# Cheenta Ganit Kendra

Passion for mathematics

## Monday, 18 November 2013

## Saturday, 15 June 2013

### Singapore Math Olympiad (Senior) 2013

- A shop sells two kind of products A and B. One day a salesman sold both A and B at the same price, $2100 to a customer. Suppose A makes a profit of 20% and B makes a loss of 20%. Then the deal

(A) make a profit of $70; (B) make a loss of $70;(C) make a profit of $175; (D) make a loss of $70; - Let f and g be functions such that for all real numbers x and y g(f(x+y)) = f(x) + (x+y)g(y).. Find the value of g(0) + g(1) + ... + g(2013)
- Each chocolate costs 1 dollar, each licorice stick costs 50 cents and each lolly costs 40 cents. How many different combinations of these three items cost a total of 10 dollars.
- Let A= {1, 2, 3, 4, 5, 6}. Find the number of distinct functions f: A → A such that f(f(f(n))) = n for all n ∈ A.
- Find the number of triangles whose sides are formed by the sides of the diagonals of a regular heptagon (7 sided polygon). (Note: the vertices of the triangle need not be vertices of the heptagon).
- Six seats are arranged in a circular table. Each seat is to be painted in red, blue or green such that any two adjacent seats have different colors. How many ways are there to paint the seats?
- ΔABC is an equilateral triangle with side length 30. Fold the triangle so that A touches a point X on BC. If BX = 6, find the value of k, where √k is the length of the crease obtained from folding.

- As shown in the figure below, circles C
_{1 }and C_{2 }of radius 360 are tangent to each other, and both tangent to the straight line l. If the circle C_{3 }is tangent to C_{1 }, C_{2 }and l, and circle C_{4 }is tangent to C_{1 }, C_{3 }and l, find the radius of C_{4 }. - Set {x} = x - ⌊x⌋, where ⌊x⌋ denotes the largest integer less than or equal to x. Find the number of real solutions to the equation {x} + {x
^{2 }} = 1 , |x| ≤ 10. - Find the value of ⌊\( ( \frac {3 + \sqrt 17 } {2} )^6 \)⌋.
- A regular dodecagon (12 sided polygon) is inscribed in a circle of radius 10. Find it's area.
- to be continued...

## Monday, 10 June 2013

### Synthesis 2013 (Reunion of Cheenta) ... revisited

The seminar began with an analysis of linear algebra (which deals with finite dimensional vector space). Sil explained why the mathematical machinery of linear algebra fails in the case of infinite dimensional vector space. Functional Analysis was developed as a discipline to address this issue and by 1970 linear partial differential equations could be understood in totality using it.

Sil mentioned that partial differential equations come up in various real life scenario. A large portion of problems are motivated by physics. The problem of non-linear partial differential equation (where partial derivatives are in multiplied form) is much more daunting than it's linear counterpart. Most mathematicians today agree that no general theory is possible to 'crack' non-linear partial differential equations.

The seminar concluded with questions from the audience.

Western Blot is one of the most important techniques for protein detection. Raikamal Paul, a master of science in microbiology from Vellore Institute of Technology, participated in extensive lab work at Indian Institute of Chemical Biology this year. She delivered a seminar on this technique on Sunday which took us to the world of practical research work in genetics.

The seminar was supplemented by presentation and board work.

The lectures covered the following topics:

The law of gravitation, an example of physical law

The relation of mathematics and physics

The great conservation principles

Symmetry in physical law

The distinction of past and future

Probability and uncertainty - the quantum mechanical view of nature

Seeking new laws

At Cheenta Reunion, we watched the first lecture (as recorded by BBC). It discussed "The law of gravitation, an example of physical law".

Anubhav Chatterjee presented a Grand Quiz Show, which included Puzzle Round, Audio/Video Round, Straight Questions and Rapid Fire Round. Subhajit and Akash won the tournament with 187 points. The puzzles were given to the teams at the very beginning of the day (at 10 AM). The rest of the rounds happened in the evening.

There were some short 'talks'. Somrik spoke on Inner Workings of L'Hopital's rule. Subhajit discussed his experience in I.S.I. B.Math interview.

Last but not the least, there was good food. Beginning with cold drinks and potato chips, we ended up with Biriyani, Chicken Tikka Masala, Chicken Tanduri Masala, Lachha Paratha, Rumali Roti, Paneer, Mutton Kasa, Fried Rice. A local restaurant, Chhota Elaichi, served us lunch.

বৃহত্তর মানচিত্র দেখুন

The Reunion concluded at 7 PM. We hope to see you at the reunion next summer (and coming winter).

## Saturday, 8 June 2013

## Wednesday, 15 May 2013

## Monday, 13 May 2013

## Sunday, 12 May 2013

### ISI 2013 B.Math and B.Stat Subjective Solutions

**1. For how many values of N (positive integer) N(N-101) is a square of a positive integer?**

**Solution:**

(We will not consider the cases where N = 0 or N = 101)

\( N(N-101) = m^2 \)

Roots of this quadratic in N is

=> \( \frac{101 \pm \sqrt { 101^2 + 4m^2}}{2} \)

The discriminant must be square of an odd number in order to have integer values for N.

Thus \( 101^2 + 4m^2 = (2k + 1)^2 \)

=> \( 101^2 = (2k +1)^2 - 4m^2 \)

=> \( 101^2 = (2k +2m + 1)(2k - 2m + 1) \)

Note that 101 is a prime number

Hence we have two possibilities

Case 1:

\( 2k + 2m + 1 = 101^2; 2k - 2m + 1 = 1 \)

Subtracting this pair of equations we get \(4m = 101^2 - 1\) or \(4m = 100 \times 102\) or m = 50*51

This gives N = 2601 (ignoring extraneous solutions)

Case 2:

\(2k + 2m + 1 = 101 ; 2k - 2m + 1 = 101 \) which gives m = 0 or N = 101. This solution we ignore as it makes N(N- 101) = 0 (a non positive square).

Hence the only solution is N = 2601 and there are no other values of N which makes N(N-101) a perfect square.

**2.**

## Saturday, 9 February 2013

### AMC 10 (2013) Solutions

12. In \(\triangle ABC, AB=AC=28\) and BC=20. Points D,E, and F are on sides \(\overline{AB}, \overline{BC}\), and \(\overline{AC}\), respectively, such that \(\overline{DE}\) and \(\overline{EF}\) are parallel to \(\overline{AC}\) and \(\overline{AB}\), respectively. What is the perimeter of parallelogram ADEF?

\(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad \)

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).

Hence perimeter = 2(AF + EF).

Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.

Hence triangle CEF is isosceles. Thus EF = CF.

Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = \(2 \times 28\) = 56.

Ans. (C) 56

\(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad \)

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).

Hence perimeter = 2(AF + EF).

Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.

Hence triangle CEF is isosceles. Thus EF = CF.

Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = \(2 \times 28\) = 56.

Ans. (C) 56

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