Take up the Challenge

Aakash chaudhary

Olympiads are highly challenging competitive examinations conducted through various independent organisations that aim to identify a child’s capability and real potential. It provides a platform to test a student’s aptitude as well as conceptual understanding of the subject. Participating in Olympiads is a matter of great pride as it provides exposure to students at national and international platforms. Each Olympiad is a separate exam with its own organising body and a common set of rules and regulations. The objective behind conducting the Olympiad exams is that they help in promoting a career and serve as a platform to challenge the brightest minds in science and mathematics across the world.

Olympiad Programme in India

The Mathematical Olympiad Programme in India, which leads to participation of Indian students in the International Mathematical Olympiad (IMO) is organised by the Homi Bhabha Centre for Science Education (HBCSE) in collaboration with the National Board for Higher Mathematics (NBHM) of the Department of Atomic Energy (DAE), Government of India. 

The first stage (PRMO) of the Olympiad is conducted by MTA (Mathematics Teachers’ Association) with assistance from IAPT, while the later stages are the responsibility of HBCSE. This programme is one of the major initiatives undertaken by the NBHM. 

Its main purpose is to spot mathematical talent among pre-university students in the country. The Mathematical Olympiad programme consists of six stages with Indian National Mathematical Olympiad (INMO)-2020 being stage three. 

Those who clear RMO are eligible to sit for the Indian National Mathematical Olympiad (INMO). 

Top 30 - 35 performers in INMO receive a certificate of merit. The INMO certificate awardees are invited to a month-long training camp (junior batch) (IMOTC) conducted in May-June each year.

— Inputs from National Academic Director (Engineering) at Aakash Educational Services Limited)

Test Pattern

A detailed, year wise analysis of INMO papers since 2010 has suggested that primarily six questions on various topics are given to be solved in four hours time. The main topics included are:

Geometry and Trigonometry

• Number Theory
• Combinatorics
• Polynomials
• Functions

At a glance

• Usually held on third Sunday of January or first Sunday of February
• Four hour written test
• Top 30 - 35 students are chosen in order of merit

Eligibility: Students who are selected in RMO in the previous year and those holding an INMO certificate of merit

Preparation strategy

As we move higher up the stages of this Olympiads, the difficulty level of the questions increases, so INMO is certainly a tougher exam than RMO. Students need to develop a deeper understanding of the concepts. 

• Solving 10 to 12 of previously asked problems in INMO daily will keep you in good stead. 

Trying your hand at about 500 similar problems form the books suggested will not only make you exam ready but enable you develop analytical thinking and enhance your problem solving skills. 

• Remember, passion, patience and skill are the virtues you need at this level to excel. 

At the end, we would like to wind up by quoting Bertrand Russel, “The true spirit of exaltation, the sense of being more than man, which is the touchstone of highest excellence is to be found in Mathematics as surely as poetry.”

Topic wise Preparation Tips For INMO

Geometry & Trigonometry: Apart from deep understanding of similar triangles and cyclic quadrilaterals, a workmanlike approach about the basic trigonometry is a must. To this end, basic trigonometry by SL loney is ideal. 

For geometry, The Challenge And Thrill of Pre-College Mathematics and Mathematical Circles (Russian experience) could be very helpful.

Number Theory: Basic divisibility in integers, properties of prime numbers, basic properties and results on congruences especially linear congruences and squares and cubes are the bulk of the topics upon which one should concentrate. 

The list of books that one should follow are: Elementary Number Theory by David M Burton, Mathematical Olympiad Challenges by Titu Andreescu and R?zvan Gelca.

Combinatorics: Apart from a good book for IIT JEE, The Challenge And Thrill of Pre-College Mathematics as well as combinatorics by S Muralidharan (AIMER) can be quite handy.

Polynomials: Again Challenge and thrill of Pre-College Mathematics and Problem Solving Strategies by Arthur Engel are good reads on this topic.

Functions: For Functions, Functional Equations A problem solving approach by B J Venkatachala is the most dependable source.

Previous Year’s Questions (For Practice)

