Preparing for
the National Olympiad takes a lot of efforts. You must have a solid
grasp of some basic problem solving skills. In this note, I'll be talking about category-wise preparation.
Junior, Secondary and Higher Secondary
Juniors, first go through all the math textbooks of high school, and learn any basic algebra you don’t know. Here are some useful books that you can follow to prepare your best for the NMO.
Geometry
Theory And Exercise Books
Problem Book
1. Geometry Camp 2009 Problem Set.
You can download the problem set from here.
Number Theory
Theory Books
1. Number Theory by S G Telang.
2. Number Theory By Marie A Jones And Joseph A Jones.
Problem Book
1. 104 Number Theory Problems
Combinatorics
1. A Path To Combinatorics For Undergraduates.
(Don't worry if you have little background in Combinatorics, this book will help you a lot)
Miscellaneous Excellent Books
1. An Excursion in Mathematics.
2. The Art and Craft of Problem Solving.
বাংলা বই
১। অলিম্পিয়াড সমগ্র
২।যারা গনিত অলিম্পিয়াডেযাবে
৩। বহুপথে পিথাগোরাস ওঅন্যান্য সমস্যা
৪।প্রাণের মাঝে গণিত বাজে (জ্যামিতির জন্য ভালবাসা)
৫। বাংলাদেশ গণিতঅলিম্পিয়াডের প্রশ্ন
Online Resources
Contributor
Mutasim Mim.
Honorable Mention, International Mathematics Olympiad 2014.
Theory And Exercise Books
1. The two Geometry text books of Class 9-10. These two books have very rich content and very good for starting geometry.
2. Geometry Revisited, Bengali translated version available as ‘জ্যামিতির দ্বিতীয়পাঠ’
Problem Book
1. Geometry Camp 2009 Problem Set.
You can download the problem set from here.
Number Theory
Theory Books
1. Number Theory by S G Telang.
2. Number Theory By Marie A Jones And Joseph A Jones.
Problem Book
1. 104 Number Theory Problems
Combinatorics
1. A Path To Combinatorics For Undergraduates.
(Don't worry if you have little background in Combinatorics, this book will help you a lot)
Miscellaneous Excellent Books
1. An Excursion in Mathematics.
2. The Art and Craft of Problem Solving.
বাংলা বই
১। অলিম্পিয়াড সমগ্র
২।যারা গনিত অলিম্পিয়াডেযাবে
৩। বহুপথে পিথাগোরাস ওঅন্যান্য সমস্যা
৪।প্রাণের মাঝে গণিত বাজে (জ্যামিতির জন্য ভালবাসা)
৫। বাংলাদেশ গণিতঅলিম্পিয়াডের প্রশ্ন
Online Resources
1. We have our own online forum here. Meet and discuss problems with math enthusiasts of all over the country here. There is even a sub forum for National Olympiad.
2. Have a look here for various math Olympiad books reviews, links, notes etc.
3. To begin number theory, have a look at this note. It contains some typos, I’ll try to upload the edited version soon.
Contributor
Mutasim Mim.
Honorable Mention, International Mathematics Olympiad 2014.
Another Article:
Mathematics has so many divisions, but you do not have to be expert in all parts for Mathematical Olympiads. If you check the past papers of any Mathematical Olympiad, especially International Mathematical Olympiad, then you’ll see that it basically covers Number Theory, Algebra, Geometry, Combinatorics and Inequality. To solve the problems you’ve to increase your problem solving skills, know some tricks and enhance the capability of doing problems using the best method.
Here are some Mathematics Olympiad Sample Problems:
III. Consider a right-angled triangle ABC, where AC=15, CB=20 and AĈB= 90’. Let CD be a perpendicular on AB. Let P and Q be in centre of triangle ACD and BCD. Find the length of PQ.
V. Given that, (TWO)²=(THREE) where TWO is a 3 digit number where different letters represent different digits. Find the five digit number THREE.
