Regular Morley tetrahedra from $p$-secting planes

Introduction

Morley’s theorem says that the adjacent angle trisectors of a triangle meet in the vertices of an equilateral triangle [3, 4]. Tran [5] carried the construction to a tetrahedron $ABCD$ by trisecting its six dihedral angles, and proved that the resulting Morley tetrahedron of an isosceles tetrahedron is again isosceles. In the same paper he replaced the trisecting planes by $p$-secting planes [5 Theorem 6]. We use the following form of the definition.

Since $0<\theta_{XY}/p<\theta_{XY}$, the half-plane $H$ lies inside the dihedral angle at $XY$. For $p=3$ this is the trisecting plane of $XY$ that is closer to $F$. For numbers $e,f,g>1$ we attach the parameter $e$ to the edges $AB$ and $CD$, $f$ to $BC$ and $AD$, and $g$ to $AC$ and $BD$, as in [5 Theorem 6]. The tetrahedron

is defined as follows: $A'$ is the common point of the three planes that are $p$-secting towards the face $BCD$ through the edges $CD$, $BD$ and $BC$, with $p$ equal to the parameter of the edge, so $p=e$, $p=g$ and $p=f$ respectively; $B'$, $C'$ and $D'$ are obtained in the same way from the faces $ACD$, $ABD$ and $ABC$. Proposition 2.3 below shows that the three planes always meet in exactly one point. For $e=f=g=3$ this is Tran’s Morley tetrahedron.¹

Tran [6] recently showed that both converse conjectures of [5] are false: there are tetrahedra $T_1$ (not isosceles) and $T_2$ (isosceles but not regular) whose Morley tetrahedra are regular. In both, some pair of opposite edges $AB$, $CD$ has its four cross edges $AC$, $AD$, $BC$, $BD$ of equal length. He conjectured that a regular Morley tetrahedron always forces this [6 Conjecture 8.1], proved it for every tetrahedron with a nontrivial symmetry [6 Corollary 11.6], and stated the analogue for $p$-secting planes.

Tran notes that his methods do not apply to parameters other than $e=f=g=3$, and he lists the decision of Conjecture 1.2 among his open problems [6 Section 18.5 and Problem 8 of Section 19]. We show that the conjecture is false, already for isosceles tetrahedra, where it holds for $e=f=g=3$ [6 Theorems 10.1 and 10.3].

A tetrahedron is isosceles if its opposite edges have equal lengths. Every isosceles tetrahedron is congruent to one with vertices

that is, to a tetrahedron inscribed in the box $[-a,a]\times[-b,b]\times[-c,c]$ [2, 1]; Lemma 3.1 below gives a short proof, with the labels of the vertices preserved. Its edge lengths are

Let $O$ be the origin, the centre of the box, and let $\ell_1,\ell_2,\ell_3,\ell_4$ be the lines through $O$ in the directions $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, $(-1,-1,1)$.

Distinct points $X$ in part (b) give distinct triples, and the interior points of $\ell_1$ already form an open segment through $O$ (§ 3), so there are infinitely many such triples. Figure 1 shows one of the regular Morley tetrahedra.

The parameters can be written down in closed form. Here is an example in which all three parameters lie strictly between 2 and 4, and one in which two of them are equal to 4.

Then $2<e,f,g<4$, and $\mathcal{M}_{e,f,g}(ABCD)$ is the regular tetrahedron with vertices $\frac{2}{7}(-1,-1,-1)$, $\frac{2}{7}(-1,1,1)$, $\frac{2}{7}(1,-1,1)$, $\frac{2}{7}(1,1,-1)$.

When the three parameters are equal, Theorem 1.3 still applies, but the three conditions on $A'$ become harder to satisfy. For $e=f=g=3$, Tran showed that an isosceles tetrahedron with a regular Morley tetrahedron is regular or similar to $T_2$ [6 Theorem 10.3]. For quadrisecting planes the second possibility disappears.

