Download pdf - Issue 1

Transcript
Page 1: Issue 1

THIS IS A CIRCLE!Curious?

Page 2: Issue 1

1 CHEENTA GANIT KENDRA

Reason, Debate and Story

য�ক্ত�, তক্কো আর গপ আর গক্কো আর গপ�Mostly mathematics...

sprinkled with a little bit of computer science and physics

May, 2012 Year 1, Issue 1

Ashani DasguptaEditorCalcutta, IndiaPhone: +91 9804005499

Sudip MondalCo-EditorSanta Clara, CAPhone: +001 4084767585

Arkabrata DasCo-EditorCalcutta, IndiaPhone: +91 8984655870

Content DevelopmentPratyush Kumar MishraNiran Meher

Cheenta Ganit Kendra, center of mathematics, strives to bring out the mathematical talent in middle school and high school students. Math Olympiads and challenging Math Entrances of Indian Statistical Institute, Chennai Mathematical Institute, Institute of Mathematics and Applications (in India) and STEP or Sixth Term Examination Paper (in Cambridge, UK) act as motivation of our course work. Our offline activities (classes, summer camps etc.) are mostly implemented in Calcutta (India) and San Francisco, Bay Area (USA). We try to reach out to as many as possible through our online courses, online math forum and blog and pages in facebook and twitter.

REACH US

E-Mail: [email protected]: B-37, C.I.T. Buildings,

Calcutta 700007, West Bengal, India

Page 3: Issue 1

2 CHEENTA GANIT KENDRA

A few words“No mathematician should ever allow himself to forget that mathematics, more than any other art or science, is a young man's game.”G.H. Hardy, A Mathematician's Apology

'Reason, Debate and Story' is a journal of mathematics. High School level math is sufficient for accessing most of it's contents. A considerable portion of the journal require much less sophistication in formal math. However, that does not mean the mathematics presented here is 'easy'. Elementary ideas can actually motivate deceivingly simple looking challenging problems.

Cheenta Ganit Kendra (literally meaning 'center of mathematical thoughts' in Bengali) engages with bright mathematically inclined students every year in and outside classroom. We have math olympiad courses and other advanced mathematics programs for high school and middle school students. Young men and women join us from diverse backgrounds and nationality (we have online as well as offline courses). Although our work is mostly concentrated around Calcutta, we are always looking out to communicate with mathematically inclined students; currently, we are working with students from USA too.

While engaging with the students, we have realized that there is a serious lack of popular math literature in India. We do have more than one widely distributed math magazines here; unfortunately they mostly publish questions and solutions from engineering entrances.

Journal on school level mathematics can generate considerable amount of interest among young students. The celebrated journals like Komal (Hungary) and Kvant (Russia) have been extremely successful in motivating innovation in the young mind. The other magazines of similar scope include Crux Mathematicoram (Canadian Mathematical Society), Plus Magazine (University of Cambridge), Die Wurzel (Germany) etc. We are inspired by these initiatives.

Apart from publishing interesting and motivating articles on mathematics (and allied sciences) , problems and discussions from challenging math contests, we wish to use this journal as a vehicle of communication with our student community and evolve it as their first step towards mathematical exploration.

We are sharing the beauty of truth through mathematics. The motivation is to appreciate the Eureka moments of knowledge and try and experience some of them.

Ashani DasguptaEditor

May, 2012

Page 4: Issue 1

3 CHEENTA GANIT KENDRA

Cheenta Awardees1. Krishnendu (Class XI) for

excellent performance in I.S.I. Entrance Model Test 10 .The story of Astronomy (Lloyd Motz )

2. Abhra Abir (Class XII, external canditate) for scoring maximum points in I.S.I. Entrance Model Test 9.Feynman Lectures on Physics, Volume I (Richard Feynman).

3. Shobhan (Class XII) for being the most interactive student of the week.One two three Infinity (George Gamow).

4. Shubhojit (High School Graduate) for scoring maximum points in I.S.I. Entrance Model Test 10. Feynman Lectures on Physics, Vol I (Richard Feynman)

______________

ProblemsINDIAN NATIONAL MATH OLYMPIAD 2012 (INMO 2012)

1. GEOMETRYLet ABCD be a quadrilateral inscribed in a circle. Suppose AB = √2+√2 and it

subtends an angle of 1350 at the center of the circle. Find the maximum possible area of ABCD.

2. NUMBER THEORYLet p1<p2< p3< p4 and q1<q2<q3<q4 be two sets of prime numbers such that p4 − p1=8 and q4−q1=8 . Suppose p1>5 and q1>5 . Prove that 30 divides p1−q1.

3. FUNCTIONAL EQUATIONDefine a sequence f⟨ n(x) n⟩ ∈ℕ0 of functions as f0(x )=1, f1(x )=x, (f n(x))2 -1= f n−1(x ) f n+1( x) , for n≥1 . Prove that each f n(x ) is a polynomial with integer coefficients.

4. COMBINATORICSLet ABC be a triangle. An interior point P of ABC is said to be good if we can find exactly 27 rays emanating from P intersecting the sides of the triangle ABC such that the triangle is divided by these rays into 27 smaller triangles of equal area. Determine the number of good points for a given triangle ABC.

5. GEOMETRYLet ABC be an acute angled triangle. Let D, E, F be points on BC, CA, AB such that AD is the median, BE is the internal bisector and CF is the altitude. Suppose that ∠FDE = C and∠ DEF = A and EFD = B. Show that ∠ ∠ ∠ ∠ ABC is

equilateral

Page 5: Issue 1

4 CHEENTA GANIT KENDRA

6. FUNCTIONAL EQUATIONLet f : be a function satisfying f (0)≠0 , fℤ→ℤ (1)=0 and1. f(xy) + f(x)f(y) = f(x) + f(y)2. (f(x-y) – f(0)) f(x)f(y) = 0for all x , y simultaneously.∈ℤ1. Find the set of all possible values of the function f.2. If f (10)≠0 and f(2) = 0, find the set of all integers n such that f (n)≠0 .

HINTS FOR INMO 2012

1. GEOMETRY1ST HINT: Find out radius by applying cosine rule in triangle AOB where O is the center. The radius is computed to be 1 unit. Apply sine rule of area1 to find the areas of rest of the 3 triangles. Apply Jensen's Inequality2.

2ND HINT:

2. NUMBER THEORY1ST HINT: First check for the possible remainders of p1 when divided by 6 and conclude that p1 must be of the form 6k – 1.

1 See Trigonometry section for sine rule of the area of a triangle.

2 See Algebra section for Jensen's Inequality

Note that any prime is of the form 6k + 1 or 6k -1 (the converse is not true).

2ND HINT:

3. FUNCTIONAL EQUATION1ST HINT: Compute the first few terms to 'search' for a pattern. In the given recurrence relation we do not have fn(x) in an explicit form.