Solution
TWO is a 3-digit number less than 317 because 317² is more than 10000, but THREE is a 5-digit number.
T can be 1, 2 or 3.
Suppose that T = 2; observing that 200*200 = 40000, whose first digit is 4 and not 2 we can say that T=1.
Now O²= E. So E can be 1, 4 or 9, but as T=1 E can only be 4 or 9.
If we multiply TWO*TWO we'll notice that O²=2OW. As O² is even, E is even, so E=4.
Since (TWO)² < 20000, we must have W ≤ 4. And since T = 1, E = 4, we get W = 2 or 3.
Since the units digit of (TWO)² is 4, the units digit of TWO must be 2 or 8. That leaves us with TWO= 128 or 132 or 138.
For only 138² ones and tens digit are same. So THREE=19044.
Solution
X. Find all integer values of N for which N²+ 25N+158 is a perfect square.
Solution
Let’s assume N²+25N+158=y²
Or, N²+25N+(158-y²) =0
We know, N= {-b ±√( b²-4ac)}/2a
I’m giving you some book suggestion to start your preparation for Mathematical Olympiad. Books are given in order of difficulty. So, if you are a beginner, you should start with the first book of each part.
Number theory
(i) Beginning Number Theory
(ii) Number Theory
(iii) Adler’s Number Theory
(iv) A Pathway into Number Theory
(v) Elementary Number Theory
(vi) A Primer of Analytic Number Theory
Algebra
(i) Concrete Abstract Algebra
Geometry
(i) Plane Euclidean Geometry Theory & Problems
(ii) Geometry Revisited
(iii) Geometry Unbound
(iv) Complex number & Geometry
(v) The Algebra of Geometry
(vi) Transformation Geometry
Combinatorics
(i) Combinatorics
(ii) Principles and Techniques in Combinatorics
(iii) A Path to Combinatorics for Undergraduates
Inequality
(i) Inequalities
(ii) Algebra Inequalities
(iii) Secrets in Inequality
Problem Books
(i) 101 Problems in Algebra
(ii) 102 Combinatorial Problems
(iii) 103 Trigonometry Problems
(iv) 104 Number Theory Problems
(v) USSR Olympiad Problem Book
(vi) The IMO compendium
Other books
(i) The Art & Craft of Problem Solving - Paul Zeitz
(ii) Problem Solving Strategies
(iii) Discrete Mathematics & It’s Application
(iv) Proofs that really count: The Art of Combinatorial Proof
Here are some Mathematics Olympiad Sample Problems:
 |
| Figure 1 |
I. The Figure 1 shown is a square with quarter circles drawn from two adjacent corners. If the side of the square is 16, find the area of the largest circle that can be inscribed in the blue region.
Solution
Let O be the center of the circle and r be the radius of the circle.
The circle must touch the arc AC and BD. Let extended DO touch the arc AC at P.
DP= 16 as it is the radius of the quarter circle ACD.
OP=r, DO=10-r.
Radius from center O is perpendicular bisector of DC.
So by Pythagoras theorem we can write,
r²+8²=(10-r)²
20r=36
r=1.8
So area of the circle=π(1.8)²
The circle must touch the arc AC and BD. Let extended DO touch the arc AC at P.
DP= 16 as it is the radius of the quarter circle ACD.
OP=r, DO=10-r.
Radius from center O is perpendicular bisector of DC.
So by Pythagoras theorem we can write,
r²+8²=(10-r)²
20r=36
r=1.8
So area of the circle=π(1.8)²
II. There are 20 questions in an exam and you must answer 12 questions including at least 5 questions from the first 6. How many different combinations of questions can you choose to answer?
Solution
Let's imagine two scenarios. In first scenario you chose to answer all of the first 6 questions and in second scenario you chose 5 questions out of the first 6 questions.
