Indian National Math Olympiad 2013

1     Let \(\Gamma_1\) and \(\Gamma_2\) be two circles touching each other externally at R. Let \(O_1\) and \(O_2\) be the centres of \(\Gamma_1\) and \(\Gamma_2\), respectively. Let \(\ell_1\) be a line which is tangent to \(\Gamma_2\) at P and passing through \(O_1\), and let \(\ell_2\) be the line tangent to \(\Gamma_1\) at Q and passing through \(O_2\). Let \(K=\ell_1\cap \ell_2\). If KP=KQ then prove that the triangle PQR is equilateral.  

Sketch of the solution

2     Find all \(m,n\in\mathbb N\) and primes \(p\geq 5\) satisfying
\(m(4m^2+m+12)=3(p^n-1)\).    

Sketch of the solution

3     Let \(a,b,c,d \in \mathbb{N}\) such that \(a \ge b \ge c \ge d\). Show that the equation \(x^4 - ax^3 - bx^2 - cx -d = 0\) has no integer solution.   

Sketch of the solution

4     Let N be an integer greater than 1 and let \(T_n\) be the number of non empty subsets S of \(\{1,2,.....,n\}\) with the property that the average of the elements of S is an integer.Prove that \(T_n - n\) is always even.    

Sketch of the solution

5     In an acute triangle ABC, let O,G,H be its circumcentre, centroid and orthocenter. Let \(D\in BC, E\in CA\) and \(OD\perp BC, HE\perp CA\). Let F be the midpoint of AB. If the triangles ODC, HEA, GFB have the same area, find all the possible values of \(\angle C\).    

6     Let a,b,c,x,y,z be six positive real numbers satisfying x+y+z=a+b+c and xyz=abc. Further, suppose that \(a\leq x<y<z\leq c\) and a<b<c. Prove that a=x,b=y and c=z.

British Mathematics Olympiad (BMO) Round 1 2012

Regional Mathematics Olympiad Region 2 Questions

RMO 2012 solution to Question No. 6

6. Find all positive integers n such that \(3^{2n} + 3 n^2 + 7 \) is a perfect square.

Solution:

We use the fact that between square of two consecutive numbers there exist no perfect square. That is between \(k^2 \) and \((k+1)^2 \) there is no square.

Note that \(3^{2n} = (9^n)^2 \) and  \((9^n + 1)^2 \) are two consecutive perfect square and \(3^{2n} + 3 n^2 + 7 \) is always a number between them for n > 2 (easily proved by induction).

Hence the only solution is n = 2.

RMO 2012 solution to Question No. 5

5. Let ABC be a triangle. Let D, E be points on the segment BC such that BD = DE = EC. Let F be the mid point of AC. Let BF intersect AD in P and AE in Q respectively. Determine the ratio of triangle APQ to that of the quadrilateral PDEQ.

Solution:

Applying Menelaus' theorem to ΔBCF with AD as the transversal, we have
 \(\frac {BD}{DC} \frac{CA}{AF} \frac {FP}{PB} = 1 \)
But BD/DC = 1/2 (as BD = DE = EC) and CA/AF = 2/1 (as CF = FA).
Hence we have BP = PF.
Again applying Menelaus' Theorem to ΔBCF with AE as the transversal we have
\(\frac {BE}{EC} \frac{CA}{AF} \frac {FQ}{QB} = 1 \)
But BE/EC = 2/1 and CA/AF = 2/1
Hence 4FQ = QB.

Suppose FQ= x unit. The QB = 4x unit. That is BF = 5x unit. Since BP = PF hence each is 2.5x unit.
Thus PQ = 2.5x - x = 1.5x unit

Hence  \(\frac{ΔAPQ}{ΔABF} = \frac{1.5x}{5x}\)
Also \(\frac{ΔABF}{ΔABC} =\frac{1}{2}\)

Thus  \(\frac{ΔAPQ}{ΔABF} \frac{ΔABF}{ΔABC}= \frac{1.5x}{5x} \frac{1}{2}\)
=> \(\frac{ΔAPQ}{ΔABC} = \frac{1.5}{10}\) ...(1)

Again \(\frac{ΔADE}{ΔABC} = \frac{1}{3}\) (as DE/BC = 1/3)
Thus \(\frac{ΔADE}{ΔABC} - \frac{ΔAPQ}{ΔABC}  = \frac{1}{3} -  \frac{1.5}{10}\)
=> \(\frac{PQED}{ΔABC}  = \frac{5.5}{30}\) ...(2)

Using (1) and (2) we have 

 \(\frac{ΔAPQ}{PQED} = \frac{4.5}{5.5} = \frac{9}{11}\)

RMO 2012 solution to Question No. 4

4. Let X = {1, 2, 3, ... , 10}. Find the number of pairs {A, B} such that A ⊆ X, B ⊆ X, A ≠ B and A∩B = {5, 7, 8}.

Solution:

First we put 5, 7, 8 in each of A and B.