2ND HINT:

4. COMBINATORICSANS: 3251ST HINT: First we note that 3 of the 27 rays join P and the vertices of the triangle (otherwise we do not have small triangles; instead we have quadrilaterals near the vertices). Rest of the 24 rays 'impinge' on the three sides. For every position of P there is exactly one type of impingement.3

3 See Combinatorics section for a formula to count non-integer solutions

Area of each of the other triangle comes out to be ½sin x1 ,½sin x2 etc. where x1, x2, x3 are the angles AOD, COD,

BOC. Apply Jensen's inequality to maximize the sum sinx1 + sin x2 + sin x3

Since p1 and q1 are both of the form 6k – 1 their difference

is divisible by 6. Check for the possible last digits of k and k' where p1 =6k -1 and q1=6k' – 1 and conclude that they

are 2 or 7.

Check till n = 5 and 'guess' that fn(x) = xfn-1(x) – fn-2(x). Complete the proof by

induction.

Page 6: Issue 1

5 CHEENTA GANIT KENDRA

2ND HINT:

5. GEOMETRY1ST HINT: Use property of nine point circle4 to conclude that D and F are on that circle. The circle passing through DEF could be the nine point circle provided it's radius should be equal to the radius of the nine point circle which happens to be half the circum radius of a triangle. Also it should cut or touch AC at at least one point in order to be the nine point circle (which it does).

2ND HINT:

6. FUNCTIONAL EQUATION1ST HINT: Put x=y=0 to conclude f(0) = 1. By assuming f(x) = 1 and replacing y by m conclude that if f(x) = 1 so is f(mx) for all integer m. Use Bezout's Theorem5 to establish the case of f(x) = 1.

4 See Geometry for nine point circle theorem5 See Number Theory section for Bezout's Theorem.

2ND HINT

REFERENCE POINTS OF THE HINT-SECTION

Trigonometry

Area of a triangle = 12

a bsin C where A, B, C

denote the three angles and a, b, c denote the sides opposite to the three angles respectively. You may remember this as half of the product of (any two sides)

and (sine of the angle between them).

Algebra

Jensen's Inequality

Let f be a convex function on an interval I and let w1, ... , wn, be n be non negative real numbers whose sum be 1. Then for all

x1 , ... , x n∈ I , w1 f (x1)+w2 f ( x2)+...+wn f (xn)≥ f (w1 x1+...+wn xn)

Combinatorics

The number of non-negative integer solutions

As postion of P determines how many rays fall on each side on the triangle on the other hand the number of rays falling on each side uniquely determines the position of P. Say x, y, z number of rays fall on AB, BC, CA respectively. Then each nonnegative integer solution of x+y+z=24 determines exactly one 'good point'.

BD=DC=DF as they are the radii of triangle BFC (as angle BFC is right and D is the midpoint of hypotenuse making D the center of the circumcircle of BFC). Use sine rule to establish radius of circumcircle of DEF equals half of the radius of circumcircle of ABC. Hence E is either foot of perpendicular or midpoint of AC. Take cases.

Put y = 0 to conclude that if f(x) is not 1 it must be 0. Use f(10) not equal to 0 and f(2) equal to 0 to conclude f(5) = 1 and hence f(5m)=1 for all integer m. Use Bezout's theorem to prove that these are the only numbers for which f(x) = 1.

Page 7: Issue 1

6 CHEENTA GANIT KENDRA

to the equation x1+x2+x3+...+xn=m is

(m+n−1m ) or

(m+n−1)!m!(n−1) ! .

Geometry

The nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant points defined from the triangle. These nine points are:

• The midpoint of each side of the triangle

• The foot of each altitude• The midpoint of the line segment

from each vertex of the triangle to the orthocenter (where the three altitudes meet; ).

The nine-point circle is also known as Feuerbach's circle, Euler's circle, Terquem's circle, the six-points circle, the twelve-points circle, the n-point circle, the medioscribed circle, the mid circle or the circum-midcircle.

Number Theory

Bezout's Theorem

In number theory, Bézout's identity for two integers a, b is an expression: ax+by=d , where x and y are integers (called Bézout coefficients for (a,b)), such that d is a common divisor of a and b. Bézout's lemma states that such coefficients exist for every pair of

nonzero integers (a,b). In addition d > 0 is their greatest common divisor and the smallest positive integer that can be written, in this form, for any integers x,y. This value of d is therefore uniquely determined by a and b, but the Bézout coefficients are not unique.

Computer scienceAt Cheenta, we are trying to integrate concepts from computer science into our early bird (middle school) math Olympiad course. This article is prepared by Sudip Mondal. It forms the basis of our effort in computer science.

What is a computerSudip Mondal

It is a “system” which has the following important properties:

a) It can accept inputs.b) It can deliver outputs.c) It can follow instructions to “generate”

the outputs from the inputs.

What are inputs?

Anything that can be measured and defined unambiguously (A piece of text can be measured; by its number of words, numerical value of each word, position of each word in the text, etc etc).

Page 8: Issue 1

7 CHEENTA GANIT KENDRA

What are outputs?

Anything that can be measured and defined unambiguously. That’s same as inputs! Yes, it is. We use computers to “convert” inputs to outputs.

Think: The instructions which the computer follows to “convert” the inputs to the outputs are also inputs to the computers.

Example 1:

Computer = YouInput 1: Look up your math book to find an integer. (This is an instruction).Input 2: An integer P. (this is the actual integer you looked up from the book).Input 3: Look up your math book to find an integer. (This is another instruction).Input 4: An integer Q. (this is the second integer you looked up from the book).Input 5: Add them (this is another instruction).Input 6: Write the sum into your exercise copy (yet another instruction)Output 1: Sum (This is the sum that you wrote in the exercise copy).Input 7: Take 5. (This is an input, with a fixed value).Input 8: Add that to the sum (Another instruction).Input 9: Write the result into the exercise copy (another instruction).Output 2: Sum + 5 (The result you wrote in the exercise copy).

Example 2:

Computer = YouInput 1: Look up your math book to find an integer. (This is an instruction).Input 2: An integer P. (this is the actual integer you looked up from the book).Input 3: Look up your math book to find an integer. (This is another instruction).Input 4: An integer Q. (this is the second integer you looked up from the book).Input 5: Find GCD of P and Q (this is another instruction)Input 6: Write the GCD into your exercise copy (yet another instruction)Output 1: GCD (This is the GCD that you wrote in the exercise copy).So you see that in a sense, input consists of both data and instructions. Same data and different instructions will generally result in different outputs; so will same instructions and different data.

What does a computer need to have?

1. It must have a way to accept inputs (A blind person cannot look up an integer in the math book).

2. It must have a memory (If you forget the integers after looking them up, you cannot add them).

3. It must “KNOW” how to add, or find the GCD (else, it simply cannot carry out the instruction). Actually, it must also “KNOW” how to look up an integer in a math book. Try to make a 3 year old “look up an integer in a math book”.

Page 9: Issue 1

8 CHEENTA GANIT KENDRA

Believe it or not, computers “KNOW” nothing. Computer designers design electronic circuits, which “can” do certain things. It means those electronic circuits are bound to do those things according to the laws of physics, they cannot choose to do anything else. Think: If I throw a stone at a window pane with enough force, it will hit the pane and will break it. Now think: A stone thrown with enough force “can” break a window pane.Actually, the stone has no choice but to break the window pane. Similarly, electronic circuits can be designed which have no choice but to add the two numbers given to them. Now, let us draw a computer with the blocks that we discussed. It will look like something in Figure 1.

