#nine-point circle

[sozeugs]

[squareallworthy]

Triangle Tuesday 8: the nine-point circle, the deep lore, and did I say nine? I meant, like, fifty-six.

Today we are looking at the nine-point circle, a famous construction with connections to other triangle geometry that we've looked at before. But first, I'd like to re-introduce the centers that we've met so far. You should imagine that they're members of a super team each with their own little sequence that the cartoon always shows when they go into action.

And imagine that they're appearing with one of these dramatic cartoon backgrounds like this. I would go ahead and do that, but it would look really messy on the little screen you're staring at. It will look better in your imagination.

Anyway, here we go. Here they are, now each with its own thematic color.

G, the Centroid!

Secret origin: intersection of the medians

Trilinear coordinates: 1/a : 1/b : 1/c

Affiliations: Euler line, isogonal conjugate of the symmedian point

Super powers: center of gravity of triangle area • center of gravity of triangle vertices • divides triangle into equal parts ABG, BCG, and ACG • divides medians 2/3 of the distance from vertex to midpoint

O, the Circumcenter!

Secret origin: intersection of the perpendicular bisectors

Trilinear coordinates: cos A : cos B : cos C

Affiliations: Euler line, isogonal conjugate of the orthocenter

Super powers: center of circumcircle • circumcircle is locus of points with degenerate pedal triangles • relates to incenter via Euler's theorem in geometry

H, the Orthocenter!

Secret origin: intersection of the altitudes

Trilinear coordinates: sec A : sec B : sec C

Affiliations: Euler line, isogonal conjugate of the circumcenter

Super powers: constitutes an orthocentric system along with the vertices • incenter of orthic triangle • sides of orthic triangle are antiparallel to sides of reference triangle • concyclic with a vertex and two adjacent feet of altitudes

I, the incenter!

Secret origin: intersection of the angle bisectors

Trilinear coordinates: 1 : 1 : 1

Affiliations: entourage of excenters

Super powers: center of the incircle • isogonally self-conjugate • forms orthocentric system along with excenters • relates to circumcenter via Euler's theorem in geometry • always lies in interior of orthocentroidal disc

K, the symmedian point!

Secret origin: intersection of the symmedians

Trilinear coordinates: a: b: c

Affiliations: isogonal conjugate of centroid

Super powers: perspector of reference and tangential triangles • parallels through K intersect sides in six concyclic points • antiparallels through K intersect sides in six concyclic points, with K at the center of the circle • circles through K and two vertices intersect sides in six concyclic points • lines joining midpoints of altitudes and sides pass through K • centroid of its own pedal triangle • just very pretty

So now that we've got the super team together, it's time to introduce them to someone new. I know, you're thinking we've got this established cast of characters, and they relate to each other in so many varied and intimate ways. How could a new triangle fit into these relationships? Wouldn't it feel like a superfluous add-on at this point in the show?

Well, let's see. Let's go back and look at the midpoints of the sides. We find the midpoints by drawing perpedicular bisectors on each side. Those lines meet at the circumcenter, which makes the midpoints the feet of the circumcenter. The triangle formed by the midpoints is the pedal triangle of the circumcenter.

Any three points also determine, naturally, a circle. The circle formed by the feet of a point is called the pedal circle of the point. And every point has a pedal circle with respect to a given triangle.

The orthocenter, for instance, has a pedal circle defined by the feet of the altitudes, and guess what? It's the same circle! What's more, this circle bisects the segments that join the orthocenter with the vertices. For this reason, the circle is called the nine-point circle, and its center is the nine-point center, N. Let's prove that it exists.

Theorem: in any triangle the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the orthocenter to the vertices lie on a common circle.

Here's our triangle ABC, with midpionts of sides Ma, Mb, Mc, feet of the altitudes Ha, Hb, Hc, and midpoints of segments joining H and the vertices Sa, Sb, Sc. We will rely on the fact that a line segment joining the midpoints of two sides (a midline) is parallel to the third side and half as long.

In green we have horizongal segments MbMa, a midline of ABC, and SaSb, a midline of ABH. Both of these midlines are parallel to side AB and half its length, so they are parallel to each other and equal. Vertical green segments MbSa and MaSb are midliness of ACH and BCH respectively, so they are also parallel and equal. Therefore the green segments form a rectangle.

