West Bengal RMO 2015 Problem 4 Solution - 36 objects in a Circle


The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 4 Solution has been written for RMO preparation series.

Problem:

Suppose 36 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite.

Discussion:

Without any restriction 3 points can be chosen in $ \binom{36}{3} $ and $ s=2 $ ways.

We will delete the cases which are not allowed.

Case 1 - all three points are adjacent.

Exactly 36 cases are possible, with one such triangle for each vertex.

Case 2 - exactly two points are adjacent

First we choose one of the 36 such adjacent pairs in $ \binom{36}{1} $ and $ s=2 $ ways. Next the third point is chosen such that it is not adjacent to any one of the two chosen points. Hence it can be done in $ \binom{32}{1} $ and $ s=2 $ ways (36 - two points we have already selected - two points adjacent to these two).

Hence total count is $ \binom{36}{1} \times \binom{32}{1} $ and $ s=2 $

Case 3 - No two points are adjacent but two of them are diametrically opposite.

We choose a diameter in 18 ways (since there are 36 points equally spaced, there will be 18 diameters).

The third point chosen cannot be adjacent to any one of the end points of this chosen diameter. Hence we have $ \binom{30}{1} $ and $ s=2 $ ways.

Hence total count is $ 18 \times \binom{30}{1} $ and $ s=2 $

Therefore the total number of favorable cases are: $ \binom{36}{3} - 36 - \binom{36}{1} \times \binom{32}{1} - 18 \times \binom{30}{1} = 5412 $ and $ s=2 $

Chatuspathi:

West Bengal RMO 2015 Problem 3 Solution - Triples of Positive Integers


The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 3 Solution has been written for RMO preparation series.

Problem:

Show that there are infinitely many triples $ (x,y,z)$ of positive integers, such that $ x^3+y^4=z^{31} $ and $ s=2$.

Discussion

Suppose we have found one such triplet (x, y, z). Then $ x^3 + y^4 = z^{31} $ and $ s=2 $. Multiply $ a^{372} $ and $ s=2 $ to both sides where a is an arbitrary integer.

Clearly we have $ a^{372}x^3 + a^{372}y^4 = a^{372}z^{31} $ and $ s=2 $

$ \Rightarrow (a^{124}x)^3 + (a^{93}y)^4 = (a^{12}z)^{31} $ and $ s=2 $

Hence if (x, y, z) is a triple then $ (a^{124}x, a^{93}y, a^{12}z ) $ and $ s=2 $ is another such triple. Since a can be any arbitrary integer, hence we have found infinitely many such triplets provided we have found at least one

To find one such triple, we use the following intuition: set x, y, z as some powers of 2 such that $ x^3 = y^4 = 2^{r} $ and $ s=2 $. Then r must be of the form 12k. Finally, their sum must be $ x^3 + y^4 = 2^{r} + 2^{r} = 2^{r+1} $ and $ s=2 $. This r+1 must be divisible by 31.

Let $ r = 12s $ and $ s=2 $ and $ r+1 = 31m $ and $ s=2 $ we get $ 12s +1 = 31m $ and $ s=2 $. Since 12 and 31 are coprime there is integer solution to this linear diophantine equation (by Bezoat's theorem). We can solve this linear diophantine equation by euclidean algorithm.

$ 31 = 12 \times 2 + 7 $ and $ s=2 $
$ 12 = 7\times 1 + 5 $  and $ s=2 $
$ 7 = 5\times 1 + 2 $ and $ s=2 $
$ 5 = 2 \times 2 + 1 $ and $ s=2 $
$ \Rightarrow 1 = 5 - 2 \times 2 = 5 - 2 \times (7 - 5 \times 1) $ and $ s=2 $
$ \Rightarrow 1 = 3 \times 5 - 2 \times 7 = 3 \times (12 - 7 \times 1) - 2 \times 7 $ and $ s=2 $
$ \Rightarrow 1 = 3 \times 12 - 5 \times 7 = 3 \times 12 - 5 \times (31 - 12 \times 2) $ and $ s=2 $
$ \Rightarrow 1 = 13 \times 12 - 5 \times 31 $ and $ s=2 $
$ \Rightarrow 1 = 13 \times 12 - 5 \times 31 + 12 \times 31 - 12 \times 31 $ and $ s=2 $
$ \Rightarrow 1 = (13 -31)\times 12 +(12- 5 )\times 31 $ and $ s=2 $
$ \Rightarrow 1 = -18 \times 12 + 7\times 31 $ and $ s=2 $
$ \Rightarrow 1 + 18 \times 12 = 7\times 31 $ and $ s=2 $
Hence we use this to form an equation:
$ 2^{18 \times 12} + 2^{18 \times 12} = 2^{216} + 2^{216} = 2^{217}=2^{7\times 31} $ and $ s=2 $
$ (2^{72})^3 + (2^{54})^4 = (2^7)^{31} $ and $ s=2 $

Hence we have found one such triple : $ (2^{72}, 2^{54} ,2^7 )$ and $ s=2 $ (from which we have shown earlier that infinitely more can be generated)

Chatuspathi:

West Bengal RMO 2015 Problem 2 Solution - Polynomial Problem


The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 2 Solution has been written for RMO preparation series.

