Orthocentric tetrahedra (Part 3)

This part is the third and final post of the series about orthocentric tetrahedra. In this post we will investigate a special case of orthocentric tetrahedra, which is trirectangular tetrahedra. The main focus will be their properties and equalities.

Definition 1. A tetrahedron $OABC$ is called trirectangular at $O$ if $OA$, $OB$ and $OC$ are pairwise perpendicular.

Remark 2.

a) It can be implied directly from the definition that a trirectangular tetrahedron is also orthocentric. In this case the orthocenter is $O$, the right vertex.

b) Again to check if a orthocentric tetrahedron $OABC$ is trirectangular at $O$, we just need to check any two of three pairs between $OA$, $OB$, $OC$ are perpendicular. In that case third pair is also perpendicular. Note that if $OABC$ is not orthocentric, then the perpendicularity of any two of three above pair does not imply the perpendicularity of the last pair. Readers can check this by themselves.

When a tetrahedon $OABC$ is trirectangular at $O$, we call the faces $OAB$, $OBC$ and $OCA$ the right-angled faces and the face $ABC$ the hypotenuse (or the base).

Next we will move on to the first main theorem about the equalities in a trirectangular tetrahedron.

Theorem 3. Given a trirectangular tetrahedron $OABC$ with the right vertex $O$. Let $H$ be the foot of the altitude starting from $O$. Let $S(ABC)$ denote the area of the triangle $ABC$. The following statements are true:

(i) The triangle $ABC$ has three acute angles;

(ii) $\dfrac{1}{OH^2}=\dfrac{1}{OA^2}+\dfrac{1}{OB^2}+\dfrac{1}{OC^2}$;

(iii) $S^2(OAB)+S^2(OBC)+S^2(OCA)=S^2(ABC)$ (Pythagorean equality);

(iv) $S^2(OAB)=S(HAB).S(ABC)$, $S^2(OBC)=S(HBC).S(ABC)$, $S^2(OAC)=S(HAC).S(ABC)$;

(v) Let $\varphi_A$, $\varphi_B$, $\varphi_C$ are the angle between $(ABC)$ and $(OBC)$, $(OCA)$ and $(OAB)$ respectively. Then $\cos^2{\varphi_A}+\cos^2{\varphi_B}+\cos^2{\varphi_C}=1$.

Figure 1

Proof. (i) Let $OA=a$, $OB=b$ and $OC=c$. Then by the Pythagorean theorem we have $AB^2=a^2+b^2$, $AC^2=a^2+c^2$ and $BC^2=b^2+c^2$. Since

$$AB^2+AC^2-BC^2=2a^2>0,$$

we conclude that $\angle A<90^\circ$. Similarly $\angle B<90^\circ$ and $\angle C<90^\circ$.

(ii) (see Figure 1) Let $D$ be the orthogonal projection of $D$ onto $AB$, then it is not hard to show that the altitude $CH$ of the triangle $ABC$ passes through $D$. Thus

$$\dfrac{1}{OH^2}=\dfrac{1}{OD^2}+\dfrac{1}{OC^2}=\dfrac{1}{OA^2}+\dfrac{1}{OB^2}+\dfrac{1}{OC^2}.$$

(iii) Multiplying two sides of (ii) by $9V^2$ where $V$ is the volume of $OABC$, we get

$$\left(\dfrac{3V}{OH}\right)^2=\left(\dfrac{3V}{OA}\right)^2+\left(\dfrac{3V}{OB}\right)^2+\left(\dfrac{3V}{OC}\right)^2.$$

By the formula of the volume of a tetrahedron, this equality implies (iii).

(iv) Use the notion of part (v). Since the orthogonal projection of $C$ to $(OAB)$ is $O$ and the orthogonal projection of $O$ to $(ABC)$ is $H$,

$$\cos \varphi_A=\dfrac{S(OAB)}{S(CAB)}=\dfrac{S(HAB)}{S(OAB)}. (1)$$

Thus $S^2(OAB)=S(HAB).S(ABC)$. By the same argument we have the remaining two equalities.