We are left out with 7 elements of X.

For each of these 7 elements there are three choices:

a) it goes to A

b) it goes to B

c) it goes to neither A nor B

Hence there are total \(3^7\) = 2187 choices. From these 2187 cases we delete that one case where all of the seven elements goes to neither A nor B as A≠ B thus giving 2187 -1 = 2186 cases.

Since A and B is unordered (that is A= {5, 7, 8, 1, 2} , B = {5, 7, 8, 4} is the same as B= {5, 7, 8, 1, 2} , A = {5, 7, 8, 4} ) we take half of these 2186 cases that is 1093 cases.

Hence there are 1093 such pairs.

RMO 2012 solution to Question No. 3

3. Let a and b are positive real numbers such that a+b = 1. Prove that \(a^a  b^b + a^b b^a \le 1\)

Solution:

We use the weighted A.M.-G.M. inequality which states that:

\(\frac {w_1 a_1 + w_2 a_2 }{w_1 + w_2} \ge ({a_1}^{w_1} {a_2}^{w_2})^{\frac{1}{w_1 + w_2}} \)

First we put \(w_1 = a , a_1 = a , w_2 = b, a_2 = b \)
Hence we get \(\frac {a a + b b }{a + b} \ge (a^a b^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(a^2 + b^2 \ge (a^a b^b) \) ....(1)

Similarly we put \(w_1 = a , a_1 = b , w_2 = b, a_2 = a \)

Hence we get \(\frac {a b + b a }{a + b} \ge (b^a a^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(2ab \ge (b^a a^b) \) ....(2)

Adding (1) and (2) we have 

\(a^2 + b^2 + 2ab \ge a^a b^b + b^a a^b \)
=> \((a+b)^2 \ge a^a b^b + b^a a^b \)
As a+b =1 we have the desired inequality
\(1 \ge a^a b^b + b^a a^b \).

RMO 2012 Solution to Question No. 2

2. Let a, b, c be positive integers such that a divides \(b^5\) , b divides \(c^5\) and c divides \(a^5\). Prove that abc divides \((a+b+c)^{31}\).

Solution:

A general term of the expansion of \((a+b+c)^{31}\) is \(\frac {31!}{p!q!r!} a^p b^q c^r\) where p+q+r = 31 (by multinomial theorem; this may reasoned as following: from 31 factors (a+b+c), choose p factors and from those chosen p factors take out 'a'. From remaining 31-p factors choose q factors and from these chosen q factors take out 'b'. From the remaining r factors take out 'c'.)

Now the terms of the expansion can be of three types:

Case 1

p, q and r are all non-zero. These terms are straight away divisible by abc as all of a, b, and c are present in them.

Case 2

Exactly one of p, q, r is zero and rest two are non-zero. Let us examine the subcase where r=0 and p,q are non zero. Other two subcases will be similar.

Then p+q+0 = 31 or p+q=31

Term of the expansion will have :

\(a^p b^q = ab(a^{p-1} b^{q-1})\)

We will show that \(a^{p-1} b^{q-1}\) is divisible by c where p+q=31

Suppose \((p-1)\ge 5 , (q-1)\ge 5\) then c divides \(a^{p-1}\) as it contains \(a^5\)  and by problem c divides \(a^5\)

Again if p-1 < 5 then \((q-1) \ge 25 \) as p-1 + q-1 = 29 (as p+q = 31)

Now a divides \(b^5 \) or \(a^5\) divides \(b^{25}\). As c divides \(a^5\) and \(a^5\) divides \(b^{25}\) hence c divides \(b^{25}\) implying c divides \(b^{q-1}\) as \((q-1)\ge 25 \)

Case 3.

Exactly two of p, q and r are zero. Let us again examine of the three subcases where q=0, r=0 and p nonzero. Other two subcases will be similar.

Then p = 31.

