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Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE:

Important Announcement [Updated:14-Sept-2020]

The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad (IMO) 2021 and European Girls’ Mathematical Olympiad (EGMO) 2022 has been disrupted by the COVID -19 pandemic in the country. In view of the prevailing situation, the following decisions are announced.

❖ In a departure from the usual four-stage procedure for the selection of the teams to represent India at the IMO 2021 and EGMO 2022 that has been followed in previous years, the selection procedure for the 2020-2021 cycle has been condensed to a three-stage process as an exception only for this year. The three stages will be:

1. A three-hour examination called the Indian Olympiad Qualifier in Mathematics (IOQM) organised by the Mathematics Teachers’ Association (India) (MTA(I)).
2. The Indian National Mathematical Olympiad (INMO) organised by the Homi Bhabha Centre for Science Education – Tata Institute of Fundamental Research (HBCSE – TIFR).
3. The International Mathematical Olympiad Training Camp (IMOTC) organised by HBCSE.

❖ The IOQM will have 30 questions with each question having an integer answer in the range 00-99. The syllabus and standard of this examination will be the same as that of PRMO of previous years.

❖ The INMO will be a four-hour examination with 6 questions. The syllabus and standard of this examination will be the same as that of the INMO of previous years.

❖ The IOQM will be conducted by the MTA(I) with support from Indian Association of Physics Teachers (IAPT) and HBCSE.

❖ The INMO and the subsequent stages of the Olympiad programme will be carried out by HBCSE.

❖ Online enrolment for IOQM is expected to start on the IAPT website (iaptexam.in) by October 15, 2020. Further confirmation regarding this and detailed information regarding eligibility and enrolment for IOQM will be announced shortly. These are expected to be similar to previous years.

❖ The tentative schedule for IOQM is Sunday, January 17, 2021 from 9:00 – 12:00 hrs.

❖ The tentative schedule for INMO is Sunday, March 7, 2021 from 12:00 – 16:00 hrs.

❖ Please note that the schedules are tentative, and are subject to change at short notice, depending on the prevailing pandemic situation in the country.

❖ There will not be any examination equivalent to the Regional Mathematical Olympiad for the 2020-2021 cycle. The detailed criteria for qualification from IOQM to INMO and from INMO to IMOTC will be announced soon.

❖ The programme for stages beyond INMO will be announced at an appropriate later time.

HBCSE

Apply for Indian Olympiad Qualifier in Mathematics – IOQM: iaptexam.in

Categories

Triangle Inequality – Mathematical Circles – Problem No. 5

Try this beautiful problem from Mathematical Circles book based on Triangle inequality.

Problem :

Find a point inside a convex quadrilateral such that the sum of the distances from the point to the vertices is minimal .

Key Concepts

Triangle Inequality

Inequality

Geometry

Mathematical Circles – Chapter 6 – Triangle Inequality – Problem 6

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

Do you really need a hint ? You can start thinking about the Triangle Inequality………..

If you have already get the idea about the main concept for this sum then you can start the problem by taking a quadrilateral ABCD with diagonals that intersect at point ‘o’.

The distance from ‘o’ to all vertices is equal.

Here is the diagram where OA + OB + OC + OD = AC + BD – which is sum of the diagonals. We can consider this as one case …

As a last hint you have to take another point to o’ to compare with first case

Now o’ be another point inside the quadrilateral. If we use triangle inequality here we have ,

AO’ + OC’ > AC from Triangle AO’C

BO’ + O’D > BD from Triangle BO’D

Hence AO’ + O’C + BO’ + O’D > AB + BD

Therefore its clear from this that the point that minimizes the sum of the distances is the point of intersection of diagonals.

Categories

Cubes and Rectangles – Math Olympiad Hanoi 2018

Try this beautiful problem from Math Olympiad Hanoi, 2018 based on Cubes and Rectangles.

Find the number of rectangles can be formed by the vertices of a cube.

• 6
• 8
• 18
• 12

Key Concepts

Geometry

Permutation

Combination

Geometry Vol I to IV by Hall and Stevens

Try with Hints

First hint

There are 6 squares on 6 faces on the cube.

Second Hint

There are 4 diagonals of the cube that have the same length

and pass through the center of the cube. Every two diagonals intersect at the midpoint and form a rectangle.

Final Step

Then there are 6+ $(\frac{4!}{2!2!})$ =12 rectangles.