(v) We have

$$\cos^2{\varphi_A}+\cos^2{\varphi_B}+\cos^2{\varphi_C}=\dfrac{S^2(OAB)+S^2(OAC)+S^2(OBC)}{S^2(ABC)}.$$

By (iii) the numerator and denominator of the right-hand side above are equal. Thus the right-hand side is $1$. $\square$

We know that the converse of the Pythagorean theorem for right triangles is true, therefore we may ask that if there is a similar version in solid geometry. Unfortunately, only the equality (iii) is not enough to conclude that a tetrahedron is trirectangular; following is an example:

Example 4. (see Figure 2) Given a cube $ABCD.A'B'C'D'$ whose edge length is $1$. Let $M$ be the midpoint of the segment $CD$. Consider the tetrahedron $MAB'D'$, we have:

Figure 2

(a) The triangle $AB'D'$ is equilateral with edge length $\sqrt{2}$, so $S(AB'D')=\dfrac{\sqrt{3}}{2}$;

(b) The triangle $MAD'$ is right-angled at $D'$ and has the area $S(MAD')=\dfrac{1}{2}.\sqrt{2}.\dfrac{1}{2}=\dfrac{\sqrt{2}}{4}$;

(c) The triangle $MB'D'$ has the area $S(MB'D')=\dfrac{1}{2}.\dfrac{1}{2}.1=\dfrac{1}{4}$;

(d) The triangle $MAB'$ is not of any special form (equilateral, right-angled, etc...). To compute $S(MAB')$, we will compute the length of its attitude beginning from $M$. To do this, let $N$ be the midpoint of the segment $A'B'$, then draw $H$ on the segment $AB'$ such that $NH \perp AB'$. It is then not difficult to prove that $MH \perp AB'$. Thus,

$B'H=\dfrac{B'N.B'A'}{B'A}=\dfrac{\sqrt{2}}{4}$, $MB'=\sqrt{MC'^2+C'B'^2}=\dfrac{\sqrt{5}}{2}$, $MH=\sqrt{MB'^2-B'H'^2}=\dfrac{3\sqrt{2}}{4}$,

and $S(MAB')=\dfrac{1}{2}.AB'.MH=\dfrac{3}{4}$.

It is now not difficult to verify that $S^2(MAD')+S^2(MB'D')+S^2(MAB')=S^2(AB'D')$. However, $MAB'D'$ is obviously not right-angled at any vertex. $\square$

In the above example, we see that the tetrahedron is even not orthocentric. Therefore, the orthocentric condition is necessary to check the trirectangular property of a tetrahedron. Moreover, from part (i) of Theorem 3 we see that the face opposite to the right vertex is an acute triangle, so we also add this property as a condition for the Pythagorean theorem's converse version for tetrahedra. We will now show that these two conditions are enough for the theorem.

Theorem 5. Let $OABC$ be an orthocentric tetrahedron with $ABC$ an acute triangle. Then $OABC$ is right-angled at $O$ if and only if one of the four equalities in Theorem 3 is satisfied.

Proof. See Figure 3 below. In this figure $AD$, $BE$ and $CF$ are the attitudes of the triangle $ABC$, $H$ is the orthocenter of this triangle, and $H'$ is the orthocenter of $OABC$. It is not hard to see that the attitude $OH$ is the intersection of three planes $(OAD)$, $(OBE)$ and $(OCF)$. Let $A'=AH' \cap OD$ and $D'=DH' \cap OA$, it is again not hard to see that $DD' \perp OA$, $AA' \perp (ABC)$ and $A'$ is the orthocenter of the triangle $OBC$. Define $B'$, $C'$, $E'$ and $F'$ in the similar way, we also get the similar claims.

Figure 3

Before proving the theorem, let us discuss some properties that might be useful along the way. Firstly, because the triange $ABC$ is given to have three acute triangle, its orthocenter $H$ lies inside the triangle, and lies in the attitude segments. Hence we can fix the triangle $ABC$, and thus the vertex $O$ runs through the line $\ell$ passing through $H$ and being perpendicular to $(ABC)$ to guarantee that the tetrahedron $OABC$ is orthocentric. Secondly, because $\overline{DA'}.\overline{DO}=\overline{DH}.\overline{DA}>0$, the points $O$ and $A'$ lie in the same side with respect to $D$; the similar claim is true for $O$,$B'$,$E$ and $O$,$C'$,$F$. Thus $OA=OD' \Leftrightarrow A \equiv D'$, $OB=OE' \Leftrightarrow B \equiv E'$, and $OC=OF' \Leftrightarrow C \equiv F'$

