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Let's say I have 3 points, A B and C. Each point has a fixed distance to every other point.

If I want to find the right representation for those distances, I have to place A and B according to their distance, draw two circles respectively with center A and B and d(A, C) radius and d(B, C) radius.

The two intersections of the circles are the possible place where I can put C and have the right distance between every point.

Pretty simple. My problem is the following: How do I do the exact same thing with 4 points A B C and D ?

The two options for placing my C point never allow me to place D with the right distances from every other point.

(I am sorry if this is very badly explained, I'm not a native english speaker !)

Thanks you ! Gael

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The distance between each point is the same? –  PEV Jan 14 '11 at 16:08
    
no, every point to point distance can be different –  Gael Jan 14 '11 at 16:10
    
So basically you are given the distances in advance and want to construct points that match those distances? –  PEV Jan 14 '11 at 16:12
    
Yes, exactly !! –  Gael Jan 14 '11 at 16:19
    
Perhaps consider a square as a special case. –  PEV Jan 14 '11 at 16:22

3 Answers 3

up vote 3 down vote accepted

In general, you cannot place four points in the plane at a given set of distances. Even with three points, you need to satisfy the triangle inequality. If the distances do, your construction works. But then given distances from D to A, B, and C, it is unlikely there is a point that satisfies the distances. If the distances do allow it, you can just do what you propose for A, B, and C (using either position for C-it is just a mirror image triangle), then locate D relative to A and B and check each of the two choices for being the correct distance from C. But it will not work unless the distance are carefully chosen. If you move into $\mathbb{R^3}$ you will be able to do it in general as long as the triangle inequalities are satisfied, and the construction is the same.

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Reminds me of those stats problems where you have to compute the probability that randomly generated distances (according to a given process, like "breaking a stick") give rise to a certain geometrical figure. –  Raskolnikov Jan 14 '11 at 16:52

See this. This is the construction of a square (a special case). Essentially, we are constructing a square of sides of given length $s$. A priori, from the Pythogorean Theorem, we know the distances between two given points.

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The general problem of determining the positions of n points given their mutual distances crops up in metric dimension reduction techniques, where one asks if the distances between a set of points may be preserved by a mapping into a lower dimensional space.

If a solution exists for n points, it can be embedded in not more than n-1 dimensions. Obviously any solution can be transformed by reflection, translation, and rotation, so at best one hopes for uniqueness up to an orthogonal transformation.

Nor does a solution always exist. As Ross Millikan points out, if a solution exists the distances must satisfy triangle inequalities. This is sufficient for three points (to determine a triangle in the plane), but it is not sufficient for placing four points.

Consider having placed points A,B,C in an equilateral triangle with side length 1. Then we can try to place a fourth point D. If the distances from D to each of A,B,C are slightly more than 1/2, then all triangle inequalities are satisfied yet there is no feasible location for D, even in three dimensions.

Indeed for four points the question amounts, having placed the first three points, whether a feasible location for the fourth point exists. The distances to the first three points give spheres embedded in three dimensions, the intersection of any two of which forms a circle, a point, or the empty set. The answer is easily found by cartesian geometry.

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