Philosophy of Cheenta

I think it is important to discuss the philosophical orientation of our organization. Whether you (that is the reader) are a parent or a student or a mathematician, this note will help you understand our organization.
Cheenta Ganit Kendra - center of mathematics has three objectives:-

• Innovate and cultivate mathematical and allied ideas (like logic, computer science etc.)
• Impart Bharatvarsha class mathematical education to middle school, high school and other students of mathematics.
• Create a cutting-edge services for our clients.

INNOVATE AND CULTIVATE MATHEMATICAL AND ALLIED IDEAS (LIKE LOGIC, COMPUTER SCIENCE ETC.)

Cheenta Ganit Kendra aspires to become a cultivation center of new mathematical ideas. We are preparing a journal as a spearhead project for this purpose. The journal is mainly targeted at school students who have a strong appetite for mathematical ideas. We wish to incorporate logic, computer science and other topics in our journal in the coming days. Apart from the journal, we are also on our way to create mathematical circles in different parts of India (Kolkata to begin with). Mathematical Circles are discussion forums of young minds who meet once or twice a week to discuss mathematics. They were formulated successfully in USSR. Any student or teacher of mathematics interested to join us in this mission may contact us without any hesitation.


IMPART BHARATVARSHA CLASS MATHEMATICAL EDUCATION TO MIDDLE SCHOOL, HIGH SCHOOL AND OTHER STUDENTS OF MATHEMATICS.

Teaching is our source of inspiration and money. We need inspiration to innovate, we need money to operate. Students aspiring for mathematical olympiads or challenging math entrances like those of I.S.I., C.M.I., I.M.A., MIT Putnam, or Cambridge STEP exams are our target audience. We put our fullest and sincerest dedication to help our students because we know that they are our oxygen (so true!). And mind it, this is not only a money making proposition, it is a question of inspiration too (and perhaps most importantly so).
We have long class hours (usually 4 hours), in-between 24/7 student support both online and over phone. Our policy is to make the student feel comfortable to ask questions. What is more, we never pick up the phone calls of our students, WE CALL THEM BACK. This is infact one of those policies which allows students to be comfortable asking questions (without worrying about anything, even call charge). Even at high school level we put definite homework assignments, checked meticulously my the faculty. Our forums are open for any-time, any-where collaboration between students or faculties or both. Bharatvarsha Class is defined as the standard of education which does not care about the material aspects of learning. We encourage fullest dedication to the art of mathematics (and not it's ourcome in the scorecard). Indeed when you put your honest effort in learning, score-cards seem happy automatically.

CREATE CUTTING-EDGE SERVICES FOR OUR CLIENTS

Cheenta Ganit Kendra believes in utmost professionalism and efficiency as far as client dealings are concerned. For us business means people and not sales-figure.

IIT Mathematics: Why it is important and which book is better?

IIT Mathematics books are important for I.S.I. preparation because they give a comprehensive problem collection on high school mathematics. I.S.I. entrance preparation has two layers.

Layer 1- High School Mathematics - Calculus, Trigonometry, Analytical Geometry, Algebra
Layer 2- Olympiad Mathematics - Plane Geometry, Combinatorics, Number Theory, Inequalities, Polynomials, Functional Equations

For high school mathematics portion, IIT Mathematics books are useful. Ofcourse you will need typical books like Maron for Calculus, Loney for Trigonometry and analytical geometry, Hall and Knight or Bernard Child for Higher Algebra, but to enhance problem solving ability you must have an IIT book as your companion.

There several such books available. Some of the most renowned are:

• M.L. Khanna,
• Tata Mcgraw Hill (TMH)
• Arihant
• Pearson
• Dinesh
• MTG

Each book has it's upside and downside but I have found that M.L. Khanna and TMH are the two most important books that serve our purpose. I will give M.L. Khanna some more points because it has a good mix of easier and difficult problems. It allows you to build your skill. On the other hand TMH is good for the already skilled one.

The bottom line is get M.L. Khanna IIT Math as a problem book. If you can finish it, you may move over to TMH.

Complex Number Random Problem

Problem: \(z^6 + z^3 + 1 = 0\) has one root whose argument is between \(\frac{\pi}{2}\) and \(\pi\). What is the argument?

