Group theoretic solution to an IMO problem Is there a (strictly) group theoretic interpretation (and possibly a solution) to this problem (taken from the 27th IMO)?
"To each vertex of a regular pentagon an integer is assigned in such a way that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y <0$, then the following operation is allowed: the numbers $x,y,z$ are replaced by $x+y$, $-y$, $z+y$, respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps". 
 A: The group theoretic interpretation is as the numbers game for the Coxeter group of type $\widetilde{A}_4$, with generators $s_1,s_2,s_3,s_4,s_5$, where the relations are 
$$s_1^2=s_2^2=s_3^2=s_4^2=s_5^2=1\mathrm{,}$$
$$(s_is_j)^2=1$$
if $i\not\equiv j\pm 1\pmod{5}$, and
$$(s_is_{i+1})^3=1$$
for all $i$, where $i+1$ is taken modulo $5$. This is an infinite group. This group acts on the vector space of all functions from the vertices $v_1,v_2,v_3,v_4,v_5$ of the pentagon to $\mathbb{R}$. If $f$ is such a function, then
$$(s_if)(v_j)=f(v_j)$$ 
if $j\not\equiv i\pm 1\pmod{5}$,
$$(s_if)(v_j)=-f(v_j)$$
if $i=j$, and 
$$(s_if)(v_j)=f(v_i)+f(v_j)$$
if $j\equiv i\pm 1\pmod{5}$.
This problem is given as an exercise in the textbook Combinatorics of Coxeter Groups by Bjorner and Brenti. The reader is asked to use the theory presented in the book to show that the game always terminates after the same number of steps and in the same position regardless of how it is played. The fact that the game eventually terminates is not so easy to determine from the theory; the reader is first asked to find an elementary proof of this. One proof is given here. Once we know that the game always ends up in a positive position, we can use the fact that the element of the group that was used to bring the start position to the end position is uniquely determined by the start and end positions, and the steps that were taken always decrease the length of the element when it is expressed in terms of the generators $s_1,s_2,s_3,s_4,s_5$, eventually reaching the identity.