INPUT

OUTPUT

MEMORY

CPU

PROCESS

Figure 1: A computer with its partsINPUT: This is the place where all the inputs exist. Think of the inputs in Example 1.

Someone must have prepared those inputs and kept it somewhere (like written in a piece of paper) for you to see. The box INPUT in the figure is that piece of paper. Only, for a computer, the piece of paper must be a bunch of electronic circuits storing the information. It may be a keyboard with instructions being given with keystrokes, or may be a USB stick containing the instructions, or may be a hard disk, CDROM (long list).

What is the big block called PROCESS with the smaller blocks MEMORY and CPU inside? Well, this is the guy in example 1, who was actually following all those instructions; thus it is you. Now we have already explained MEMORY; you need it to remember the numbers you looked up (maybe this is another piece of paper where you write down the two integers you looked up from your maths book). But what is this CPU? Well, this is a bunch of those electronic circuits, which can be forced to add two numbers etc. This is also the guy, which actually follows the instructions given to it. Thus, CPU is the one which adds the two numbers or computes their GCD, or even look up a place called “Maths Book” to get an integer.

Finally, OUTPUT is the place where the result is shown. This is the exercise copy in the examples. It may be a printer or a monitor or a USB stick or a Hard Disk or anything (another long list) where the result can be written/dumped in some electronic form.

Now let us think about this block called

Page 10: Issue 1

9 CHEENTA GANIT KENDRA

PROCESS for a while. Someone prepared an INPUT (a list of instructions and data) and stored it in the box called INPUT. If the PROCESS block has to process these inputs, it has to be able to connect with the INPUT block. (You cannot follow instructions unless the instruction booklet is given to you). So, we need to establish a connection between the INPUT and the PROCESS block. How do we establish connection? We use something called a BUS. Similarly we need a BUS to connect to the output. Let us add these two buses to the figure.

INPUT

OUTPUT

MEMORY

CPU

PROCESS

BUS 1 BUS 2

Figure 2: Better picture of a computer

So, now we have a way for the PROCESS block to access the INPUTs and to write to the OUTPUTs. Now, for a moment, think about Example 1 again. It is repeated below:

Computer = YouInput 1: Look up your math book to find an integer. (This is an instruction).Input 2: An integer P. (this is the actual

integer you looked up from the book).Input 3: Look up your math book to find an integer. (This is another instruction).Input 4: An integer Q. (this is the second integer you looked up from the book).Input 5: Add them (this is another instruction).Input 6: Write the sum into your exercise copy (yet another instruction)Output 1: Sum (This is the sum that you wrote in the exercise copy).Input 7: Take 5. (This is an input, with a fixed value).Input 8: Add that to the sum (Another instruction).Input 9: Write the result into the exercise copy (another instruction).Output 2: Sum + 5 (The result you wrote in the exercise copy). Think of yourself as the PROCESS block, (your mind being the CPU, and a small piece of scratch paper as the memory). Suppose the instructions above are written on a piece of paper titled “GENERAL APTITUDE TEST”. As long as you do not have this piece of paper you cannot do anything. Suppose the “GENERAL APTITUDE TEST” is now given to you. So now you can see the instructions (i.e. the BUS 1 is established). What will you do?

Obviously, you will start reading into your mind Input 1. If your mind is distracted by something else, you will not be able to read it into your mind. What does reading into your mind mean? It means converting the piece of text into known sequence of actions.

Page 11: Issue 1

10 CHEENTA GANIT KENDRA

Once you have read it into your mind you will know that you are being asked to:

1. Open your maths book.2. Look up an integer.3. Write that integer in the scratch paper

(so that you remember it).Thus, understanding the instruction, and actually following it are two different things.The first step of reading into your mind or understanding involves bringing the instruction from the “GENERAL APTITUDE TEST” into your mind (you do this by merely seeing it, a computer is stupid, it needs electronic circuits at every stage). Now, remember, your mind is the CPU. So, the first step is to read in the instruction into the CPU. This is called a “Fetch”, i.e. to bring an input into the CPU to decipher its meaning (or decode it).Once a “Fetch” is done, the instructions is deciphered or decoded (with the help of electronic circuits again). Now it is time to do the job (i.e. open the maths book, look up the integer…). This part is called the “Execute”.Thus, following instruction 1 involves “Fetch”ing it, “Understanding” it or “Decode”ing it, and finally, “Execut”ing it. Once these are done, we can move to Input 2. This is how all operations on a computer is done, by following a “Fetch-Decode-Execute” cycle. In short, it is written as the “Fetch-Execution” cycle, with the decoding being implicitly implied.Think about Input 7. This instruction contains some information or data (the number 5). Thus, data can also be fetched (but data cannot be executed).Things to remember:

A computer needs a block called INPUT.A computer needs a block called OUTPUT.A computer needs a block called PROCESS (MEMORY + CPU)The CPU fetches the instructions and data from INPUT, and executes the instructions after decoding them. This is called the Fetch-Execution cycle.This is enough guys. Digest this information. Come up with questions (as many as you can). (...to be continued... send your comments and questions to [email protected])

More ProblemsUNITED STATES OF AMERICA JUNIOR MATH OLYMPIAD 2012

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≥3 such that among any n positive real numbers

a1 , a 2 , ... , an withmax ( a1 , a 2 , ... , an )≤n min (

a1 , a 2 , ... , an ) there exist three that are the side lengths of an acute triangle.

Page 12: Issue 1

11 CHEENTA GANIT KENDRA

3. Let a, b, c be positive real numbers. Prove that

a3+3b3

5a+b+b3+3c3

5b+c+c 3+3a3

5c+a≥ 2

3(a 2+b2+c2) .

4. Let α be an irrational number with 0<α<1, and draw a circle in the plane whose circumference has length 1. Given any integer n≥3, define a sequence of points P1 ,P 2 ,... , Pn as follows. First select any point P1 on the circle, and for 2≤k≤n define Pk as the point on the circle for which the length of the arc Pk−1 Pk is α, when travelling counterclockwise around the circle from Pk−1 to Pk . Suppose that Pa and Pb are the nearest adjacent points on either side of Pn . Prove that a+b≤n.

5. For distinct positive integers a, b < 2012, define f(a, b) to be the number of integers k with 1≤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 γ be a line passing through P. Let A', B', C' be the points where reflections of the lines PA, PB, PC with respect to γ intersect lines BC, AC, AB, respectively. Prove that A', B' and C' are collinear.

