Ana, Beto, Carlos, Diana, Elena and Fabian are in a circle, located in that order. Ana, Beto, Carlos, Diana, Elena and Fabian each have a piece of paper, where are written the real numbers $a,b,c,d,e,f$ respectively. At the end of each minute, all the people simultaneously replace the number on their paper by the sum of three numbers; the number that was at the beginning of the minute on his paper and on the papers of his two neighbors. At the end of the minute $2022, 2022$ replacements have been made and each person have in his paper it´s initial number. Find all the posible values of $abc+def$. $\textbf{Note:}$ If at the beginning of the minute $N$ Ana, Beto, Carlos have the numbers $x,y,z$, respectively, then at the end of the minute $N$, Beto is going to have the number $x+y+z$.
Problem
Source: 2022 Centroamerican and Caribbean Mathematical Olympiad, P2
Tags: combinatorics, algebra
01.12.2022 11:19
$\textbf{Claim 1:} \;a+b+c+d+e+f = 0$ Total sum is every time tripled. So after $2022$ replacements we have total sum equal to $3^{2022}(a+b+c+d+e+f)$ meaning that $a+b+c+d+e+f = 0$. $\textbf{Claim 2:} \;a+d = b+e = c+f = 0$ After one replacement opposite players add up to zero. This property holds until the end so it must hold at the beginning. Now, \[ abc+def = abc+(-a)(-b)(-c) = 0. \]
02.12.2022 05:31
Why didn't it just ask for a general solution for $(a,b,c,d,e,f)$? First, note that the end result after $2022$ minutes will be the same if the paper is passed to the right after each minute. Next, note that this is equivalent to taking the sum of the numbers on one's paper and the two papers to the right. Thus we can use the generating function $(1+x+x^2)^{2022}$, where the sum of the coefficients of all $x$ with the same exponent modulo $6$ gives us the amount of each number on the final paper after $2022$ minutes. Roots of unity filter (lots of tedious algebra left as an exercise to the reader, I hope I did it right :p) gives us \[(3^{2022} + 2^{2023} - 5)a + (3^{2022} + 2^{2022} - 1)(b+f) + (3^{2022} - 2^{2022} + 1)(c+e) + (3^{2022} - 2^{2023} - 1)d = 0,\]and a symmetric equation for the other $5$ variables. Sum the equations cyclically to get that the six variables sum to $0$. Sum the equations for $a$ and $d$ to get that $a=-d$. Simplify each equation to get $2a+b+f=0$, and symmetric equations for the other variables. We get $a-(-f) = (-b)-a$ and symmetric equations for the other 5 variables. Thus $a,-b,c,-d,e,-f,a$ form an arithmetic sequence, so $(a,b,c,d,e,f)=(n,-n,n,-n,n,-n)$ for any real $n$. Thus $abc+def=n^3-n^3=\boxed{0}$.
03.12.2022 23:17
I'm going to use slightly sophisticated methods, but by doing so we will not only prove that $abc+def=0$ but also we will characterize all solutions. We have the following lemma: Lemma: If $(a,b,c,d,e,f)$ and $(a',b',c',d',e',f')$ are periodic with period 2022, then $(a+a',b+b',c+c',d+d',e+e',f+f')$ also is periodic with period 2022. Proof: It follows from the linearity of the operation.$\square$ As $(a,b,c,d,e,f)$ works, then $(b,c,d,e,f,a)$, $(c,d,e,f,a,b)$, $(d,e,f,a,b,c)$, $(e,f,a,b,c,d)$ and $(f,a,b,c,d,e)$ also work. By the lemma \[(a+b+c+d+e+f,a+b+c+d+e+f,\dots, a+b+c+d+e+f)\]also works. After every minute the numbers triple, thus $a+b+c+d+e+f=0$ Applying the lemma to $(a,b,c,d,e,f)$ and $(d,e,f,a,b,c)$ we have that $(a+d,e+b,c+f,a+d,e+b,c+f)$ also works . After a minute it becomes $(0,0,0,0,0,0)$ that stays constant. Thus $d=-a$ ,$e=-b$ and $f=-c$. Now we know that the configuration has the form $(a,b,c,-a,-b,-c)$. If we apply the lemma on lists $(a,b,c,-a,-b,-c)$ and $(b,c,-a,-b,-c,a)$, we get that \[(a+b,b+c,c-a,-a-b,-b-c,a-c)\]also works. After each minute the numbers double, and therefor $a+b=b+c=c-a=0$. This implies that the configuration is of the form $(a,-a,a,-a,a,-a)$ for some real number $a$ and we are done.