 45 \%$ In essence saying that,                              $ \cfrac {x}{100} = x \%$ Since any ratio can also be expressed as a division, it can also be represented as a percentage.  For ex., a ratio of $\cfrac{1}{2}$ or $1:2$ can be converted to a percent.  $ \cfrac {1}{2}$ $=$ $ \cfrac {1 * 50}{2 * 50} = \cfrac {50}{100} = 50$ Per Cent $=  50\%$  Expressing $x \%$ as a Fraction Any percentage can be expressed as a decimal fraction by dividing the percentage by $100$.  As $x \% = x$ out of $100 = \cfrac {x}{100}$ $75 \% = 75$...
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Progressions

Let us observe the following pattern of numbers.     $(i)$   $5, 11, 17, 23, ............$    $(ii)$   $6, 12, 24, 48, ............$   $(iii)$   $4, 2, 0, -2,-4, ............$   $(iv)$   $\cfrac {2}{3}, \cfrac{4}{9}, \cfrac{8}{27}, \cfrac {16}{81},......... $ In example, $(i)$, every number (except 5) is formed by adding $6$ to the previous numbers. Hence a specific pattern is followed in the arrangement of these numbers. Similarly, in example $(ii)$, every number is obtained by multiplying the previous number by $2$. Similar cases are followed in examples $(iii)$ and $(iv)$.  SEQUENCE A systematic arrangement of numbers according to a given rule is called a sequence.     The numbers in a sequence are called its terms. We refer the first term of a sequence as $T_{1}$, the second term as  $T_{2}$ and so on. The $n$th terms of a sequence is denoted by  $T_{n}$, which may also be...
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Polynomials: Key Vocabulary

Let us look at some key terms that pertain to Polynomials before we deep dive into the study of Polynomials.  Constant: A number having a fixed numerical value Example: $3, \cfrac{4}{5}, 4.2, 6.\overline{3}$ Variable:  A number which can take various numerical values Example:   $ x,  y,  z$ Algebraic Expression: A combination of constants and variables connected by arithmetic operators Example: $2x^2 + 7, 5x^3 + 4xy + 2xy^2 + 7,$ etc   Terms: Several parts of an algebraic expression separated  by $+$ or $-$ signs are called the terms of the expression.  Example: In the expression $9x + 7y + 5$, $9x$, $7y$, and $5$ are terms.   Coefficient of a Term:  In the term $8x^2$, $8$ is the numerical coefficient of $x^2$ and $x^2$ is said to be the literal coefficient of 8.  Like Terms: Terms having the same literal coefficients are called Like Terms.  Example: $8xy$, $9xy$, and $10xy$ are Like Terms ...
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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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Circumradius of a Triangle

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In $\bigtriangleup ABC$, bisect the two sides $BC$ and $CA$ in $D$ and $E$ respectively and draw $DO$ and $EO$ perpendicular to $BC$ and $CA$.  Thus, $O$ is the center of the circumcircle. Join $OB$ and $OC$.  The two triangles $BOD$ and $COD$ are equal in all respect, so  $ \angle BOD = \angle COD$ Therefore, $ \angle BOD = \cfrac{1}{2} \angle BOC =  \angle BAC = \angle A$ Also, $BD = BO \sin \angle BOD$ Therefore, $\cfrac{a}{2} = R \sin \angle A$ or $R =  \cfrac{a}{2 \sin \angle A}$ Now, we know that sin $A = \cfrac{2}{bc}$ $\sqrt{s(s-a)(s-b)(s-c)} = \cfrac {2S}{bc}$ where $S$ is the area of the triangle.   Therefore, substituting the above values of sin $\angle A$ in eq $(1)$, we get $R = \cfrac {abc}{4S}$ giving the radius of the circumcircle in terms of the sides. 
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Simson Line

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In geometry, given a triangle $ABC$ and a point $P$ on its circumcircle, the three closest points to $P$ on lines $AB$, $AC$, and $BC$ are collinear. The line through these points is the Simson line of $P$, named for Robert Simson. Prove that the feet of perpendiculars drawn from a point on the circumcircle of a triangle on the sides are collinear. Solution: Let $D, E, F$ be the feet of perpendiculars drawn from a point $P$ on the circumcircle of $ \bigtriangleup ABC$ on the sides $BC, CA, AB$ respectively. We shall prove that the points $D, E, F$ are collinear by showing that $ \angle PED + \angle PEF = 180 ^\circ $ We start by noting that $ \angle PEA + \angle PFA = 180 ^\circ $ Therefore, the points $P, E, A, F$ are concyclic. Consequently, $ \angle PEF + \angle PAF$                 (angles in the same segment)   $...........(1) $ Since $ \angle PEC = \angle PDC = 90^\circ $, therefore $P, E, D, C$ are concyclic...
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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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Incenter/Excenter Lemma

Image
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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Sequences and Series

 Let's take a look at some interesting sequences and series problems. 1. (USAMTS) Evaluate the value of  $$S = \sqrt{1+\cfrac{1}{1^2}+\cfrac{1}{2^2}} + \sqrt{1+\cfrac{1}{2^2}+\cfrac{1}{3^2}} + ... + \sqrt{1+\cfrac{1}{1999^2}+\cfrac{1}{2000^2}}$$ Solution Notice that $$1+\cfrac{1}{n^2}+\cfrac{1}{(n+1)^2} = \cfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2} = \cfrac{n^2+n+1)^2}{n^2(n+1)^2} = \Bigg(1+\cfrac{1}{n(n+1)} \Bigg)^2$$ Thus, $$S= \Bigg(1+\cfrac{1}{1 \cdot 2} \Bigg) + \Bigg(1+\cfrac{1}{2 \cdot 3} \Bigg) + \Bigg(1+\cfrac{1}{3 \cdot 4} \Bigg) + ... + \Bigg(1+\cfrac{1}{1999 \cdot 2000} \Bigg)$$ $$= 1999 + \Bigg(1 - \cfrac{1}{2} \Bigg) + \Bigg(\cfrac{1}{2} - \cfrac{1}{3} \Bigg) + ... + \Bigg(\cfrac{1}{1999} - \cfrac{1}{2000} \Bigg) = 1999 + 1 - \cfrac{1}{2000} = \boxed{\cfrac{3,999,999}{2000}}$$ 2. (HMMT) Find the value of  $$S = \cfrac{1}{3^2+1} + \cfrac{1}{4^2+2} + \cfrac{1}{5^2+3} + ...$$ Solution For this problem, we can use partial fraction decomposition and try to find an alt...
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