Solution:
\(z^6 + z^3 + 1 = 0\)
\(\Rightarrow (z^3 - 1)(z^6 + z^3 + 1) = 0\)
\(\Rightarrow (z^9 - 1) = 0\)
\(\Rightarrow z =  \cos\frac {2k\pi}{9} + \sin\frac{2k\pi}{9}\)
Required argument is found by putting values for k (=3)

GEOMETRY RANDOM PROBLEM

Given two triangles ABC and DEF where AC = DF, AB = DE, angle B is a right angle, angle E is obtuse, and angle F < angle C, is BC greater than, equal to, or less than EF. Support your answer with a proof.



Given: ABC be any triangle right angled at B. DEF be another triangle such that AB = DE, AC = DF and ∠DEF is obtuse.

To Prove: BC > EF or BC = EF or BC < EF whichever is true

Construction: Apply ΔDEF on ΔABC in the following manner: put D on A and DF along AC. As DF = AC (given) hence F falls on C. Join BE.

Proof:
In ΔBAE; AB = AE (since AE = DE and DE = AB).
Hence ∠ABE = ∠AEB
Since ∠ABC < ∠AEC (∠ABC = 90; ∠AEC is obtuse).
Therefore ∠ABC - ∠ABE < ∠AEC - ∠AEB
That is in ΔBCE
∠EBC < ∠CEB.
As side opposite to the bigger angle is bigger, BC > EF. (proved)

NEXT 30 DAYS for TARGET ISI+OLYMPIAD '12 - a minimalistic strategy

What to study and how much to study is a relative question (depends much on the student). We present here a minimalistic strategy which is good for someone who has a fair idea about high school math (and who is interested to crack I.S.I. or C.M.I. undergraduate math/stat entrance in 2012). The strategy is minimalistic because we suggest the must-do stuff only. The more work the better you are off.

Regular (problems that you must to every day, every season)

• I.S.I. 10+2 - subjective 100 problems and objective 300 problems (that is just 13 problems per day)
• Geometry - Challenges and Thrills of pre college math - 10 riders + theorems per day
• Calculus - TMH IIT math - 10 problems per day

Specialization (special topics that you may cover in the upcoming 30 days given that you have 1 year to prepare for the entrance)

• High School centric topic - Trigonometry - TMH IIT math + Loney (Height and Distance harder + Properties of triangle harder) - 10 problems per day
• Olympiad centric topic - Inequalities + Functional Equation + RMO (4 papers), INMO (2 papers) + IMO (2 papers) =  300 problems i.e. about 10 problems per day.

That is about 50 problems per day. Earlier we suggested in a post that you should work on 100 problems per day. That is very much true. However in this post we have suggested the bare minimum work that you must do in order to crack I.S.I. For support and further discussion you may call us at 9804005499. If you are outside Kolkata add '0' in the beginning. If you are outside India add 091 in the beginning.

I.S.I. 10+2 Subjectives Solution

P148. Show that there is no real constant c > 0 such that \(\cos\sqrt{x+c}=\cos\sqrt{x}\) for all real numbers \(x\ge 0\).
Solution:

If the given equation holds for some constant c>0 then,

f(x) = \(\cos\sqrt{x}-\cos\sqrt{x+c}=0\) for all \(x\ge 0\)
\(\Rightarrow 2\sin\frac{\sqrt{x+c}+\sqrt{x}}{2}\sin\frac{\sqrt{x+c}-\sqrt{x}}{2}=0\)
Putting x=0, we note
\(\Rightarrow\sin^2\frac{\sqrt{c}}{2}=0\)
As \(c\not=0\)
\(\sqrt{c}=2n\pi\)
\(\Rightarrow c=4n^2\pi^2\)
We put n=1 and x=\(\frac{\pi}{2}\) to note that f(x) is not zero.
Hence no c>0 allows f(x) =0 for all \(x\ge 0\). (proved)

The INEQUALITY quest!

