Projective Geometry Lesson 6
Harmonic Mirroring
Continuing with our play with the Harmonic Point: We discovered last time that the placement of the Harmonic Point only depends on the placement of the first three points. Now let’s look at that relationship in a little more detail to see something else truly remarkable.
Draw another line and choose your three points as before. You might be getting good enough to be able to predict where the fourth harmonic point will appear based on your choices. I’m going to label the two points that connect to all the sides of my quadrilaterals as S1 and S2, and the point between them that connects to a diagonal as D1. So, the Harmonic Point by this labeling scheme will be the second diagonal, D2.
By looking at the way I’ve chosen the placement, do you already know where D2 will appear? Try it yourself, of course, but you may be getting good enough at this that you can just look at my drawings and see it.
Now, we are going to try to imagine something: starting from D1’s current position, we will slide it slowly to the left toward S1. We will keep S1 and S2 stationary, so what we will be imagining is how D2 responds to this movement. Right now, D1 and D2 are both close to S2. Now, imagining that D1 slides to the left, I’m going to use another color to construct a new quadrilateral and find the new location for D2:
Moving D1 a short distance to the left causes D2 to move by a larger amount to the right. They are kind of like mirror images, but D2’s movement is more dramatic than D1.
Now, taking a third snapshot, as it continues to slide to the left, I’m going to intentionally choose the midpoint between S1 and S2 for my new position for D1. Do you already know what D2 will do?
The next three pictures are just showing you how I found the midpoint. You could use a ruler, of course, and calculate what is halfway, but this is more direct. I actually got this idea from my brilliant wife, Jenny, who tried something similar on the first drawing entirely by accident. I marked the distance from S1 to S2, then folded the paper so the marks matched. That gave me exactly half the distance so that is where I marked D1’s new position!
Now, using a third color, find D2 again, the Harmonic Point:
Whoops, I labeled it D3 by accident! It’s actually still D2, but where has it gone? This is exactly the same construction that we first did for the Projective Net with equidistant points, and now D2 (mislabeled as D3) is way, way out there. In fact, it’s so far out there that we use the language that “D2 is at infinity”! We will play a lot more with that language, but go with me for now!
Choosing one more position for D1 in which it still slides left, and it is now much closer to S1, what has happened?
(I fixed my mistake and every D2 is correctly labeled now!). It has “returned” to the visible line . . . from the other side!!!!
Can you now link up the motion smoothly in your imagination that is depicted here? (If you can’t then actually doing these drawings helps a lot!) D1 starts close to S2, and so D2 is also close on the other side. (Can you see that if D1 slid up to S2, that D2 would join it and all three points would coincide?) Now D1 moves slowly to the left, heading for S1. D2 responds by shooting out quickly to the right, so far and so fast in fact that D2 reaches “infinity” (again, we’ll discuss that a lot more later). When D1 is exactly halfway between S1 and D2, D2 is “at infinity”. Then D1 slides further left and approaches S1; and D2, following it, returns “from the other side”. And again, if D1 were to slide to be on top of S1, D2 would join it.
What does this mean, or what can it tell us about our exploration of Projective Geometry and infinity? Well, there are quite a few things, but for now I want to just highlight a couple: The points D1 and D2 are like “projective mirror images” of each other. But they are not like mirror images we are accustomed to in a flat mirror where both image and object are the same size. Rather, the movement of D2 in response to D1 is more like a convex mirror in which the slow constant motion of D1 to the left is “distorted” into the much more vigorous motion of D2 to the right. In fact, you can’t really say which point is the “real one” and which is the mirror image. They are both entirely equivalent. And that means that the line segment from S1 to S2, which appears to us to be a short segment that we can see on our paper, is the perfect, projected image of the line segment that goes from S2 to the right, through infinity and back to S1 on the other side. While one appears super long and passes through infinity, it is entirely equivalent to the apparently short finite segment we see on the paper! And, the midpoint between S1 and S2 is a projective mirror image of the point at infinity!!!
Wow, we are really into the thick of it now, aren’t we? I welcome all of your reactions. Am I being clear enough here? It would help if I labelled my points consistently! We will go even deeper into these ideas, especially the idea of the point at infinity, next time.
Here is something to provide an aid if you’re having trouble imagining it. I used a free online geometry program called Geogebra to make a version of this drawing in which you can watch the mirrored points smoothly move along the line. Check it out and let me know what you think. It’s mesmerizing!
The link was cool to try out.
Either we are in the thick of it or I am kinda thick. It makes sense...I guess.