Conjecture 1.2 for $e=f=g$ remains open; for $e=f=g=3$ it is Tran’s Conjecture 8.1 in [6].

§ 2 describes the construction through the distances to the faces; Proposition 2.3 shows that $A'$ runs once through the interior of the tetrahedron as $(e,f,g)$ runs through $(1,\infty)^3$. § 3 proves Theorem 1.3, Corollary 1.4 and the two examples, and § 4 proves Theorem 1.7.

The construction in terms of distances

Let $ABCD$ be a nondegenerate tetrahedron with volume $V$. For a vertex $X$ let $F_X$ be the opposite face, $S_X$ its area, $\mathbf{n}_X$ its inward unit normal, and $d_X(P)$ the signed distance from a point $P$ to the plane of $F_X$, positive on the side of $X$. Each $d_X$ is an affine function. Splitting $ABCD$ into four tetrahedra with apex $P$ gives

for $P$ in the tetrahedron, and hence for every $P \in\mathbb{R}^3$, since both sides are affine. The interior of $ABCD$ is the set where all four distances are positive. The interior dihedral angle between the faces $F_U$ and $F_W$, along the edge they share, satisfies $\cos\theta= -\mathbf{n}_U \cdot\mathbf{n}_W$.

For $\theta\in(0,\pi)$ and $p>1$ put

Proof. Write $\varphi=\theta/p$, which decreases from $\theta$ to $0$ as $p$ increases from $1$ to $\infty$. Then

Since $\sin\theta>0$ and $\cot$ decreases on $(0,\pi)$, this is a strictly decreasing function of $\varphi$; it tends to $0$ as $\varphi\to\theta$ and to $\infty$ as $\varphi\to0$. This gives the bijection and the formula for $p$. Finally

and $\cos(\theta/2)>0$, so $R_p(\theta)>1$ exactly when $\varphi<\theta/2$, that is, when $p>2$. $\square$

Proof. Let $\Pi$ be the $p$-secting plane and $\theta=\theta_{XY}$. The line $XY$ cuts $\Pi$ into two half-planes, and one of them lies inside the dihedral angle at $XY$. A point $P$ of that half-plane at distance $r$ from the line $XY$ has $d_U(P)=r\sin(\theta/p)$ and $d_W(P)=r\sin(\theta-\theta/p)$, because the half-plane makes the angles $\theta/p$ and $\theta-\theta/p$ with $F_U$ and $F_W$ (Figure 2). So the affine function $d_W-R_p(\theta)d_U$ vanishes on this half-plane and on the line $XY$. It is not constant, because $\mathbf{n}_U$ and $\mathbf{n}_W$ are linearly independent, so its zero set is a plane, and this plane is $\Pi$. $\square$

For the vertex $A'$ the three edges are $CD=F_A\cap F_B$, $BD=F_A\cap F_C$ and $BC=F_A\cap F_D$, with parameters $e$, $g$ and $f$. By Lemma 2.2, $A'$ is the common point of the three planes

and $B'$, $C'$, $D'$ are described in the same way.

Proof. The four normals $\mathbf{n}_A,\ldots,\mathbf{n}_D$ span $\mathbb{R}^3$, so $P\mapsto(d_A(P),d_B(P),d_C(P),d_D(P))$ is an injective affine map from $\mathbb{R}^3$ to $\mathbb{R}^4$. By (3) its image lies in the hyperplane $H=\{t\in\mathbb{R}^4:S_A t_A+S_B t_B+S_C t_C+S_D t_D=3V\}$, and since both are three-dimensional, the image is all of $H$. Write $\rho_e=R_e(\theta_{CD})$, $\rho_g=R_g(\theta_{BD})$ and $\rho_f=R_f(\theta_{BC})$; these are positive by Lemma 2.1. A point lies on the three planes (4) exactly when its distance vector has the form $(\tau,\rho_e\tau,\rho_g\tau,\rho_f\tau)$, and such a vector lies in $H$ for exactly one $\tau$, namely