Inequalities are important for olympiads and I.S.I., C.M.I. entrances. Sometimes they appear as standalone problems. However mostly they are clubbed with other problems (geometry, calculus etc). The Cheenta Inequality course is a 7-session-super-course to achieve excellence and efficiency in solving inequalities. The course work consists of 600 problems on inequalities with interdisciplinary treatment in certain special areas. For example while discussing triangular inequality or Euler's inequality we bring geometry in action. Similarly we discuss Lagrange's mean value theorem, idea of limit while discussing sequential inequalities, convexity of function and so on.
The 600 problems are sourced from:

• 70 problems from Little Mathematical Library
• 125 problems from Arthur Engel
• 100 problems from Excursion in Math and Challenges and Thrills of Precollege Math
• 300 problems from Inequalities: An Approach Through Problems By Venkatchala, International Math Olympiads and other national olympiads.

The course is resumed with certain trivial-looking ideas (natural numbers can be counted, ordered, square of real number is always non-negative, positive times positive is always positive). In the first stroke we cover inequalities related to Means, Bernoulli, Cauchy–Bunyakovsky–Schwarz, Holder. We also cover the simple cases of approximating sequential sums.
Then we move over to Jensen (convexity etc.) Rearrangement, Schur, Murihead and some other not-so-trivial-looking inequalities. Triangular inequality and application of calculus in understanding inequalities is discussed in the final phase.
The Kiran Kedlaya Inequality pack and similar courses available freely online were helpful. We tried to incorporate the best part of all of them. A course on inequality must be holistic as well as efficient. We must not loose our focus from problem-solving techniques. At the same-time the course must bear an elegance of the art-form.

I.S.I. 10+2 Subjectives Solution (2 problems)

P164. Show that the area of the bounded region enclosed between the curves \(y^3=x^2\) and \(y=2-x^2\), is \(2\frac{2}{15}\).

Solution:

Note that \(y=x^{\frac{2}{3}}\) is an even function (green line).


P165. Find the area of the region in the xy plane, bounded by the graphs of \(y=x^2\), x+y = 2 and \(y=-\sqrt {x}\)

Solution:



The parabola and straight line intersects at (1,1) (we find that by solving the \(y=x^2\) and x+y=2)
Thus the area is found by adding area under parabola (from 0 to 1) and area under straight line (from 1 to 2).
\(\int^1_0 x^2\,dx=\left[\frac{x^3}{3}\right]^1_0=\frac{1}{3}\) (area under parabola)
area under straight line above 'x' axis is the triangle with height 1 unit and base 1 unit (from x=1 to x=2, area under x+y=2)
that area = \(\frac{1}{2}\times 1\times 1=\frac{1}{2}\)
Thus total area above x axis (of the required region) is \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Now we come to the region below 'x' axis.
x+y = 2 and \(y=-\sqrt{x}\) intersect at (4, -2) (found by solving the two equations). We calculate the area under the curve \(y=-\sqrt{x}\) from x=0 to x=4 and subtract from it the area of the triangle with base from x=2 to x=4 and height =2 (hence the area of the triangle to be subtracted is 2 sq unit).
Area under the square root curve is
\(|\int^4_0 -\sqrt{x}\,dx|=\int^4_0 \sqrt{x}\,dx= \left[\frac {x^{\frac{1}{2}+1}} {\frac{1}{2}+1}\right]^4_0\).
=\(\frac {2}{3}\times 8=\frac {16}{3}\)
Delete 2 square unit from this and add the area computed before (above 'x' axis).
Area = \(\frac {16}{3} - 2 + \frac{5}{6} = \frac{25}{6}\) (ANS)

So you have got a year?!!