\(a^{31} = a\times a^5 \times a^{25}\). c divides \(a^5\) and b divides \(c^5\) which divides \(a^{25}\).

Hence we have checked all possible terms and have shown than abc divides each of them.

RMO 2012 solution to Question No. 1

1. Let ABCD be a unit square. Draw a quadrant of a circle with A as the center and B, D as the end points of the arc. Similarly draw a quadrant of a circle with B as the center and A, C as the end points of the arc. Inscribe a circle Γ touching the arcs AC and BD both externally and also touching the side CD.

Solution:

Suppose O is the center of Γ. The point of tangency of Γ with CD be P, with arc AC be R and with BD be Q. 

By symmetry DP = 1/2 unit.

Also note that AQO is a straight line. (If we draw a tangent line through Q, AQ will be perpendicular to the tangent line as A is the center of quadrant ABD. Also OQ will be perpendicular to the tangent line as O is center of Γ. OQ, AQ are perpendicular to same line - the tangent line - at the same point (Q). Hence they are the same straight line).

Length of AQO = AQ + QO = 1 + r (AQ = AB = 1 and let the radius of Γ be r)

Drop a perpendicular OM on AD.

Then OM = DP = 1/2

AM = 1-r.

Applying Pythagoras' Theorem on triangle AQM we have

\( (1-r)^2 + (\frac{1}{2})^2 = (1+r)^2 \)

=> r = 1/16.

Answer: Radius of smaller circle is \(\frac{1}{16}\) unit.

Regional Mathematics Olympiad (RMO) 2012

Complex Numbers versus Projective Geometry - One problem, Two solutions

The Problem

Suppose ABC is any triangle. D, E, F are points on BC, CA, AB respectively such that \(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB}\). Prove that the centroids of triangles ABC and DEF coincide.

A little Complex Number

Let A, B, C be points on the Complex plane with complex coordinates a, b, c.

The origin of the plane is made to coincide with the centroid of triangle ABC (the complex coordinates of the vertices adjusted accordingly).

Hence the complex coordinate of the centroid G is 0.

\(G = \frac{a+b+c}{3} = 0\) implying a+b+c= 0.

Next we find the complex coordinates of D, E, F in terms of a, b, c and the given ratio \(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = k\) (say)

Using the section formula we find that

\(d = \frac{ck+b}{k+1}\)
\(e = \frac{ak+c}{k+1}\)
\(f = \frac{bk+a}{k+1}\)

where d, e, f are the complex coordinates of D, E, F respectively.

Hence the centroid of triangle DEF is given by \(\frac{d+e+f}{3} = \frac{\frac{ck+b}{k+1} + \frac{ak+c}{k+1} + \frac{bk+a}{k+1} } {3} = 0\) using (a+b+c) = 0.

Thus the centroid of triangle ABC and DEF are same.

A little Projective Geometry

We project triangle ABC into a plane T such that the projection is an equilateral triangle.

(to be continued)


 

Is it a prime number?

353 is a prime number. So is 7919 (in fact it is the 1000th prime). There are 25 primes between 1 and 100. From 1 to 1000 there are 168 of them.

It is difficult to check whether a number is prime or not. One simple method is to try and divide the number with smaller numbers. For example to find out whether 351 is a prime or not, you start checking by dividing the number by small numbers such as 2. Of course, it is an odd number so we immediately know that it is not divisible by 2. Next we try by 3 and it works! Indeed 117 times 3 is 351.

But suppose we working with 899. To check if it is a prime or not we start dividing it small numbers. 2 will not divide as the number is odd. But 3 will not divide either. If we keep on checking, we will see that up till 28 no number divides it (899 is 29 times 31).

How far should we check to be sure whether a number is prime or not? Suppose n is the number. Then we should try and divide 'n' with all prime numbers up till \( \sqrt n \). Indeed if n is not a prime number then it must have a divisor smaller than or equal to \( \sqrt n \) apart from 1. This fact has a simple proof.

Suppose n is not prime. Then it is composite (that is it has a divisor apart from 1 and it self).
Let p be a divisor of n. then \( q = \frac {n}{p}\) is also a divisor of n.
That is n = pq
Now we claim that one of p or q is smaller than or equal to \( \sqrt n \).
If our claim is false then both of them are greater than \( \sqrt n \).
p > \( \sqrt n \) ; q > \( \sqrt n \)
This implies \(p \times q > \sqrt n \times \sqrt n \ = n \).
But p*q is not greater than n.
Hence our claim is correct.

