Elementary problems in Ramsey number theory for RMO

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let $\{ A, B, C, D, E, F\}$ be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) $R(p,1)=1$ , and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that $R(3,3)=6$ .

Solution 5:

Question 1 and its proof in this blog article imply that $R(3,3) \leq 6$ .

To prove that $R(3,3)>5$ , it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

$R(m,n) \leq R(m-1,n) + R(m,n-1)$ .

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let $p \equiv R(m-1,n)$ , $q=R(m,n-1)$ and $r \equiv p + q$ . Consider a group $\{ 1,2, 3, \ldots, r\}$ of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have $r-1$ people, so either L has at least p people or M has at least q people. (a) If L has $p \equiv R(m-1,n)$ people, then, by definition, it contains a subset of $(m-1)$ people known to one another or it contains a subset of n people unknown to one another. In the former case, the $(m-1)$ people and person 1 constitute m people known to one another.

Thus, in their case, a group of $R(m-1,n) + R(m,n-1)$ people necessarily includes m mutual acquaintances or n mutual strangers. That is, $R(m,n) \leq R(m-1,n) + R(m,n-1)$ .

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then $R(m,n) \leq { {m+n-2} \choose {m-1}}$ — a non-recursive upper bound.

Solution 7:

When $m=2$ , or $n=2$ , (i) holds with equality (see problem 4 in this blog article). The proof is by induction on $k=m+n$ . As we have just seen, the result is true when $k=4$ . Assume the result true for $k-1$ . Then,

$R(m-1,n) \leq {{m+n-3} \choose {m-2}}$ and $R(m,n-1) \leq {{m+n-3} \choose {m-1}}$

Now, Pascal’s identity gives:

${{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1}$ so that $R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}$

But, from the previous question and its solution, we get $R(m,n) \leq R(m-1,n) + R(m, n-1)$

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

How to find the number of proper divisors of an integer and other cute related questions

Posted by Nalin Pithwa

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, $441000 = (2^{3})(3^{2})(5^{3})(7^{2})$ . Any divisor, proper or improper, of the given number must be of the form $(2^{a})(3^{b})(5^{c})(7^{d})$ where $0 \leq a \leq 3$ , $0 \leq b \leq 2$ , $0 \leq c \leq 3$ , and $0 \leq d \leq 2$ . In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be $(4)(3)(4)(3)-2=142$ .

Question 2:

Count the proper divisors of an integer N whose prime factorization is: $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}$

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is $(\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2$ , where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that $m>1, n>1$ , and the GCD of m and n is 1.

Solution 3:

Consider the set $A = {2^{3}, 3^{2}, 5^{3}, 7^{2}}$ associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

$A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\}$ ,

and $A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}$

Hence, the required answer is $4+3=2^{4-1}+1=7$ .

Question 4:

Generalize the above problem by showing that any integer has $2^{k-1}-1$ factorizations into relatively prime pairs m, n ( $m>1, n>1$ ).

Solution 4:

Proof by mathematical induction on k:

For $k=1$ , the result holds trivially.

For $k=2$ , we must prove that a set of k distinct elements, $Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\}$ has $2^{k-1}-1$ sets. Now, one partition of Z is

$Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W$

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in $2^{k-2}-1$ ways — and then, including $a_{k}$ in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

$1 + (2^{k-2})(2)=2^{k-1}-1$ . QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case.

Cheers,

Nalin Pithwa.

Practice questions based on combinatorics for RMO Training and IITJEE Mathematics

Posted by Nalin Pithwa

Question 1:

Prove that if n is an even integer, then

$\frac{1}{(1!)(n-1)!} + \frac{1}{3! (n-3)!} + \frac{1}{5! (n-5)!} + \ldots + \frac{1}{(n-1)! 1!} = \frac{2^{n-1}}{n!}$

Question 2:

If ${n \choose 0}$ , ${n \choose 1}$ , ${n \choose 2}$ , …. ${n \choose n}$ are the coefficients in the expansion of $(1+x)^{n}$ , when n is a positive integer, prove that

(a) ${n \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^{r}{n \choose r} = (-1)^{r}\frac{(n-1)!}{r! (n-r-1)!}$

(b) ${n \choose 0} - 2{n \choose 1} + 3{n \choose 2} - 4{n \choose 3} + \ldots + (-1)^{n}(n+1){n \choose n}=0$

(c) ${n \choose 0}^{2} - {n \choose 1}^{2} + {n \choose 2}^{2} - {n \choose 3}^{2} + \ldots + (-1)^{n}{n \choose n}^{2}=0$ , or $(-1)^{\frac{n}{2}}{n \choose {n/2}}$ , according as n is odd or even.