Theorem 3 shows the necessity of the equalities, so it suffices to show their sufficiency. We will start with (iv). We have

$S^2(OAB)=S(HAB).S(ABC) \Leftrightarrow (OF.AB)^2=(HF.AB).(CF.AB)$

$\Leftrightarrow FO^2=FH.FC \Leftrightarrow \angle FOC=90^\circ \Leftrightarrow OF \perp OC$

$\Leftrightarrow OC \perp (OAB)$ (since $OC \perp AB$ and $(OAB)$ contains $AB$ and $OF$)

$\Leftrightarrow OC \perp OA$ and $OC \perp OB \Leftrightarrow$ $OABC$ is right-angled at $O$.

Here we have $FO^2=FH.FC \Leftrightarrow \angle FOC=90^\circ$ because $H$ lies on the segment $CF$. The proof for the remaining two equalities are similar.

Next we move on to (iii) and (v) and prove the conclusion simultaneously for these two identities. Reuse the notation $\varphi_A$, $\varphi_B$ and $\varphi_C$ as defined in Theorem 3, we have

$\cos \varphi_A=\frac{S(HBC)}{S(OBC)}=\frac{S(A'BC)}{S(ABC)} \Rightarrow S(OBC).S(A'BC)=S(ABC).S(HBC)$
$\cos \varphi_B=\frac{S(HAC)}{S(OAC)}=\frac{S(AB'C)}{S(ABC)} \Rightarrow S(OAC).S(AB'C)=S(ABC).S(HAC)$
$\cos \varphi_C=\frac{S(HAB)}{S(OAB)}=\frac{S(ABC')}{S(ABC)} \Rightarrow S(OAB).S(ABC')=S(ABC).S(HAB)$

Adding the identities side-by-side, we get

$$\sum S(A'BC).S(OBC)=S^2(ABC),$$

here the sum takes all the cyclic permutation of $A,B,C$. So

$OABC$ is right-angled at $O$ $\Leftrightarrow$ $OA$, $OB$ and $OC$ are pair-wise perpendicular

$\Leftrightarrow O \equiv A' \equiv B' \equiv C' \Leftrightarrow OA'=OD, OB'=OE, OC'=OF$

$\Leftrightarrow S(OBC)=S(A'BC), S(OAC)=S(AB'C), S(OAB)=S(ABC')$

$\Leftrightarrow \sum S^2(A'BC)=\sum S^2(OBC)=S^2(ABC)$

$\Leftrightarrow \sum S^2(OBC)=S^2(ABC) \text{ and } \sum \frac{S^2(A'BC)}{S^2(ABC)}=1$

$\Leftrightarrow \sum S^2(OBC)=S^2(ABC) \text{ and } \cos^2\varphi_A+\cos^2\varphi_B+\cos^2\varphi_C=1$

Finally we will show that (ii) is also a necessary and sufficient condition. Multiplying both sides of (ii) by $OH^2$, we get

$$1=\frac{OH^2}{OA^2}+\frac{OH^2}{OB^2}+\frac{OH^2}{OC^2}.$$

Denote $\psi_A$, $\psi_B$, $\psi_C$ respectively the angle between the plane $(ABC)$ and the plane $(H'BC)$,  $(H'AC)$ and $(H'AB)$. We have

$$\cos\psi_A=\cos \angle HDH'=\cos \angle AOH=\frac{OH}{OA},$$

and similarly $\cos\psi_B=\frac{OH}{OB}$ and $\cos\psi_C=\frac{OH}{OC}$. Thus (ii) is equivalent to

$$\cos^2\psi_A+\cos^2\psi_B+\cos^2\psi_C=1.$$

Note that the tetrahedron $OABC$ is orthocentric and has the orthocenter $H'$, so the tetrahedron $H'ABC$ is also orthocentric and has the orthocenter $O$ (the proof is left as an easy exercise). Therefore (ii) holds if and only if $H'ABC$ is right-angled at $H'$, i.e if and only if $H' \equiv O$, i.e if and only if $OABC$ is right-angled at $O$. $\square$