Why Triangles Are Easy and Tetrahedra Are Hard

Do you think there’s a triangle whose angles measure 41, 76 and 63 degrees?

At first, answering this may seem easy. From geometry class we know that the sum of the measures of the interior angles of a triangle is 180 degrees, and since 41 + 76 + 63 = 180, the answer must be yes.

But there’s more to this question than meets the eye. The triangle angle sum theorem tells us that, given a triangle in flat Euclidean geometry, the sum of the measures of the interior angles is 180 degrees. But our problem doesn’t give us a triangle. Instead, we are asked if one exists. The triangle angle sum theorem doesn’t directly answer that question. But it can help us construct the triangle we want.

To satisfy the triangle angle sum theorem, each angle in a triangle has to be less than 180 degrees, which means we can always arrange two of them on the same side of a line segment. Let’s put the 41-degree angle and the 76-degree angle at either end of line segment AB, like this.

The rays extending from A and B can’t be parallel, because in Euclidean geometry this would require these “same-side interior” angles to be “supplementary” — that  is, to sum to 180 degrees. These angles don’t, so the rays aren’t parallel, and if they aren’t parallel, they must intersect.

Call their intersection C. Now we’ve got a triangle, and now we can apply the triangle angle sum theorem. The third angle must measure 180 − (41 + 76) = 63, and so $latex \triangle A B C$ is exactly the triangle we were looking for.

This argument can be generalized to show that any three angle measures that sum to 180 degrees can make a triangle, and one immediate consequence is that it’s easy to find triangles whose angle measures (in degrees) are all rational numbers. Just start with any two positive rational numbers whose sum is less than 180; call them x and y. Then 180 − (x + y) is also a rational number, and since x + y + (180 − (x + y)) = 180, you can make a triangle with those three rational angles.

As easy as it is to make triangles with rational angles, a similar problem in three dimensions proved so challenging that it took decades for the world’s best mathematicians to resolve. What makes this kind of problem so much harder one dimension up? To understand this is to appreciate the triangle angle sum theorem even more.

The related problem in three dimensions involves tetrahedra — four-sided shapes with triangular faces. You could think of them as the three-dimensional version of triangles. In two dimensions the triangle is the simplest closed shape you can make with flat sides, and you need three line segments to do it. In three dimensions, the tetrahedron is the simplest closed shape you can make with flat sides, and it takes four triangular faces to do it.

The four triangular faces of a tetrahedron are like the three sides of the triangle. But how should we think about the angles? You could imagine a “solid angle” at each of the four vertices of the tetrahedron. But the question we’re interested in involves the “dihedral angles” formed by the intersecting faces.

If you sketch two intersecting planes, you’ll discover many different angles you could measure. Which one should you choose to represent the dihedral angle?

The answer is to rotate the intersecting faces until they look like a two-dimensional angle.

It’s this angle that we associate with the dihedral angle.

In a tetrahedron each of the four faces intersects every other face, making six edges and six dihedral angles. For decades, mathematicians wondered what kinds of tetrahedra have six rational dihedral angles. As before, an angle is considered rational if its degree measure is a rational number. This is equivalent to its radian measure being a rational multiple of π (To see the equivalence, note that to convert from degrees to radians you multiply the degree measure by $latex \frac{\pi}{180^{\circ}}$, so if the degree measure is rational, then the radian measure is a rational multiple of π, and vice versa.)

We’ve seen how easy it is to make triangles with rational angles, but the problem is much trickier for tetrahedra. Consider the following simple tetrahedron $latex O A B C$ that is formed from a corner sliced off a cube.

Right away we see that three of the dihedral angles in this tetrahedron are right angles, since they were formed by the faces of the cube. It’s convenient to identify each dihedral angle with its edge, so in this tetrahedron the dihedral angles on $latex OA$, $latex OB$ and $latex OC$ are all right angles.