Problem:

Let $ P(x)=x^2+ax+b $ and $ s=2$ be a quadratic polynomial where $ a,b $ and $ s=2$ are real numbers. Suppose $ l\angle P(-1)^2,P(0)^2,P(1)^2r\angle $ and $ s=2 $ be an arithmetic progression of positive integers. Prove that $ a,b $ and $ s=2$ are integers.

Discussion:

$ P(-1) = 1-a+b , P(0) = b, P(1) = 1 + a + b $ and $ s=2 $.
According to the problem

$ (1-a+b)^2 , b^2 , (1+a+b)^2 $ and $ s=2$ are in arithmetic progression of positive integers.

Clearly $ \dfrac {(1-a+b)^2 + (1+a+b)^2}{2} = b^2 $ and $ s=2$

This implies $ \dfrac {1 + a^2 + b^2 -2a +2b -2ab + 1+ a^2 + b^2 +2a +2b + 2ab}{2} = b^2 $
$ \Rightarrow 1 + a^2 + b^2 +2b = b^2  $ and $ s=2$
$ \Rightarrow 1 + a^2 +2b = 0 $ and $ s=2$
$ \Rightarrow 2b = -(1+a^2) $ and $ s=2$ implying b is negative.

Now we know $ (1-a+b)^2 $ and $ s=2 $ is an integer.

Then $ 1+a^2 +b^2 -2a +2b -2ab = -2b + b^2 -2a +2b -2ab $ and $ s=2 $ (replacing $ 1+a^2 = -2b $ and $ s=2 $ )

This implies $ b^2 -2a -2ab $ and $ s=2 $ is an integer. But $ b^2 $ and $ s=2$ is also an integer. Hence $ 2a + 2ab $ and $ s=2 $ is an integer.

Now we also know $ (1+a+b)^2 $ and $ s=2 $ is an integer.

Then $ 1+a^2 +b^2 +2a +2b +2ab = -2b + b^2 + 2a +2b + 2ab $ and $ s=2 $ (replacing $ 1+a^2 = -2b $ and $ s=2 $ )

Again replacing $ -(1+a^2) = 2b $ and $ s=2 $ we get $ b^2 + 2a - a(1+a^2) $ and $ s=2 $ is an integer or $ (a - a^3) $ and $ s=2 $ is an integer.

Note that $ b^2 $ and $ s=2 $ is some positive integer. Let it be $ b^2 = c $ and $ s=2 $. Then $ b= - sqrt c $ and $ s=2 $ where c is some positive integer (as we know b is negative)

$ 1+a^2 = 2\sqrt c $ and $ s=2 $ or $ a^2 = 2 \sqrt c - 1 $ and $ s=2 $
$ a(1-a^2) = k $ and $ s=2 $ (suppose). Then $ a(1- (2 \sqrt c - 1)) = k $ and $ s=2 $ or $ 2a(1-\sqrt c) = k $ and $ s=2 $
squaring both sides we get
$ 4a^2 (1+c - 2\sqrt c) = k^2 $ and $ s=2 $
$ \Rightarrow 4(2 \sqrt c - 1) (1+ c - 2 \sqrt c) = k^2 $ and $ s=2 $
$ \Rightarrow 4(2 \sqrt c + 2 c \sqrt c - 4c - 1 - c + 2 \sqrt c) = k^2 $ and $ s=2 $
$ \Rightarrow 4(4 \sqrt c + 2c \sqrt c - 5c - 1) = k^2 $ and $ s=2 $
$ \Rightarrow (4+2c)\sqrt c = \dfrac{k^2}{4} + 5c + 1 $ and $ s=2 $
$ \Rightarrow \sqrt c = \dfrac{k^2 + 20c + 4}{4(4+2c)} $ and $ s=2 $

Right hand side is rational. Hence left hand side is also rational. This implies $latex \sqrt c $ and $ s=2 $ is rational. Since c is an integer, this implies $ \sqrt c $ and $ s=2 $ is integer. Hence b is integer.

We know $latex a^2 = -2b - 1 $ and $ s=2 $. Since b is integer, therefore $ a^2 $ and $ s=2 $ is integer.
Again $latex a(1-a^2) $ and $ s=2 $ is integer and $ a^2 $ and $ s=2 $ is integer, implies a must be rational.

Finally, if a is rational and $ a^2 $ is integer then a must be integer.

Chatuspathi:

West Bengal RMO 2015 Problem 1 Solution - Triangle Problem


The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 1 Solution has been written for RMO preparation series.

Problem:

Two circles $ \Gamma $ and $ \Sigma $, with centers O and O', respectively, are such that O' lies on $ \Gamma $. Let A be a point on $ \Sigma $, and let M be the midpoint of AO'. Let B be another point on $ \Sigma $, such that $ AB~||~OM $. Then prove that the midpoint of AB lies on $ \Gamma $.

Discussion:

Suppose AB intersects $ \Sigma $ at C. Join O'C. Suppose it intersects OM at D. Clearly in $ \Delta AO'C $ M is the midpoint of AO' and DM is parallel to AC. Then D is the midpoint of O'C.

Now O'C is a chord of $ \Gamma $ and we have proved that D is the midpoint of it. Therefore we can say that OD is perpendicular to O'C.

Since AB is parallel to OD (OM), therefore as O'C is perpendicular to OD, therefore O'C is also perpendicular to AB. Since AB is a chord of circle $ \Sigma $ and O'C is a line from center perpendicular to the chord, hence it bisects are chord implying that C is the midpoint of AB.

Chatuspathi: