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In a regular pentagon, all diagonals are drawn. Initially, each vertex and each point of intersection of the diagonals is labeled by the number $1$. In one step it is permitted to change the signs of all numbers of a side or diagonal. Is it possible to change the signs of all labels to $−1$ by a sequence of steps?

My strategy was to make the regular pentagon as a graph and then partition the vertices into sets $A,B$ and $C$ where the vertices in each set are not adjacent to one another. Thus if we consider the quantity $$Q=(S_a-S_b)(S_b-S_c)(S_a-S_c)$$ then it is an invariant. Here $S_a$ is the sum of the numbers associated to each vertex in the set $A.$ Depending on how you come up with the partition one might get different values but I got $S_a=5,S_b=3$ and $S_c=2$ which on addition must give $10$ which makes sense as initially we had placed $1$ on each vertex. Thus $Q=6.$ However we want $Q=-6$ which happens when $S_a=-5,S_b=-3$ and $S_c=-2,$ which is a contradiction as $Q$ is an invariant.

Is this solution correct?

Student
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1 Answers1

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First a hint: Consider the parity of the number of $-1$s on the pentagon (not including the intersections on diagonals).

Now for your solution, suppose a side is flipped from $1$ to $-1$, so WLOG say $S_a, S_b $ decrease by $2$, while $S_c$ remains unchanged. $Q$ clearly doesn’t seem invariant...

Macavity
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  • Thanks for your comment. Please check out this post. https://math.stackexchange.com/questions/2585622/trying-to-generalize-a-contest-math-problem – Student Dec 30 '17 at 16:11