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In two dimensional Euclidean space, it is not hard to calculate the radius of the circumscribed circle of an arbitrary triangle when all the side lengths are given. We can use Heron's formula to calculate the area of the triangle, then immediately obtain the requested radius since there is $$S=abc/4R$$where R is the radius.
So what about the case in three dimensional Euclidean space? Now there is a tetrahedron whose edge lengths are $a,b,c,d,e,f$ such that they can construct a tetrahedron. So far I have proved that for any possible given edge lengths there always exist one and only one circumscribed sphere. Let its radius be $R$, then it is clear that there exists an unique function $F$ such that$$R=F(a,b,c,d,e,f)$$ Since I haven't seen $F$ in any books I read, I started my exploration then, trying to work out $F$. But every time I had to give up because the calculation was scarily daunting.I have tried many different ways, but none of them seemed to remove the pain of huge amount of calculation which I just don't want to spend too much time and energy in. Further observation tells that $F$ should be symmetrical as to all the $a,b,c,d,e,f$, but that's, till now, all the knowledge I have about it. So I am wondering if there is a simple way, like in the two dimensional case, to get the result I want. I don't expect $F$ to take a simple form. I am just seeking an intuitive (and simple, if possible) METHOD that shows how to deduce $R$ from $a,b,c,d,e,f$.

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    $\begingroup$ Pages 279 and 280 give formulas, but without proof: archive.org/stream/jstor-2973351/2973351#page/n1/mode/2up $\endgroup$ – Edward Jiang Dec 31 '14 at 20:33
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    $\begingroup$ Problem 70, "100 Great Problems of Elementary Mathematics" [Heinrich Dörrie] reprinted by Dover Publications. $\endgroup$ – Senex Ægypti Parvi Dec 31 '14 at 22:26
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    $\begingroup$ Don't forget that how the skew lines comprising the tetrahedron are paired makes a difference. For example, is $a$ opposite to $d$, $b$ to $e$, usw? $\endgroup$ – Senex Ægypti Parvi Dec 31 '14 at 23:40
  • $\begingroup$ Oh , thanks for reminding me of that. $\endgroup$ – Vim Jan 1 '15 at 4:05
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    $\begingroup$ A note for anyone using the formula in the document provided by Edward Jiang: that document oversimplifies in an important way. It says that a,a1, b,b1, c,c1 are to be three opposite pairs of sides; but there are different ways to pick those, and only one works. You must take a1,b1,c1 to be the edges meeting at one vertex, and a,b,c to be the sides of the "opposite" face. $\endgroup$ – Gareth McCaughan Aug 17 '17 at 12:30
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Instead of tetrahedron, let us work out a general formula for $n$-simplex first.

Given any non-degenerate $n$-simplex $S$ with vertices $v_1, \ldots, v_{n+1}$. Let

  • $V$ be the volume of $S$.
  • $\ell_{ij} = \| v_i - v_j \|$ be the edge lengths.
  • $c$ and $R$ be the center and radius for the circumsphere.
  • $\lambda_1, \ldots, \lambda_{n+1}$ be the barycentric coordinates of $c$ with respect to $S$.
    i.e the list of $n+1$ real numbers such that $$c = \sum_{i}\lambda_i v_i\quad\text{ with }\quad \sum_{i}\lambda_i = 1$$

Recall for any two points $u, v$ with barycentric coordinates $\mu_i, \nu_i$ with respect to $S$.
Their distance is given by the formula:

$$\| u - v \|^2 = -\sum_{i < j}\ell_{ij}^2(\mu_i - \nu_i)(\mu_j - \nu_j)$$

Apply this to circumcenter $c$ and vertex $v_k$, we obtain

$$R^2 = -\sum_{i<j}\ell_{ij}^2 (\lambda_i - \delta_{ki})(\lambda_j - \delta_{kj}) = \sum_{j}\ell_{kj}^2 \lambda_j - \frac12 \sum_{i,j} \ell_{ij}^2\lambda_i\lambda_j$$ Multiply by $\lambda_k$ and sum over $k$, we obtain

$$R^2 = \frac12\sum_{i,j} \ell_{ij}^2 \lambda_i\lambda_j \quad\implies\quad 2R^2 = \sum_{j}\ell_{kj}^2 \lambda_j\;\;\text{ for all }k\tag{*1}$$

To proceed, rewrite everything in matrix form. Let

  • ${\bf \theta}$ be the $(n+1)\times 1$ column vector with all entries $1$.
  • ${\bf \lambda}$ be the $(n+1)\times 1$ column vector with entries $\lambda_i$.
  • $\Lambda$ be the $(n+1)\times(n+1)$ matrix with entries $\ell_{ij}^2$.
  • $\Delta$ be the $(n+2)\times(n+2)$ matrix $\begin{bmatrix}0 & {\bf \theta}^T \\ {\bf \theta} & \Lambda\end{bmatrix}$

In terms of them, $(*1)$ becomes $2R^2{\bf \theta} = \Lambda {\bf \lambda}$.

