Competitive Math Made Easy

 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...

Two triangles are called similar , if we can get two congruent triangles after enlarging or compressing the sides of one of them according to an equal ratio. That is, two triangles are similar means they have a same shape but may have different sizes. Criteria for Similarity of Two Triangles  (I) Each pair of corresponding angles are equal (A.A.A.);  (II) All corresponding sides are proportional (S.S.S.);  (III) Two pairs of corresponding sides are proportional, and the included corresponding angles are equal (S.A.S.);  (IV) For two right triangles, a pair of two corresponding acute angles are equal (A.A.);  (V) Among the three pairs of corresponding sides two pairs are proportional (S.S.). Basic Properties of Two Similar Triangles (I) For two similar triangles, their corresponding sides, corresponding heights, corresponding medians, corresponding angle bisectors, corresponding perimeter are all proportional with the same ratio.  (II) Consider the similarit...

When solving problems relating to clocks, keeping the following two points in mind will be helpful. The Minute hand covers $360^{\circ}$ in 60 minutes, or $6^{\circ}$ per minute. The Hour hand covers $360^{\circ}$ in 12 hours, or   $30^{\circ}$ per hour. Using these facts, the we can easily derive the results below: In a period of 12 hours, the hour and minute hand coincide 11 times. In a period of 12 hours, the hour and minute hand form a $180^{\circ}$ 11 times. In a period of 12 hours, any other angle between the minute and hour hand is formed 22 times. The time gap between any two coincidences is $\cfrac{12}{11}$ hours or $65\cfrac{5}{11}$ minutes. One common problem involving clocks is to find the angle between the minute and hour hand at a given time. To do so, one can use the formula  $$\theta = |5.5m - 30h |$$ Where $\theta =$ desired angle, and the time can be written as $h:m$. Let's use what we have learned to solve a couple problems. 1. What is the angle between the ...

Let $ABCD$ be a convex quadrilateral with $AB=BC=AC$. Suppose that a point P lies in the interior of the quadrilateral such that $AD=AP=DP$ and $\angle PCD=30^{\circ}$. Given that $CP=2$ and $CD=3$, find that length of $AC$.  The first step in solving this question is to draw the segment $PB$. Doing this allows us to see that $\triangle APB \cong \triangle ADC$ because $AP=AD$, $AB=AC$, and $\angle DAC = \angle PAB$ because $\angle DAP = \angle CAB = 60^{\circ}$. Thus, $PB=3$.  Next, let's set $\angle ACD = x$. We see that $\angle PCA = 30-x$, and $\angle BCP = 30 +x$. Due to the triangle congruency we established earlier, $\angle PBA = x$ as well. This means that we can write $\angle PBC = 60 - x$. Now,  $$\angle CPB = 180^{\circ} - \angle BCP - \angle PBC = 180^{\circ} - (30+x) - (60-x) = 90^{\circ}$$.  Thus, $\triangle BPC$ is a right triangle, and we know that $CP=2$ and $PB=3$, so we can easily find $CB$ to be  $$CB = \sqrt{2^2+3^2} = \boxed{\sqrt{13}}$$

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 ...

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...

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}{...

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...

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}...

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]}$        ...

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...

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 ...

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...