Therefore when we have a number to check whether it is a prime or not, it is sufficient to try and divide it with all the prime numbers smaller than or equal to \( \sqrt n \) (since if it is composite then it will have a divisor which smaller than or equal to \( \sqrt n \) and this divisor is either itself prime or is divisible by a still smaller prime number).

Hence to check whether 641 is prime or not it is sufficient to check with all primes smaller than \( \sqrt {641} \) ~ 25. That is we try to divide it with 2, 3, 5, 7, 11, 13, 17, 19, 23. None of them divides 641. Thus we conclude that it must be a prime.

Regional Math Olympiad Camp in India

An application of Menalaus' theorem

Given: AB is the diameter of a circle with center O. C be any point on the circle. OC. is joined. Let Q be the midpoint of OC. AQ produced meet the circle at E. CD be perpendicular to diameter AB. ED and CB are joined.

R.T.P. : CM = MB

Construction: AC and BD joined.

Proof: In triangle BOC, AQF is the transversal. Applying Menalaus' Theorem we have CF/FB = 1/2.
Now ΔABC is similar to ΔBDH and ΔACF is similar to ΔDGH (right angles equal and angle subtended by the same segment equal in both cases).
Hence HG/GB = CF/FB. But CF/FB =1/2. Hence HG/GB = 1/2.
Now applying Menalaus' Theorem in ΔBCH with transversal DGM we have CM/MB=1.

Q.E.D.

24th June at Cheenta

We have three events coming up today.

1. The 2013 graduating batch will attend class at 7:30 AM
2. The 2014 graduating batch will attend class at 11 AM
3. Mock interview session for Indian Statistical Institute B.Math aspirants

The first class will begin in an hour. We will focus on two themes:

• Geometry - Stewart's Theorem and an introduction to mass point geometry
  
• Complex Numbers - Elementary inequalities involving complex numbers and further discussion on their geometry

The two problem sheets on geometry and two problem sheets on complex numbers will aid the discussion. We will resume the two-test structure from today. That is to say the class will begin with a d-test (diagnosis test of 15 minutes) and end with an e-test (effect test of 15 minutes).

Rittinkar did the maximum amount of homework last week. He will be awarded with a book. I like his diligence very much. Even Krishnendu has worked hard (he submitted some INMO problem solutions yesterday).

The discussion on geometry will introduce mass-point technique (the lever of Archimedes :)). In the long run I want to link barycenter and complex number application. In fact the discussion on complex numbers is mainly motivated from geometry of Vector Algebra. We have considered some problems on geometry in the last class (and applied complex numbers to solve them). We want to continue the journey.

The second class is equally interesting.

We will again focus on two themes:

• Inequalities
• Differential Calculus

The concept of inequality and calculus are intertwined historically. In fact, quite a few calculus books begin from this perspective. The one I like most however attacks limiting processes through the discussion on sequences (the book is Pre-calculus by Tarasov which happens to be our prize book too).

We have already covered some grounds in inequality, We are freely talking about Arithmetic and geometric means, Cauchy Schwarz, Jensen's inequality etc. In today's class we want to focus on some problem solving (from problem sheet 1 and 2).

The discussion on differentiation, continuity etc. will follow. We will discuss Rolle's Theorem, Lagrange's Mean Value Theorem (with geometric intuition). Presently we are following I.A. Maron's Calculus in One Variable. The Art and Craft of Problem Solving by Paul Zeitz is also a nice book (it has a section on calculus).

The Mock Interview Session will follow the classes.

Subhojit and Shalmali will attend the session. I have also invited Abhra Abir. There will be board-work, problems and interactions. I am looking forward meet them one more time.

C.M.I. ENTRANCE 2012

CHENNAI MATHEMATICAL INSTITUTE
B.SC. MATH ENTRANCE 2012

ANSWER FIVE 6 MARK QUESTIONS AND SEVEN OUT 10 MARK QUESTIONS.