Question 3:

If $s_{n}$ denotes the sum of the first n natural numbers, prove that

(a) $(1-x)^{-3}=s_{1}+s_{2}x+s_{3}x^{2}+\ldots + s_{n}x^{n-1}+\ldots$

(b) $2(s_{1}s_{2m} + s_{2}s_{2n-1} + \ldots + s_{n}s_{n+1}) = \frac{2n+4}{5! (2n-1)!}$

Question 4:

If $q_{n}=\frac{1.3.5.7...(2n-1)}{2.4.6.8...2n}$ , prove that

(a) $q_{2n+1}+q_{1}q_{2n}+ q_{2}q_{2n-1} + \ldots + q_{n-1}q_{n+2} + q_{n}q_{n+1}= \frac{1}{2}$

(b) $2(q_{2n}-q_{1}q_{2m-1}+q_{2}q_{2m-2}+\ldots + (-1)^{m}q_{m-1}q_{m+1}) = q_{n} + (-1)^{n-1}{q_{n}}^{2}$ .

Question 5:

Find the sum of the products, two at a time, of the coefficients in the expansion of $(1+x)^{n}$ , where n is a positive integer.

Question 6:

If $(7+4\sqrt{3})^{n} = p + \beta$ , where n and p are positive integers, and $\beta$ , a proper fraction, show that $(1-\beta)(p+\beta)=1$ .

Question 7:

If ${n \choose 0}$ , ${n \choose 1}$ , ${n \choose 2}$ , …,, ${n \choose n}$ are the coefficients in the expansion of $(1+x)^{n}$ , where n is a positive integer, show that

${n \choose 1} - \frac{{n \choose 2}}{2} + \frac{{n \choose 3}}{3} - \ldots + \frac{(-1)^{n-1}{n \choose n}}{n} = 1 + \frac{1}{2} + \frac {1}{3} + \frac{1}{4} + \ldots + \frac{1}{n}$ .

That’s all for today, folks!

Nalin Pithwa.

Combinatorics for RMO : some basics and examples: homogeneous products of r dimensions

Posted by Nalin Pithwa

Question:

Find the number of homogeneous products of r dimensions that can be formed out of the n letters a, b, c ….and their powers.

Solution:

By division, or by the binomial theorem, we have:

$\frac{1}{1-ax} = 1 + ax + a^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-bx} = 1+ bx + b^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-cx} = 1 + cx + c^{2}x^{2} + c^{3}x^{3} + \ldots$

Hence, by multiplication,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \times \ldots$

$= (1+ax + a^{2}x^{2}+a^{3}x^{3}+ \ldots)(1+bx + b^{2}x^{2} + b^{3}x^{3}+ \ldots)(1+cx + c^{2}x^{2} + c^{3}x^{3}+ \ldots)\ldots$

$= 1 + x(a + b + c + \ldots) +x^{2}(a^{2}+ab+ac+b^{2}+bc + c^{2} + \ldots) + \ldots$

$= 1 + S_{1}x + S_{2}x^{2} + S_{3}x^{3} + \ldots$ suppose;

where $S_{1}$ , $S_{2}$ , $S_{3}$ , $\ldots$ are the sums of the homogeneous products of one, two, three, … dimensions that can be formed of a, b, c, …and their powers.

To obtain the number of these products, put a, b, c, …each equal to 1; each term in $S_{1}$ , $S_{2}$ , $S_{3}$ , …now becomes 1, and the values of $S_{1}$ , $S_{2}$ , $S_{3}$ , …so obtained give the number of the homogeneous products of one, two, three, ….dimensions.

Also,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \ldots$

becomes $\frac{1}{(1-x)^{n}}$ , or $(1-x)^{-n}$

Hence, $S_{r} =$ the coefficient of $x^{r}$ in the expansion of $(1-x)^{-n}$

$= \frac{n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}= \frac{(n+r-1)!}{r!(n-1)!}$

Question:

Find the number of terms in the expansion of any multinomial when the index is a positive integer.

Answer:

In the expansion of $(a_{1}+ a_{2} + a_{3} + \ldots + a_{r})^{n}$

every term is of n dimensions; therefore, the number of terms is the same as the number of homogeneous products of n dimensions that can be formed out of the r quantities $a_{1}$ , $a_{2}$ , $a_{3}$ , … $a_{r}$ , and their powers; and therefore by the preceding question and solution, this is equal to

$\frac{(r+n-1)!}{n! (r-1)!}$

A theorem in combinatorics:

From the previous discussion in this blog article, we can deduce a theorem relating to the number of combinations of n things.

