In Part 1 we discussed the meaning of the adjective "orthocentric" and several properties of orthocentric tetrahedron. In this post, we continue the topic with two objects called the Euler line and the Euler sphere (its definition will be stated later).
We have known from plane geometry that given a triangle $ABC$, then the orthocenter $H$, the centroid $G$ and the circumcenter $O$ satisfy the equality $\overrightarrow{HG}=2\overrightarrow{GO}$. Moreover, let $S$ be the center of the Euler circle of $ABC$ (i.e. the circle passing through the midpoints of the three edges), then $S$ divides $OH$ into two equal-length segments. Again, when we transfer from two-dimensional to three-dimensional, we only know that a tetrahedron $ABCD$ always has the centroid $G$, the intersection of four medians, and the circumcenter $O$. The tetrahedron has the orthocenter $H$ if and only if its three pairs of opposite edges are perpendicular. We will now show that in this case $H$, $G$ and $O$ are also collinear and, furthermore, find the ratio $\overline{HG}:\overline{GO}$.
Before doing so we will state and prove the following lemma:
Lemma 1. Given a tetrahedron $ABCD$ and its centroid $G$. Then $ABCD$ is orthocentric if and only if $G$ is equidistant to the midpoints of all the edges of $ABCD$.
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| Figure 1 |
Proof. Define the midpoints as in Figure 1. It is not difficult to show that $MPNQ$, $MRNS$ and $PRQS$ are parallelograms and $MN$, $PQ$, $RS$ are concurrent at $G$. Therefore
$ABCD$ is orthocentric $\Leftrightarrow AB \perp CD, AC \perp BD, AD \perp BC$
$\Leftrightarrow RQ \perp RP, MR \perp RN, MQ \perp MP$
$\Leftrightarrow$ $MPNQ$, $MRNS$, $PRQS$ are rectangles
$\Leftrightarrow GM=GN=GP=GQ=GR=GS$.
Theorem 2. Given an orthocentric tetrahedron $ABCD$. Let $H$, $G$ and $O$ are respectively the orthocenter, the centroid and the circumcenter of $ABCD$. Then $H$, $G$, $O$ are collinear and $\overline{HG}:\overline{GO}=1$.
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| Figure 2 |
Proof. (see Figure 1 and 2) Define $H_A$, $G_A$, $O_A$ respectively the orthocenter, the centroid and the circumcenter of the triangle $BCD$. Define $H_B$, $G_B$, $O_B$ similarly. Since the lines $AH_A$ and $OO_A$ are both perpendicular to the plane $(BCD)$, they are parallel. Therefore $O$ and $H$ are in $(AH_AO_A)$. On the other hand $H_A$, $G_A$, $O_A$ are collinear, therefore $G_A \in (AH_AO_A)$. This implies $AG_A \subset (AH_AO_A)$ and $G \in (AH_AO_A)$. We have successfully proved that $H,G,O$ lie in the plane $(AH_AO_A)$. By the same argument, we also get $H,G,O \in (BH_BO_B)$. Thus $H,G,O$ lies in the intersection line of two planes $(AH_AO_A)$ and $(BH_BO_B)$.
Next we will prove that $\overline{HG}:\overline{GO}=1$. Let $S_A$ be the orthogonal projection of $G$ to the plane $(BCD)$, then $S_A \in H_AO_A$. Applying the Thales theorem, $\overline{HG}:\overline{GO}=\overline{H_AS_A}:\overline{S_AO_A}$. Keep the name of the midpoints of the edges of $ABCD$ as Figure 1. Lemma 1 tells us that $GS=GQ=GN$, and since $GS_A \perp (BCD)$ the point $S_A$ is the circumcenter of the triangle $SQN$. But the circumcircle of the triangle $SQN$ is the Euler circle of the triangle $BCD$. Therefore $\overline{H_AS_A}:\overline{S_AO_A}=1$. The proof is now completed. $\square$
Theorem 2 states that if a tetrahedron $ABCD$ is orthocentric, then its orthocenter lies in the line passing through its centroid and its circumcenter. We call this lines the Euler line of $ABCD$.
Also, from the definition of the Euler circle of a triangle and Lemma 1, we see that when $ABCD$ is orthocentric, then the Euler circles of four faces of the tetrahedron lies in a spheres whose center is the centroid $G$ of $ABCD$ and whose radius is the distance from $G$ to the midpoint of any edge of $ABCD$, and vice versa, We call this sphere the Euler sphere of $ABCD$.