6 mark questions

1. Find the number of real solutions of x = 99 sin \(\pi \) x
2. Find \(\displaystyle\lim_{x\to\infty}\dfrac{x^{100} ln(x)}{e^x tan^{-1}(\frac{\pi}{3} + sin x)}\)
3. (part A)Suppose there are k students and n identical chocolates. The chocolates are to be distributed one by one to the students (with each student having equal probability of receiving each chocolate). Find the probability of a particular student getting at least one chocolate.
   (part B) Suppose the number of ways of distributing the k chocolates to n students be \(\dbinom{n+k-1}{k}\). Find the probability of a particular student getting at least one chocolate.
4. Show that \(\dfrac{ln 12}{ln 18}\) is an irrational number.
5. Give an example of a polynomial with real coefficients such that \(P(\sqrt{2} + i)=0\). Further given an example of a polynomial with rational coefficients such that \(P(\sqrt{2} + i)=0\).
6. Say f(1) = 2; f(2) = 3, f(3) = 1; then show that f'(x) = 0 for some x (given that f is a continuously differentiable function defined on all real numbers).

10 mark questions

1. (part A) Suppose a plane has 2n points; n red points and n blue points. One blue point and one red point is joined by a line segment. Like this n line segments are drawn by pairing a red and a blue point. Prove that each such scheme of pairing segments will have two segments which do not intersect each other.
   (part B) Suppose the position of the n red points are given. Prove that we can put n blue points in such a way that there are two segments (produced in the manner described in part A) which do not intersect each other.
2. (part A) Let ABCD be any quadrilateral. E, F, G and H be the mid points of the sides AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram whose area is half of the quadrilateral ABCD.
   (part B) Suppose the coordinates of E, F, G, H are given:  E (0,0) , F(0, -1), G (1, -1) , H (1, 0). Find all points A in the first quadrant such that E, F, G, H be the midpoints of quadrilateral ABCD.
3. Let f be a function whose domain and co-domain be non negative natural numbers such that f(f(f(n))) < f(n+1). Prove that:
   (a) If f(n) = 0 then n = 0.
   (b) f(n) < n+1
   (c) If f(x) < n then x<n
   Using the above prove that f is an identity function, that is f(n) = n.
4. Consider a sequence \(c_{n+2} = a c_{n+1} + b c_n\) for \(n \ge 0\) where \(c_0 = 0\). If gcd(b, k) = 1 then show that k divides n for infinitely many n.
5. Find out the value of \(x^{2012} + \dfrac{1}{x^{2012}}\) when \(x + \dfrac{1}{x} = \dfrac{\sqrt{5} + 1}{2}\).
   Hint
   (a) Show that \(|{r + \dfrac{1}{r}}|\ge 2\) for all real r.
   (b) Prove that \(\sin \dfrac{\pi}{5} < \cos \dfrac{2\pi}{5} < \sin \dfrac{3\pi}{5}\).
6. A polynomial P(x) takes values \(prime^{positive number}\) for every positive integer n,then show that p(x) is a constant polynomial.
   If such a polynomial exist then show that there also exists a polynomial g(x)= \(prime^l\) where l is a fixed number.
7. Consider a set A = {1, 2, ... , n}. Suppose \(A_1 , A_2 , ... , A_k \) be subsets of set A such that any two of them consists at least one common element. Show that the greatest value of k is \(2^{n-1}\). Further, show that if they any two of them have a common element but intersection of all of them is a null set then the greatest value of k is \(2^{n-1}\).
8. Suppose \(\displaystyle x = \sum_{i=1}^{10} \dfrac{1}{10 \sqrt 3} . \dfrac{1}{1+ (\dfrac{i}{10 \sqrt 3})^2}\) and \(\displaystyle y = \sum_{i=0}^9 \dfrac{1}{10 \sqrt 3} . \dfrac{1}{1+ (\dfrac{i}{10 \sqrt 3})^2}\)
1.  Show that \(x < \dfrac{\pi}{6} < y \) 
2. \(\dfrac{x+y}{2} < \dfrac{\pi}{6} \)

[b]OTHERS PLEASE CONTRIBUTE THE REST OF THE QUESTIONS (AND SOLUTIONS). WE ARE TRYING ON OUR END TO DO THE SAME[/b]

Solutions to I.S.I. 2012 Subjective (B.Stat, B.Math)

Q7. Consider two circles with radii a, and b and centers at (b, 0), (a, 0) respectively with b<a. Let the crescent shaped region M has a third circle which at any position is tangential to both the inner circle and the outer circle. Find the locus of center c of the third circle as it traverses through the region M (remaining tangential to both the circle.


Discussion:

Join AC and BC. AC passes through, \(T_1\) the point of tangency of the smaller circle with the circle with center at (a, 0) and BC when extended touches \(T_2\) which is the other point of the tangency.