So the planes meet in exactly one point, and all its distances to the faces are positive. Conversely, for an interior point $P$ the three ratios in (5) are positive, and by Lemma 2.1 each of them equals $R_p$ of the corresponding dihedral angle for exactly one $p>1$. With these parameters $P$ lies on the three planes (4), so $P=A'$; and no other triple gives $P$, again by Lemma 2.1. □

Two vertices of $M_{e,f,g}(ABCD)$ can coincide only if a parameter equals 2. For instance, a point $P=A'=B'$ satisfies $d_B(P)=R_e(\theta_{CD})d_A(P)$ and, since $B'$ lies on the $e$-secting plane of $CD$ towards $F_B$, also $d_A(P)=R_e(\theta_{CD})d_B(P)$; so $R_e(\theta_{CD})=1$ and $e=2$ by Lemma 2.1.

The symmetries we need permute the vertices by elements of the Klein group

Every element of $K$ maps each of the three pairs of opposite edges to itself; for instance $(AB)(CD)$ maps $AB$ and $CD$ to themselves, interchanges $AC$ and $BD$, and interchanges $AD$ and $BC$. So it preserves the parameter attached to every edge.

Proof. Write $d_X^1$ for the signed distances of $T_1$. The isometry $h$ maps the plane of $F_X$ to the plane of the face of $T_1$ opposite $\pi(X)$, and the side of $X$ to the side of $\pi(X)$, so $d_{\pi(X)}^1(h(P))=d_X(P)$ for every point $P$. It maps the inward normals of $ABCD$ to those of $T_1$, so it also preserves the dihedral angles: the angle of $T_1$ between the faces opposite $\pi(U)$ and $\pi(W)$ equals $\theta$ at $F_U\cap F_W$. By Lemma 2.2, $X'$ satisfies $d_Y(X')=R_p(\theta)d_X(X')$ for the three vertices $Y\ne X$, where $\theta$ is the angle at $F_X\cap F_Y$ and $p$ is the parameter of that edge. Hence $h(X')$ satisfies $d_{\pi(Y)}^1(h(X'))=R_p(\theta)d_{\pi(X)}^1(h(X'))$, with the same $p$ and the same angle. By Lemma 2.2 and Proposition 2.3 for $T_1$, $h(X')$ is the vertex of $M_{e_1,f_1,g_1}(T_1)$ with label $\pi(X)$.

We use Lemma 2.4 in two ways: with $T_1=ABCD$, the same parameters and $\pi\in K$; and with $\pi$ the identity, when $h$ moves a tetrahedron into a standard position.

Remark 2.5. If the angle $\theta_{XY}/p$ in Definition 1.1 is measured from the other face through $XY$, the plane used for a given parameter $p$ is the one we use for $p/(p-1)$. The map $p\mapsto p/(p-1)$ is an involution of $(1,\infty)$, so the family of tetrahedra $M_{e,f,g}(ABCD)$ is the same, with the parameters renamed. If instead one always takes the $p$-secting plane closer to the face, only parameters $p\ge2$ occur. In Examples 1.5 and 1.6 all parameters exceed 2, so these examples disprove Conjecture 1.2 under each of these readings.

