Yitzi2

Even if we interpret "bandolier" as indicating a visit to Scadrial, I'm not sure that visiting one other world counts as "cosmere aware".

1. (It's possible the answer to this is known already, but I wasn't able to find it.) If a circle has something at some valid bind points but not others (e.g. three of the four points of a four-pointer), does that weaken it? If not, why is a two-pointer considered a separate type, given that those are two of the points of a four- or six-pointer? 2. If the Rithmatist had been cosmere-related, would it be a shardworld? If so, which Shard(s) would be present there? 3. The height of a Line of Forbiddance is proportional to its width. Does that mean that with a thin enough chalk "pencil" you can make LoFs you can reach over? If so, would this be permitted in a duel?

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

There can only be so many Rithmatists (presumably because there are only so many Shadowblazes), and the difference between Joel's usefulness to the cause as another Rithmatist and his usefulness as support for Melody is far less than for most candidates.

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.

Thanks. I did try resending it, and recovering the old one would be nice, but not a big deal. The real goal was just that you be aware of the issue in case there's something that needs to be done.

I had a previous account, signed up with my hotmail email, but I forgot my password and when I tried to have a password reset link sent, it didn't arrive (not in my inbox, not in my junk folder). I signed up with this one instead, but I thought you should know.