Assume the radius of the moving (and growing circle) to be r at a particular instance. Then AC = a+r and BC = b-r.

Then AC+BC = a+b which is a constant for any position of C. Hence C is a point whose some of distances from two fixed points at any instant is a constant. This is the locus definition of an ellipse with foci at (a, 0) and (b, 0).

Q8. Let S = {1, 2, ... , n}. Let \(f_1 , f_2 , ... \)  be functions from S to S (one-one and onto). For any function f, call D, subset of S, to be invariant if for all x in D, f(x)  is also in D. Note that for any function the null set and the entire set are 'invariant' sets. Let deg(f) be the number of invariant subsets for a function.
 a) Prove that there exists a function with deg(f)=2.
 b) For a particular value of k prove that there exist a function with deg(f) = \(2^k\)


Discussion:


(a)

Consider the function defined piecewise as f(x) = x - 1 is \(x \ne 1\) and f(x) = n if x = 1

Of course null set and the entire sets are invariant subsets. We prove that there are no other invariant subsets.

Suppose D =  {\(a_1 , a_2 , ... , a_k \)} be an invariant subset with at least one element.

Since we are working with natural numbers only, it is possible to arrange the elements in ascending order (there is a least element by well ordering principle).

Suppose after rearrangement D = {\(b_1 , b_2 , ... , b_k \)} where \(b_1\) is the least element of the set

If \(b_1 \ne 1\) then \(f(b_1) = b_1 -1\) is not inside D as \(b_1\) is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.

This \(b_1\) must equal to 1.

As D is invariant subset \(f(b_1) = n \) must belong to D. Again f(n) = n-1 is also in D and so on. Thus all the elements from 1 to n are in D making D=S.

Hence we have proved that degree of this function is 2.

(b)

For a natural number 'k' to find a function with deg(f) = \(2^k\) define the function piecewise as

f(x) = x for \(1\le x \le k-1\)
      = n for x=k
      = x-1 for the rest of elements in 'n'

To construct an invariant subset the 'k-1' elements which are identically mapped, and the entirety of the 'k to n' elements considered as a unit must be considered. Thus there are total k-1 + 1 elements with which subsets are to be constructed. There are \(2^k\) subsets possible.

I.S.I. B.Stat and B.Math Entrance 2012 Subjective Paper

USAJMO 2012 questions

1. Given a triangle ABC, let P and Q be the points on the segments AB and AC, respectively such that AP = AQ. Let S and R be distinct points on segment BC such that S lies between B and R, ∠BPS = ∠PRS, and ∠CQR = ∠QSR. Prove that P, Q, R and S are concyclic (in other words these four points lie on a circle).
2. Find all integers \(n \ge 3 \) such that among any n positive real numbers \( a_1 , a_2 , ... , a_n \) with
   max \((a_1 , a_2 , ... , a_n) \le n\) min \(( a_1 , a_2 , ... , a_n)\) there exist three that are the side lengths of an acute triangle.
3. Let a, b, c be positive real numbers. Prove that \(\frac{a^3 + 3 b^3}{5a + b} + \frac{b^3 + 3c^3}{5b +c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2)\).
4. Let \(\alpha\) be an irrational number with \(0 < \alpha < 1\), and draw a circle in the plane whose circumference has length 1. Given any integer \(n \ge 3 \), define a sequence of points \(P_1 , P_2 , ... , P_n \) as follows. First select any point \(P_1\) on the circle, and for \( 2 \le k \le n \) define \(P_k\) as the point on the circle for which the length of the arc \(P_{k-1} P_k\) is \(\alpha\), when travelling counterclockwise around the circle from \(P_{k-1} \) to \(P_k\). Suppose that \(P_a\) and \(P_b\) are the nearest adjacent points on either side of \(P_n\). Prove that \(a+b \le n\).
5. For distinct positive integers a, b < 2012, define f(a, b) to be the number of integers k with \(1\le k < 2012\) such that the remainder when ak divided by 2012 is greater than that of bk divided by 2012. Let S be the minimum value of f(a, b), where a and b range over all pairs of distinct positive integers less than 2012. Determine S.
6. Let P be a point in the plane of triangle ABC, and \(\gamma\) be a line passing through P. Let A', B', C'  be the points where reflections of the lines PA, PB, PC with respect to \(\gamma\) intersect lines BC, AC, AB, respectively. Prove that A', B' and C' are collinear.

I.S.I. Entrance Model Test at Cheenta Ganit Kendra