Isosceles tetrahedra

Proof. Put $O=\frac{1}{4}(A+B+C+D)$ and

Then $A=O+u_1+u_2+u_3$, $B=O+u_1-u_2-u_3$, $C=O-u_1+u_2-u_3$ and $D=O-u_1-u_2+u_3$. Since $A-B=2(u_2+u_3)$ and $C-D=2(u_2-u_3)$, we get $|AB|^2-|CD|^2=16u_2\cdot u_3$; in the same way $|AC|^2-|BD|^2=16u_1\cdot u_3$ and $|AD|^2-|BC|^2=16u_1\cdot u_2$. So $u_1,u_2,u_3$ are pairwise orthogonal. A direct expansion gives $(B-A)\cdot((C-A)\times(D-A))=-16u_1\cdot(u_2\times u_3)$, which is not zero, so none of the $u_i$ vanishes. With $a=|u_1|$, $b=|u_2|$, $c=|u_3|$, the vectors $u_1/a$, $u_2/b$, $u_3/c$ form an orthonormal basis, and the isometry $h(P)=((P-O)\cdot u_1/a,(P-O)\cdot u_2/b,(P-O)\cdot u_3/c)$ sends $A$ to $(a,b,c)$, $B$ to $(a,-b,-c)$, $C$ to $(-a,b,-c)$ and $D$ to $(-a,-b,c)$.

From now on let $ABCD$ be the tetrahedron (1), and write $\alpha=1/a$, $\beta=1/b$, $\gamma=1/c$ and $N=(\alpha^2+\beta^2+\gamma^2)^{1/2}$. For a point $P=(x,y,z)$ put

Each $L_X$ vanishes at the three vertices other than $X$ and equals 4 at $X$; hence $d_X=L_X/N$, and the inward normals are $n_A=(\alpha,\beta,\gamma)/N$, $n_B=(\alpha,-\beta,-\gamma)/N$, $n_C=(-\alpha,\beta,-\gamma)/N$, $n_D=(-\alpha,-\beta,\gamma)/N$. The dihedral angles at the edges of the face $F_A$ are therefore given by

The three half-turns about the coordinate axes,

map the tetrahedron to itself and induce the permutations $(AB)(CD)$, $(AC)(BD)$ and $(AD)(BC)$.

Proof of Theorem 1.3. (a) Let $A'=(x,y,z)$. Lemma 2.4, applied to $h_1$, $h_2$ and $h_3$, gives

Hence

If $A'=O$, the four points coincide. Otherwise the six distances are equal if and only if $x^2=y^2=z^2$, that is, if and only if $A'$ lies on one of the lines $\ell_i$; and then $A',B',C',D'$ are alternate vertices of the cube $[-|x|,|x|]^3$, which form a regular tetrahedron.

(b) This is Proposition 2.3 applied to the interior point $X$. □

For $0<t<1/(\alpha+\beta+\gamma)$ the point $-t(1,1,1)$ of $\ell_1$ is interior: $L_A(-t,-t,-t)=1-t(\alpha+\beta+\gamma)>0$, and $L_B$, $L_C$, $L_D$ exceed $L_A$ there by $2t(\beta+\gamma)$, $2t(\alpha+\gamma)$ and $2t(\alpha+\beta)$. By part (b) each of these points is the vertex $A'$ of a regular Morley tetrahedron, and distinct points give distinct triples because the map of Proposition 2.3 is injective.

Proof of Corollary 1.4. If $a$, $b$, $c$ are pairwise different, so are $b^2+c^2$, $a^2+c^2$ and $a^2+b^2$, and by (1.2) the tetrahedron has three different edge lengths, each shared by a pair of opposite edges. Under any labeling, the four cross edges of a pair of opposite edges are the two other pairs, so they have two different lengths. On the other hand, by Theorem 1.3 and the paragraph above, $M_{e,f,g}(ABCD)$ is regular for the infinitely many triples obtained from the points $-t(1,1,1)$ with $0<t<1/(\alpha+\beta+\gamma)$. □

For a point $X$ on $\ell_1$ the ratios in (2.3) take a simple form. If $X=-t(1,1,1)$ with $t>0$, then

so that, for $0<t<1/(\alpha+\beta+\gamma)$ and with $\lambda=t/L_A(X)>0$,

For $p=4$ the identity $\sin3u=\sin u(1+2\cos2u)$ gives

Since $p\mapsto R_p(\theta)$ is increasing (Lemma 2.1), a parameter $p$ with $R_p(\theta)=\rho$ satisfies $p<4$ as soon as $(\rho-1)^2<2(1+\cos\theta)$, and $p>2$ as soon as $\rho>1$.