In the same way, the blue segments also form a rectangle. These two rectangles have a common diagonal MbSb, and so have the same circumcircle. And McHcSc is a right triangle, and its hypotenuse McSc is also a diagonal of the blue rectangle. Therefore Hc is on the same circle, and by analogous argument so are Ha and Hb.

Since now we can be sure the nine-point circle and its center exist, it should get its own introduction scene and signature color.

N, the nine-point circle!

Secret origin: pedal circle of circumcenter and orthocenter

Trilinear coordinates: cos(B-C) : cos(C-A) : cos(A-B)

Affiliations: Euler line

Super powers: passes through nine points • many many others which we will get to next

So those are some nice ways that connect the nine-point circle to the circumcenter and orthocenter. And it's cool that this new character N has some connections to the established characters O and H, but what about the other points? Can we find some connections there?

Yes, we can, because it turns out the nine-point center is connected to everything. Would you like to see some of them?

(This is me right now.)

Because there are kind of a lot.

The nine-point circle is tangent to the incircle and the excircles. If you are keeping track, that's now 13 points of interest on the nine-point circle.

The radius of the nine-point circle is half that of the circumcircle. This is easy to see, because the circle is the circumcircle of the medial triangle, which is half the size of the reference triangle.

The nine-point center lies on the Euler line midway between the circumcenter and the orthocenter. The centroid is one-third of the way from the circumcenter to the orthocenter. These positions give the orthocenter and centroid special status with regard to the nine-point circle.

The nine-point circle bisects any line segment from the orthocenter to the circumcircle, making the fact that it bisects the tops of the altitudes into a special case of a general property.

The centroid trisects any line segment extending from a point on the nine-point circle through G and on to the circumcircle. Another way of saying this is that G is a center of dilation between the circumcircle and nine-point circle with scaling factor -1/2. In triangle geometry, this particular dilation through G is called taking the complement of a point, so a third way of describing this relationship is to say that the nine-point circle is the complement of the circumcircle.

The nine-point center is also the center of gravity of A, B, C, and H.

As we have seen previously, the vertices and the orthocenter form an orthocentric system, meaning that any one of the points A, B, C, and H is the orthocenter of the other three. For all four of the triangles ABC, ABH, ACH, and BCH, the feet of the altitudes are the same, so they all share the same orthic triangle and the same nine-point circle.

But each of these four triangles has its own incircle and set of excircles, and the nine-point circle is tangent to all of them, making sixteen points of tangency. That's now 29 points of interest on the nine-point circle.

The incenter and the excenters form another orthocentric system, and the nine-point center of this system is the circumcircle of the original triangle. The reference triangle ABC is the orthic triangle of the excentral triangle JaJbJc.

And speaking of the orthic triangle, it defines three other triangles in the corners that are similar to the reference triangle. The Euler lines of these corner triangles coincide at a point on the nine-point circle. That brings us to 30 points of interest on the nine-point circle.

Let's go back to pedal triangles, and remember that if a point is on the circumcircle, its feet are colinear and its pedal triangle has collapsed to a line segment, called a Simson line. The envelope of these line segments form a shape called the Steiner deltoid.

The nine-point center is also the center of the Steiner deltoid, and the circle is tangent to the deltoid at three points. There are four different circumcircles in an orthocentric system, so there are twelve points of tangency in all, so we now have 42 points of interest on the nine-piont circle.

The Simson lines of two opposite points on the circumcircle intersect perpendicularly on the nine-point circle.

You may have noticed that in citing all these connections I haven't mentioned the symmedian point K so far. As far as I can tell, K is just not that very strongly connected to N. However, here's one: the circle through K, N, and O is orthogonal to the circle with diameter GH (the orthocentroidal circle). That's five members of our super team tied into one relation. But that's all I've got. Sometimes members of a team don't interact that much.

And this is the kind of thing you look for in a successful series, right? You have an established cast of characters, with a rich, dense lore linking them all together in various ways. Then, as the story develops and someone new is introduced, new retlationships make the lore even richer. That that's why I'm in the triangle fandom.

With that, I think the only question left to answer is "Hey, Square, do you happen to have a group portrait of the whole team together in one clear and not at all confusing picture?" And of course I do.

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.

An index of all posts in this series is available here.