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I.S.I. and C.M.I. Entrance

AM-GM Inequality Problem | ISI Entrance

Here is a video solution for ISI Entrance Number Theory Problems based on AM-GM Inequality Problem. Watch and Learn!

Here goes the question…

a, b, c, d are positive real numbers. Prove that: (1+a)(1+b)(1+c)(1+d) <= 16.

We will recommend you to try the problem yourself.

Done?

Let’s see the proof in the video below:

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AMC 10 USA Math Olympiad

Combination of Equations | SMO, 2010 | Problem No. 7

Try this beautiful problem from Singapore Mathematical Olympiad, SMO, 2010 – Problem 7 based on the combination of equations.

Problem – Combination of Equations (SMO Entrance)


Find the sum of all the positive integers p such that the expression (x-p) (x – 13) + 4 can be expressed in the form (x+q) (x+r) for distinct integers q and r.

  • 26
  • 27
  • 16
  • 20

Key Concepts


Basic Algebra

Combination of Terms

Generator of a group

Check the Answer


Answer: 26

Singapore Mathematical Olympiad

Challenges and Thrills – Pre College Mathematics

Try with Hints


If you got stuck in this problem start this problem using this hint :

Start with the given hint

(x-p) (x-13) +4 = (x+q)(x+r)

Let’s try to minimize the expression by taking x= -q

so , (-q -p)(-q -13) = -4 , which becomes (q+p) (q+13) = -4

As it is already given p and q are integers we can come up with many cases .

Try to find out the different cases we can have ………………………..

Starting after the last hint :

p+q = 4 and q+13 = -1 ………………………………..(1)

q+p = -4 and q +13 = 1……………..(2)

p+q = 2 and q+13 = -2 and …………………………(3)

p+q = -2 and q+13 = 2 …………………………(4)

For 1st case its simple calculation that we get q = -14 and p = 8

The initial expression becomes (x-p) (x-13) +4 = (x-14) (x-17)

For 2 nd case :

q= -12 and p = 8

so the initial expression becomes : (x-p)(x-13)+4 = (x-9)(x-12)

try the rest of the cases…………

Now let’s talk about 3rd case ,

q= -15 , p = 17 and

hence (x – p) (x – 13) +4 = \(( x – 15)^2\) which is not true to this problem .

For last case , we obtain q = -11, q = 9 so the initial

(x- p) (x-13) +4 = \((x-11)^2\)

So p is 8 and 18 which add upto 8 +18 = 26 -(Answer)

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AMC 10 USA Math Olympiad

Trigonometry Simplification | SMO, 2009 | Problem 26

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Trigonometry Simplification.

Problem – Trigonometry Simplification (SMO Entrance)


If \(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ} = tan x^\circ \)

Find \( x^\circ \) ?

  • 12
  • 95
  • 46
  • 28

Key Concepts


Trigonometry

Geometry

Check the Answer


Answer: 95

Singapore Mathematical Olympiad

Challenges and Thrill – Pre College Mathematics

Try with Hints


If you really got stuck into this sum we can start from here

\(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ}\)

= \(\frac {cos 100^\circ}{1-2sin 50^\circ cos 50^\circ}\)

Now let’s check with some basic values in trigonometry

\( Cos 2 A = cos^2 A – sin^2 A \) and

\(2 sin A cos A = sin 2 A\)

Now try the rest of the sum by using these two above mentioned values………………

Let’s continue from the last hint :

\( cos 100^\circ = cos^2 50^\circ – sin^2 50^\circ \)

\( 2 sin 25^\circ cos 25^\circ = sin 50^\circ\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ}{2sin 50^\circ cos 50^\circ}\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ }{(cos 50^\circ – sin 50^\circ)^2}\)

Using \(a^2 – b^2 = (a+b) (a-b)\) formula

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

Do the rest of the steps ……………..

Starting from right after the last hint:

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

= \(\frac {1+ tan 50^\circ}{1-tan 50^\circ}\)

= \(\frac {tan 45^\circ + tan 50^\circ}{1-tan 45^\circ tan 50 ^\circ}\)

= \( tan 95^\circ\) – Answer

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AMC 10 USA Math Olympiad

Problem on Trigonometry | SMO, 2008 | Problem – 22

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

Problem on Trigonometry | SMO, 2008 |Problem 22

Find the value of \(\frac {tan 40^\circ tan 60^\circ tan 80^\circ}{tan40^\circ + tan 60^\circ + tan 80^\circ}\)

  • 1
  • 15
  • 6
  • 0

Key Concepts


Trigonometry

Tan Rule

Check the Answer


Answer: 1

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre College Mathematics

Try with Hints


If you got stuck in this sum how to get started you can start by consider a general case where \(40^\circ = A\) , \(60^\circ = B\) and \(80^\circ = C\).