Proof of Example 1.5. Here $(\alpha,\beta,\gamma)=(1,\frac{1}{2},\frac{1}{3})$ and $N^2=\frac{49}{36}$, so (6) gives $\cos\theta_{CD}=-\frac{23}{49}$, $\cos\theta_{BD}=\frac{31}{49}$ and $\cos\theta_{BC}=\frac{41}{49}$, with sines $\frac{12\sqrt{13}}{49}$, $\frac{12\sqrt{10}}{49}$ and $\frac{12\sqrt{5}}{49}$. Take $X=-\frac{2}{7}(1,1,1)$. Then $L_A(X)=\frac{10}{21}>0$, so $X$ is interior, $\lambda=\frac{3}{5}$, and (7) gives the ratios $2$, $\frac{13}{5}$ and $\frac{14}{5}$. By Lemma 2.1 the parameters are $\theta/\varphi$ with $\tan\varphi=\sin\theta/(\rho+\cos\theta)$:

These are the stated values of $e$, $g$ and $f$; the three angles $\varphi$ lie in $(0,\pi/2)$, so the arctangents are the right branches. The three ratios exceed $1$, and $2(1+\cos\theta)$ equals $\frac{52}{49}$, $\frac{160}{49}$ and $\frac{180}{49}$ for the edges $CD$, $BD$ and $BC$, which exceed $1^2$, $(\frac{8}{5})^2$ and $(\frac{9}{5})^2$; so $2<e,f,g<4$. By Proposition 2.3, $A'=X$, and by Theorem 1.3 and its proof the other vertices are $h_1(X)$, $h_2(X)$ and $h_3(X)$. □

Proof of Example 1.6. Here $(\alpha,\beta,\gamma)=(\frac{1}{4},\frac{1}{2},1)$ and $N^2=\frac{21}{16}$, so $\cos\theta_{CD}=\frac{19}{21}$, $\cos\theta_{BD}=\frac{13}{21}$ and $\cos\theta_{BC}=-\frac{11}{21}$. By (8), $R_4(\theta_{CD})=1+4\sqrt{\frac{5}{21}}$ and $R_4(\theta_{BC})=1+2\sqrt{\frac{5}{21}}$. By (7), with $\beta+\gamma=\frac{3}{2}$ and $\alpha+\beta=\frac{3}{4}$, both equations $e=4$ and $f=4$ say $\lambda=\frac{3}{4}\sqrt{\frac{5}{21}}$. For this $\lambda$ the point $X=-t(1,1,1)$ with $t=\lambda/(1+\frac{7}{4}\lambda)=\frac{5}{2}-\frac{9}{2}\sqrt{\frac{5}{21}}$ is interior, and the third ratio is $1+\frac{5}{2}\lambda=1+\frac{10}{3}\sqrt{\frac{5}{21}}$. With $\sin\theta_{BD}=\frac{4\sqrt{17}}{21}$, Lemma 2.1 gives

which is the stated value of $g$. The third ratio exceeds $1$, and $(\frac{10}{3})^2\cdot\frac{5}{21}=\frac{500}{189}<\frac{612}{189}=2(1+\cos\theta_{BD})$, so $2<g<4$. Finally $s>0$ because $\frac{5}{21}<\frac{5}{81}$. As before, $A'=X$ and the Morley tetrahedron is regular. □

Quadrisecting planes

Proof of Theorem 1.7. By Lemma 3.1 there is an isometry $h$ that maps $A$, $B$, $C$, $D$ to the points (1), in this order. By Lemma 2.4, with $\pi$ the identity and all parameters equal to $4$, $h$ maps $\mathcal{M}_{4,4,4}(ABCD)$ onto $\mathcal{M}_{4,4,4}$ of the image, and an isometry preserves regularity. So we may assume that $ABCD$ is the tetrahedron (1). By (2) it is regular if and only if $a=b=c$, that is, $\alpha=\beta=\gamma$.

