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Given a closed spatial polygon with fixed edgelengths $r_i, i=1..4$ (that is a cyclicly ordered 4-tuple of vectors $v_i\in \mathbb R^3$ with $|v_i|=r_i$ such that $\sum v_i=0$) one can associate a pair of numbers to each diagonal.

Namely, for a choice of a number $i$ from $1..4$ one has

  1. Length of diagonal, that is $l=|v_i+v_{i+1}|$.
  2. Dihedral angle at the diagonal, that is, $\theta=\angle (v_i\times v_{i+1},v_{i+2}\times v_{i+3})$.

This pair of numbers defines the polygon uniquely up to rigid motions (once the diagonal is nonzero). One can also write coordinates with respect to another diagonal.

Is there any simple relation between $(l,\theta)$ with respect to two diagonals?

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  • $\begingroup$ Do you mean to say that “this pair of numbers together with the edge lengths defines the polygon uniquely…”? This is what the word “fixed” suggests in the first sentence but it’s a little confusing as written. $\endgroup$ Commented Aug 14 at 12:38
  • $\begingroup$ @JackEdwardTisdell yes, exactly $\endgroup$ Commented Aug 14 at 15:29

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For $i=1,2$, let $l_i$ be the length of a diagonal, $\theta_i$ the corresponding dihedral angle, and $A_i,B_i$ the areas of the two faces meeting at this diagonal. Then the following relation holds: $$4l_1^2 l_2^2-(r_1^2+r_3^2-r_2^2-r_4^2)^2=16(A_i^2+B_i^2-2A_iB_i \cos \theta_i).$$ Note that we can express $A_i$ and $B_i$ in terms of $l_i$ and $r_1,r_2,r_3,r_4$ using Heron's formula.

The procedure for changing coordinates is now quite simple. Given $(l_1,\theta_1)$, we can find $l_2$ using the relation for $i=1$ (which is linear in $l_2^2$). Once we have $l_2$, we can find $\theta_2$ using the relation for $i=2$ (which is linear in $\cos \theta_2$).

Here is a proof for $i=1$. We will refer to segments with the same labels as their lengths. Let $k_2$ the projection of $l_2$ onto a plane $P_1$ orthogonal to $l_1$. Let $\lambda$ be the angle between the projections of the diagonals onto a mutually parallel plane. Working in this plane, we see that $k_2=l_2 \sin \lambda$. Now working in $P_1$, apply the law of cosines to the triangle whose vertices are the endpoints of $k_2$ and the intersection of $P_1$ and $l_1$. The angle opposite $k_2$ is $\theta_1$, and the two other sides have the same lengths as the altitudes dropped to $l_1$ of the faces with areas $A_1$ and $B_1$, namely $2A_1/l_1$ and $2B_1/l_1$. Hence $$k_2^2=\left(\frac{2A_1}{l_1}\right)^2+\left(\frac{2B_1}{l_1}\right)^2-2\left(\frac{2A_1}{l_1}\right)\left(\frac{2B_1}{l_1}\right)\cos \theta_1.$$ Now just equate the two expressions for $k_2$ and use the fact that $$\sin \lambda = \sqrt{1-\frac{(r_1^2+r_3^2-r_2^2-r_4^2)^2}{4l_1^2l_2^2}},$$ which is well known (see here, for instance).

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  • $\begingroup$ Thanks! However this is gives only one relation, not a change of coordinates between $(l_1,\theta_1)$ and $(l_2,\theta_2)$ $\endgroup$ Commented Aug 24 at 13:48
  • $\begingroup$ OK, I replaced the relation in my original answer with a different pair of relations that should do what you want. $\endgroup$ Commented Aug 24 at 19:06
  • $\begingroup$ @N M Thank you, this is cool! Can you hint at how you got this relation? [my hope was that there might exist some (may be highly nonlinear in $l$ and $\theta$) complex coordinates $C$ and $C'$ on the space of such 4-gons such that change of coordinates has the form $C\cdot C'=$some function of $r$] $\endgroup$ Commented Aug 25 at 23:50
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    $\begingroup$ @DmitriiKorshunov I added a proof. $\endgroup$ Commented Aug 26 at 4:13

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