Categories

FERMAT POINT

ABC is a Triangle and P be a Fermat Point Inside it.draw three equilateral triangle based on the three sides i.e$\triangle ABA’$, $\triangle ACC’$, $\triangle BCB’$ respectively.Join $AB’$,$BC’$ and$CA’$ .Show that $AB’$,$BC’$ and $CA’$ pass through a single piont i.e they are concurrent.

Key Concepts

Rotation

Geometry

shortest distance

Challenges and thrills of pre college mathematics

Try with Hints

Rotation:

ABC is a Triangle . Let P Be any point join $AP,BP$ and $CP$. Now if we rotate the $\triangle ABP$ about the point at B $60 ^{\circ}$ anti clockwise we will get $\triangle BP’A’$.

SHORTEST DISTANCE:

Join the point P and P’.Now In the triangle BPP’ we have

BP-BP’

$\angle PBP’=60 ^{\circ}$, SO $\triangle BPP’$ is a equilateral triangle. so $BP=BP’=PP’$

and also $AP’=AP$ (Length remain unchange after Rotation).

So from the point $A’$ to $C$ the path is $A’P’+PP’+PC$.This path will be Shortest distance if $A’P’+PP’+PC$ i.e A’C be a straight line. and also $AP+PB+AB=A’P’+PP’+PC$

the shortest path betwween two points is a straight line and so $PA+PB+PC$ reaches its minimum if and only if the point $p$ and $P’$ lie on the line $A’C$

By symmetry it follows that $P$ must also lie on the line $BC’$ and $AB’$.

So the point of intersection of these lines is a fermat point of a $\triangle ABC$.

EQUILATERAL TRIANGLE :

Now the triangle $AA’B$ we have

$A’B=AB$ (length remain unchange due to rotation)

$\angle A’BA =60^{\circ}$. so the triangle $AA’B$ is a equilateral triangle .

similarly for the other two triangles $AC’C$ and $BB’C$

Categories

Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

Key Concepts

Triangle Inequality

Perimeter

Geometry

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , $\frac {perimeter}{2} > a$

$\frac {perimeter}{2}$ = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

Categories

Math Olympiad in India | A Comprehensive Guide

The Math Olympiad Program in India aims to encourage the students interested in Mathematics. It nurtures their talent through healthy competition among pre-university students in India. This programme is one of the major initiatives undertaken by the National Board of Higher Mathematics (NBHM) and is organized by Homi Bhabha Centre for Science Education (HBCSE).

In India, there are 25 regions designated for training and selection of students in these stages. Each of these regions is assigned to a Regional Coordinator (RC). The Indian National Mathematics Olympiad (INMO) leads the Indian Students to participate in the International Math Olympiad (IMO).

Candidates born on or after August 1, 2001 and studying in Class 8, 9, 10, 11 or 12 are eligible to write PRMO 2020. Further, the candidates must be Indian citizens. Provisionally, students with OCI cards are eligible to write the PRMO, subject to this condition: https://olympiads.hbcse.tifr.res.in/how-to-participate/eligibility/mathematical-olympiad/

Stages of Math Olympiad in India

Math Olympiad in India is a six-stage process:

• Pre Regional Mathematics Olympiad (Pre RMO): This is the first phase of the International Mathematics Olympiad, conducted in August. In this exam, there are 30 questions that the students need to solve in two and a half hours of examination. Answers are marked on a machine-readable OMR Sheet. Students from Class 8 can enroll.
• Regional Mathematical Olympiad (RMO): This is the second phase of the International Mathematics Olympiad, conducted in October. The question paper contains six problems and the duration of the exam is three hours. The difficulty level of the exam is higher than in the first phase.
• Indian National Mathematical Olympiad (INMO): This is the third phase of the International Mathematics Olympiad, conducted in January. The students selected (approximately 900) in the RMO are eligible for the Indian National Mathematical Olympiad (INMO). It is held at 28 centers across the country. Successful candidates get direct entry in Indian Statistical Institute and Chennai Mathematical Institute interview pool for their prestigious bachelor’s program.
• IMO Training Camp – Approximately best 35 students are invited to take training at HBCSE from April to May. In this camp, stress is laid on building the concepts and problem-solving skills of the students. Also, orientation is provided for the IMO. Students need to take several selection tests during this training period. Taking the performances of the students into consideration, only six students are selected to go for the IMO Examination.
• Pre Departure Camp – The selected students go through rigorous training of 8-10 days before the main Olympiad.
• IMO – A team of 6 candidates accompanied by 4 teachers or mentors represents India in the greatest mathematics contest of the world.

What is not an actual Math Olympiad?

Many private organizations use the name of IMO to conduct their own contests. These include booksellers who are trying to sell books to coaching centers who use it as a marketing trick. Parents and teachers have requested them to not use the name of IMO.