Consider n letters a, b, c, d, ….; then, if we were to write down all the homogeneous products of r dimensions, which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice, …up to r times.

Therefore, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n letters, and therefore equal to $\frac{(n+r-1)!}{r!(n-1)!}$ , or ${{n+r-1} \choose r}$ .

That is, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of combinations of $n+r-1$ things r at a time when repetitions are NOT allowed.

We conclude this article with a few miscellaneous examples:

Example 1:

Find the coefficient of $x^{r}$ in the expansion of $\frac{(1-2x)^{2}}{(1+x)^{3}}$

Solution 1:

The expression $= (1-4x+4x^{2})(1+p_{1}x+p_{2}x^{2}+ \ldots + p_{r}x^{r}+ \ldots)$ , suppose.

The coefficients of $x^{r}$ will be obtained by multiplying $p_{r}$ , $p_{r-1}$ , $p_{r-2}$ by 1, -4, and 4 respectively, and adding the results; hence,

the required coefficient is $p_{r} - 4p_{r-1}+4p_{r-2}$

But, with a little work, we can show that $p_{r} = (-1)^{r}\frac{(r+1)(r+2)}{2}$ .

Hence, the required coefficient is

$= (-1)^{r}\frac{(r+1)(r+2)}{2} - 4(-1)^{r-1}\frac{r(r+1)}{2} + 4 (-1)^{r-2}\frac{r(r-1)}{2}$

$= \frac{(-1)^{r}}{2}\times ((r+1)(r+2) + 4r(r+1) + 4r(r-1))$

$= \frac{(-1)^{r}}{2}(9r^{2}+3r+2)$

Example 2:

Find the value of the series

$2 + \frac{5}{(2!).3} + \frac{5.7}{3^{2}.(3!)} + \frac{5.7.9}{3^{3}.(4!)} + \ldots$

Solution 2:

The expression is equal to

$2 + \frac{3.5}{2!}\times \frac{1}{3^{2}} + \frac{3.5.7}{3!}\times \frac{1}{3^{3}} + \frac{3.5.7.9}{4!}\times \frac{1}{3^{4}} + \ldots$

$= 2 + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times \frac{2^{2}}{3^{2}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times \frac{2^{3}}{3^{3}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times \frac{2^{4}}{3^{4}} + \ldots$

$= 1 + \frac{\frac{3}{2}}{1} \times \frac{2}{3} + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times (\frac{2}{3})^{2} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times (\frac{2}{3})^{3} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times (\frac{2}{3})^{4} + \ldots$

$= (1-\frac{2}{3})^{\frac{-3}{2}} = (\frac{1}{3})^{-\frac{3}{2}} = 3^{\frac{3}{2}} = 3 \sqrt{3}$ .

Example 3:

If n is any positive integer, show that the integral part of $(3+\sqrt{7})^{n}$ is an odd number.

Solution 3:

Suppose I to denote the integral and f the fractional part of $(3+\sqrt{7})^{n}$ .

Then, $I + f = 3^{n} + {n \choose 1}3^{n-1}\sqrt{7} + {n \choose 2}3^{n-2}.7 + {n \choose 3}3^{n-3}.(\sqrt{7})^{3}+ \ldots$ …call this relation 1.

Now, $3 - \sqrt{7}$ is positive and less than 1, therefore $(3-\sqrt{7})^{n}$ is a proper fraction; denote it by $f^{'}$ ;

Hence, $f^{'} = 3^{n} - {n \choose 1}.3^{n-1}.\sqrt{7} + {n \choose 2}.3^{n-2}.7 - {n \choose 3}.3^{n-3}.(\sqrt{7})^{3}+ \ldots$ …call this as relation 2.

Add together relations 1 and 2; the irrational terms disappear, and we have

$I + f + f^{'} = 2(3^{n} + {n \choose 2}.3^{n-2}.7+ \ldots ) = an even integer$

But, since f and $f^{'}$ are proper fractions their sum must be 1;

Hence, I is an odd integer.

Hope you had fun,

Nalin Pithwa.

Science Lives: Laslo Lovasz: Discrete Maths, Combinatorics and Computer Science

Posted by Nalin Pithwa

Thanks to Simon Foundation : a youtube video.

