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

But try the problem first…

Answer: 26

Source

Suggested Reading

Singapore Mathematical Olympiad

Challenges and Thrills – Pre College Mathematics

Try with Hints

First hint

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 ………………………..

Second Hint

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…………

Final Step

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)

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

But try the problem first…

Answer: 1

Source

Suggested Reading

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre College Mathematics

Try with Hints

First hint

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………………

Second Hint

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……………..

Final Step

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……………….

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 :

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…

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 ]

Now again considering the formula as 2n such that :

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.

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 :

First Problem …………………….

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 )

Second Problem ……………………

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

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 .

468

456

654

400

Key Concepts

Area of Triangle

Area of Rectangle

2-D Geometry

Check the Answer

But try the problem first…

Answer: 468

Source

Suggested Reading

Singapore Mathematics Olympiad,

Challenges and Thrills – Pre – College Mathematics

Try with Hints

First hint

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………………………

Second Hint

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

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

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.

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).

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

But try the problem first…

The answer is 43 m

Source

Suggested Reading

AMC – 2006 – 10 B – Problem 10

Secrets in Inequalities.

Try with Hints

First Hint……..

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 you really need another hint try this out …………….

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 ………..

Final Step

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.