That is a good start. And a demanding one. All good starts are demanding by birth-right. They ask you to do more in the subsequent days. This article is mainly targeted at class 12 pass-outs who are targeting I.S.I. 2012 (or those 12th graders who are able to devote some serious time to mathematics).
Target to solve 100 problems a day (no less). You DO NOT HAVE 365 DAYS. Actually it's about 360 days (counting 2012 to be a leap year and assuming I.S.I. entrance to be held in first week of May). So really we are looking at some 36000 problems in the coming year and trust me that is sufficient to get you there.
Wake up at 5 in the morning. Do about 5 hrs of mathematics. In the afternoon devote 2-3 hours. Same in the evening. That is sufficient for 100 problems a day. That is sufficient as far as time management is concerned.
The books you need to solve->

1. TMH IIT Math and M.L. Khanna IIT Math books make about 20000 problems together. They are important because they are comprehensive.
2. I.S.I. 10+2 Test of Mathematics has about 1500 problems (along with Test Papers)
3. Excursion in Mathematics + Challenges and Thrills in Pre college mathematics together have about 2000 problems.
4. Algebra - Hall and Knight (or better Barnard Child), Complex Numbers from A to Z, (2000 problems)
5. Trigonometry - Loney (about 1000 problems)
6. Coordinate Geometry - Loney (about 1000 problems)
7. Calculus - A combined effort of Apostle, Maron, Piscunov, Tarasov (about 4000 problems)
8. Mathematical Circles, Combinatorics by Brualdi, Selected Problems by yaglom, Problem Solving Strategies by Arthur Engel, IMO compendium, Functional Equation by Venkatchala, Inequalities of Little Mathematical Library and Venkatchala, Number Theory by Burton or Zuckermann should account for the rest 4000-5000 problems.

Note that some problems will need superb intellectual effort, some will ask for tidy computation. So the key is to distribute all kinds of problems in a day. Try about 20 challenging problems per day. Rest 80 should be normal (or subnormal) stuff.

All the best and keep working!

IMO 1959 solution of Q4

Statement: Construct a right angled triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.


1st Solution:

Let |AB| = x and |AC| = x'

Then AD² = x.x'

Also x² + x'² = c² (since ABC is right angled at A).

Note that |AD| = c/2 since AD is the median of a right angled triangle with hypotenuse c units. (application of Apolonius Theorem)

Thus \(\frac{c^2}{4}=x\sqrt{c^2 - x^2}\) 

Hence we find 'x' in terms of c.

We draw a circle with diameter c. Cut off x = f(c) length from 'c' and construct the desired triangle.

15th May 2011 class preview

On 15th May 2011 (Sunday) we focus on inequalities once more. We have already covered Problem Sheet 1 and Problem Sheet 2 on inequalities which covered the Little Mathematical Library booklet on inequalities. Problem Sheet 3 will conclude the LML book and assist a migration to the Arthur Engel- Problem Solving Strategies collection on inequalities.

• Diagnosis Test (covering Problem Sheet 1 and 2 on inequalities)
• Problem Set 3 (on inequalities) with a marathon revision of previous two sheets.
• Regional Mathematics Olympiad 1993 paper discussion
• International Math Olympiad 1959 paper discussion
• Effect Test (covering Problem Sheet 3 with special stress on Holder's inequality)

IMO and RMO related discussion will need prior preparation in the part of the student.In fact that is of the homework of current week. This class is open to all. Students from class 9 to 12 may attend it. Practice Problems for next week includes 100 problems from Arthur Engel inequalities, Inequality chapter from Excursion in Math, Challenges and Thrills of Precollege Math, Mathematical Circles (especially triangular inequalities) and IMO, INMO, RMO level inequality problem collection.

In May will have 3-4 classes (since it is summer vacation in most schools), which are open to all. We want to complete Functional Equation (the book by Venkatchala, and problem collection in Arthur Engel). We enter June to work on Combinatorics and Geometry.

In December the Regional Math Olympiad is due. Summer Vacation is really the prime time for preparation. Look out for more updates in this blog.

RMO 1992, question no. 4

ABCD is a quadrilateral and P , Q are mid-points of CD, AB respectively. Let AP , DQ meet
at X, and BP , CQ meet at Y . Prove that
area of ADX + area of BCY = area of quadrilateral PXQY 

Dysfunctional: LaTeX using MathJax on Blogger

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Chinese Remainder Theorem

I want to discuss the 'ideai behind the famous Chinese Remainder Theorem. Let us leave the jargon and start our exploration by a problem.


Find a number that leaves remainder 1, when divided by 5, 7 and 13.

Clearly such a number can be found by trial. A stupid method is to check out all the numbers (at least some them) which leaves 1 as remainder when divided by 5 (we choose 5 because it is the smallest).

