Given the vertices of a triangle that forms the base of a trirectangular tetrahedron, is the right angle vertex (the vertex opposite the base where all three face angles are right angles) uniquely defined (except, perhaps for a mirror reflection)?

I am guessing this can be expressed in vector notation as: Given a set of three vectors ($\vec{a},\vec{b},\vec{c}$), is there only one vector $\vec{x}$ that does not lie on the plane defined by the vertices of ($\vec{a},\vec{b},\vec{c}$), and satisfies the following relation?

$$ \lvert (\vec{a}-\vec{x}) \cdot (\vec{b}-\vec{x})\lvert \ + \ \lvert (\vec{b}-\vec{x}) \cdot (\vec{c}-\vec{x}) \lvert \ + \ \lvert (\vec{c}-\vec{x}) \cdot (\vec{a}-\vec{x})\lvert \ =0\ $$

(except perhaps for the vector obtained by reflecting the vertex of $\vec{x}$ about the plane defined by ($\vec{a},\vec{b},\vec{c}$))

Apologies if this is a bit elementary. Its been a while since college geometry!

  • $\begingroup$ Thanks for the great responses! Here's why the question was interesting to me in the first place: Given two points $A$ and $B$ on the two-dimensional Cartesian plane, there are an infinite number of orthogonal coordinate axes (origin $O$) such that $\vec{OA}$ and $\vec{OB}$ form orthogonal basis vectors (ie, such that $A$ and $B$ lie on orthogonal axes with origin $O$). The locus of $O$ is the "Thales' circle" (thanks to @ChristianBlatter for the term!). $\endgroup$ – Sridharan Devarajan Nov 2 '13 at 19:42
  • $\begingroup$ On the other hand, if the conjecture posed in my question is true, then the following has to be true: Given three points $A$, $B$ and $C$ in three-dimensional Cartesian space, there is only one orthogonal coordinate system (with origin $O$) such that ($\vec{OA}, \vec{OB}, \vec{OC}$) form orthogonal basis vectors (ie, such that $A$, $B$ and $C$ lie on orthogonal axes with origin $O$). (the other one generated by mirror reflecting $O$ about the plane $ABC$ can be discounted by adding a "handedness" constraint). I'm surprised I've never encountered this result (or anything like it) before! $\endgroup$ – Sridharan Devarajan Nov 2 '13 at 19:43

The vertex is unique (up to mirror reflection in the base).

You can prove this with vector methods, but here's a more synthetic approach.

Let $PABC$ be a tri-rectangular tetrahedron with "right corner" at $P$, and let $D$, $E$, $F$ be the feet of altitudes from $A$, $B$, $C$ in $\triangle ABC$.

Consider plane $PAD$ (that is, the plane of $\triangle PAD$). Since $AD \perp BC$, we must have $PAD \perp ABC$; likewise $PBE \perp ABC$ and $PCF \perp ABC$. The three altitudes of a triangle meet at a common point called the orthocenter; consequently, our three planes meet at a common line, $\ell$, through the orthocenter, $Q$, of $\triangle ABC$, and this $\ell$ is perpendicular to $ABC$. One observes that $P$ (on $\ell$) must be a specific distance away from $ABC$[*] in order to create right angles $\angle BPC$, $\angle CPA$, $\angle APB$; as a result, $P$ is in one of two mirrored locations, as claimed. $\blacksquare$

[*] A "continuity" argument shows this: When $P$ is in plane $ABC$, clearly the angles are too large; as $P$ moves away from the plane, the angles get continually smaller until they're too small. Somewhere in there, and only once (on a given side of the plane), the angles are (ahem) just right.

For a more-constructive justification, let's write $$a := |BC| \quad b := |CA| \quad c := |AB| \quad u := |PA| \quad v := |PB| \quad w := |PC| \qquad h := |PQ|$$

Since $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ are all right triangles, we have $$v^2 + w^2 = a^2 \qquad w^2 + u^2 = b^2 \qquad u^2 + v^2 = c^2$$ We can solve this system for $u$, $v$, $w$. The values happen to be $$u^2 = \frac{1}{2}\left(-a^2+b^2+c^2\right) \qquad v^2 = \frac{1}{2}\left(a^2-b^2+c^2\right) \qquad w^2 = \frac{1}{2}\left(a^2+b^2-c^2\right)$$ but that's a bit beside the point. Now, simply note that we can compute the volume of the tetrahedron in two ways $$V = \frac{1}{3} h \; |\triangle ABC|= \frac{1}{6} u v w$$ Since $u$, $v$, $w$ are determined by $a$, $b$, $c$, as is area $|\triangle ABC|$, height $h$ must have a unique value.

  • $\begingroup$ I think the second part of your response nails it. Just to clarify further, the volume of a general tetrahedron is given by: $$V = \frac{\mathcal{C}}{6} uvw \\ \mathcal{C} = \sqrt{1 + 2 \cos \alpha \cos \beta \cos \gamma - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma}$$ where $\alpha, \beta, \gamma$ are the three angles at the vertex $P$ (intersection of $u, v, w$). As these are $90^{\circ}$ for the trirectangular tetrahedron, $\mathcal{C}=1$ and, the formula for the volume simplifies to: $$V = \frac{1}{6} uvw$$ $\endgroup$ – Sridharan Devarajan Nov 2 '13 at 18:59
  • $\begingroup$ I am not sure I am convinced by your geometric argument (the first part of your response). Even if $AD \perp BC$ it remains to be shown that plane $PAD \perp ABC$ for a trirectangular tetrahedron (either rigorously or with an intuitive argument). $\endgroup$ – Sridharan Devarajan Nov 2 '13 at 19:06
  • $\begingroup$ @SridharanDevarajan: Perhaps it's easier to think of the perpendicularity thing the other way: The plane through $PA$ that is perpendicular to $BC$ necessarily contains altitude $AD$ in $\triangle ABC$ (as well as altitude $PD$ in $\triangle PBC$). $\endgroup$ – Blue Nov 2 '13 at 20:03
  • $\begingroup$ @SridharanDevarajan: Your volume formula is correct. For the purposes of my argument, however, the simpler formula is $$V=\frac{1}{3} K k$$ where $K$ is the area of a face, and $k$ is the length of the altitude to that face. For "leg face" $\triangle PBC$, for instance, we have $K = \frac{1}{2} vw$ and $k = u$; for the "hypotenuse face", $K = |\triangle ABC|$ and $k = |PQ|$. $\endgroup$ – Blue Nov 2 '13 at 20:09

The locus of points in ${\mathbb R}^3$ from which the segment $AB$ is seen under a right angle is the "Thales' sphere" with center the midpoint of $AB$ and diameter $AB$.

Therefore we have to consider the three Thales' spheres over the three sides of the base triangle. The two spheres over $AB$ and $AC$ intersect in a circle $\gamma$ passing through $A$. This circle intersects the sphere over $BC$ in at most two points $P_i$, which then have to lie symmetrically with respect to the base plane. From each of these two points the three sides of the base are seen under a right angle.

The circle $\gamma$ need not intersect the third sphere. In this case the problem has no solution. Consider as an example the case where the angle at $A$ is almost $180^\circ$ and $|AB|=|AC|$.

  • $\begingroup$ Christian, I had the same geometric intuition and hence made the conjecture. I did not know that the spheres were called "Thales' spheres". Thank you for sharing this information! $\endgroup$ – Sridharan Devarajan Nov 2 '13 at 17:56

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