• Let AB be a diameter of a circle G and let C be a point on G different from A and B. Let D be the foot of perpendicular from C on to AB. Let K be a point of the segment CD such that AC is equal to the semiperimeter of the triangle ADK. Show that the excircle of triangle ADK opposite A is tangent to G.
• (Geometry and Trigonometry) Let ABC be a triangle with ÐA = 90o and AB < AC. Let AD be the altitude from A on to BC. Let P, Q and I denote respectively the incentres of triangles ABD, ACD and ABC. Prove that AI is perpendicular to PQ and AI = PQ.
• Let n ³ 1 be an integer and consider the sum
• x=?_(k?0)?(?(n@2k))  2^(n-2k) 3^k=(?(n@0)) 2^n+(?(n@2)) 2^(n-2).3+(?(n@4)) 2^(n-4).3^2+?.
• Show that 2x-1,2x,2x+ 1 form the sides of a triangle whose area and inradius are also integers.
• (Number Theory) For any natural number n > 1, write the infinite decimal expansion of 1/n (for example, we write 1/2=0.49 ? as its infinite decimal expansion, not 0.5). Determine the length of the non-periodic part of the (infinite) decimal expansion of 1/n.
• Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number n ? 3, prove that there is a regular n-sided polygon all of whose vertices are blue.
• (Combinatorics) Written on a blackboard is the polynomial x^2+x +2014. Calvin and Hobbes take turns alternatively (starting with Calvin) in the following game. During his turn, Calvin should either increase or decrease the coefficient of x by 1. And during his turn, Hobbes should either increase or decrease the constant coefficient by 1. Calvin wins if at any point of time the polynomial on the blackboard at that instant has integer roots. Prove that Calvin has a winning strategy.
• (Polynomials) Let a,b,c,x,y,z be positive real numbers such that a+b+c = x+y+z and abc = xyz. Further, suppose that a ? x < y < z ? c and a < b < c. Prove that a = x,b = y and c = z.

Mock Test For INMO

Let A and B be the common points of two circles. A line passing through A intersects the circles at C and D. Let P and Q be the projections of B onto the tangents to the two circles at C and D. Prove that PQ is tangent to the circle of diameter AB.

Show that for each positive integer n,

n!=?_(i=1)^n?lcm{1,2,…,?n/i?}.

(Here lcm denotes the least common multiple, and ?x? denotes the greatest integer ?x.)

For each integer m, consider the polynomial

P_m (x)=x^4-(2m+4)x^2+?(m-2)?^2.

For what values of m is P_m (x) the product of two non-constant polynomials with integer coefficients?

Let f(x) be a polynomial with integer coefficients. Define a sequence a_0,a_1,… of integers such that a_0=0 and a_(n+1)=f(a_n) for all n ? 0. Prove that if there exists a positive integer m for which a_m  = 0 then either a_1=0 or a_2=0.

Given a regular n-gon and M a point in its interior, let x_1,x_2,...,x_n be the distances from M to the sides. Prove that

1/x_1 +1/x_2 +?+1/x_n >2?/a,where a is the side length of the polygon.

Prove that the number of triangles that can be formed with the sides of lengths a, b and c where a, b, c are integer such that a?b?c is 

(i)  (c(c+2))/4 if c is even (ii)  ?(c+1)?^2/4 if c is odd.

Solution to Mock test for Maths Olympiad

Let T be the intersections of the tangents at C and D.  The angles ÐABD and ÐADT are equal, since both are measured by half of the arc (AD) ?.  Similarly, the angles ÐABC and ÐACT are equal, since they are measured by half of the arc (AC) ?. This implies that 

ÐCBD = ÐABD + ÐABC = ÐADT + ÐACT = 180o - ÐCTD

where the last equality follows form the sum of the angles in triangles TCD. Hence the quadrilaterals TCBD is cyclic.

The quadrilateral TPBQ is also cyclic, since it has two opposite right angles. This implies that ÐPBQ = 180o - ÐCTD; thus ÐPBQ = ÐDBC as they both have ÐCTD as their supplement. Therefore, by subtracting ÐCBQ, we obtain ÐCBP = ÐQBD.

The quadrilateral BMCP and BMQD are cyclic, since ÐCMB = ÐCPB = ÐBQD = ÐDMB = 90o. Hence

ÐCMP = ÐCBP = ÐQBD = ÐQMD,

which shows that M lies on PQ. Moreover, in the cyclic quadrilateral QMBD,

ÐMBD = 180o ? ÐMQD = ÐQMD + ÐQDM = ÐQMD + ÐABD,

because ÐQDM and ÐABD are both measured by half of the arc (AD) ?. Since ÐMBD = ÐMBA + ÐABD, the above 

equality implies that ÐQMD = ÐMBA; hence MQ is tangent to the circle.

First solution: It is enough to show that for each prime p, the exponent of p in the prime factorization of both sides is the same. On the left side, it is well-known that the exponent of p in the prime factorization of n! is 

?_(i=1)^n??n/p^i   .?

(To see this, note that the i-th term counts the multiples of p^i among 1,...,n, so that a number divisible exactly by p^i gets counted exactly i times.) This number can be reinterpreted as the cardinality of the set S of points in the plane with positive integer coordinates lying on or under the curve y=np^(-x)  : namely, each summand is the number of points of S with x=i. 

On the right side, the exponent of p in the prime factorization of 1cm(1,...,?n/i?) is ?log_p ?n/i?? = ?log_p (n/i)?. However, this is precisely the number of points of S with y = i. Thus

?_(i=1)^n???log_p ?n/i??= ? ?_(i=1)^n??n/p^i ?   ,

and the desired result follows.