If you choose to answer the first 6 questions, there are 14ℂ6=3003 ways to answer the remaining 6.
If you choose to answer 5 questions from the first 6 questions, there are 6ℂ5=6 ways to choose them.
Remaining 7 questions can be chosen in 14ℂ7=3432 ways.
So, 3003+3432×6=23595 ways of choosing 12 questions.
If you choose to answer 5 questions from the first 6 questions, there are 6ℂ5=6 ways to choose them.
Remaining 7 questions can be chosen in 14ℂ7=3432 ways.
So, 3003+3432×6=23595 ways of choosing 12 questions.
III. Consider a right-angled triangle ABC, where AC=15, CB=20 and AĈB= 90’. Let CD be a perpendicular on AB. Let P and Q be in centre of triangle ACD and BCD. Find the length of PQ.
Solution
AB=25 (By Pythagoras Theorem)
sin B=15/25= 3/5
3/5=CD/20
CD=15
DB=16
Area of ΔBCD= 0.5*12*16= semi perimeter* in-radius= 0.5*(12+16+20)*r
r=4
AD=9
Area of ΔACD= 0.5*9*12= 0.5*(15+12+9)*r̅
r̅=3
r+r̅=7
PQ=7.
sin B=15/25= 3/5
3/5=CD/20
CD=15
DB=16
Area of ΔBCD= 0.5*12*16= semi perimeter* in-radius= 0.5*(12+16+20)*r
r=4
AD=9
Area of ΔACD= 0.5*9*12= 0.5*(15+12+9)*r̅
r̅=3
r+r̅=7
PQ=7.
IV. ABCD is a square. P is a point inside the square.∠PAB=∠PBA=15°. What is the measure of ∠CPD in degrees?
Solution
Let length of square is x
Triangle PAB is an isosceles triangle.Draw a perpendicular bisector PZ from P on AB.
tan 15°= PZ/(x/2) = 2PZ/x
PZ= xtan15°/2
Notice triangle PDC is also an isosceles triangle.
The perpendicular bisector from P to DC= x-(xtan15/2)
tan (PDC)= (2x-xtan15)/x
2-tan(60°-45°)=tan (PDC)
tan(PDC)= 2- (tan 60°-tan 45°)/(1+tan 45°tan 60°)
tan (PDC)=√3
angle PDC=60°=angle PCD
So angle CPD= 180°-120°=60°
Triangle PAB is an isosceles triangle.Draw a perpendicular bisector PZ from P on AB.
tan 15°= PZ/(x/2) = 2PZ/x
PZ= xtan15°/2
Notice triangle PDC is also an isosceles triangle.
The perpendicular bisector from P to DC= x-(xtan15/2)
tan (PDC)= (2x-xtan15)/x
2-tan(60°-45°)=tan (PDC)
tan(PDC)= 2- (tan 60°-tan 45°)/(1+tan 45°tan 60°)
tan (PDC)=√3
angle PDC=60°=angle PCD
So angle CPD= 180°-120°=60°
Solution
TWO is a 3-digit number less than 317 because 317² is more than 10000, but THREE is a 5-digit number.
T can be 1, 2 or 3.
Suppose that T = 2; observing that 200*200 = 40000, whose first digit is 4 and not 2 we can say that T=1.
Now O²= E. So E can be 1, 4 or 9, but as T=1 E can only be 4 or 9.
If we multiply TWO*TWO we'll notice that O²=2OW. As O² is even, E is even, so E=4.
Since (TWO)² < 20000, we must have W ≤ 4. And since T = 1, E = 4, we get W = 2 or 3.
Since the units digit of (TWO)² is 4, the units digit of TWO must be 2 or 8. That leaves us with TWO= 128 or 132 or 138.
For only 138² ones and tens digit are same. So THREE=19044.
 |
| Figure 2 |
VI. In the Figure 2, area of DQBP is 18unit². Given that BP=PC and AQ=QB. Find the area of ABCD.
Solution
Let AQ be x and BP be y.