These private contests are no way similar to the actual math olympiad which is organized by the department of atomic energy and Indian Statistical Institute. In fact, they can sometimes be counterproductive as they promote rote learning of formulas.

In the actual International Math Olympiad, students have 9 hours to solve 6 problems. These problems require special problem-solving skills and creativity. The fake olympiads usually kill this creative spirit and replace them with rote formula learning.

Here is a public petition against fake math olympiads in India.

Taste some Real & Mind-Bending Mathematics Problems with us!

How to prepare for Math Olympiad?

Math Olympiad has four core topics:

• Number Theory
• Geometry
• Algebra
• Combinatorics

The preparation involves months of problem-solving and concept building. The real math olympian is regarded very highly by the prestigious universities of the world. For example, an IMO medal is almost a sure shot entry to Ivy League universities like Harvard, MIT, Yale. Even an INMO level medal, lets the student take the interview of Indian Statistical Institute’s B.Stat – B.Math Program, and Chennai Mathematical Institute’s B.Sc. Math Program, bypassing the entrance tests.

Cheenta has a rigorous program for Maths Olympiad candidates starting as early as class 1. Our team consists of Olympians and researchers from leading universities in the world.

The unique feature of Cheenta program is that it has both one-on-one and group classes for every student, every week.

Are you in Serious love with Mathematics? Participating in Math Olympiad is a great way to measure your love and who else is a better choice than Cheenta for Math olympiad. Cheenta has an experience of teaching Advanced Mathematics to Outstanding Kids since 10 years from India and abroad.

Don’t believe us? Take the Trial Class for free and decide yourself.

Miscellaneous

• Challenges and Thrills of Pre-College Mathematics by Venkatchala
• Excursion in Mathematics by Bhaskaracharya Pratishthana
• Problem Solving Strategies by Arthur Engel
• Test of Mathematics at 10+2 Level by East West Press
• IMO Compendium

Number Theory

• Elementary Number Theory by David Burton
• Elementary Theory of Numbers by W. Sierpinsky

Combinatorics

• Principles and Techniques in Combinatorics by Chen Chuan Chong and Koh Khee Meng
• Graph Theory by Harary
• Notes by Yufei Zhao

Algebra

• Polynomials by Barbeau
• Inequality by Little Mathematical Library
• Secrets in Inequalities by Pham Kim Hung
• Complex Numbers from A to Z by Titu Andreescu

Geometry

• Lines and Curves by Vasiliyev (something else)
• Geometric Transformation by Yaglom
• Notes by Yufei Zhao
• Trigonometric Delights by El Maor
• Trigonometry by S.L. Loney
• 101 Problems in Trigonometry by Titu Andreescu

Number Theory I

This is the first course in elementary number theory:

• NT.I.1 Primes, Divisibility
• NT.I.2 Arithmetic of Remainders
• NT.I.3 Bezout’s Theorem and Euclidean Algorithm
• NT.I.4 Theory of congruence
• NT.I.5 Number Theoretic Functions
• NT.I.6 Theorems of Fermat, Euler, and Wilson
• NT.I.7 Pythagorean Triples
• NT.I.8 Chinese Remainder Theorem

Combinatorics I

This is the first course in combinatorics and elementary counting techniques:

• Com.I.1 Multiplication and Addition rules
• Com.I.2 Bijection Principles
• Com.I.3 Combinatorial Coefficients
• Com.I.4 Inclusion and Exclusion Principles
• Com.I.5 Pigeon Hole Principle
• Com.I.6 Recursions
• Com.I.7 Shortest Route Problems

Algebra I

This is the first course is school algebra. (We assume that the student is familiar with algebraic expressions, and elementary algebraic identities)

• Alg.I.1 Algebraic identities (Sophie Germain, Cube of three etc.)
• Alg.I.2 Mathematical Induction
• Alg.I.3 Binomial Theorem
• Alg.I.4 Linear Equations
• Alg.I.6 Remainder Theorem
• Alg.I.7 Theorems related to roots of an integer polynomial

Geometry I

• Geo.I.1 Locus visualization
• Geo.I.2 Straight Lines
• Geo.I.3 Triangles
• Geo.I.4 Geometric Constructions
• Geo.I.5 Circles

Trigonometry I

• Trig.I.1 Angle and rotation
• Trig.I.2 Half arcs and Half chords – Genesis of trigonometric ratios
• Trig.I.3 Elementary ratios and associated angles
• Trig.I.4 Trigonometric identities
• Trig.I.5 Geometry and trigonometry
• Trig.I.6 Basic properties of Triangles
• Trig.I.7 Compound Angles
• Trig.I.8 Multiple and Submultiple Angles
• Trig.I.9 Trigonometric Series
• Trig.I.10 Height and Distance

Inequality I

This first course in inequality must be preceded by a basic course in algebra.