Together with ${\bf \theta}^T{\bf \lambda} = \sum_{i} \lambda_i = 1$, we obtain

$$\Delta \begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix} = \begin{bmatrix}0 & {\bf \theta}^T \\ {\bf \theta} & \Lambda\end{bmatrix} \begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix} = \begin{bmatrix} 1 \\ 0\end{bmatrix} \quad\implies\quad \begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix} = \Delta^{-1}\begin{bmatrix} 1 \\ 0\end{bmatrix} $$ Extracting the first component of both sides and notice $\Delta^{-1} = (\det\Delta)^{-1} {\rm adj}\Delta$, we get $$-2R^2 = \frac{\det\Lambda}{\det\Delta}$$

Notice $\det\Delta$ is simply the Cayley-Menger determiant for $S$ which satisfies $$V^2 = \frac{(-1)^{n+1}}{2^n n!^2} \det\Delta$$ This means the volume $V$, the circumradius $R$ and $\det\Lambda$ of a $n$-simplex are related by following formula:

$$\bbox[border:1px solid blue;padding: 16px;]{(-1)^n 2^{n+1}(n!)^2 (VR)^2 = \det\Lambda}\tag{*2}$$

For example, when $n = 2$, the $n$-simplex becomes a triangle. If $a, b, c$ are the side lengths of a triangle. $(*2)$ reduces to the familiar relation among its area, circumradius and side lengths.

$$32V^2R^2 = \det \begin{bmatrix} 0 & a^2 & b^2\\ a^2 & 0 & c^2\\ b^2 & c^2 & 0 \end{bmatrix} = 2a^2b^2c^2 \quad\iff\quad 4VR = abc $$

Back to the original problem of tetrahedron which corresponds to $n = 3$. Let $a, b, c$ be the lengths of the three edges attached to a vertex. Let $a_1, b_1, c_1$ be the lengths of corresponding opposite edges. $(*2)$ becomes $$-576(VR)^2 = \det\begin{bmatrix} 0 & a^2 & b^2 & c^2\\ a^2 & 0 & c_1^2 & b_1^2\\ b^2 & c_1^2 & 0 & a_1^2\\ c^2 & b_1^2 & a_1^2 & 0 \end{bmatrix} $$ Reproducing the complicated formula appeared in the link in Edward Jiang's comment.

To simplify further, we first multiply $2^{nd}/3^{rd}/4^{th}$ row/column of $\Lambda$ by $\frac{bc}{a}, \frac{ac}{b}, \frac{ab}{c}$ and then divide the $1^{st}$ row/column or resulting matrix by $abc$. At the end, we find $$\det\Lambda = \det\begin{bmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & (cc_1)^2 & (bb_1)^2\\ 1 & (cc_1)^2 & 0 & (aa_1)^2\\ 1 & (bb_1)^2 & (aa_1)^2 & 0 \end{bmatrix} $$ This has the form of a Cayley-Menger determinant for a "triangle" with sides $aa_1$, $bb_1$ and $cc_1$. Recall Heron's formula for computing area of triangle, $\det\Lambda$ can be factorized as follows:

$$\det\Lambda = -16 p(p-aa_1)(p-bb_1)(p - cc_1)\quad\text{ where }\quad p = \frac{aa_1 + bb_1 + cc_1}{2}$$

As a result, we obtain a much simpler relation for tetrahedron.

$$\bbox[border:1px solid blue;padding: 16px;]{6VR = \sqrt{p(p-aa_1)(p-bb_1)(p-cc_1)}}\tag{*3}$$

Update

I recently come across an elegant proof of formula $(*3)$ using sphere inversion (the 3d-counterpart of circle inversion in 2d). It explains why the RHS is the area of a triangle with sides $aa_1$, $bb_1$, $cc_1$.

Let $OABC$ be a tetrahedron with sides $OA = a, OB = b, OC = c, BC = a_1, CA = b_1, AB = c_1$. Let $V$ and $R$ be its volume and circumradius. Let $A', B', C'$ be the image of $A, B, C$ under a sphere inversion with respect to the sphere centered at $O$ with radius $\rho = \sqrt[3]{abc}$.

Notice tetrahedron $OA'B'C'$ is sharing a trihedral angle at $O$ with $OABC$. Since $$|OA'||OB'||OC'| = \frac{\rho^2}{a}\frac{\rho^2}{b}\frac{\rho^2}{c} = abc = |OA| |OB| |OC|$$ The volume of tetrahedron $OA'B'C'$ also equals to $V$.

Under the sphere inversion, the circumsphere of $OABC$ get mapped at a plane at a distance $\frac{\rho^2}{2R}$ from $R$. The points $A'$, $B'$, $C'$ belong to this plane and they are forming a triangle with sides $B'C' = \frac{a_1\rho^2}{bc}, C'A' = \frac{b_1\rho^2}{ac}, A'B' = \frac{c_1\rho^2}{ab}$. If $\Delta(u,v,w)$ is the area of a triangle of sides $u,v,w$, then $$V = \frac{1}{3}\frac{\rho^2}{2R}\Delta\left( \frac{a_1\rho^2}{bc}, \frac{b_1\rho^2}{ac}, \frac{c_1\rho^2}{ab} \right) = \frac{1}{6R} \Delta\left( \frac{a_1\rho^3}{bc}, \frac{b_1\rho^3}{ac}, \frac{c_1\rho^3}{ab} \right) = \frac{1}{6R} \Delta(aa_1,bb_1,cc_1) $$ Together with Heron's formula, $(*3)$ follows.

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  • $\begingroup$ very interesting reply, thanks for giving it $\endgroup$ – G Cab Sep 1 '17 at 22:42

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