We can write Area of ABCD- Area of AQD- Area of PCD= Area of DQBP
So 4xy- 0.5*x*2y- 0.5*2x*y=18
2xy=18
4xy= 36
So area of ABCD is 36unit²
We can write Area of ABCD- Area of AQD- Area of PCD= Area of DQBP
So 4xy- 0.5*x*2y- 0.5*2x*y=18
2xy=18
4xy= 36
So area of ABCD is 36unit²
VII. Let a, b, c and d be distinct positive integers such that ab+cd=60, ac+bd=75, ad+bc=68. Find the value of a+b+c+d.
Solution
ac+bd+ad+bc=75+68=143
a(c+d)+b(c+d)=143
(a+b)(c+d)=11*13
As 11 and 13 are prime numbers and (a+b) and (c+d) are integers, we can write (a+b)=11 and (c+d)=13
So a+b+c+d=24.
a(c+d)+b(c+d)=143
(a+b)(c+d)=11*13
As 11 and 13 are prime numbers and (a+b) and (c+d) are integers, we can write (a+b)=11 and (c+d)=13
So a+b+c+d=24.
VIII. Looking at pascal's triangle (figure 3) from left to right, find the 2015th number on the 2017th row of pascal's triangle. Consider the very top row as the 1st row.
 |
| Figure 3 |
Solution
From 5th row if you look at the 3rd number from right for every row, you'll notice a pattern. The 3rd number from right for each consecutive rows are : 6, 10, 15, 21, 28, 36, 45 and so on.
If you look closely you will notice that the 3rd number from right in nth row is the summation of all integers from 1 to (n-2)
So the 3rd number from right in the 2017th row is:
2015*2016/2= 2031120
If you look closely you will notice that the 3rd number from right in nth row is the summation of all integers from 1 to (n-2)
So the 3rd number from right in the 2017th row is:
2015*2016/2= 2031120
IX. How many two digit numbers are there so that the sum of the number and its reverse is a perfect square?
Solution
Let one number be 10x+y. So its reverse is 10y+x.
So 10x+y+x+10y=a*a
11x+11y= 11(x+y) = a*a
So (x+y) needs to be 11 or 11b*b.
As 1<x,y <=9, (x+y) <=18
So (x,y) can be (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)
So there are 8 such numbers.
So 10x+y+x+10y=a*a
11x+11y= 11(x+y) = a*a
So (x+y) needs to be 11 or 11b*b.
As 1<x,y <=9, (x+y) <=18
So (x,y) can be (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)
So there are 8 such numbers.
Solution
Let’s assume N²+25N+158=y²
Or, N²+25N+(158-y²) =0
We know, N= {-b ±√( b²-4ac)}/2a
Let, D= b²-4ac and D needs to have an integer root for N to be an integer.
25²-4×(158-y²)= 625-632+4y²= -7+4y²
So, (-7+4y²) needs to have an integer root
Let, -7+4y² = x²
x²- (2y)² =-7
(x+2y)(x-2y)=-7
As both “a” and “x” are integers, (x+2y) can be 1 (x-2y) can be -7 or (x+2y) can be 7 and (x-2y) can be -1.
Solving these equations for “a” we get a=-2 or 2.
So D= 9
N= (-25+3)/2 or (-25-3)/2
So N= -11 or -14
25²-4×(158-y²)= 625-632+4y²= -7+4y²
So, (-7+4y²) needs to have an integer root
Let, -7+4y² = x²
x²- (2y)² =-7
(x+2y)(x-2y)=-7
As both “a” and “x” are integers, (x+2y) can be 1 (x-2y) can be -7 or (x+2y) can be 7 and (x-2y) can be -1.
Solving these equations for “a” we get a=-2 or 2.
So D= 9
N= (-25+3)/2 or (-25-3)/2
So N= -11 or -14
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