If you slice the cube so that  $latex OA$ = $latex OB$ = $latex OC$, the dihedral angles at $latexAB$, $latexAC$ and $latexBC$ will all be congruent. Let’s cut the cube so that $latex OA$ = $latex OB$ = $latex OC$ = 1, and then calculate the measure of the dihedral angle at $latex BC$. The key to measuring this dihedral angle is to draw segments from $latex O$ and $latex A$ to the midpoint of $latex BC$. Let’s call that point $latex M$.

If we rotate the tetrahedron to look at the dihedral angle at $latex BC$ from the side, we’ll see $latex \angle A M O$, which has the same measure. To measure $latex \angle A M O$ we need the lengths $latex OA$ and $latex OM$. We already know that $latex OA$ = 1, and to find $latex OM$ we just need to take a closer look at triangle $latex \Delta O C B$.

Since $latex \angle B O C$ is a right angle, this is a right triangle, so we can use the Pythagorean theorem to find that $latex BC$ = $latex \sqrt{2}$. And because $latex M$ is the midpoint of $latex BC$, we know $latex MC$ = $latex \frac{\sqrt{2}}{2}$. But in addition to being a right triangle, $latex \Delta O C B$ is also isosceles, because $latex OB$ = $latex OC$. This makes it a 45-45-90 triangle, which means that $latex \angle OBC$ and $latex \angle OCB$ both measure 45 degrees. The fact that $latex \Delta O C B$ is isosceles guarantees that $latex OM$ is perpendicular to $latex BC$, which makes $latex \Delta O M C$ a right triangle as well. But if the measure of $latex \angle O M C$ = 90° and the measure of $latex \angle O C B$ = 45°, then the triangle angle sum theorem tells us that the measure of $latex \angle M O C$ = 45°. That makes the smaller triangle $latex \Delta O M C$ isosceles, so $latex OM$= $latex MC$ = $latex \frac{\sqrt{2}}{2}$.

Now we’re finally ready to find the measure of $latex \angle A M O$.

In $latex \Delta A M O$ we know $latex AO$ = 1 and $latex OM$ = $latex \frac{\sqrt{2}}{2}$, and since $latex \angle A O M$ is a right angle, we can apply trigonometry. In a right triangle the tangent of an angle is the ratio of the opposite side and the adjacent side:

$latex\tan \angle A M O$ = $latex\frac{1}{\frac{\sqrt{2}}{2}}$ = $latex \frac{2}{\sqrt{2}}$ = $latex\sqrt{2}$

So the measure of $latex \angle A M O$ is the inverse tangent, or arctangent, of $latex \sqrt{2}$. This turns out to be an irrational number, so this is not an example of a rational tetrahedron, as three of its angles aren’t rational. But even though it’s not what we’re looking for, this irrational tetrahedron can tell us something important about the search for rational tetrahedra.

To see this, let’s find the approximate sum of all the dihedral angles in our irrational tetrahedron. Using a calculator or a trig table, we find the approximate measure of $latex\angle A M O$ to be around 54.74 degrees.

We can now sum the six dihedral angles of tetrahedron $latex OABC$: Three are right (and measure 90 degrees), and the other three are all congruent to the angle we just found. Thus, the sum of the six dihedral angles in this tetrahedron is approximately 3 × 90° + 3 × 54.74°≈ 434.22°.

Here’s where things take a turn. Let’s go back to the cube, and instead of cutting it so $latex OA$ = $latex OB$ = $latex OC$, imagine taking a very thin slice off the corner.

This new tetrahedron still has three 90-degree dihedral angles at $latex OP$, $latex OC$, $latex OB$, but the other dihedral angles have changed. The angle at $latex BC$ now looks quite small, and the angles at $latex PB$ and $latex PC$ look more like the angles at $latex OB$ and $latex OC$.

In fact, if you continue to take thinner and thinner slices, $latex P$ will get closer to $latex O$, the dihedral angle $latex BC$ will approach 0 degrees, and the dihedral angles at $latex PB$ and $latex PC$ will each approach 90 degrees. Notice the approximate sum of the angles:

90° + 90° + 90° + 90° + 90° + 0° = 450°.

