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A step-by-step guide to the Irregular Nonagon and beyond


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First of all, I'm really sad there hasn't been a post on here in a long time. Rithmatist needs more love! Anyhow, I made this today to teach myself something and I feel like sharing it, even though I doubt many people check here any more :)

So, the irregular nonagon, AKA the Line of Warding with 9 bind points. The important thing to remember is that while the 2, 4, and 6 bind point versions all look the same (apart from rotation), ones made with 9 bind points are much more flexible. Now, let's get down to business.

First, draw your circle. Then, pick 3 points on the circle and mark them. These are your first 3 bind points. The only restriction is: they can't all be in the same 180 degree section of the circle. You'll see why in a moment.

Now, you're going to need to mark down 3 lines to make a triangle where the midpoint of each of the lines is one of the bind points you marked in the last step. There's only one way to do this, so don't worry about messing it up! You'll probably notice that the lines of your triangle intersect the circle in more than one place. The extra spots where they intersect are your next 3 bind points! I forgot to mark them in the picture though. The only issue would be if you put all 3 of your original bind points on the same half of the circle - then your triangle would be obtuse, and you'd probably feel pretty obtuse too! Obtuse triangles aren't allowed here, so keep some healthy distance between those points, mister.

The final step now is to draw a line from each corner of your triangle to the extra bind points you marked in the last step. In order to do that you'll have to intersect your circle again, and you know what that means! That's your last 3 bind points, done! See, that wasn't so hard.

Now, it's pretty well known (thanks to the hard work of people like KalynaAnne) that you can make the 6-point, 4-point and 2-point circles with the irregular nonagon, and you can even create top-secret 5 and 8 point circles that aren't discussed in the books. How exactly you can do this I will show after I update my post tomorrow (or the day after), as I'm currently working on some swanky animations to get my point across better



Edited by Anchpop
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Delivering now! I'll walk you guys through the creation of all the bindpoint arrangements - 6, 4, and 2, followed by the top secret 5 and 8 not yet discovered in the books!

Animation - So, we'll do 6 first because it's the easiest. Simply start with your initial 3 bindpoints equidistant from each other, so the triangle they create will be equilateral. The reason that this creates 6 bindpoints instead of 9 is because 3 of them overlap. 

Animation - Next up, 4. to make this one, imagine a square inscribed in your circle, and then take three of the corners (doesn't matter which) and make them bindpoints. The big triangle you draw will have one of its corners resting right on the circle! 

Animation - Now, 2! This one is the hardest to wrap your head around. You have to make 2 of your bindpoints overlap, and put the other one on the exact opposite side of the circle, which will make a straight line instead of a triangle! Well, you can think of it as a triangle with 2 sides that are twice the diameter of your circle and one side with no length at all. Now you might be thinking, "Why does the last point have to be exactly opposite of the other two? Why can't I put it wherever I want since I'd still get a straight line?" The answer to that is a little in-depth, so I've put it in a spoiler tag. Read it at your own peril! 


A triangle with one side having no length doesn't really make sense, the only way to interpret it in a way that's consistent with more advanced parts of rithmatics is that there are two points there infinitely close together. And if those two points weren't directly opposite of the last one, the triangle drawn would be obtuse and I said not to do that earlier. Now why are obtuse triangles not allowed? Well, turns out that's more like a guideline than a general rule. With the exception of the 2 and 4 pointed ones, you can get any possible line of warding using either obtuse or acute triangles. When you use an acute triangle the circle has to it at 6 points (or 3 in the case of the 6-pointed circle, or 5 in the case of the yet-to-be-revealed 8-pointed circle that we won't be considering here), but an issue arises when you try with an obtuse angle. If you start with your bindpoints making an obtuse triangle, you can get a triangle that only intersects your circle at 3 or 4 points! But that's only where the trouble begins - when your triangle is obtuse, the 3rd point on the triangle is inside the circle. That's bad news, because it means you'll only have 1 extra point where your triangle intersects the circle, and therefore only 1 extra bindpoint. If you followed the rules I said above, you'd end up with an impossible circle with only 4 bindpoints. But the solution is not too difficult, luckily. All you need to do is continue the lines of your triangle into infinity, and they'll intersect the circle two more times. After that go to the corners like you're used to and draw lines to the extra intersections that have now appeared (two more bindpoints down), and then continue those as well until they hit the circle again, getting the last 3 you need. But if you continued your lines into infinity like I told you to, you'll notice something interesting. The extra lines you drew to connect to where the triangle re-intersects make a new, acute triangle, and the new triangle conveniently has bindpoints at the midpoints of all of its edges! And now if you redrew the extra lines where your new triangle intersects the circle, you'd be remaking your old triangle again! Basically, it means that any set of bindpoints you get from an obtuse triangle you can also get from an acute one, but obtuse ones are much harder to draw so there's no point to them.

For extra credit, you can notice that for every acute triangle, is not one but three obtuse triangles you can make that will have the same bindpoints.

Animation - Okay, to make a 8-pointer couldn't be easier. Just start our with your initial bindpoints equidistant, like you did with the 6-pointer. This makes the triangle equilateral. Now, move two of the bindpoints along the circle, one clockwise and one anticlockwise, making the triangle an isosceles. Basically you've stretched the triangle upwards. The strange thing is that Rithmatists don't use this shape, even though you can make it into a nice regular octagon. Off the top of my head, the Eskridge Defense could be modified to used 4 defensive walls instead of 3, possibly offering greater maneuverability and offering stronger defense.

Animation - Last but not least, the 5-pointer. It's very similar to the 8-pointer, except you start with a 4-pointer instead of a 6-pointer. Jeez, pointer doesn't even sound like a word anymore. Because of the lopsided nature of this one, I can't really imagine it being useful in the Melee. 

That's a wrap, folks! I know all of this was old news thanks to advanced Rithmatists like kalynaanne (except for the last line in the obtuse triangle section), but in my next post I'll be going a little farther into the unknown. Why a nonagon? How does the circle "know" to be weaker if it's bindpoints are wrong? I mean, surely there isn't a shadowblaze doing the math in its head every time a rithmatist draws a bindpoint. Tune in next time to find out!

Edited by Anchpop
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  • 2 months later...
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There's actually a (usually) easier way to construct a nine-pointer, without even drawing any triangles.

1. Start with your circle.

2. Mark three points anywhere on it (they can all be in the same 180-degree arc, or even have two or three in the same location).  These will be the feet of the altitudes.

3. The three points you just marked divide the circle into three arcs, and any two adjacent arcs can be combined to form another arc.  Bisect these six arcs; the points bisecting the three basic arcs will be the other altitude crossings, and the points bisecting the combined arcs will be midpoints. 

3a. If two of the arcs are the same size, the bisector of the arc combining them will lie on one of the points from step 2, resulting in an 8-pointer.  If all three are the same size, all three combined-arc bisectors will lie on the points from step 2, resulting in the 6-pointer.

3b. If two of the points are in the same location, the arc between them is of size 0, so the bisector of that arc is in that location as well.  Two of the combined arcs also become identical to basic arcs, so there are only three bisectors, resulting in a five-pointer (or four-pointer, if the third point lies on the bisector of the combined arc).

3c.  If all three points were in the same location, three of the bisectors will be there as well, and the other three will be on the opposite side of the circle (making a 2-pointer).

3d. The bisector of a combined arc will be exactly opposite the circle from the bisector of the opposite basic arc; as such, it may be easier to only bisect the arcs less than or equal to 180 degrees, and then mark the points opposite them.

Edited by Yitzi2
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