The Monge Point

This blog post corresponds to my recent camper talk at MBL Poland on tetrahedra¹.

Today, we shall take a look at the three-dimensional analogue of the triangle — the tetrahedron. On the way, we shall answer the question of whether the standard triangle centers also exist for tetrahedra, with particular attention on the orthocenter.

Throughout our work, we consider a tetrahedron in the three-dimensional Euclidean space.

Recall that the locus of points equidistant to a line segment , is the perpendicular-bisector plane of — the plane passing through the midpoint of the segment, which is perpendicular to it. Now, the perpendicular bisector planes of segments , and have a unique common point , which is equidistant to each of the vertices. This, is the circumcenter of tetrahedron .

Now, what about the orthocenter. Sure, we could try defining it to be the point of concurrence of the four altitudes from each vertex to the opposite face. The only issue is that this point does not necessarily exist.

Before we get into the gory details of how to fix this, it is best to have a quick refresher on how perpendicularity works in three dimensions.

We know that a line is said to be perpendicular to another line if the two lines intersect at a right angle, and it is fairly intuitive that two planes in space are said to be perpendicular if the dihedral angle at which they meet is a right angle. Moreover, a  line is said to be perpendicular to a plane if it is perpendicular to every line in the plane that it intersects (play around with this if it feels awkward).

Our escape route shall be to define a new object called the Monge plane as the plane perpendicular to one edge of a tetrahedron which contains the midpoint of the opposite edge. In what follows, assume the circumcenter of the tetrahedron lies at the origin.

Proof. We shall begin by making the seemingly unrelated observation that the segments connecting the midpoints of opposite edges of a tetrahedron bisect each other. Indeed, the midpoint of the segment joining the midpoints of and has position vector,

which is symmetric in , , and , settling the claim.

Now, let denote the perpendicular-bisector plane of side . Consider the reflection of across point . This plane must be parallel to and pass through the reflection of the midpoint of across , which is the midpoint of segment . Thus, is precisely the Monge plane of side .

Since the perpendicular bisector planes concur at the circumcenter of , it follows that the Monge planes concur at the reflection of across , the Monge point of the tetrahedron.

It also happens that the point above is the centroid of the tetrahedron, and the proof is similar to the two-dimensional case. Thus, the circumcenter, centroid and Monge point of a tetrahedron lie on a straight line known as the Euler line of the tetrahedron.

Moreover, the Monge point has position vector

.

To see why this point is considered the three-dimensional analogue of the orthocenter, note that the definition of a Monge plane reduces directly to an altitude for a triangle. Moreover, whenever the crude definition for the orthocenter exists, it coincides with the Monge point of the tetrahedron.

The next step obviously is to determine when the orthocenter of a tetrahedron does exist.

Proof. Suppose the altitudes concur at . Since , line is perpendicular to side . Similarly, line is perpendicular to side . It then follows that line is perpendicular to plane so it is also perpendicular to side . The proof for the other pairs is similar.

Repeat the same argument for the other direction.

Moreover, it is also true that if any two pairs of opposite sides are perpendicular, then so is the last pair.

The next problem provides an alternative characterization for orthocentric tetrahedra.

It is not hard to see that

.

Thus,

.

Repeating the argument for the other sides and applying the above theorem solves the problem.

When the orthocenter of a tetrahedron does exist, it is related to the orthocenters of the faces as follows.

Proof. Orthogonally project line to side . Point stays fixed, and maps to . Thus, (confirm why!). Repeating the argument for the two other sides implies that is the orthocenter of .

Problem 3 from the 1983 IMO Longlist shows the converse result that if at least one altitude passes through the orthocenter of the opposite face, then the tetrahedron must be orthocentric.

You can also take this one step further and show that the tetrahedral Euler line maps to the Euler line of triangle under orthogonal projection at . Hint: first try showing that the orthogonal projection of the centroid is the nine-point center of the respective face.

The similarities with the two-dimensional case continue with the following result which shows that the orthocenter of a tetrahedron forms a five-point orthocentric system when it exists.

Proof. For example, consider tetrahedron . Since the altitude from passes through . Furthermore, since and , it follows that is the altitude from to . The other vertices work identically, implying that is this tetrahedron’s orthocenter.

Let us wrap up with an application of the above observations to a recent Olympiad problem. This is in a sense the three-dimensional analogue of the famous ‘Reflecting the orthocenter’ lemma from configurational geometry.

We shall use several results about circumspheres that you should feel free to look up if you have not encountered them before.

We know that must in fact be the orthocenter of face , so let hit again at , and let hit and again at and respectively.

Since lies on the circumsphere of (why?) we may consider the power of a point at with respect to (think about why this is well defined) to note,

since and are symmetric about .

Now, let hit the circumcircle of again at . Since is the orthocenter of (why?), due to the power of a point at with respect to we may similarly note,

since and are symmetric about . Combining these two relations solves the problem.

1. Here’s the link to the original slides which include diagrams. ↩︎