Theory: All rithmatic circles are nine-pointers.

[Alfa]

Warnings. TONS of Geometry.

 

The theory shows, that each known rithmatic circle, is, in fact, a special case of a nine-point circle. Also, this theory postulates a five-point and an eight-point circle.We start with the construction of a basic nine-point circle, as discribed on the graphic "the nine-point circle". For this, we need a sharp-angled triange cornered between the points A, B and C. We construct the middle-points of each side. Those we call M_a, M_b and M_c. We draw a circle through those three points.

 

 

 

We construct the heights on the sides a,b and c and call them h_a, h_b and h_c. Note, that the height meet our rithmatic circle d exactly at the intersections of d with a, b and c. We name the crossings of the h-lines with k with the names H_a, H_b and H_c as well as L_a, L_b and L_c. The M, H and L points now build the anchor points of the basic nine-point circle.

 

 

As far we have the usual nine-pointer. The only requirement for it was a triangle ABC with sharp angles.

Now let's make a step from general to special. Let us assume, that each of the angles of ABC would measure exactly 60°.

 

 

In this case, the points M and H on each side merge together, so we basically loose three points, leaving us with a six-pointer. Note, that a six-pointer absolutely must be regular under this constraints.

Anotther special case: let us assume, that one of the angles of ABC is exactly 90° and the other two 45°.

 

 

Here we have a lot of merging: the points L and M on the sides a and b merge togehter (note: L_a merges with M_b and vice versa), H_c and M_c merge with each other; all the other points merge with C.

The two-pointer is an even more special case: now we bring the points A and B together, making all points merging with either them or C. My graphical program is not capable of this construction, but theoretically it underlays the same laws as the nine-pointer construction.

 

Now: predictions.

 

The construction of a five-pointer requires ABC to have one 90° angle, but the other two to be different from each other. The Idea is the same as with the four-pointer, but M_c and H_c separate from each other. The pentagon in the five-pointer is irregular, but has a rectangle structure between four of it's points (all except H_c)

 

 

Also, you can construct an eight-point circle. It requires for ABC to have two similar angles, none of them 45° or 90°. In this situation only the points H_c and M_c merge. The octagon is irregular, but symmetrical around h_c.

 

 

Other forms might exist, but i didn't found a way to create three-pointers and seven-pointers. Properties, usefullness and rithmatic specialities of this circles are yet to discover.

Edited December 4, 2015 by Alfa

[Purelake Earthquake]

I've done a lot of the geometry for this too, but came to some different conclusions.

 

I don't think a two-pointer would work, if my theory is correct.  Here's why, but it requires a big theoretical geometry warning as well:

 

So, if you bring points A and B together, then line AB will have no direction, since it is defined by only one point.  Since it has no direction there can be no lines orthogonal to it.  No orthogonal lines means no altitude.  Two of the nine significant points are defined by the altitude, following your naming the L point merges with C, but the H point  either exists anywhere or doesn't exist at all

 

You also can't define a circle based on two points around it's edges, you can't assume it's a diameter.  The other defining point would be Hc, which can't be defined, therefore the nine point circle for this triangle can't be defined.  In fact, the two points cannot be the diameter.

Actually here's something that may make more sense.  Let's look at the midpoints, BC and AC are the same, so their midpoints will be the same.  We've got three points now, 1: The merge between A and B, 2: the merged midpoints, and 3: point C.  The merged midpoint doesn't merge with A+B or C, it's the midpoint of those two points,  That gives us three co-linear points.  All this means that we do have a third point that defines the circle.  Now when you've got three points on the edge of a circle, and those three points are distinct and co-linear, that means your circle is big, infinitely big circle.  Infinitely big circles are basically lines.

 

Except there's also something wrong with this.  I was assuming that at least one of the nine points merged, but none do, except for Lc, which is still kind of undefined right now.  Lc does merge there, so it does become a point and we do wind up with an infinite circle.  The circle line end up existing on top of both AC and BC, and the altitude for AB intersects C at a line perpendicular to the circle which is also a line.  I believe this means that we now have a place to put Hc, on the line at infinity (it's infinity far away from the triangle).  Now, I may be wrong about several steps on this paragraph, and I'm not sure how to prove everything here, but I'm pretty sure this is true.

 

Either way the other points hold up, namely that Hc is undefined, and two midpoints don't merge with A,B, or C. 

 

 

Basically a lot of things start to break down when you start dealing with one dimension triangles.  Someone correct me if I messed up anywhere here.

 

That said, there is still some possibility for my theory.