6, 11, 16, 21, 26, 31, 36, 41,...

The above sequence of numbers leave 1 as a remainder when divided by 5. Now the second condition is that, our required number must generate 1 as a remainder when divided by 7. Clearly that number is one of the numbers of the above sequence. We check out the remainders when divided by 7.

Number	Remainder when divided by 7
6	6
11	4
16	2
21	0
26	5
31	3
36	1 (WHOA!)

So 36 is the first number that fulfills the second condition. It leaves remainder 1 when divided by 5 as well as 13. Note that 36 is the first number with such a character. 71, 106 etc. are other numbers with such characteristic.

Anyways, we move to the third condition now. The number must leave a remainder 1 when divided by 13. A little introspection will bring the idea of product + 1 to the forefront. Note that 36 = 5*7 + 1; 70=5*7(*2)+1; 106 = 5*7(*3)+1; i.e. each of the numbers which leaves remainder 1 when divided by 5 and 7, contains 5 and 7 (as prime factors) and 1 more (rightly so!).

Therefore our required number is 13*5*7*(any number from 1 toward infinity) +1. In short we write 455k + 1 where k is any natural number.

Now we move forward with the second problem.


Find all integers that leave a remainder of 3 when divided by 5, a remainder of 5 when divided by 7, and a remainder of 7 when divided by 11. 

A brain-breaker idea will be to check out all the numbers that produce a remainder 3 when divided by 5. The following is the least of such numbers.

3, 8, 13, 18, 23, 28, 33, 38,

Then we shoot the numbers of the above sequence by 7 and find that 33 is a number that produces remainder 5 when divided by 7 (33=4*7 + 5). Can we guess a second number that produces remainder 5 when divided by 7? Clearly 68 is a second number with the same feature. Lets try 1 more (and while we do this lets us ask our brain the method it is using to compute the number). 103! Yes, we add 35 to the preceeding number (we still need to REASON why this works). But before that let us assure ourselves that 103 = 5*20 + 3 and 103 = 7*14 + 5. Also we need to be convinced that when we jumped 35 we did not miss any number with 3 and 5 as remainders when divided by 5 and 7 respectively.

Number	Remainder when divided by 7
3	3
8	1
13	6
18	4
23	2
28	0
33	5
38	3
43	1
48	6
53	4
58	2
63	0
68	5
73	3
78	1
83	6
88	4
93	2
98	0
103	5

So we did not miss any one number.Hmm, that is good.And one more observation. The remainders are repeating (3164205 3164205 ... so on). We will learn later that is a beautiful (and logical) consequence of the concept of divison.

So far so good. Our brain tells us: Find the first number with the character (remainder 3, 5, 7 when divided by 5, 7, 11) and add 5*7*11 to that to find all such numbers. The challenge is however to find that FIRST NUMBER.

Halayudha and Meru Prastara

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INDIAN MATH IVY

Indian Statistical Institute (I.S.I.), Chennai Mathematical Institute (C.M.I.) and Institute of Mathematics and Application (I.M.A.) can be regarded as three Indian institutions that provided world class mathematics course at undergraduate level. The B.Stat Course at I.S.I. is also world famous. The courses at C.M.I. and I.M.A. have computer science as second major.

Each of these three institutes conduct entrance test at the month of May. The prospective students are expected to know more rigorous mathematics than the IIT's. Apart from High School math, the entrance test syllabus includes Math Olympiad curriculum- plane geometry, combinatorics, number theory, inequalities, functional equation, polynomials and more.

Indian National Math Olympiad qualified students are not required to take the Entrance test. They will directly appear for the interview. Yes, there is an interview after the entrance test where students are asked math related questions.

The entrance tests are on Mathematics only. There is no age bar and non science students may also sit in the test provided he/she had pure maths in High School. The alumni of these three institutes are working in the most prestigious labs, universities and corporate sectors of the world. India hopes to regenerate it's mathematical fervor through state-of-art institutes like these.

Preparatory guidance for I.S.I., C.M.I. and I.M.A. entrances (and Math Olympiads) are difficult to find. Check out our website for some help.