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Given: four points in the Euclidean plane and the distances to all but one pair of them. Find the remaining distance.

My solution attempts to be exhaustive--I consider each case individually: four collinear points, all but one point are collinear, etc.

I would really appreciate it if someone could check the example I came up with below for analyzing the troublesome case (at least for me) where no three points are collinear. Did I overlook anything?

Essentially, the law of cosines takes center stage in most of my cases. I've come up with an example which is making me rethink several of my cases. Consider the distances $$|XY|=100,|ZY|=109,|XZ|=11,|WY|=75,|WZ|=136.$$

I suspect I have--what I'm now calling--an ambiguous case. Since $$\frac{109^2+136^2-75^2}{2(109)(136)}=\frac{109^2+11^2-100^2}{2(11)(109)}$$ (law of cosines) there are congruent angles which means, to make a long story short, two segments that could coincide, taking me back to a case where there are three collinear points. Is this right or did I make a mistake along the way?

I apologize in advance if my question is unclear in any way.

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Do you have four or five points? In your example I only see four points. The first three distances fix the triangle $X$, $Y$, $Z$. The last three fix the triangle $X$, $Z$, $W$, however, you may reflect this at the line $ZY$ and get two possibilities for the distance from $X$ to $W$... –  Dirk Nov 16 '11 at 20:42
    
Ahhh! Four points, I meant four points everywhere. (It appears I don't have the privileges to edit my title. Can a mod edit it for me?) –  sasha Nov 16 '11 at 20:46
    
@Dirk What you say confirms my suspicion in general. But in the example, are there in actuality two possibilities for $|XW|$ or one? –  sasha Nov 16 '11 at 20:50
    
@sasha: two possibilities. For a simpler case, let $XY=XZ=YZ=3, WY=WZ=2$. Then XYZ is equilateral and WYZ is isoceles on one side of XYZ. But it can be inside or outside XYZ. This will often be the case. –  Ross Millikan Nov 16 '11 at 21:35
    
@RossMillikan Of course! Thank you. –  sasha Nov 16 '11 at 22:00
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1 Answer

up vote 2 down vote accepted

This is not an answer to your question, but I wanted to post this approach anyway.

One can solve the problem using the Cayley-Menger determinants. The points $X, Y, Z, W$ lie on a plane if and only if the determinant $$ \left|\begin{matrix} 0 & d(X, Y)^2 & d(X, Z)^2 & d(X, W)^2 & 1 \\ d(Y, X)^2 & 0 & d(Y, Z)^2 & d(Y, W)^2 & 1 \\ d(Z, X)^2 & d(Z, Y)^2 & 0 & d(Z, W)^2 & 1 \\ d(W, X)^2 & d(W, Y)^2 & d(W, Z)^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ \end{matrix} \right| $$ vanishes.

Plugging in the given values, we get $$ \left|\begin{matrix} 0 & 100^2 & 11^2 & t & 1 \\ 100^2 & 0 & 109^2 & 75^2 & 1 \\ 11^2 & 109^2 & 0 & 136^2 & 1 \\ t & 75^2 & 136^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ \end{matrix} \right| = 0, $$ where $t = |XW|^2$. Evaluating this using Wolfram|Alpha, we get $$ -2(t - 15625)(11881 t - 207183025) = 0. $$ Using this, either $|XW| = \sqrt{15625} = 125$, or $|XW| = \sqrt{\frac{207183025}{11881}} = \frac{35 \sqrt{169129}}{109}$.

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TIL! And by the way, you've confirmed that my hunch was correct, so your comment is indeed an answer to my question. Also, thanks for editing the post title. –  sasha Nov 16 '11 at 22:11
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