Suppose first that $\alpha=\beta=\gamma$. By (6) the edges $CD$, $BD$ and $BC$ have the same dihedral angle $\theta$, and since $d_X=L_X/N$, (4) says that $L_B=L_C=L_D=R_4(\theta)L_A$ at $A'=(x,y,z)$. Now $L_B-L_C=2\alpha(x-y)$ and $L_C-L_D=2\alpha(y-z)$, so $x=y=z$ and $A'$ lies on $\ell_1$. If $A'$ were $O$, then $L_A(A')=L_B(A')=1$ would give $R_4(\theta)=1$, while $R_4(\theta)>1$ by Lemma 2.1. So $A'\ne O$, and $\mathcal{M}_{4,4,4}(ABCD)$ is regular by Theorem 1.3.

Conversely, suppose that $\mathcal{M}_{4,4,4}(ABCD)$ is regular. By Theorem 1.3, $A'=X$ is a point $t\delta\ne O$, where $\delta$ is one of $\delta_1=(1,1,1)$, $\delta_2=(1,-1,-1)$, $\delta_3=(-1,1,-1)$, $\delta_4=(-1,-1,1)$. Put $\mu=t/L_A(X)\ne0$ and

By (6), $1+\cos\theta_{CD}=2q_{CD}/N^2$, and similarly for the other two edges. A direct computation from the formulas for $L_X$ gives

where $(w_{CD}, w_{BD}, w_{BC})$ equals

for $\delta=\delta_1,\delta_2,\delta_3,\delta_4$ respectively. Since all parameters equal $4$, (5) and (8) give, for each of the three edges $E$ of $F_A$,

In particular $w_E\ne0$, since $q_E>0$, and the three quotients $q_E/w_E^2$ are equal. For $\delta=\delta_2$ this is impossible: $q_{CD}/w_{CD}^2=(\beta^2+\gamma^2)/(\beta+\gamma)^2<1$, while $q_{BD}/w_{BD}^2=(\alpha^2+\gamma^2)/(\alpha-\gamma)^2>1$. In the same way the edges $BD$ and $CD$ rule out $\delta_3$, and the edges $BC$ and $CD$ rule out $\delta_4$. So $\delta=\delta_1$, and

Since $(u^2+v^2)/(u+v)^2=1-2r(u,v)$ with $r(u,v)=uv/(u+v)^2$, this says $r(\gamma,\beta)=r(\gamma,\alpha)$ and $r(\alpha,\gamma)=r(\alpha,\beta)$. For positive $u,v,w$,

so the two equations say

Suppose $\alpha\ne\beta$. Then $\gamma^2=\alpha\beta$. If $\gamma=\beta$, this gives $\beta^2=\alpha\beta$, so $\alpha=\beta$; otherwise $\alpha^2=\beta\gamma$, and dividing $\gamma^2=\alpha\beta$ by $\alpha^2=\beta\gamma$ gives $\gamma^3=\alpha^3$, so $\gamma=\alpha$ and then $\alpha^2=\alpha\beta$, again $\alpha=\beta$. Both cases contradict $\alpha\ne\beta$. Hence $\alpha=\beta$, and the second equation becomes $-\beta(\gamma-\beta)^2=0$, so $\gamma=\beta$. Thus $a=b=c$, and $ABCD$ is regular. $\square$

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Funding

Funding. No funding was received for conducting this study.

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Competing interests. The author has no competing interests to declare that are relevant to the content of this article.

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Use of generative AI

Use of generative AI. During the preparation of this work the author used Claude (Anthropic) to search the literature, to find and check the proofs and examples, and to draft the text. The author checked every step of the proofs and every reference, and takes full responsibility for the content.

Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL), Bhubaneswar, Odisha 751030, India

Email address: [email protected]

Email address: [email protected]

URL: https://orcid.org/0000-0003-0466-750X