Triangle Tuesday 2: The Circumcenter, Pedal Triangles, Degeneracy, And What Even Is A Triangle Anyway?

Triangle Tuesday 2: The circumcenter, pedal triangles, degeneracy, and what even is a triangle anyway?

The circumcenter is almost as simple an idea as the centroid, which we looked at before. To define it, you start the same way. Take triangle ABC, find the midpoints of the sides Ma, Mb, and Mc. Then instead of drawing lines to the midpoints from the vertices, draw perpendicular lines through the midpoints. These lines all coincide at a point O, which is the center of a circle that you can draw through the vertices. The circle is called the circumcircle, and that's why the point is called the circumcenter.

I say almost as simple, but in a sense the circumcenter is simpler than the centroid, because you could easily discover it by accident in the process of simply finding the midpoints. Drawing that perpendicular line, the perpendicular bisector, is the standard way of finding the midpoint of a line segment. It's covered all the way back in Book 1, Proposition 10 of Euclid's Elements, and it's simply this:

So if you find the midpoint of all three sides of a triangle with this method, you've already identified the circumcenter. But that doesn't prove that the perpendicular bisectors always coincide, nor that their point of crossing is the center of the circumcircle. For that, let's return to Euclid (Elements, book 4, proposition 5). Euclid's proof is very straightforward, and leads nicely into something interesting, so we'll follow that, but I will state the theorem differently.

Theorem: the perpendicular bisectors of a triangle coincide and their point of intersection is the center of a circle that meets all three vertices.

Let ABC be a triangle with midpoints of the sides Ma opposite A, similarly for Mb and Mc. Draw perpendiculars to sides AC and BC from their midpoints to meet at point O. Connect three segments from O to A, B, and C.

Consider the two blue triangles. Their sies AMb and CMb are equal, since Mb is the midpoint of AC. They also have OMb in common. Their angles at Mb are right angles, and therefore equal. So they have two sides and one angle the same, making them congruent, and therefore OA = OC.

The same argument applied to the green triangles shows that OB and OC are equal. By transitivity, OA = OB and O is equidistant from the three vertices. The radii of a circle are all equal, so a circle centered at O passing through A also passes through B and C.

Finally, draw a line from O perpendicular to AB. This creates two white triangles with sides OA and OB equal, side OZ in common, and equal right angles at Z. The two triangles are then congruent and the two sides AZ and BZ are equal. So Z is the midpoint Mc, showing that the perpendicular bisectors all meet.

And the same argument works when ABC is obtuse. The circumcenter lands outside the triangle, and in this coloring the white triangles are no longer white, but all the relationships between the segments are the same.

(What Euclid didn't prove is that the perpendicular bisectors of AC and BC do in fact meet somewhere, that is, that they aren't parallel. It's not difficult, but I'm not going to prove that either, at least not yet, for reasons.)

Let's develop another idea. We located the circumcenter by drawing the perpendicular bisectors, but now consider doing this construction in reverse. That is, pick a point, and then draw perpendiculars to the three sides. The intersection of the perpendicular and the side is called the foot of that point with respect to that side. If you do that with with the circumcenter, the feet are of course the midpoints, but you can find the feet for any point.

And if we connect those three feet, we get a triangle. In this case, the medial triangle, which we have seen before. For a point in general, the triangle formed by its feet is called the pedal triangle of that point. ("Pedal" meaning "related to feet," and yes, that is why a lever operated with your foot is also called a pedal.)

So let's draw the pedal triangle for an arbitrary point, move it around, and see what happens. The point is going to sometimes be outside the triangle, but that's all right. With extended sides (dashed lines) we will still be able to draw a perpendicular to find a foot, no matter where the point is.

So there's something interesting -- the three feet become colinear and the pedal triangle flattens out into a straight line when the point is on the circumcircle. Does that always happen?

Looks like it does! So let's prove that. Below is a drawing of the flattened-out pedal triangle of a point on the circumcircle, all labeled up. I've also added a couple dashed lines to make following the proof easier. What we would like to show is that ∠JKP + ∠PKL = 180°.

We're going to extract some information from this drawing based on two facts: a) in a cyclic quadrilateral (meaning it has all vertices on the same circle), opposite angles sum to 180° and b) if two right triangles have the same hypotenuse, the triangles have the same circumcircle. I'm not going to prove either of those here because this post is long enough already, but both of these results follow straightforwardly from the inscribed angle theorem.

Theorem: For a point P on the circumcircle of a triangle ABC, the feet J, K, and L with respect to ABC are colinear.

Okay. PCBA is a cyclic quadrilateral, so

1) ∠BAP + ∠PCB = 180°.

And ∠BAP is the same as ∠LAP, so

2) ∠LAP + ∠PCB = 180°.

The two triangles AKP and ALP are right triangles with the same hypotenuse (the dashed segment AP), so all four points are on the same circle and ALKP is a cyclic quadrilateral. Therefore,

3) ∠LAP + ∠PKL = 180°,

4) ∠PKL = ∠PCB.

Quadrilateral PKCJ is also cyclic (again because of right triangles sharing the same hypotenuse), so

5) ∠JKP = ∠JCP

by the inscribed angle theorem. ∠PCB is supplemental to ∠JCP, so

6) ∠JKP = 180° - ∠PCB

and then combining 4) and 6),

7) ∠JKP + ∠PKL = ∠PCB + (180° - ∠PCB) = 180°,

which means that the pedal triangle of a point on the circumference of a circle is flattened to a line segment. Can we consider such a figure to be a triangle?

Now we can return to Euclid's omission in the existence proof of the circumcircle. Proving that the perpendicular bisectors aren't parallel is equivalent to proving that no two sides of a triangle are parallel, or that the three vertices of a triangle aren't colinear. Euclid didn't do that, but it's pretty simple, so he could have. And then he would simply have said that such an arrangement of line segments isn't a triangle. Modern geometers working with projective geometry can answer differently, and might say that this is a degenerate triangle, but we haven't gotten into that yet.

Let's do one more thing. We can extend the flattened line segment into a line, called the Simson line, after Robert Simson, who never wrote anything about it. It was actually discovered by William Wallace, but not named for him, because that's how things work in math.

The set of all Simson lines from all points on the circumcircle form an envelope in the shape of a deltoid, the Steiner deltoid, named for Jakob Steiner, who for all I can tell was its actual discoverer.

The deltoid is tangent to the sides of the triangle at three points where the Simson line coincides with the sides. I'll have more to say about this lovely deltoid later, but for now, please just enjoy this gif. It took me several hours to figure out how to make it, so if people reading this could spend a collective several hours staring at it, that would be great.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.