• Ineq.I.1 Geometric Inequalities
• Ineq.I.2 Arithmetic and Geometric Mean Inequality
• Ineq.I.3 Cauchy Schwarze Inequality
• Ineq.I.4 Titu’s Lemma

Complex Number I

• Complex.I.1 Geometry of Screw Similarity
• Complex.I.2 Field Properties of complex Number
• Complex.I.3 nth roots of unity and Primitive roots
• Complex.I.4 Basic applications to geometry

Intermediate Curriculum

Number Theory II

• NT.II.1 Mobius Inversion Formula
• NT.II.2 Greatest Integer Function
• NT.II.3 Elementary Group Theory
• NT.II.4 Primitive roots and indices
• NT.II.6 Representation of Integers as sum of squares
• NT.II.7 Perfect Numbers

Combinatorics II

• Com.II.1 Chu Shih Chieh’ Identity (Hockey Stick)
• Com.II.2 Multinomial Coeffiecients
• Com.II.3 Advanced Pigeon Holes and Ramsay numbers
• Com.II.4 Catalan Numbers (and advanced bijection)
• Com.II.5 Stirling numbers of second kind
• Com.II.6 Generating functions
• Com.II.7 Non-linear recurrance

Algebra II

• Alg.II.1 Elementary ring and field theory
• Alg.II.2 Eisenstein’s criterion

Geometry II

• Geo.II.1 Barycentric Coordinates
• Geo.II.2 Miquel Point Configuration
• Geo.II.3 Translation
• Geo.II.4 Rotation
• Geo.II.5 Screw Similarity

Inequality II

• Ineq.II.1 Schur’s Inequality
• Ineq.II.2 Rearrangement Inequality
• Ineq.II.3 Jensen’s Inequality
• Ineq.II.4 Bernoulli’s Inequality and Power means

Complex Number II

• Complex.II.1 Cyclotomic Polynomials
• Complex.II.2 Nine Point theorem and other geometric investigations using complex numbers

Number Theory III

• NT.III.1 Thue’s Theorem
• NT.III.2 Square Free Numbers
• NT.III.3 Diophantine Analysis of second and higher degrees
• NT.III.4 Arithmetic Progression whose terms are primes.
• NT.III.5 Trinomial of Euler
• NT.III.6 Scherk and Richart’s Theorem
• NT.III.7 Amicable Numbers
• NT.III.8 Liouville function
• NT.III.9 Roots of polynomials and roots of congruences
• NT.III.10 Numeri Idonai

Combinatorics III

• Com.III.1 Graph Theory
• Com.III.2 Invariance and Extremal Principles
• Com.III.3 Combinatorial Geometry

Algebra III

• Alg.III.1 Polynomials

Geometry III

• Geo.III.1 Inversive Geometry
• Geo.III.2 Advanced Application of complex numbers
• Geo.III.3 Projective Geometry

Inequality III

• Ineq.III.1 Holder and Minkowski’s inequality

Categories

Understand the problem

Given a circle $\Gamma$, let $P$ be a point in its interior, and let $l$ be a line passing through $P$.  Construct with proof using a ruler and compass, all circles which pass through $P$, are tangent to $\Gamma$, and whose centres lie on $l$.

Source of the problem

RMO 2019 Maharashtra and Goa region

Geometry

Easy

Try these problems first before watching the video or reading the hints:

(Send it to support@cheenta.com. Our priority response is for internal students, however we occasionally try to respond to external students as well). 1. How do you infer that a parallel line needs to be drawn through the center (to the given line AB (L)?  2. Can you find any isosceles triangle in the picture (once one of the little circles is drawn)? 3. How is the second small circle drawn?

Watch the video

Do you really need a hint? Try it first!

Consider an inversion with respect to a circle with centre $P$. Call this map $f$. Note that, given any point $X$, $f(X)$ is constructible using ruler and compass. Construct the circle $f(\Gamma)$.

Suppose $\Gamma'$ is one of our solutions. Then $f(\Gamma')$ is a line perpendicular to $l=f(l)$ and tangent to $f(\Gamma)$

There can be no more than two lines perpendicular to $l$ and tangent to $f(\Gamma)$. Thus these two lines are the images of our solution circles.

Invert the lines back to get the solution circles.

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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