Second solution: We prove the result by induction on n, the case n = 1 being obvious. What we actually show is that going from n - 1 to n changes both sides by the same multiplicative factor, that is,

n=?_(i=1)^(n-1)?(lcm{1,2,…,?n/i?})/(Icm{1,2,…,?n-1/i?})

Note that the i-th term in the product is equal to 1 if n/i is not an integer, i.e., if n/i is not a divisor of n. It is also equal to 1 if n/i is a divisor of n but not a prime power, since any composite number divides the 1cm of all smaller numbers. However, if n/i is a power of p, then the i-th term is equal to p. 

Since n/i runs over all proper divisors of n, the product on the right side includes one factor of the prime p for each factor of p in the prime factorization of n. Thus the whole product is indeed equal to n, completing the induction.

By the quadratic formula, if P_m (x) = 0, then x^2=m± 2?2m+2, and hence the four roots of P_m are given by S={±?m±?2}. If P_m factors into two nonconstant polynomials over the integers, then some subset of S consisting of one or two elements form the roots of a polynomial with integer coefficients.

First suppose this subset has a single element, say ?m±?2; this element must be a rational number. Then ?(?m  ± ?2)?^2=2+m ± 2?2m is an integer, so m is twice a perfect square, say m=2n^2. But then ?m± ?2=(n±1)?2 is only rational if n = ±1, i.e., if m = 2.

Next, suppose that the subset contains two elements; then we can take it to be one of {?m±?2},{?2±?m} or {±(?m+?2)}. In all cases, the sum and the product of the elements of the subset must be a rational number. In the first case, this means 2?m?Q, so m is a perfect square. In the second case, we have 2?2?Q, contradiction. In the third case, we have ?(?m+?2)?^2?Q, or m+2+2?2m?Q, which means that m is twice a perfect square. We conclude that P_m (x) factors into two nonconstant polynomials over the integers if and only if m is either a square or twice a square.

Recall that if f(x) is a polynomial with integer coefficients, then m-n divides f(m)-f(n) for any integers m and n. In particular, if we put b_n=a_(n+1)-a_n, then b_n divides b_(n+1) for all n. On the other hand, we are given that a_0=a_m=0, which implies that a_1=a_m+1 and so b_0=b_m. If b_0=0, then a_0=a_1=?=a_m and we are done. Otherwise, |b_0 |=|b_1 |=|b_2 |=? , so b_n=±b_0 for all n. 

Now b_0+ • • •+b_(m-1)=a_m-a_0=0, so half of the integers b_0,...,b_(m-1) are positive and half are negative. In particular, there exists an integer 0 < k < m such that b_(k-1)=-b_k, which is to say, a_(k-1)= a_(k+1). From this it follows that a_n=a_(n+2) for all n ? k - 1; in particular, for m = n, we have

a_0=a_m=a_(m+2)=f(f(a_0))=a_2.

The area of the regular n-gon of side a is equal to (?na?^2)/(4 tan?/n). This area can also be written as the sum of the areas of the triangles formed by two consecutive vertices and the point M. Therefore,

a(x_1+x_2+?+x_n)=?na?^2/(2tan ?/n) .

We use two well-known inequalities:  tanx > x for x ? (0,?/2) and the arithmetic-harmonic mean inequality to obtain

(x_1+x_2+?+x_n )(1/x_1 +1/x_2 +?+1/x_n )?n^2.

We have

1/x_1 +1/x_2 +?+1/x_n =(1/x_1 +1/x_2 +?+1/x_n ) ?(x?_1+x_2+?+x_n)(2tan ?/n)/?na?^2  >2?/?na?^2  (1/x_1 +1/x_2 +?+1/x_n ) ?(x?_1+x_2+?+x_n)?2?/a  

Let c=2m+1 where m is a positive integer (i.e., when c is odd).

Since, c<a+b and a?b, we get

c<2b?b>1/2 c=m+1/2

Thus, b can take values from m+1 to 2m+1

If b= 2m+1, a can be 1,2,…,(2m+1). Thus, there are (2m+1) values.

If b = 2m, a can be 2,3,……,2m. In this case, a can take (2m-1) values and so on.

When b = m+1, a can take just one value viz. (m+1). Thus, there are (2m+1)+(2m-1)+?+1=?(m+1)?^2

= 1/4 (c+1)^2 Such triangles.

Now, let c=2m where m is a positive integer (i.e., when c is even)

When b=2m, a can be 1,2,……,2m

When b=2m-1, a can be 2,3,….,2m-1

When b=2m-2, a can be 3,4,……,2m-2 and so on.

When b = m+1, a can be m,m+1.

For b=m and any value ?m, for a will not work.

Thus, required number of triangles is 

2m+(2m-2)+?+2

= 2[1/2 m(m+1) ]=(c/2)(c/2+1)

= 1/4 c(c+2)