So , A+B+C = \( 180 ^\circ\)

\( A+B = 180^\circ – C\)

(tan (A+B) = tan \(180^\circ – C)\)……………………….(1)

Now try to implement the basic formula and try to do this sum………………

In this we can continue from the last hint:

the formula of tan (A + B) = \(\frac {tan A + tan B}{1- tan A . tan B}\)

From the equation (1) …….

tan (A+B) = tan (180 – C)

\(\frac {tan A + tan B}{1- tan A . tan B} = tan (180^\circ – c)\)

(frac {tan A + tan B}{1- tan A . tan B} = -tan C )

Now just rearrange this expression and you will get the final answer……………..

Here is the final solution:

{tan A + tan B} = -tan C {1- tan A . tan B}

tan A + tan B = – tan C + tan C tan A tan B

tan A + tan B + tan C = tan A tan B tan C

\(\frac {tan A tan B tan C}{tan A + tan B + tan C} = 1\)

Which is the given question. It can be a proof also……………….

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AMC 10 USA Math Olympiad

Problem on HCF | SMO, 2013 | Problem 35

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2013 based on HCF.

Problem on HCF | SMO, 2013 | Problem 35

What is the smallest positive integer n,where \( n \neq 11\) such that the highest common factor of n-11 and 3n +20 is greater than 1?

  • 62
  • 65
  • 66
  • 60

Key Concepts


HCF and GCD

Number Theory

Check the Answer


Answer: 64

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


If you got stuck in this sum we can start from here:

Let d is the highest common factor of n-11 and 3n +20 which is greater than 1.

So d|(n-11) and d|(3n + 20) .

If we compile this two then d|(3n +20 -3(n-11) when d|53 .

Now one thing is clear that 53 is a prime number and also d >1

so we can consider d = 53.

Now try the rest………………

Now from the previous hint

n-11 = 53 k (let kis the +ve integer)

n = 53 k +11

So for any k 3n +20 is a multiple of 53.

so 3n + 20 = 3(53k +11) +20 = 53(3k+1)

Finish the rest………..

Here is the final solution :

After the last hint :

n = 64 (if k = 1) which is the smallest integer. as hcf of (n – 11,3n +20)>1(answer)

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AMC 10 USA Math Olympiad

Sequence | Arithmetic and Geometric | Learn with Problems

What we say is definition:


Sequence : A sequence is an arrangement of objects or a set of numbers in a particular order followed by some rule . In other words we can say that each sequence has a definite pattern. For example :

Example 1 : {1,2,3,4,5,…………………………} – here if we add 1 with the previous term then we are getting the next term as 1 , 1+1 = 2 , 2+1 = 3, and so on.

Again in a sequence the terms can repeat itself such as :

{0,1, 0, 1 , 0 , 1 ,……………} – here 1’s and 0’s are alternately repeating itselves.

Series : A “series” is what you get when you add up all the terms of a sequence; the addition, and also the resulting value, are called the “sum” or the “summation”.

For an example if we say there is a sequence of {1,2,3,4} then the corresponding series is {1+2+3+4} and the sum of this series is 10.

Know Something More About Sequence :


In a sequence each number is called TERM or ELEMENT or MEMBER .

Sequences can be of two types (primarily ) :

(1) Finite Sequences : These are the sequences where the last term is defined in other words. We can say it has a finite number of terms . For an example we can say :

{1,2,3,4,5} – here the last term is already defined so this is a finite sequence .

{4,3,2,1} – We can apply the same logic and can say this is a finite sequence as well (only its in backward )

(2) Infinite Sequences : Thee are the sequences where the last term is not defined .In other words we can say it has an infinite number of terms. For example :

{1,2,3,4,…………………….} – here we have used some dots after 4 instead of any number . The only reason for this is to tell you it can continue till infinity. Huh! funny…….

For this reason these types of sequences are called infinite sequences.

Apart from these two there are some commonly used sequences we have :

  1. Arithmetic Sequences: In these sequences every term is created by adding or subtracting a definite number to the preceding number. Example : {1,5,9,13,17,21,25,…} – where the difference of (5-1) = 4 , (9-5) = 4 and so on…
  2. Geometric Sequences : In these sequences every term is obtained by multiplying or dividing a definite number with the preceding number. Example : { 6, 12, 24, 48 ,…} -where if we divide the next term by the previous term then \(\frac {12}{6} = 2\) again \(\frac {24}{12} = 2 \) and so on………………….

Some examples for better understanding :

Before starting with an example lets try to find the importance of formula to represent one sequence :

Let’s start with a sequence : {3,5,7,9,…………………..}

Now from this sequence we can understand that

1st term is = 3

2nd term is = 5

3rd term is = 7

4th term is = 9 and so on .