The Story�������� ������ �� ������� ���� ��� �� �� ������ ���� ������ �����

�������� �������� � ��� !"##�$�%� &�'� ��� �(����� $%�� ��� &�'� ���� )� �*+������ ,-��� ��� � �+ �� ,����� . / 0 0 0� ����� $�%� �1 � �)��23���� %�� �� ��� 4#!4�*�� � �������� ������� 5���� %�� /� !"##�6 � �7�� ����/� ��� �8� �� '3�� 9� 5 � ���� ������ ����� �+ ��� %��� � 5 ��*� �6 � 1���� �� � :���� �;� ������ � '3�� 9� ���� ��8� :/�� ����� �<�=���� �> � � �?�� +�� �*���� ������ /� �*� ��� ��� ���� @��� AB���=�� ����<�=���� ���C3�� �� � D��� ������ ����� ��7 ������)� ��� ��� +���� E ����� ����� ��F'� �G���H� ��D����� ���'� ���� ����� ���� ���@� �%��� 56 ��������� ���� ���� ��� ��� ���� A��� ��� /� �*�1��� �%�� � ���%�� �> � � �?���� ������� ����� �&@��� ������ ����� ,�D� "###��� ���C3� ���������� �*��7�-� ��� � �I�$%-� ��*��� /�> ��� *�� J��� � �� ��� ������ ������ � � ��:*��� ���� �K � � � ��L� � ��� D��� ���� %�� M� ��� ��*�� ��1��-� ��N-��� �� � ���� �1��� %��� ���� /O

Page 13: Issue 1

12 CHEENTA GANIT KENDRA

O D�� M���� ��� @F���� &� � �13���� P"#�� � ���� �P"#��� �� *���� ����� ���� D��� �+ �� /� D��� Q ����� ��� %�R� ��� (�ST�� ��� � �*��U�� �U�� �*� ���� .0 0 0����� P"#��%���6 � &� � @�V���� �� '3�� �9�� ����� /� �1� *��<� *������ � � ��B�� ��� ������� ����� ������� �B��?�� ���� I��� %�� �1��� /� WP"��� D���� WP"�6 � ����� � '3�� �9� ������/� ��I���� � ��� ��� ��/ O�> � � �?�� ��C� X��� � ������� ������ ���� ���� �� � � � �� �� ����+� ����� � �O��� ���� &�� �P"#� D��� ������� WP"��� D�� /� % � � ��� !!4"�� � D��/� ����� ��� %�� MOO ��� MOO ���� �1� ������� �B��?���� 6 � ���� �� � ����� ����� ��� :YI��� : �I��� ����� ������� �� /� ����� ���I���� �+ ���� M:O�*� ������ ����� ��� � Z�� ���� ���� �� /� ���9 ����� �%�� � �1�� ���� �%��� ����� 6 � �� ���� ����) ������ ���� �+� ��� ��)��� 1��� /OO��� /� ��L� ���� D��� %�R� ��� � ��� ���� � �� �� ������7� ��� ��� �� 5��� ��� ���� ��� MO��� ���@� J��� �����-� �I�$%-� +��� ��[��� ��< ��� /� � ����� %����������� ��BY\�= �� %� � � � ��O)��� ����>� %�R��� � ����� ��� � ���� � ����� D�� �%��� � � ��*�� ��1��-� ����� �+������ ��� /O� ��L ��I�$%-� �*��7�-� ���� ��� ��BY\�= ��� �� ��� ��� �

���������� �,��C� C��� ������ ������ �ID]� �� /��*� ���� ����� � ��� � � ��O������ ����� �� �@��'�� ���� � �� ���� D��� %�� /� �*�� ��1��-� ��� �������� ����� ���R/ O�> � � �?�� ��� � 8��*-�� ����+� %��� A* ��O��� ���� � ������ � ��� ���������� D����� �� ��� ������ 1�� /� ���� ����� D��� 1��� %��� 1��� ��%� � �������� ���� ���/� ������ ���� ��� I������ ���/ �����<�� ��^� � %)���� �'����� ���� �*�� ��� ������ ��� � � ������ ����� �*��/� �������� �,�D� ��� %�� ��)� ����� 00 0O�*��7�-� I���� +��� ��� ���������� D����� ��� � ��\����� 1��R� � �����<�� �� � � ����� *�,�[� ���� /� ������ �G?�[� J��� �*� � ��O �*��7�-I�� ����� � �1 ��� �%��� ��� ��� � �!_``����� ����� D�� �� '3�� �9� %���� ������ /� ��D�a O���� ��� ������ ��B����� � � � � ��O!_``���� � � D��� )� %��� ����� �� /� �*��7�-���� ����/� O�*�� ���� /� ����� ���� � ��C� WP"���� 6 � �+ ��� ���� ��� �Z�� % � ��/� ��%� � ���� D��� %��� ��� ���� � �� ������/ O�> � � �?���� ���� %��� 1�)��� ���+��� �$�%D ����� ���� � ��O�� ���� *%,� �%�*�� I�� /� ��� �1��� �%����� ���� D��� %��� !_``���/� ����� � � �� ������� ����� �� /O�> � � �?�� ��� ������ ��� � �� � �O�*������ ���B<� ��

Page 14: Issue 1

13 CHEENTA GANIT KENDRA

1�� ,���*� ��� ��� ����� ��� E � ��� �%�*��� � ��* �!_``MOO*%,� �%�*��� ���� /� !"##�� /� ����� 6 � �� '3�� 9� �,���/� ����� !_``��� 6 � %����/� &���� � D���� ������ �%���� 6 /� !_``��� �%���� 6 � ��� �!_``��� D��/� ����� ����� �� � �����)� � � �� EO��*��7�-�� +� ���� %�B /� ���� ��� � �b@13� %�� �� ��O�K� � �� M��R�� @�� ����)� ��� ���� �� '3�� �9� !"##��� D��� % /� �1�%� � ����)� D��� �Q� %�R� �� � ����� ���� !"##� D���� !"##���� �%��� �6 � ����� /� ��L� ��� � % � ����� 6 � ������ �%�� ���� � ��� �+ ��� ����� ��/� ������� ���� 8 ��� ����� ����� @�� � %�� 5 ������ 6 � ������� %���� ����� /� ��L� � ��� ,���� �1� � '3�� �9�� ���,�� �� 6� ����)� D���� %�����/� ���� ��� '3�� �9� ����C� � ��� !_``��� �%���� 6 �� ������ ���� /������ !"##� D��� %��� ������ ��� ������/ O�*��7�-� ���� �E �O�� I���� I��� � ��� ���� ��� MOO� � � � ��� �� � ,��� � ��� ��� O������ ��� 5�������� %)���� �� � /O�*� ��<� �� � � ��� ���� I��� � ���� � /OO@�� �� � ���R� �,��C��� �'��� ��� /� ��L ��,��C�� �� � �I���� ,�<��� �<��� ���� �1 ������� +�� �c/� �� � 1��� ��D� ��� �R� �,��C� ��� ��1�� �� &� �� �� �� �R6�� *��� ����� ��7 ��d�-��� ������ *%, ����%� �� ��e�� �� /�

