distances

king.of.dragon

Let n points ( $n\ge4$ )such that distance of any two points is in Z.Prove that minimum is $\frac{1}{6}$ distances which are multiples of 3

**Posted:** Wed Sep 16, 2009 2:19 am

king.of.dragon

No one can do it?
I think it seems very difficult .Who can solve it?

**Posted:** Tue Sep 22, 2009 4:55 am

mavropnevma

Assume we have proven it for the base case $n = 4$ . The simple approach of arguing

**Posted:** Wed Sep 23, 2009 1:05 pm

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mavropnevma

For the base case $n=4$ , by choosing convenient coordinates we can assume the points are $O(0,0), A(a,0), B(b\cos\beta, b\sin\beta)$ and $C(c\cos\gamma, c\sin\gamma)$ , with the squares of their pairwise distances given by $a^2, b^2, c^2, r^2 = a^2+b^2-2ab\cos\beta, s^2 = a^2+c^2-2ac\cos\gamma$ and $t^2 = b^2+c^2-2bc\cos(\beta -\gamma)$ , where $a,b,c, r, s, t \in \mathbb{Z}_+^*$ .

Assume none is divisible by $3$ , so $a^2 \equiv b^2 \equiv c^2 \equiv r^2 \equiv s^2 \equiv t^2 \equiv 1 \pmod{3}$ .
We have $\cos\beta = \frac {a^2+b^2-r^2} {2ab} = \frac {p} {q}$ with $(p,q) = 1$ , $\cos\gamma = \frac {a^2+c^2-s^2} {2ac} = \frac {u} {v}$ with $(u,v) = 1$ , so we also have $p^2 \equiv q^2 \equiv u^2 \equiv v^2 \equiv 1 \pmod{3}$ .
Applying a dilation of ratio $qv \not \equiv 0 \pmod{3}$ to the whole configuration, the points become $O(0,0), A'(aqv,0), B'(bpv, \pm bv\sqrt{q^2-p^2})$ and $C'(cqu, \pm cq\sqrt{v^2-u^2})$ , so the squares of the distances become $1, 1, 1, (aqv - bpv)^2, (aqv - cqu)^2$ and $(bpv - cqu)^2$ modulo $3$ .

But among the three integer abscissa $aqv, bpv, cqu$ , congruent to $\pm 1$ modulo $3$ , there must exist two with their difference divisible by $3$ , so one of the latter three distances will be $0$ modulo $3$ , contradiction.

**Posted:** Thu Sep 24, 2009 7:13 am

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king.of.dragon

**Posted:** Sun Nov 08, 2009 9:13 pm

mavropnevma

1. The dilation of ratio $qv$ (a homothety of center $O$ ) is only needed because the distances between the points $O,A,B,C$ are only known to be rational. By doing this homothety, we get the points $O,A',B',C'$ with integer distances between them.

2. I never said that from $a, q, v$ , congruent to $\pm 1$ modulo $3$ , it must follow that $aqv \equiv 1 \pmod{3}$ . If you read the last part, I only said that $aqv, bpv, cqu$ , are congruent to $\pm 1$ modulo $3$ . This in turn implies that two of them must have same congruence (pigeonhole principle: three numbers, and only two congruences $-1$ and $1$ ).

**Posted:** Sun Nov 08, 2009 10:45 pm

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king.of.dragon

Can u tell me solution with n>4? Mr. Green

I haven't already proved with n>4

**Posted:** Tue Nov 10, 2009 6:06 am