How many altitudes of a tetrahedron can lie outside the tetrahedron?
Let the dihedral angle between the faces $ABC$ and $ABD$ of the tetrahedron $ABCD$ be an obtuse angle. The perpendicular from $D$ to the plane $S$ of the face $ABC$ intersects the plane at $D'$ (1st figure). Since the two faces form an obtuse angle, $D'$ lies on the other side of the line $AB$ in the plane $S$ compared to $C$. Therefore, the altitude $DD'$ - except for the point $D$ - lies outside the tetrahedron. A similar statement can be made about the altitude from vertex $C$. If the faces meeting at the edge $DC$ also form an obtuse angle, then the altitudes from vertices $A$ and $B$ also lie outside the tetrahedron. Such a tetrahedron exists. In the 2nd figure, let the opposite edges $AB$ and $CD$ of the tetrahedron $ABCD$ be 6 units long, and both are parallel to the plane of the drawing, with a distance of 1 unit between them, and their projections form a square. The dihedral angle $\alpha$ at the edges $AB$ and $CD$ is such that $\operatorname{tg} \frac{\alpha}{2} = \frac{3}{1}$, so the dihedral angles are obtuse.

If one of the dihedral angles at the opposite edges $AB$ and $CD$ is an obtuse angle and the other is an acute angle, then exactly two altitudes lie outside the tetrahedron. We get such a tetrahedron if, in the previous construction, we take $AB$ to be less than 2 units, keeping all other data unchanged.

We will show that exactly three altitudes can also lie outside the tetrahedron. In the 3rd figure, $ABC$ is an equilateral triangle, and the projection of vertex $D$ is the center of the triangle, with $F$ being the midpoint of edge $BC$. The altitude from $D$ has a segment inside the tetrahedron. If the angle $ADF$ is an obtuse angle, the other three altitudes lie outside the tetrahedron.

In a regular tetrahedron, no altitude lies outside the tetrahedron.

It remains to examine the case where exactly one altitude can lie outside the tetrahedron. We give a negative answer to this. Based on the 1st figure, if $D'$ is a point outside the triangle $ABC$, then at least one dihedral angle is an obtuse angle, and there will be at least one more altitude that lies outside the tetrahedron.

Summarizing, we can conclude that 0, 2, 3, or 4 altitudes of the tetrahedron can lie outside the tetrahedron.

Based on the work of Máthé András (Bp., Apáczai Csere J. Gym., 10th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_288d249bbe054ff07ab6g-1.jpg?height=1242&width=438&top_left_y=1081&top_left_x=824)