�7��,-�� �f� :����,��� : ���� ��7 �� &���f �,������ �� � ����� �f��� �D�%��� �� � /� Y6� �R ������ %�� 1+�� gh� �� �� ��R�� ��6� *��*D� �,�D� ���*��� ��R�� ��6� ����C� *�*D��� :�8 :�� ��� ����� ��7� �d�-�� ��R�� &����� *��*D�� ���6� �i� :�� : �&� � ��R� ����� /� � ���� %� �&� � )� �d�-�� ��R�� *�*D� *7+D�� * ��� %��/ ������ &� � ��R� ��,�� ���� ������ �c� %� ��d�-���R�� *�*D*7+D�� ��� �� � ��/� @��� 1�� ����� ���� � ��� ����� A��� ���/� �,�� ���� ����� ��c� ���'� ����� *�� �j��� �� � ��<��R/� �� � 1�� ���k ��� %�� ��%� � � ��� ��� ���� ������ ���6 �,���� ���� �'�/� �+���� &� � �R� ( ��C . �5���&��6� *��*D�� ��6 �i� �8� (��l�i� ,��� .��d�-�� ��R� (,��� . ���� ����7� ��7� �d�-� ���R�� ����)� ��6� *�*D� (����)� ��6� ,���� .5��� ��� (,����� �� � .���,��� (����� ,��� �Y,���� %��� ������ . /� Y6� �R� ���� ����� ������ ����� ��� �K��� ��� � ��k� +������ %��/� 1�� �1�c� ����� & �'� ���� 1��� �1� ����� ����� ���� �� ����� *%�,� ������� �m��� +����/ OO��% E ���� :���� :5�� @����� ���� +�� ���,� /���) �����)� �l�i� ��� &������� ���� � ��� ����� MOO ��n� /� @�� ��� ��� ����� ��-�� �<��� o��- ��* � ��� ���� !4� ��*/ O�*��7�-� ��� �> � � �?�� ��*��� �,�I�� , �

Page 15: Issue 1

14 CHEENTA GANIT KENDRA

���� � � �O ������ O/���� ���� %�* ��Op 0 0 0 ���� ������ ��*� �* � �� �� � � ��� ��� ���� I������ �@D� I��� ���� ���� ���) /� * ��� * ��� ��)� ����� �����/� � ���� ����) ���� I������� ����)� ��� ����� �����/� &q� %�R ���� I��� ����� ��,��� ��I���� ���� 1��/ O�*��7�-� ��� ������ ��� ����� �%�*� � �O�*��D �I��� � �I��� ����� ���� I������ ������� /� ��� ��* � ��� ���� ��-�� �<�/� � ��� ���� �� ���� ��8����� ���<� *��@� ��� ��� � ������ ��*/� ���* ����*�� �*���e� � ���-�� �<��� �* � �� � ��=�<� �, I�� � ���� ���� �� � � ���r�������� �*s�� �0 0 08 * 00 08 * 00 0 � /O ����� ��� ��� I�� � � :��%� � J@� ��^�� ��� ���� �� �� / : ������ �

Some More ProblemsINDIAN INTERNATIONAL MATH OLYMPIAD TRAINING CAMP 2011

DAY 1

1. Let ABC be a triangle each of whose angles is greater than 300. Suppose a circle centered with P cuts segments BC in T, Q; CA in K, L and AB in M, N such that they are on a circle in counterclockwise direction in that order.

Suppose further PQK, PLM, PNT are equilateral. Prove that: a) The radius of the circle is

2abca2+b 2+c2+4√3 S

where S is the

area. b) a.AP = b.BP = c.PC

2. Let the real numbers a, b, c, d satisfy the relations a+b+c+d = 6 and a2+b2+c2+d2 = 12. Prove that

36≤4 (a3+b3+c3+d 3)−(a4+b 4+c4+d 4)≤48 3. A set of n distinct integer weights

w1 ,w2 ,... , wn is said to be balanced if after removing any one of weights, the remaining (n - 1) weights can be split into two subcollections (not necessarily with equal size) with equal sum. a) Prove that if there exist balanced

sets of sizes k ; j then also a balanced set of size k + j 1.

b) Prove that for all odd n≥7 there exist a balanced set of size n .

DAY 2

1. Find all positive integers n satisfying the conditions a) n2=(a+1)3−a 3 . b) 2n + 119 a perfect square

2. Suppose a1 , ... , a n are non-integral real numbers for n≥2 such that

a1k+...+a n

k is an integer for all integers 1≤k≤n . Prove that none of a1 , ... , a n is rational.

3. Let T be a non-empty finite subset of positive integers greater than or equal

Page 16: Issue 1

15 CHEENTA GANIT KENDRA

to 1. A subset S of T is called good if for every integer t∈T there exists an s in S such that gcd(t,s) > 1. Let A = (X , Y ) | X ⊆T , Y ⊆T ; gcd (x, y ) = 1 for all x∈X , y ∈Y .Prove that : a ) If X 0 is not go o d then the number of pairs (X 0 ; Y ) in A is even . b ) the number of good subsets of T is o dd .

DAY 3

1. Let ABCDE be a convex pentagon such that BC || AE, AB = BC + AE, and ∠ ABC = CDE. Let M be the midpoint of∠ CE and let O be the circumcenter of triangle BCD. Given that DMO = 90∠ 0

prove that 2 BDA= CDE.∠ ∠

2. Prove that for no integer n is n7 + 7 is a perfect square.

3. Consider a n×n grid which is divided into n2 unit squares (think of a chess board). The set of all unit squares intersecting the main diagonal of the square or lying under it is called an n-staircase. Find the number of ways in which an n-staircase can be partitioned into several rectangles, with sides along the grid lines, having mutually distinct areas.

DAY 4

1. Let ABC be an acute angled triangle. Let AD, BE, CF be internal bisectors, with D, E, F on BC, CA, AB respectively.

Prove that EFBC

+ DEAB

+ DFAC

≥1+ rR .

2. Find all pairs (m, n) of nonnegative integers for which m2 + 2.3n = m(2n+1 – 1).

3. Let {a0 , a1 , ....} and {b0 , b1 , ....} be two infinite sequences of integers such that

(a n−a n−1)(an−a n−2)+(bn−bn−1)(bn−bn−2)=0for all integers n≥2 . Prove that there exist a positive integer k such that

a k+2011=a k+20112011 .

(Send the solutions of these questions to [email protected]. You may also send them by post. )

Topic

NUMBER THEORY

1. Given the Well Ordering Principle (any

For some of our students the month of May is special. There are three mathematics entrances coming up in this month which are relevant for Indian students: I.S.I. B.Stat and B.Math Entrance, C.M.I. B.Sc. Math & Comp. Sc and B.Sc. Math & Physics entrance,IoM.A., B.Sc. Math and Comp. Sc. Entrance. The following collection of results are useful for students appearing for these entrances. If you have not prepared for them over years, you won't find these results useful. They are NEITHER COMPREHENSIVE, NOR SELF SUFFICIENT. They can only act as a reminder of what you have learnt (or discovered) while doing intensive mathematics over years. ALL THE BEST!

Page 17: Issue 1

16 CHEENTA GANIT KENDRA

non-empty subset of non-negative integers has a smallest element) prove the Principles of Mathematical Induction: a) If a subset S of positive integers

contains 1, and contains n+1 whenever it contains n, then S contains all the positive integers.

b) (Strong form) If a subset S of positive integers contains 1 and contains n+1 whenever it contains 1, 2, 3, …, n, then S contains all the positive integers.

