Optimal 4-Point Configuration

Geometry Level 5

Let { x 1 , x 2 , x 3 , x 4 } \{x_1,x_2,x_3,x_4\} be 4 distinct points on the Euclidian plane such that the distance between any two points is at least 1. We want to minimize the average distance beween 2 points: min { x 1 , x 2 , x 3 , x 4 } R 2 1 6 x i , x j i < j dist ( x i , x j ) . \min_{\begin{array}{c}\scriptstyle \{x_1,x_2,x_3,x_4\} \in {\mathbb{R}}^2 \\[-4pt] \end{array} } \frac{1}{6} \sum_{\begin{array}{c}\scriptstyle x_i,x_j\\[-4pt] \scriptstyle \space i < j \end{array} } \text{dist}(x_i,x_j). If the minimum average distance is expressed as a + b c , \large \frac{a+\sqrt{b}}{c}, where a a and c c are coprime and b b is square-free, submit a + b + c a+b+c .


The answer is 14.

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1 solution

Mark Hennings
Jan 18, 2018

The optimal configuration has the four points at the vertices of two equilateral triangles of side 1 1 , so that the distances are 1 , 1 , 1 , 1 , 1 , 3 1,1,1,1,1,\sqrt{3} , and the average is 5 + 3 6 \frac{5 + \sqrt{3}}{6} for a solution of 5 + 6 + 3 = 14 5+6+3 = \boxed{14} .

I got the answer, but how do you prove that that is the optimal configuration?

Nick Turtle - 3 years, 4 months ago

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Considering all rhombuses with side length 1, you get a minimal sum of diagonals in the configuration stated above.

Maria Kozlowska - 3 years, 4 months ago

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