As $latex P$ gets closer to $latex O$ the sum of the six dihedral angles of the tetrahedron gets closer to 450°. Which means the sum of the angles is changing! In our original tetrahedron $latex OABC$ the dihedral angle measures added up to around 432°, but when we change the angles, the overall sum changes. The tetrahedron may be the 3D version of the triangle in some ways, but in one way it’s quite different: There’s no tetrahedron dihedral angle sum theorem guaranteeing that the sum of the angles is constant.

It turns out the best we can do is guarantee that the sum of the measures of the six dihedral angles of a tetrahedron lies between 360 and 540 degrees. And if you’re searching for tetrahedra with rational dihedral angles, this is a problem. You can’t just pick five rational angles and be sure the sixth is automatically rational because, unlike in a triangle, you have no idea what the sum has to be.

Even worse, you can’t be sure any six angles can even be the dihedral angles of a tetrahedron. Consider five right angles and an acute angle. The sum of these six angles is between 450 and 540 degrees, which is inside the acceptable range for a tetrahedron. But there is no tetrahedron that has these six angle measures. If five of the six dihedral angles are right, then one of the faces must have three right dihedral angles. But if this happens, those faces can’t close up and form a tetrahedron: Like parallel lines, they’ll never meet.

Thus, the problem of finding all possible rational tetrahedra requires much more than just finding five or six rational numbers with a certain sum. Among other things, it requires solving an equation with 105 terms inspired by a 1976 paper written by John Conway and Antonia Jones. A group of mathematicians accomplished this in 2020, and the result was a complete classification of all rational tetrahedra.

The triangle angle sum theorem is one of many reasons to admire the beauty and elegance of triangles. The lack of such a theorem for tetrahedra is a reason to appreciate their beauty and complexity, one dimension up.

Exercises

1. What is the sum of the dihedral angles in a cube?

2. What is the (approximate) sum of the six dihedral angles in a regular tetrahedron?

3. Imagine a regular tetrahedron sitting on a tabletop. As you push the top vertex down toward the bottom face, what happens to the sum of the six dihedral angles?

4. Can any four angle measures that sum to 360 degrees be the angles of a quadrilateral?

Click for Answer 1:

Click for Answer 2:

All six dihedral angles are congruent, so you can use an appropriate right triangle to find the measure of one of the dihedral angles.

All the faces of the regular tetrahedron are equilateral triangles, so the median — the segment from vertex to midpoint — of a face has length $latex \frac{\sqrt{3}}{2} s$, where s is the side length. This is the hypotenuse of the desired triangle. The altitude from the top vertex of the tetrahedron meets the bottom equilateral triangular face at its centroid, which is known to lie $latex\frac{1}{3}$ of the way along the median from the midpoint, making the length of this leg of this triangle $latex\frac{1}{3}$ × $latex\frac{\sqrt{3}}{2} s$. The dihedral angle is the angle between these two sides, and so is equal to the inverse cosine of $latex\frac{\frac{1}{3} \times \frac{\sqrt{3}}{2} s}{\frac{\sqrt{3}}{2} s}$ = $latex\frac{1}{3}$. Since $latex \arccos \frac{1}{3}$ ≈ 70.53°, the sum of the six (congruent) dihedral angles of the regular tetrahedron is approximately 6 × 70.53° ≈ 423.18°.

Click for Answer 3:

Click for Answer 4:

Yes. Say the measures are a, b, c and d, with a + b + c + d = 360. Assume that a and b are each less than or equal to both c and d. Split c into c₁ and c₂ and d into d₁ and d₂ so that c = c₁ + c₂, d = d₁ + d₂, a + c₁ + d₁ = 180, and b + c₂ + d₂ = 180 (there is plenty of freedom to do this in many different ways). Use the two sets of angles to create two triangles, adjusting so that the sides opposite angles a and b are congruent. Now put them together so c₁ and c₂ merge to form c and d₁ and d₂ merge to form d, and you’ve got your quadrilateral with angles a, b, c and d.

An interesting challenge is to consider if a quadrilateral can always be constructed with the angles in a particular order.

Correction: February 1, 2022

This column has been revised to correct a typographic error. 3 × 90° + 3 × 54.74°≈ 434.22° (not 432.22).