Competitive Math Made Easy

The Pigeon Hole principle follows a simple observation that if you have 10 pigeons and 9 pigeonholes and you distribute these pigeons randomly, its quite obvious that there will be one pigeonhole with more than 1 pigeon. In other words, there are n boxes (the Pigeonholes), into which more than n objects (the Pigeons), have been placed, then at least one of the boxes must have received more than one of the objects. This principle is frequently applied in problems where we need to determine a minimal number of objects to ensure that some integral number property is satisfied. Consider the following puzzle:  A box contains red, black and white balls. The objective is to pick balls satisfying some constraints. How many balls must be taken to ensure that there is a pair of same color? The key to solve this puzzle is pigeonhole principle.  The principle says:  If $n$ pigeons are nesting in $m$ pigeonholes, where $n > m$, then at least one pigeonhole has more than one pig...

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...

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 $100 = \cfrac {x}{100}$ $75 \% = 75$...

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...

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 ...

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 ...

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. 

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...

Circular Permutations are type of arrangements where the objects are to be arranged around a circle or in a circular order. Observe that in circular permutations the order around the circle (or the relative positions ) alone need to be taken into consideration and not the actual positions. For example, suppose $5$ different things are arranged around a circle. Consider the 5 positions around a circle. $A$, $B$, $C$, $D$, $E$ can be arranged in 5 different positions in 5! ways. Consider one such arrangement say $ABCDE$ in that order around the circle. This arrangement and the $4$ new arrangements $BCDEA$, $CDEAB$, $DEABC$, and  $EABCD$ are not only really different arrangements because the same relative positions around the circle are maintained by the $5$ letters. Therefore, the number of different ways of arranging the  $5$ letters around a circle is $ \cfrac {5!}{5} = 4!$ Also, in the above we are considering the clockwise and anti-clockwise arrangements on the circle ...