2. Let a, b, c, m, x, y be integers: a) If a | b then a | bc for any integer c b) If a|b and b|c then a|c. c) If a |b and a|c then a| bx + cy for

any integers x and y. d) If a|b and b≠0 then∣a∣≤∣b∣ e) If a|b and b|a then a=±b . f) If m≠0 then a|b if and only if

ma|mb. 3. (Division Algorithm) Given any integer

a and b with a≠0 there exists unique integers q and r such that b = qa + r,

0≤r≤∣a∣ . If a∤b then r satisfies the stronger inequality 0<r<∣a∣ .

4. (Bezout's Theorem) If a, b are any integers, not both zero, then gcd(a,b) exists and there exists integers

x0, y0 such that gcd(a,b)=a x0+b y0 .

5. Let a,b be integers, not both zero. Then a, b are coprime if and only if there are integers x, y such that ax + by = 1.

6. If a, b are coprime integers then every

integer n can be expressed as n = ax + by for some integers x and y.

7. If d= gcd(a,b), then gcd ( ad

, bd) = 1.

8. (Euclidean Algorithm) Given integers b and c, c>0, we make a repeated application of division algorithm to obtain a series of equations:

b=cq+r1 ,0<r1<c ,c=r1 q1+r2,0<r2<r1 ,r1=r2 q2+r3, 0<r3<r2 ,...r j−2=r j−1 q j−1+r j , 0<r j<r j−1 ,r j−1=r j q j .

Then the gcd(b,c) of b and c is r j , the last non-zero remainder of the division process. Moreover if (b, c) =

b x0+c y0 then the values of x0, y0can be obtained by eliminating

r j−1 , ..., r2, r1 from the above set of equations.

9. (Euclid's Lemma) If (a, m) = 1 and m | ab then m| b.

10. If p is a prime and p|ab then p|a or p|b.

11. If p is a prime such that p | a1 a2 ... an , then p divides at least one

of the factor ai of the product. 12. If a|m and b|m and (a,b)=1,

then ab|m. 13. Every integer n> 1 can be

Page 18: Issue 1

17 CHEENTA GANIT KENDRA

expressed as a product of primes (with perhaps only one factor).

14. (The fundamental Theorem of Arithmetic) Every positive integer n > 1 can be expressed as a product of primes in a unique way except for the order of the prime factors.

15. If a, b are positive integers such that ab=c2 and (a, b) =1, then a and b are both perfect squares.

16. Prove that there are infinitely many primes.

17. Prove that given any positive integer n, there exist n consecutive composite integers.

18. Let a, b, c, d, x, y denote integers then:

a)a≡b(mod m) , b≡a(mod m) ,a−b≡0(mod m)

are

equivalent statements. b) If a≡b(mod m) and

c≡d (mod m) , then a x+c y≡b x+d y (modm) .

c) If a≡b(mod m) and d|m, then a≡b(mod d ).

19. Let f(x) denote a polynomial with integral coefficients. If

a≡b(mod m) , then f (a)≡ f (b)(mod m).

20. Let a, b, x, y, m, m1 , ... , mr be integers. Then, a) a x≡a y (mod m) if and only if

x≡ y (mod m(a ,m)

) .

b) If a x≡a y (mod m) and (a, m) = 1 then

21. The following holds a) a≡a(mod m) (reflexive) b) If a≡b (mod m) , b≡c(mod m)

then a≡c(mod m) (transitive) c) If a≡b (mod m) then

b≡a (mod m) . d) If a≡b (mod m) , then for any

integer k, ka≡kb (mod m) . e) If a≡b (mod m) and

c≡d (mod m) , then ac≡bd (mod m) . In general , if a i≡bi(mod m) , i = 1, 2, 3, ..., k,

then a1...ak≡b1 ...bk (mod m) . In particular, if a≡b (mod m) , then for any positive integer k,

ak≡bk (mod m) . f) We have a≡b (mod mi) ,i=1, ... , k

, kf and only if a≡b (mod lcm(m1 ,... , mk )) . In

particular, if m1 , ... , mk are pairwise relatively prime, then

a≡b (mod mi) ,i=1, ... , k , if and only if a≡b (mod m1 m2... mk )

22. Only a perfect square has an odd number of divisors.

23. The even and odd numbers have the following properties: a) an odd number is of the form 2k +

1, for some integer k; b) an even number is of the form 2m,

for some integer m;

Page 19: Issue 1

18 CHEENTA GANIT KENDRA

c) the sum of two odd numbers is an even number;

d) the sum of two even numbers is an even number;

e) the sum of odd and even number is an odd number;

f) the product of two odd numbers is an odd number;

g) the product of integers is even if and only if atleast one of its factors is even.

24. If n is a composite integer, then it has a prime divisor p not exceeding

n . 25. or any given integer m, there is

no polynomial p(x) with integer coefficients such that p(n) is prime for all integers n with n≥m .

26. The following holds: a) if p is a prime, then gcd (p, m) = p

or gcd (p, m) = 1. b) If d = gcd(m,n), m = dm' , n = dn',

then gcd(m',n') = 1. c) If d = gcd(m,n), m=d'm'', n=d'n'',

gcd(m'',n'')=1, then d'=d. d) If d' is a common divisor of m and

n, then d' divides gcd(m,n). e) If p x∥m and p y∥n , then

pmin x , y∥gcd x , y . Further more, if m= p1

1 ... p k k and

n= p11 ... pk

k , i , i≥0 , i=1,2, ... , k then gcd(m,n) = p1

min 1, 1... pkmin k ,k

f) If m = nq + r, then gcd(m,n) = gcd(n,r)

g) gcd((m,n),p) = gcd(m, (n,p)) = gcd(m,n,p).

h) If d| ai , i = 1, ... , s , then d|gcd( a1 , ... as )

i) If ai= p11i ... pk

k i ,i=1,... , s , then gcd a1 , ... , a s= p1

min 11 , ... ,1k ... pkmin k1 , ... ,kk

27. The following holds a) If lcm(s,t) = m , m = ss' = tt', then

gcd (s', t') =1 b) If m' is a common multiple of s and

t and m' = ss' = tt', gcd(s',t')=1, then m=m'.

c) If m' is a common multiple of s and t, then m|m'.

d) If m|s and n|s, then lcm(m,n) | s. e) If n is an integer, n lcm(s,t) =

lcm(ns, nt). f) If s= p1

1 ... pk k and

t= p11 ... pk

k ,i ,i≥0 , i=1, 2,... , k then lcm(s,t) = p1

max1 ,1 ... pkmax k ,k

28. For any integer m and n the following relation holds: mn = gcd(m,n)*lcm(m,n)

29. If n = p1a1 p2

a2 ... pka k is a prime

decomposition of n, the number of divisors of n, that is

taon =a11 a21...ak1 30. If n = p1

a1 p2a 2 ... pk

a k is a prime decomposition of n, then there are

2a 112 a21 ...2 ak1 distinct pairs of ordered positive integers (a, b) with lcm (a, b) = n.

31. For any positive integer n

∏d∣n

d =ntao n

2 .

32. For any positive integer n, tao(n) taon ≤2 n .

Page 20: Issue 1

19 CHEENTA GANIT KENDRA

33. If n = p1a1 p2

a 2 ... pka k is a prime

decomposition of n, then

σ (n)=∑d∣n

d=p1

α1+1−1p1−1

...pk

α k+1−1pk−1

.

34. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. Assume that S is a complete set of residue classes modulo m. The set T = aS + b = { aS + b |

s∈S } is also a complete set of residue classes modulo n.

35. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. There exists integers x such that ax≡b(mod m)and all these integers form exactly one ersidue class modulo m.

36. For any prime p, ( p−1)!≡−1(mod p) .

37. Let m be a positive integer. Let a be an integer relatively prime to m. Assume that S is a reduced complete set of residue classes modulo m. Set T = aS = { as | s∈S } is also a reduced complete set of residue classes modulo m.

38. (Euler's Theorem) Let a and m be relatively prime positive integers. Then aϕ(m)≡1(mod m)

39. (Fermat's Little Theorem) Let a be a positive integer and let p be a prime then a p≡a (mod p) .

40. Let p be a prime and let a be a positive integer. Then ϕ( pa)= pa−pa−1 .

41. Let a and b be two relatively prime positive integers. Then ϕ(ab)=ϕ(a )ϕ(b) .

42. If n=p1α1 ... pk

α k is the prime factorization of n>1, then

ϕ(n)=n(1− 1p1

)(1− 1p2

) ...(1− 1pk

)

43. For any positive integer n ∑d ∣n

ϕ(d )=n .

44. Let n=ah ah−1...a0 be a positive integer. a) Let s(n) = a0+...+ah denote the

sum of it's digits then n≡s(n)(mod 3) . In particular, n

is divisible by 3 if and only if the sum S(n) of its digits is divisible by 3.

b) We can replace 3 by 9 in (1) ; that is n≡s (n)(mod 9) . In particular, n is divisible by 3 if and only if the sum S(n) of its digits is divisible by 3.

c) Let s'(n) = a0−a1+...+(−1)h ah(alternating sum). Then n is divisible by 11 if and only if s'(n) is divisible by 11.

d) n is divisible by 7, 11, or 13 if and only if ah ah−1 ... a3−a2 a1 a0 has this property.

e) N is divisible by 27 or 37 if and only if ah ah−1... a3+a2 a1 a0 has this property.

45. N is divisible by 2k or 5k

Page 21: Issue 1

20 CHEENTA GANIT KENDRA

(k≤h) if and only if ak −1...a0 has this property.

46. Let n be a positive integer, and let S(n) denote the sum of its digits. Then a) 9|S(n) – n; b) S (n1+n2)≤S (n1)+S (n2) (subaddi

tivity property) c) S (n1 n2)≤min (n1 S (n2) , n2 S (n1)) ; d) S (n1 n2)≤S (n1)S (n2) (submultipli

cativity property)...(to be continued)

Debate

This is a discussion on I.S.I. B.Stat and B.Math entrance paper, 2012. I.S.I. or Indian Statistical Institute is one of the most prestigious institutes in India. It's mathematics and statistics courses at the undergraduate level are held at high esteem throughout the world. Along with Chennai Mathematical Institute and Institute of Mathematics and Applications, Bhubaneswar, it offers a world class mathematics course to selected few students.

Tintin: How was the paper this year?

Dhanshiri: It was easier than any I.S.I. Entrance paper I have ever encountered. The Subjective section (8 questions, 2 hours) was

hopelessly simple.

Shankar: There were a couple of nice questions though. The seventh question was like this: 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 circles.)

Dhanshiri: Yes, this was a descent one. When I first looked at it, I thought the locus is a circle.

Shankar: Same here (grin). A circle with a

center at a+b

2 . Actually this is how I

started solving the problem: the midpoint of AB (let us call it D) be joined with C. Also 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.Tintin: How can we be sure that AC passes through T 1 and BC meets the outer circle at

T 2 . They may pass through any other point. Why specifically through the point of tangency?

Dhansiri: Well.. that is a result from geometry. The proof is also simple. Join AT1

and CT1. Both are perpendicular to the tangent

Page 22: Issue 1

21 CHEENTA GANIT KENDRA

passing through T1 (tangents are perpendicular to the radius at the point of tangency). Hence AT1 and CT1 make 1800 with each other making AC a straight line. For BCT1 you can argue in a similar manner.

Shankar: Absolutely. Now I assumed the radius of the moving (and growing circle) to be r at a particular instance. Then AC = a+r and BC = b-r.

Dhansiri: So you tried to use Apollonious' Theorem to measure DC which is a median to the triangle ACB?

Shankar: Yes. Since I know AC (a+r), BC (b-

r) and AD (a+b

2−a= b−a

2 ) , and as I

guessed that the locus is a circle, I tried to

prove DC = a+b

2 . Of course the guess was

wrong.

2DC2 = (a+r)2 + (b+r)2 - 2 (b−a2 )

2

. This

makes DC dependent on r and r (or radius of the moving circle) is visibly changing it's length during the travel.

Dhansiri: The actual solution is basically a one liner. As you have already observed that 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 sum 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).

Shankar: Yes, I figured that out as soon as I looked into a+r and b-r. It was quite a simple problem put in a beautiful style :)

Tintin: I did not understand the part of Apollonious Theorem.

Shankar: Apollonious Theorem gives the measure of a median of any triangle (line joining any vertex and midpoint of the opposite side). The statement of the theorem is : in any triangle ABC, if AD is a median, then AB2 + AC2 = 2(AD2 + BD2).

(discussion will continue in the next issue. The rest of the questions from I.S.I. Subjective Paper follows)

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

Page 23: Issue 1

22 CHEENTA GANIT KENDRA

Subjective Questions

1. If x, y, z are the angles of a triangle prove that:

a) tan x2

tan y2+tan y

2tan z

2+tan z

2tan x

2=1

b) Using part (a) or otherwise prove

that tan x2

tan y2

tan z2≤ 1

3√3

2. The function f(x) = (α+∣x∣)2 e(5−∣x∣)2

a) is continuous for what value of α?

b) is differentiable for what value of α ?

3. Consider that number array

1 3 6 ..2 5 9 ..4 8 13 ..7 12 .. .... .. .. ..

In this array the position of 8 is in the 3rd row and 2nd coloumn. What is the position of 20096?

4. Prove that the polynomial x8−x7+x2−x+15 has no real root.

5. Let 'm' be a number all of whose digits are either 6 or 0. Prove that 'm' cannot be a perfect square.

6. a) Let a be the base of a triangle and

a+b be it's perimeter. Prove that the area of the triangle is maximized if the other two sides are equal.

b) Using part (a) or otherwise prove that the among all quadrilaterals with fixed perimeter square has the greatest area.

7. 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 circles).

8. 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) Find a function with deg(f)=2. b) For a particular value of k find a

function with deg(f) = 2k .

Page 24: Issue 1

23 CHEENTA GANIT KENDRA

Some More StoryPURSUING PURE SCIENCE AND

MATHEMATICSInstitutions of Excellence in India

after High School

Mathematics

• Indian Statistical Institute, Banglalore◦ B.Math course (3 years)◦ Admission through Entrance Test

and Interview.◦ Indian National Math Olympiad

qualified students are exempted from Entrance Test.

