## The Steiner ratio conjecture is true for five points.(English)Zbl 0576.05015

The long-standing conjecture of Gilbert and Pollak states that for any set of n given points in the Euclidean plane, the ratio of the length of a Steiner minimal tree and the length of a minimal spanning tree is at least $$\sqrt{3}/2$$. This conjecture has been shown to be true for $$n=3$$ by Gilbert and Pollak, and for $$n=4$$ by Pollak. Recently, the autors used a different approach to give a shorter proof for $$n=4$$. In this paper, we continue this approach to prove the conjecture for $$n=5$$. Some results are useful in obtaining bounds for the ratio of two lengths in the general case. Let L(AB) denote the Euclidean distance between two points A and B and define $$L(A_ 1A_ 2...A_ n)=L(A_ 1A_ 2)+L(A_ 2A_ 3)+...+L(A_{n-1}A_ n)$$. For a sequence of points $$A_ 1,A_ 2,...,A_ n$$, let $$\sphericalangle A_ i$$ denote the angle $$\sphericalangle A_{i-1}A_ iA_{i+1}$$, the angle from line $$[A_{i- 1},A_ i]$$ to line $$[A_{i+1},A_ i]$$. The main lemma is that suppose that $$k\cdot 180-60\leq \sum^{j+k-1}_{i=j}\sphericalangle A_ i\leq k\cdot 180+60$$, $$j=1,2,...,n-1$$, $$k=1,2,...,n-j$$, then $$L(A_ 1A_ n)\geq (\sqrt{3}/2)L(A_ 1A_ 2...A_ n).$$
By Melzak’s method, a Steiner minimal tree can be transfered into a segment and every spanning tree can be transfered into a path interconnecting two endpoints of the segment. By a case argument on angles, we proved that there always exists a spanning tree path which satisfies the condition of the main lemma, for $$n=5$$. The same idea would work for $$n=6$$. The same idea would work for $$n=6$$. However, the argument becomes too complicated to be written down.

### MSC:

 05C05 Trees 51M15 Geometric constructions in real or complex geometry 05C35 Extremal problems in graph theory

### Keywords:

ratio; Steiner minimal tree; minimal spanning tree
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### References:

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