 

My personal reasoning regarding all this is that merging doesn't work, i.e. the nine points have to be distinct.  The Rithmatic scholars probably thought about this theory and tried merging, but it didn't work.  I took the fact that they hadn't come up with five or eight point circles as evidence that they don't work.  The illustration on nine-pointers in the book mentions they can't be obtuse, but it doesn't mention, them being right triangles, isosceles, or equilateral, all of which cause merging.

 

In fact I don't know why it says they can't be obtuse, those don't cause any merging, you just have to extend the lines of the triangle passed the vertices to get some of the H points.

 

 

[Alfa]

Defense of the two-pointer. Very ugly geometrics, working with limits. Don't read it if you care for your sanity.

I also don't know if this is even legitimate, but it's the only possible way to explain the two-pointer via the nine-pointers theory.

 

Actually, the two-pointer would work if we'll look at it from the point of limits. Say, we don't bring A and B together but let the space between them shrink ad infinitum. Mathemtaically speaking: lim _{A -> B} (|AB|)=0. This allows us to avoid the "indefinitivess" mentioned before.

 

Then the angles at A and B would be 90° each (or a infinite small step from 90°). Then H_a  and L_b would merge with B and respectively H_b and L_a with A respectively, and since there is almost no space between, they will merge with each other. The point H_c  - which is identical to M_c - would fall in the not-existant space between A and B ans therefore merge with them.

Where are now M_a,  M_b and L_c? The first two are clearly between the A/B-Point and C. Here the circle d crosses h_c , giving us the position of L_c. Now, the locations of all points are clear: we have C, we have the M_a/M_b/L_c-  Mergepoint and the A/B/M_c/H_c/L_a /H_a/L_b /H_b -  Mergepoint.

Ugly.

Even more ugly is, that the two "Mergepoints" define not one, but infinite amounts of circles.

Luckyly, we have worked with a sharp triangle (NOTE: in this limit-scenario, two lim=90°-Angles still count as a sharp triangle) which is inscribing the center of the circle. So the center has to stay inside the triangle.

Now, we don't have any real triangle, but from the limit-pov we have a infite small space between a and b.

So the center of d has to stay between a and b or, in fact, lying on them. The only possible location is in the center between both "Mergepoints". Now we have a center and two points on the circle, so we can actually draw it. Unsurprisingly, both mergepoints come to rest at exactly opposing points of the circle.

Edited December 4, 2015 by Alfa

[Purelake Earthquake]

Ah, that makes sense.  It looks like that solution works.

 

I was starting to consider that sort of a scenario before, but I moved on too early.

 

Merging the points along those parameters I see it approaches a circle with A/B and the midpoints as a diameter with a removable discontinuity where they actually meet.

 

 

With that discrepancy resolved, I'd say that there's a quite good chance of this being correct.  Though, I still think it shouldn't have taken long for Rithmatic scholars to try this out for themselves.

 

 

Also, could you exploit limits like this to come up with some addition circles?  I tried, but haven't come up with anything conclusive yet.

[Alfa]

Addition circles?

[WeiryWriter]

Great job on this! Unfortunately it is a topic that has already been discovered (see here: http://www.17thshard.com/forum/topic/22801-hardcore-rithmatic-theory/). Five point and eight point circles are confirmed to be possible.

[Alfa]

damnation. And I thought I might be original for once...

[Purelake Earthquake]

Addition circles?

Oops, that's supposed to say Additional circles.

 

What I meant was if you used limits to make a one-dimensional triangle, using slightly different definitions for how it's shrinking, could you come up with any different sorts of Rithmatic circles.

I don't think you can though, at least not with any methods I could come up with.

 

You can make a five point or a regular four point from this (another way to do the same ones with regular triangles), but those would violate the non obtuse rule anyway.  So I don't think there's anything new that can be done here.

[Alfa]

I actually discovered another special case:
If the outer triangle has the angles 45°, 67.5° and 67.5° we have a regular octagon.

This is strange, for that would be a very logical and easy-to-discover circle.

[Xaladin]

Brandon Sanderson has forced me to recognize and respect that there is a lot to math that I don't know. I actually have a degree in math, but geometry is not my focus or strongpoint. I didn't know about the 9-point circle until reading The Rithmatist, and was skeptical at first. Cool stuff! I think I am a little behind on the geometry, though.

[Yitzi2]

Three-pointers and seven-pointers are in fact impossible, as are four-pointers and six-pointers that are not equally spaced.

On 12/5/2015 at 2:56 PM, Alfa said:

I actually discovered another special case:
If the outer triangle has the angles 45°, 67.5° and 67.5° we have a regular octagon.

This is strange, for that would be a very logical and easy-to-discover circle.

It is also notable for minimizing the largest distance between points; any other nine-pointer will have at least one pair of adjacent points more than 45 degrees apart.

Edited March 21, 2017 by Yitzi2