◦ www.isical.ac.in

• Indian Statistical Institute, Kolkata◦ B.Stat Course (3 years)◦ Admission through Entrance Test

and Interview◦ Indian National Math Olympiad

qualified students are exempted from Entrance Test.

◦ Math Olympiad Scholars are preferred.

• Chennai Mathematical Institute, Chennai◦ B.Sc. In Mathematics and Physics

(integrated) or Mathematics and Computer Science (integrated) ,

Mathematics and Physics Integrated course.

◦ Admission through Entrance Test◦ Indian National Math Olympiad

qualified students are exempted from Entrance Test.

◦ www.cmi.ac.in

• Institute of Mathematics and Applications, Bhubaneswar◦ B.Sc. In Mathematics and

Computing (3 years)◦ Admission through Entrance Test◦ www.iomaorissa.ac.in

• Indian Institute of Science, Bangalore◦ B.S. Course (4 years)◦ Admission through IIT and AIEEE

merit list.◦ Separate application necessary◦ www.iisc.ernet.in/

Physics, Chemistry, Biology, Mathematics

• Indian Institute of Science Education and Research (Bhopal, Kolkata, Mohali, Pune, Thiruvananthapuram )◦ 5 year BS-MS dual degree with

major in biology, chemistry, mathematics and physics.

◦ Admission through Kishore Vaigyanik Protsahan Yojana (KVPY) : SX, SA, SB, SP , IIT-JEE 2012; Rank list, Central and State

Page 25: Issue 1

24 CHEENTA GANIT KENDRA

board (CSB) : With high aggregate scores in board exam (XII) and IISER science aptitude test.

◦ http://www.iiser-admissions.in/

Astronomy

• Indian Institute of Space Science and TechnologyThiruvananthapuram ◦ http://www.iist.ac.in/

Yet Another StorySOME THOUGHTS ABOUT 4th DIMENSION.... by Arkabrata Das

Have you ever imagined how a four-dimensional being would appear to you if it comes in front of you?

I think they would look like some fleshy balloons constantly changing in size.Now as they are of four dimension and you being from three dimension,you could see (4C3) or 4 different shapes of it as you could see three dimensions one at a time. Let me elaborate it.

As we are trying to think how will a 4 dimensional object look like with respect to the 3 rd dimension, firstly let us think how we,the 3 rd dimensional objects look like to the 2 dimensional objects.Consider a two-dimensional world

resembling a sheet of paper. How would you appear to the inhabitants of such a world (if such a world exists).The 2-D creatures would only see cross-sections of you as you intersected their universe. Your finger would look like a flat disc or like a rectangular sheet of skin.They would just see irregular shapes with skin boundaries as you entered their world. Similarly, an object which belong to the fourth dimension would have a cross-section in our space that looked liked a bunch of skinny balls.

A 4-D being would be a god to us. It would see everything in our world.We cannot hide anything from it.It could even look inside our stomach and remove our breakfast without cutting through our skin, just like we could remove a dot inside a circle by moving it up into the third dimension, perpendicular to the circle, without breaking the circle.Surgery would have been very simple then.

It can effortlessly remove things before your very eyes, giving you the impression that the objects simply disappeared.It can also see inside any 3-D object and if necessary remove anything from inside. The being can look inside our intestines, or remove a tumor from our brain without ever cutting through the skin. A pair of gloves can be easily transformed into two left or two right gloves.In the 4th dimension hiding an object from us will very simple then.Many more new shapes can be formed then. So what do you think its just a superstition or can a 4-D universe exist in real?

Page 26: Issue 1

25 CHEENTA GANIT KENDRA

Cover StoryHow do you 'define' a circle? It is the

collection of all points, equidistant from a fixed point on the plane. We have played with the

definition of 'distance' in our cover story.What is distance? How can we

'understand' in mathematical terms? Given two points on the plane we

should be able to compute a positive real number which can act the 'distance' between those two points. Indeed 'distance' is nothing but a function or a rule of association which relates pair of points with a given number.

This 'rule' can be anything that satisfies certain simple properties. Let x, y be two points and d(x, y) = D be the distance between those two points. Then the properties

are:1. d(x, y) must be postive for all x, y and

0 if only if x=y.2. d(x, y) = d(y, x) that is distanc e from

x to y must be the same as the distance from y to x.

3. The function must obey the triangular inequality, that for any three points x, y and z we have

d ( x , y)+d ( y , z)≥d (z , x)

Now let us define a 'different' distance function that will make the shape below a circle.

Let d(x, y) = max{|x1 – y1|, |x2 – y2|} where the coordinates of x = (x1 , x2) and y =(y1,y 2). Now try to establish that d(x, y) is indeed a distance function (that is it satisfies all the properties that a distance function should satisfy) and the above square looking this is actually a circle (that is all the points on it has a constant distance from the point (0,0) according to the new definition of distance).

A Month @ CheentaA brief preview of our activities in coming month.

Offline Classes

2013 High School Graduation Classes • Ellipse from Coordinate Geometry• Questions from I.S.I. Entrance 2012,

2010, 2009

Page 27: Issue 1

26 CHEENTA GANIT KENDRA

2014 High School Graduation Classes

• Differential Calculus (Limit, Continuity)• Inequalities• Questions from and Indian National

Math Olympiad 2012, Regional Math Olympiad 2010, 2011

Early Bird Classes

• Geometry of Circles• Trigonometry of triangles and

quadrilaterals

ONLINE CLASSES

Advance Interdisciplinary Mathematics Program, USA (online)

• Inequalities• Calculus• Mechanics

Early Bird Online Course

• Number Theory• Algebra

2013 - 14 Indian Math Entrance Program

• Number Theory (Congruency theory)• Combinatorics

Alternate Saturday Full Length Offline Examination Program

• 4 hours I.S.I. Standard full length test• 3 hours C.M.I. Standard full length test

Other Activities• An excursion in mathematics (22nd

May, 2012)◦ A day-long program in Calcutta,

India with Early Bird students. We plan to do mathematics on the bank of river Ganges, explore technological museums and planetariums, find some picnic spots where we can have good food and do some more mathematics (preferrably geometry).

• Cheenta Reunion (2nd June)◦ A day-long program where Cheenta

Alumni, students, teachers and friends come together. We will have movies, chess, mathematics and much more!

• Summer Camp in California (12th

June to 12th July 2012)◦ A month-long summer camp will be

organized in Bay area, San Francisco, USA. This is the first overseas activity on our part. If you want to register for the program please visit www.cheenta.com/mathcamp

Page 28: Issue 1

27 CHEENTA GANIT KENDRA

Number Theory Combinatorics Geometry Inequalities Functional Equation Polynomial

MATH OLYMPIAD SUMMER CAMP12th June to 12th July

WHO CAN ATTEND?

Students of Class VI through XII who have a special interest in beautiful mathematics;

from Calcutta, India

or San Francisco, USA

TO REGISTER

Call us +91 9804005499 (INDIA) +1 4084767585 (USA)

or Register atCheenta.com/mathcamp

Published by Ashani Dasgupta on behalf of Cheenta Ganit Kendra; B 37, C.I.T. Buildings, Calcutta 700007, West Bengal India; Phone +91 9804005499; www.cheenta.com; [email protected];