Before getting into the topic of this post I want to tell a story about the mathematics curriculum in Vietnam. In the (former) national curriculum, students will spend the 11th-grade year and 12th-grade year studying solid geometry. They use the axioms of solid geometry to study the parallel and perpendicular relations in 3D space. They also study geometric properties of some solid figures, as well as a little bit of analytical geometry for three-dimensional spaces.
This post is the first one of a series about a solid figure called "orthocentric tetrahedron", i.e. a tetrahedron whose pairs of opposite edges are perpendicular. For some reason, many properties of this solid figure attract me, maybe because of an object called "Euler sphere" that I will mention in the second post of this series. But first, in this post, I will state and prove some basic properties of orthocentric tetrahedra.
Before doing that let us explain the meaning of the adjective "orthocentric". In plane geometry, we have already known that three altitudes drawn from the vertices of a triangle to the opposite sides intersect at the point called the orthocenter of the triangle. However, in solid geometry, four altitudes of a tetrahedron do not need to be concurrent (In fact, they could even be non-co-planar!). Therefore one may ask which propert(ies) a tetrahedron should have if its four altitudes are concurrent. We will prove later (Theorem 3) that the necessary and sufficient condition is the property mentioned in Definition 1.
Definition 1. A tetrahedron $ABCD$ is called orthocentric if $AB \perp CD$, $AC \perp BD$ and $AD \perp BC$.
Remark 2. In practice, to check if a tetrahedron $ABCD$ is orthocentric or not, we just need to check if it has at least two perpendicular pairs of opposite edges. The third pair is therefore perpendicular too. Indeed, suppose $AB \perp CD$ and $AC \perp BD$. Doing some calculations of scalar products of vectors,
$$\overrightarrow{AD}.\overrightarrow{BC}=\overrightarrow{AD}.(\overrightarrow{AC}-\overrightarrow{AB})=\overrightarrow{AD}.\overrightarrow{AC}-\overrightarrow{AD}.\overrightarrow{AB}$$
$$=\overrightarrow{AD}.\overrightarrow{AC}-(\overrightarrow{AC}+\overrightarrow{CD}).\overrightarrow{AB}$$
$$=\overrightarrow{AC}.\overrightarrow{AD}-\overrightarrow{AC}.\overrightarrow{AB}-\underbrace{\overrightarrow{CD}.\overrightarrow{AB}}_{0}$$
$$=\overrightarrow{AC}(\overrightarrow{AD}-\overrightarrow{AB})=\overrightarrow{AC}.\overrightarrow{BD}=0.$$
Thus $AD \perp BC$.
Now we move on to Theorem 3, which will state some properties of an orthocentric tetrahedron. These are also necessary and sufficient conditions for a tetrahedron to be orthocentric.
Theorem 3. Given a tetrahedron $ABCD$, these following statements are equivalent:
(i) $ABCD$ is orthocentric
(ii) $ABCD$ has at least two perpendicular pairs of opposite edges.
(iii) The orthogonal projection of any vertex to its opposite face is the orthocenter of that face.
(iv) Four altitudes of $ABCD$ is concurrent at a point $H$. We call $H$ the orthocenter of the tetrahedron $ABCD$.
(v) $AB^2+CD^2=AC^2+BD^2=AD^2+BC^2$.
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| Figure 1 |
(i) $\Rightarrow$ (iii): Let $H_A$ be the orthogonal projection of $A$ to the plane $(BCD)$. Since $AH_A \perp (BCD)$ we have $AH_A \perp BD$. Together with $AC \perp BD$, it implies that $BD \perp (ACH_A)$ and then $BD \perp CH_A$. By the same argument, we also have $CD \perp BH_A$. Thus $H_A$ is the orthocenter of the triangle BCD.
(iii) $\Rightarrow$ (iv): Let $AH_A$, $BH_B$, $CH_C$ and $DH_D$ be the altitudes of $ABCD$ with $H_A$, $H_B$, $H_C$ and $H_D$ are the orthocenters of the respective opposite faces, we need to prove that the four altitudes are concurrent.
Indeed, we have $AH_B \perp CD$ and $BH_B \perp CD$ (all given by (iii)). Thus $CD \perp (ABH_B)$. Additionally, $CD \perp BH_A$ since $H_A$ is the orthocenter of the triangle $BCD$. Thus $BH_A$ is parallel or lies in the plane $(ABH_B)$. But $(ABH_B)$ already consists of the point $B$, therefore $BH_A$ lies in $(ABH_B)$, which makes $AH_A$ and $BH_B$ co-planar.
By a similar argument, we can also show that any two altitudes are co-planar and they intersect. But four altitudes of the tetrahedron are not co-planar, therefore they have to be concurrent.
(iv) $\Rightarrow$ (i): Suppose that four altitudes $AH_A$, $BH_B$, $CH_C$ and $DH_D$ are concurrent at $H$. Since $AH_A \perp (BCD)$ and $BH_B \perp (ACD)$, the altitudes $AH_A$ and $BH_B$ are perpendicular to $CD$. This imples $CD \perp (ABH_AH_B)$ ($(ABH_AH_B)$ is the plane containing $AH_A$ and $BH_B$), and then $CD \perp AB$. By the same argument, we also have $BC \perp AD$ and $BD \perp AC$.
(i) $\Leftrightarrow$ (v): Let $H_A$ be the orthocenter of the triangle $BCD$. Using the Pythagorean theorem, we can easily show that $BH_A^2+CD^2=CH_A^2+BD^2=DH_A^2+BC^2$. Thus
$$AB^2+CD^2=AC^2+BD^2=AD^2+BC^2$$
$$\Leftrightarrow (\overrightarrow{AH_A}+\overrightarrow{H_AB})^2+CD^2=(\overrightarrow{AH_A}+\overrightarrow{H_AC})^2+BD^2=(\overrightarrow{AH_A}+\overrightarrow{H_AD})^2+BC^2$$
$$\Leftrightarrow AH_A^2+BH_A^2+2\overrightarrow{AH_A}.\overrightarrow{BH_A}+CD^2=AH_A^2+CH_A^2+2\overrightarrow{AH_A}.\overrightarrow{CH_A}+BD^2$$
$$=AH_A^2+DH_A^2+2\overrightarrow{AH_A}.\overrightarrow{DH_A}+BC^2$$
$$\Leftrightarrow \overrightarrow{AH_A}.\overrightarrow{BH_A}=\overrightarrow{AH_A}.\overrightarrow{CH_A}=\overrightarrow{AH_A}.\overrightarrow{DH_A}$$
$$\Leftrightarrow \overrightarrow{AH_A}(\overrightarrow{BH_A}-\overrightarrow{CH_A})=\overrightarrow{AH_A}(\overrightarrow{BH_A}-\overrightarrow{DH_A})=0$$
$$\Leftrightarrow \overrightarrow{AH_A}.\overrightarrow{BC}=\overrightarrow{AH_A}.\overrightarrow{BD}=0 \Leftrightarrow AH_A \perp BC \text{ and } AH_A \perp BD$$
$$\Leftrightarrow AH_A \perp (BCD) \Leftrightarrow ABCD \text{ is orthocentric.}$$
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