Inequalities and mathematical induction: RMO sample problems-solutions

Posted by Nalin Pithwa

Problem:

1. Prove the inequality —- $2^{n}(n!)^{2} \leq (2n)!$ for all natural numbers greater than or equal to 1.

Proof 1:

First consider the following: $2.6. 10.14 \ldots (4n-2)=\frac{(2n)!}{n!}$ . Let us prove this claim first and then use it to prove what is asked: Towards, that end, consider

$RHS = \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)\ldots 4.2.1}{1.2.3.4\ldots (n-1)n}=LHS$ , cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! 🙂 )

In the given inequality:

we need to prove $2^{n}(n!)^{2} \leq (2n)!$

consider $2.6.10.14. \ldots (4n-2)=\frac{(2n)!}{n!}$ where

LHS = $(2.1)(2.3) (2.5) (2.7) \ldots 2(n-1)$ , this is an AP with first term 2 and nth term $(4n-2)$ and common difference 4; there are n factors “2”; hence, we $2^{n}1.3.5.7 \ldots (n-1)=\frac{(2n)!}{n!}$ so we get

$2^{n} (n!) 1.3.5.7.\ldots (n-1) = (2n)!$ ; multiplying and dividing RHS of this by $2.4.6.8.\ldots n$ , we get the desired inequality. Remember the inequality is less than or equal to.

Problem 2:

Establish the Bernoulli inequality: If $(1+a) > 0$ , then $(1+a)^{n} \geq 1+na$ .

Solution 2:

Apply the binomial theorem, which in turn, is proved by mathematical induction ! 🙂

Problem 3:

For all natural numbers greater than or equal to 1, prove the following by mathematical induction:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{n^{2}} \leq 2-\frac{1}{n}$

Proof 3:

Let the given proposition be P(n).

Step 1: Check if P(1) is true. Towards that end:

$LHS=\frac{1}{1^{2}}=1$ and $latex RHS=2-\frac{1}{1}=2-1=1$ and hence, P(1) is true.

Step 2: Let P(n) be true for some $n=k$ , $k \in N$ . That is, the following is true:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} \leq 2 -\frac{1}{k}$

Add $\frac{1}{(k+1)^{2}}$ to both sides of above inequality, we get the following:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} + \frac{1}{(k+1)^{2}} \leq 2-\frac{1}{k}+\frac{1}{(k+1)^{2}}$

Now, the RHS in above is $2-\frac{1}{k} +\frac{1}{(k+1)^{2}}=2-\frac{k^{2}+k+1}{k(k+1)^{2}}$ . We want this to be less than or equal to $2-\frac{1}{k+1}$ . Now, $k \in N$ , $k>1$ , so what we have to prove is the following:

$-\frac{k^{2}+k+1}{k(k+1)^{2}} \leq -\frac{1}{k+1}$ , that is, we want to prove that

$(k+1)(k^{2}+k+1) \geq k(k^{2}+2k+1)$ , that is, we want $k^{3}+k^{2}+k+k^{2}+k+1 \geq k^{3}+2k^{2}+k$ , that is, we want $k+1 \geq 0$ , which is obviously true. QED.

Cheers,

Nalin Pithwa.

RMO Training: more help from Nordic mathematical contest

Posted by Nalin Pithwa

Problem:

32 competitors participate in a tournament. No two of them are equal and in a one against one match the better always wins. (No tie please). Show that the gold, silver and bronze medal can be found in 39 matches.

Solution:

We begin by determining the gold medallist using classical elimination, where we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners in the second round, then 2-semifinal matches and finally one match making 31 matches altogether.

Now, the second best player must have at some point lost to the best player, and as there were 5 rounds in the elimination, there are 5 candidates for the silver medal. Let $C_{i}$ be the candidate who lost to the gold medalist in round i. Now, let $C_{1}$ and $C_{2}$ play, the winner play against $C_{3}$ , and so forth. After 4 matches, we know the silver medalist; assume this was $C_{k}$ .

Now, the third best player must have lost against the gold medalist or against $C_{k}$ or both. (If the player had lost to someone else, there would be at least three better players.) Now, $C_{k}$ won k-1 times in the elimination rounds, the 5-k players $C_{k+1}\ldots C_{5}$ and if k is greater than one, one player $C_{j}$ with $j<k$ . So there are either $(k-1)+(5-k)=4$ or $(k-1)+(5-k)+1=5$ candidates for the third place. At most 4 matches are again needed to determine the bronze winners.