So if I tell you to find the 10 th term (lets say each term has a general name which is ‘n’) of this sequence then it will be easy for us to find i.e we can continue counting the terms and we can say the 10th term is 21 (HUH – that’s easy) but if I tell you to find the 100th term from this sequence then ???????????????????

Its not impossible to find but it will be a waste of time , page , ink and energy. For this if we can generate a formula from one sequence we can work at ease.

So from the above sequence {3,5,7,9,…………….}

Lets draw a table and lets start considering n as the general formula for the given sequence:

[Note = We have to match the (we want to get) column and the (reality) column ]

sequence

Now again considering the formula as 2n such that :

Sequence 2

So gain the two columns are not matching but one thing we say that the gaps between two terms are same as given in the sequence {3,5,7,9,………………………………}.So we are not far from the correct answer.

Sequence 3

Now its perfectly matches with the columns. So the desire formula of the sequence is {3,5,7,9, …..} = 2n + 1.

I hope we can generate some more formula with this method. Try to do ……..


Sequence Problems :



Calculate 4th term of the sequence :

\( a_{n} = (-n)^{n} \)

\(a_{1} = – 1^{1} \) = -1

\( a_{2} = (- 2 )^{2} \) = 4

\( a_{3} = (- 3 )^{3} \) = -27

\( a_{4} = (- 4 )^{4} \) = 256 (Answer )

For the sequence defined by \(a_{n} = n^{2} – 5n + 2 \) , what is the smallest value of n for which  \(a_{n}\) is positive ?

\(a_{n} = n^{2} – 5n + 2 \)

Therefore ,

\(a_{1} = 1^{2} – 5\) times \(1 + 2 = 1 – 5 +2 = -2 < 0 \)

\(a_{2 } = 2^{2} – 5\) times \(2 + 2 = 4 – 10 +2 = -4 < 0 \)

\(a_{3} = 3^{2} – 5\times 3 + 2 = 9 – 15 +2 = -4 < 0 \)

\(a_{4} = 4^{2} – 5\times 4 + 2 = 16 – 20 +2 = -2 < 0 \)

\(a_{5} = 5^{2} – 5\times 5 + 2 = 25 – 25 +2 = 2 > 0 \)

Therefore the smallest value of n for which \(a_{n} \) is positive is n = 5 .

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AMC 10 USA Math Olympiad

Sum of Series from SMO – 2013 – Problem Number 29

Try this beautiful problem from Sum of Series from SMO, Singapore Mathematics Olympiad, 2013.

Sum of Series from SMO, 2013


Let m and n be two positive integers that satisfy

\(\frac {m}{n} = \frac {1}{10\times 12} + \frac {1}{12 \times 14} + \frac {1}{14 \times 16} + \cdot +\frac {1}{2012 \times 2014} \)

Find the smallest possible value of m+n .

  • 10570
  • 10571
  • 16001
  • 20000

Key Concepts


Greatest Common Divisor (gcd)

Sequence and Series

Number Theory

Check the Answer


Answer: 10571

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


We can start this kind some by using the concept of series and sequence …….

In this problem we can see that the series as

\(\frac {m}{n}\) =\(\frac {1}{10 \times 12}\) +\(\frac {1}{12 \times 14}\) +\( \cdot \cdot\)+

\(\frac {1}{2012 \times 2014} \)

So sum of this series is

\(\frac {m}{n} = \frac {1}{4} \displaystyle\sum _{k = 5}^{1006} \frac {1}{k(k+1)}\)

Now do the rest of the sum ………………

If you are really stuck after the first hint here is the rest of the sum……………

From the above hint we can continue this problem by breaking the formula more we will get :

= \(\frac {1}{4} \displaystyle\sum_{k=5}^{1006} \frac {1}{k} – \frac {1}{k+1}\)

Now replacing by the values:

\(\frac {1}{4} (\frac {1}{5} – \frac {1}{1007}) \)

Please try to do the rest…………………

This is the last hint as well as the final answer….

If we continue after the last hint…

\(\frac {m}{n} = \frac {501}{10070}\)

Since gcd(501,10070) = 1

we can conclude by the values of m= 501 and n = 10070

So the sum is m+n = 10571 (Answer).

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AMC 10 USA Math Olympiad

Problem from Area of Rectangle | SMO 2012 | Junior Section

Try this beautiful problem from area of rectangle from Singapore Math Olympiad, 2012, Junior Section.

Problem – Area of Rectangle (SMO Exam)


In the diagram below , A and B (20,0) lie on the x-axis and c(0,30) lies on the y-axis such that \(\angle {ABC} = 90^\circ\).A rectangle DEFG is inscribed in triangle ABC . Given that the area of triangle CGF is 351, calculate the area of the rectangle DEFG .

