I heard about Ceva's Theorem in three dimensions. Can you give me more details and link something about that? Thanks.


1 Answer 1


Here's an analogue I devised. The key is to establish the proper counterpart of the ratio of lengths; I use the "triple-ratio" of (signed) areas.

Let $OABC$ be a tetrahedron and let lines (cevians) from the vertices meet the respective opposite faces at points $O^\prime$, $A^\prime$, $B^\prime$, $C^\prime$. Write "$|\triangle XYZ|$" for the (signed) area of $\triangle XYZ$; and write "$[x:y:z]$" for the triple ratio of $x$, $y$, $z$.

(The areas of two coplanar triangles have the same sign if the triangles are traced in the same direction, and they have opposite signs if they're traced in opposite directions, with "direction" based on a particular view of their common plane. (Which direction corresponds to "$+$" is irrelevant.) Signs of non-coplanar faces are independent. If you don't like the signed areas, just assume each cevian hits the interior of its corresponding face, so that, below, all factors are positive.)

If the four cevians of $OABC$ concur, then $$\begin{align}&[\;|\triangle A^\prime BC|:|\triangle A^\prime CO|:|\triangle A^\prime OB|\;] \\ \cdot &[\;|\triangle B^\prime AO|:|\triangle B^\prime OC|:|\triangle B^\prime CA|\;]\\ \cdot &[\;|\triangle C^\prime OA|:|\triangle C^\prime AB|:|\triangle C^\prime BO|\;]\\ \cdot &[\;|\triangle O^\prime CB|:|\triangle O^\prime BA|:|\triangle O^\prime AC|\;] \;::\; [\;1 : 1 : 1\;]\end{align} \qquad (*)$$

(Note: Triple ratios multiply component-wise.)

Proof. Since everything involves ratios, we can apply an affine transformation so that $O$ is the origin, and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$. Letting $P(x,y,z)$ be the assumed point of concurrency, we can deduce

$$A^\prime = \frac{(0,y,z)}{1-x} \qquad B^\prime = \frac{(x,0,z)}{1-y} \qquad C^\prime = \frac{(x,y,0)}{1-z} \qquad O^\prime = \frac{(x,y,z)}{x+y+z}$$

Consequently, $$\begin{align} [\;|\triangle A^\prime BC|:|\triangle A^\prime CO|:|\triangle A^\prime OB|\;] &= \left[\; \frac{1-x-y-z}{2(1-x)} : \frac{y}{2(1-x)} : \frac{z}{2(1-x)}\;\right] \\ &= \left[\; 1-x-y-z : y : z\;\right] \\ [\;|\triangle B^\prime AO|:|\triangle B^\prime OC|:|\triangle B^\prime CA|\;] &= \left[\; z : x : 1-x-y-z\;\right] \\ [\;|\triangle C^\prime OA|:|\triangle C^\prime AB|:|\triangle C^\prime BO|\;] &= \left[\; y : 1-x-y-z : x \;\right] \\[6pt] [\;|\triangle O^\prime CB|:|\triangle O^\prime BA|:|\triangle O^\prime AC|\;] &= \left[\;\frac{x\sqrt{3}}{2(x+y+z)} : \frac{z\sqrt{3}}{2(x+y+z)} : \frac{y\sqrt{3}}{2(x+y+z)} \;\right] \\ &=\left[\;x : z : y\;\right] \end{align}$$

whereupon each component of the final triple-ratio product is effectively $xyz(1-x-y-z)$, so that the result reduces proportionally to $[1:1:1]$. $\square$

Conversely ... Well, we can't expect $(*)$ to be powerful enough to guarantee concurrency of four cevians, since it only reduces the configuration's degrees of freedom from eight to six. (Note that it takes some doing to guarantee that just two lines in space intersect.) However, we can say that, when we're mostly there, $(*)$ brings us home.

If three cevians of $OABC$ concur, and if $(*)$ holds, then all four cevians concur.

Proof. Taking $P(x,y,z)$ the point of concurrency of cevians from $A$, $B$, $C$, our work above gives $A^\prime$, $B^\prime$, $C^\prime$ as well as the triple ratios involving areas of triangles determined by those points. Writing $O^\prime$ generically as $(u,v,w)$, with $u+v+w=1$, we have

$$\left[\;|\triangle O^\prime CB|:|\triangle O^\prime BA|:|\triangle O^\prime AC|\;\right] = \left[\;u\frac{\sqrt{3}}{2} : w\frac{\sqrt{3}}{2} : v\frac{\sqrt{3}}{2} \right] = \left[\; u : w : v \;\right]$$

Then $(*)$ implies

$$uyz(1-x-y-z) = xvz(1-x-y-z) = xyw(1-x-y-z)$$

so that

$$\frac{u}{x} = \frac{v}{y} = \frac{w}{z}$$

That is, $O^\prime = k P$ for some $k$, and we have that our fourth cevian, $\overleftrightarrow{OO^\prime}$, contains the point of concurrency, $P$. $\square$

In 1988, Steven Landy articulated "A Generalization of Ceva's Theorem to Higher Dimensions", stating

The $N$ cevians of an $N-1$ dimensional pyramid are concurrent if and only if a vertex mass assignment exists such that each [point where a cevian meets its opposite cell] is at the center of mass of its surrounding cell.

That "(mass assignment) implies (concurrency)" follows immediately from the observation that each cevian must pass through the center-of-mass for the entire tetrahedron. Conversely, when concurrency is assumed, Landy says that

[I]n any $N$-dimensional Ceva pyramid the mass assignment is such that the mass at any vertex is proportional to the "volume" of the opposite $N$-dimensional pyramid determined by the remaining $N$ vertices and the [point of concurrency].

It's interesting that Landy's additive center-of-mass approach, and my multiplicative result (which itself naturally extends to any-dimensional figures), both feature measurements of sub-figures each of which is determined by an auxiliary point and a facet of the original figure.


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