As promised, this is the \( 4 \)th instalment, following the 3rd.

Let us start with the solution to the second part of the problem that was left to the reader:

So remember that \( A \) is smart. She wants to avoid the above case where she loses. So she says: "Aha! I only write \( 0s \), in that way,

Iwill win!" Well, sad life A, because you don't. Because if A only writes "0s", using the same argument above we can in fact make B write \( 1998 \) times \( 0 \) and then \( 3 \) times \( 1 \), which after interpreting in binary is \( 21 \) so B wins again. □

Recall the method of **infinite descent** or more commonly known as **monovariants** introduced in the 2nd instalment.

Let us now apply infinite descent to a much more difficult problem.

*Example 3*: (IMO 1986) We assign an integer to each vertex of a regular pentagon, so that the sum of all of them is positive. We are allowed to perform a transformation if three consecutive vertices have assigned numbers \( x,y, z \) respectively and \( y < 0 \), if so, then we can change the numbers \( (x,y,z) \) to \( (x+y, -y, z+y) \). This process can be continued as long as one of the numbers is negative. Does this process always come to an end?

*Solution*: Don't freak out that this is from IMO.

We label the pentagon with \( a,b,c,d,e \) as shown.

So we need to find a certain invariance. In fact the only "basic" thing we have is the sum of all numbers. Also notice that \( (x+y) + (-y) + (z+y) = (x+y+z) \) so the sum remains constant. This does not tell us whether the process ends. We need something that decreases. Now from the sum, it is certainly quite motivated to consider the average. So if every number is very close to the average then none would be negative and the process would end. So we want to search for something that will increase as the numbers are different to the average. Naturally, we might think of \( |x-y| \) but the absolute value can be quite unpredictable at times, so we consider the "nicer" alternative **squares** \( (x-y)^2 \).

So referring back to the notation at the beginning, consider a function \( f(a,b,c,d,e) = (a-c)^2 + (c-e)^2 + (e-b)^2 + (b-d)^2 + (d-a)^2 \). Clearly \( f(a,b,c,d,e) ≥ 0 \). Suppose WLOG that \( c < 0 \) and we do the transformation in the problem. Then:

\( f(a,b+c, -c, d+c, e) = (a+c)^2 + (-c-e)^2 + (e-b-c)^2+ (b-d)^2 + (d+c-a)^2 \)

\( = [(a-c)^2 +4ac] + [(c-e)^2 + 4ce] + [(e-b)^2 - 2ec + 2bc + c^2] + (b-d)^2 + [(d-a)^2 - ac + 2cd + c^2] \)

\( = f(a,b,c,d,e) + 2c(a+b+c+d+e) \)

Since \( c<0 \) and we assumed that \( a+b+c+d+e > 0 \), we have that \( 2c(a+b+c+d+e) < 0 \) so \( f(a,b,c,d,e) \) decreases at each step. By infinite descent we are done. □

**Question for the reader**: Can you find a monovariance using \( |x-y| \) as sugeested in the solution above?

*Hint*: Consider including *pairwise sums* then testing and etc.

Look forward to the next post :)

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TopNewestHere is an alternate solution for the general case of \(n\)-gon from "Mathematical Puzzles: A Connoisseur's Collection" by Peter Winkler. This solution was devised by Bernard Chazelle. I quote:

Let \(x(0),\ldots ,x(n-1)\) be the labels, summing to \(s>0\), with indices taken modulo \(n\). Define the doubly infinity sequence \(b(\cdot)\) by \(b(0)=0\) and \(b(i)=b(i-1)+x(i\pmod{n})\). The sequence \(b(\cdot )\) is not periodic, but periodically ascending: \(b(i+n)=b(i)+s\).

If \(x(i)\) is negative, \(b(i)<b(i-1)\), and flipping \(x(i)\) has the effect of switching \(b(i)\) with \(b(i-1)\), so that they are now in ascending order. It does the same for all pairs \(b(j)\), \(b(j-1)\) shifted from these by multiples of \(n\). Thus, flipping labels amounts to sorting \(b(\cdot )\) by adjacent transpositions!

To track the progress of this sorting process, we need a finite parameter \(P\) that measures the degree to which \(b(\cdot )\) is out of order. To obtain this, let \(i^+\) be the number of indices \(j > i\) for which \(b(j) < b(i)\), and \(i^-\) the number of indices \(j < i\) for which \(b(j) > b(i)\). Note that \(i^+\) and \(i^-\) are finite and depend only on \(i\pmod{n}\). Observe also that \(\displaystyle\sum_{i=0}^{n-1}{i^+}=\sum_{i=0}^{n-1}{i^-}\); we let this sum be our magic parameter \(P\).

When \(x(i+1)\) is flipped, \(i^+\) decreases by 1, and every other \(j^+\) in unchanged. Thus, \(P\) does down by

exactly 1. When \(P\) hits 0, the sequence is filly sorted, so all the labels are non-negative and the process terminates.We have shown more than asked: The process terminates in exactly the same number (\(P\)) of steps regardless of choices, and moreover, the final configuration in independent of choices as well! The reason is that there is only one sorted version of \(b(\cdot )\); entry \(b(i)\) from the original sequence must wind up in position \(i+i^+-i^-\) when the sorting is complete. \(\Box\).

I don't understand this solution, but I hope I have enlightened all of you who do.

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I don't really understand the solution of the problem, from the last note. Wasn't A going to win if it's possible to write the sum as the sum of two perfect squares? If they end with 21, as you suggest in the solution(?), wouldn't A win?

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I'm sorry, but how is 21 a sum of 2 perfect squares?

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Of course, now afterwards it´s hard to figure out how i thougt, but probably i considered 21 as a prime number, and then noticed that it was congruent to 1 modulo 4. My bad, sorry.

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