Problem from area of rectangle
  • 468
  • 456
  • 654
  • 400

Key Concepts


Area of Triangle

Area of Rectangle

2-D Geometry

Check the Answer


Answer: 468

Singapore Mathematics Olympiad,

Challenges and Thrills – Pre – College Mathematics

Try with Hints


We can try this sum from taking

OA = \(\frac {30^2}{20} = 45\)

So the area of \(\triangle {ABC} = \frac {(20+45)\times 30}{2} = 975\)

Try to do the rest of the sum………………………

Now lets try to find height of \(\triangle {CGF}\)

Suppose height of \(\triangle {CGF}\) be ‘h’. Then

\((\frac {h}{30})^2 = \frac {351}{975} = (\frac {3}{5})^2\)

\(\frac {h}{30 – h} = \frac {3}{2}\)

Now we have almost reach the answer . Try to find the area of Rectangle DEFG……

Note that the rectangle DEFG has the same base as \(\triangle {CGF}\). Then its area is

\( 351 \times \frac {2}{3} \times 2 = 468 \) (Answer )

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AMC 10 USA Math Olympiad

Area of Circle – Singapore Mathematics Olympiad – 2013

Try this beautiful problem from Singapore Mathematics Olympiad based on Area of Circle.

Problem – Area of Circle


On the xy – plane , let S denote the region consisting of all points

(x,y ) for which \(|x+ \frac {1}{2} y | \leq 10 \) and \(|x|\leq 10 \) and \(|y|\leq 10 \). The largest circle centered at ( 0,0 ) that can be fitted in the region S has area \(k\pi\). Find the value of k.

  • 81
  • 80
  • 79
  • 84

Key Concepts


Area of Circle

2 D Geometry

Check the Answer


Answer: 80.

Singapore Mathematics Olympiad – 2013 – Senior Section – Problem No. 14

Challenges and Thrills – Pre – College Mathematics

Try with Hints


We can start the sum from here :

The region S is the hexagon enclosed by lines

\(x = \pm 10 \) ; \( y = \pm 10 \) and \( x + \frac {1}{2} y = \pm 10\).

So the largest circle contained in S must be tangent to \(x + \frac {1}{2} y = \pm 10\)

Try to find the radius from this ………………..

We can start this hint by using a diagram:

Area of circle - figure

So following the last hint we can understand that its radius is the distance from the origin (0,0) to \( x + \frac {1}{2} y = 10 \)

r = \( \frac {10}{ \sqrt {1+ ({\frac {1}{2}})^2}} = 4\sqrt {5} \).

Now it very easy to find the rest of the solutiion………

So the area of a circle is \(\pi {r}^2\)

= \(\pi {(4\sqrt 5)}^2 \)

= \(80 \pi\) (Answer).

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AMC 10 USA Math Olympiad

Triangle Inequality Theorem – Explanation

What is Triangle Inequality Theorem ?

If I want to give you a perfect definition for Triangle Inequality then I can say : –

The sum of the lengths of any two sides of a triangle is always greater than the length of the third side of that triangle.

It follows from the fact that a straight line is the shortest path between two points. The inequality is strict if the triangle is non-degenerate (meaning it has a non-zero area).

Triangle Inequality Theorem

So in other words we can say that : It is not possible to construct a triangle from three line segments if any of them is longer than the sum of the other two. This is known as The Converse of the Triangle Inequality theorem .

So suppose we have three sides lengths as 6 m, 4 m and 3 m then can we draw a triangle with this side ? The answer will be YES we can.

Suppose side a = 3 m

length of side b = 4 m

Length of side c = 6 m

if side a + side b > side c then only we can draw the triangle or

side b + side c > side a or

side a + side c > side b

So from the above example we can find that 4 m + 3 m > 6 m

But look if we try to take 4 m + 6 m \(\geq \) 3 m .

This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric spaces and analysis.

Problem using Triangle Inequality :


In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?

  • 43
  • 44
  • 45
  • 46

Key Concepts


Triangle Inequality

Inequality

Geometry

Check the Answer


The answer is 43 m

AMC – 2006 – 10 B – Problem 10

Secrets in Inequalities.

Try with Hints


This can be a very good example to show Triangle Inequality

Let ‘ x ‘ be the length of the first side of the given triangle. So the length of the second side will be 3 x and that of the third side be 15 . Now apply triangle inequality and try to find the possible values of the sides.

If we apply Triangle Inequality here then the expression will be like

\(3 x < x + 15 \)

\( 2 x < 15 \)

\( x < \frac {15}{2}\)

x < 7.5

Now do the rest of the problem ………..

I am sure you have already got the answer but let me show the rest of the steps for this sum

If x < 7.5 then

The largest integer satisfying this inequality is 7.

So the largest perimeter is 7 + 3.7 +15 = 7 + 21 + 15 = 43.

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