akasketch he/him Posted February 9, 2014 Report Share Posted February 9, 2014 (edited) Sorry for the length in advance. When I read this book, the reasoning for the nine-pointer (described via annotated depiction from what I assume is Joel's notebook on page 243) kinda blew over me, and I just wanted to experience the story for once, rather than get caught up in the world's physics. That said, I still promised myself that I would check it out after I was done. I finished the novel, then recommended it to my little brother's friend, who promptly inhaled the darn thing, then asked, "Any other suggestions?" with a huge smile on his face. After this, I finally looked it up on Wikipedia (mostly due to his own interest in Rithmatics), and the results I found then and afterwards were extremely intriguing. First: The 9-point circle is an actual discovery made by Olry Terquem, and has some significance in the geometric world. Second: (From here on out is a thought process) The 9-point circle doesn't quite work for equilateral and right triangles; what do the look like; what are their Rithmatic equivalents? Third: Equilaterals would lend themselves to the six-point circle due to their nature of fusing together three different pairs of significant triangle points that would be fully represented in the 9-pointer (kind of shown in a picture on the top of pages 94-95; I just discovered this disproof of my originality in thought, as well as another in the history section of the aforementioned Wikipedia article). Fourth (fittingly): If both the 9- and 6-point circles can be represented as a relationship between the circle and a single triangle, what about the 4-point circle? The four points form an inscribed square when connected, and a square is essentially two equilateral right triangles stuck together at the Hypotenuse. On a whim, I drew this on a piece of graph paper: I noticed that all nine points were represented, and several at once in the peak (I had the hypotenuse on the bottom), and was then feeling nearly satisfied with my pursuit of Sanderson's use of Trig relations in his novel. Fifth: What about ellipses? The first thing that came to mind was Isosceles triangles, and thus I drew this on the same piece of graph paper: I'm pretty sure that the points at which the ellipse passes through the sides are their midpoints. Obviously, all nine significant points are NOT represented by said ellipse, but it does pass through at least two, probably four, of them. Sixth: This one is best described through simply showing a picture: I was messing around with isosceles triangles, so naturally, I wanted to see what their complement circle would look like point-wise, so I essentially drew up this diagram on my graph paper. That's it for the thought process, but I'm having trouble with a couple. For one, although Lines of Vigor are made from a sine or cosine graph, where did lines of revocation come from? And, I'm completely at a loss as to where the spiraling sound-sucking line comes from. Comments, further proofs, or disproofs? EDIT: sorry for the small pics. Edited February 9, 2014 by akasketch 6 Quote Link to comment Share on other sites More sharing options...

Haelbarde he/him Posted February 9, 2014 Report Share Posted February 9, 2014 (edited) Lines of Revocation are the Sawtooth function with a line through them. Try graphing y=(-2/Pi)arctan(cot(Pi*x)) or y= t - floor(t). Sawtooths are a form of triangle waves which are themselves integrals of the square wave, which is formed by getting the sign of cos or sin Ie y = cos(x)/|cos(x)| or y = sin(x)/|sin(x)| give square waves. I'm not sure about the spirals, though maybe thinking fractals? The Julia Set can do something similar (though waaaaaaay more complex). EDIT 2 - Re Fractals - http://fractalfoundation.org/resources/what-are-fractals/ This is awesome Edited February 9, 2014 by Haelbarde 2 Quote Link to comment Share on other sites More sharing options...

akasketch he/him Posted February 9, 2014 Author Report Share Posted February 9, 2014 (edited) First of all, that IS cool, and I was thinking along the same lines for the spirally thingy, but am wondering how it relates to trig functions, as everything else does. Thanks for the Sawtooth; I'll check that out. EDIT: OK, so the vertical lines in the line of Revocation are asymptotes, and the line through the middle is the midpoint of all the triangles formed. Edited February 9, 2014 by akasketch 0 Quote Link to comment Share on other sites More sharing options...

WeiryWriter he/him Posted February 9, 2014 Report Share Posted February 9, 2014 A note on ellipses, if I remember correctly they only have two bindpoints, at the point where they curve the most. 1 Quote Link to comment Share on other sites More sharing options...

akasketch he/him Posted February 9, 2014 Author Report Share Posted February 9, 2014 Ok. That means that they don't pass through the midpoints of the isosceles, nor the feet of the altitudes; just the top and the bottom. 0 Quote Link to comment Share on other sites More sharing options...

Tarontos he/him Posted February 28, 2014 Report Share Posted February 28, 2014 (edited) First i need to provide i link to my thread on this that was posted quite a while a go.. http://www.17thshard.com/forum/topic/3600-triangles/ second the lines of silencing are likely entirely unrelated to the circle or could possibly be related to another shape we have not seen yet., i have hypothesized in the past that they would absorbs lines of vigor, and the lines of revocation as they absorb sound. this is based off of the waves. also look at none isosceles right triangles using the nine point system you will come up with five point circles. on elipses us a triangle with angle 180,0,0 Edited February 28, 2014 by Tarontos 0 Quote Link to comment Share on other sites More sharing options...

Shardlet he/him Posted February 28, 2014 Report Share Posted February 28, 2014 Second: (From here on out is a thought process) The 9-point circle doesn't quite work for equilateral and right triangles; what do the look like; what are their Rithmatic equivalents? I think you have a mistake here. A 9-point circle is produced using an inscribed circle within a right triangle so long as the triangle is not isosceles. 0 Quote Link to comment Share on other sites More sharing options...

akasketch he/him Posted April 3, 2014 Author Report Share Posted April 3, 2014 I think you have a mistake here. A 9-point circle is produced using an inscribed circle within a right triangle so long as the triangle is not isosceles. No, actually: http://en.wikipedia.org/wiki/Nine-point_circle My thought here was that right triangles and the like did not have nine points relating to their circle, but I was wrong, for there are multiple points converged into one. 0 Quote Link to comment Share on other sites More sharing options...

Shardlet he/him Posted April 3, 2014 Report Share Posted April 3, 2014 (edited) No, actually: http://en.wikipedia.org/wiki/Nine-point_circle My thought here was that right triangles and the like did not have nine points relating to their circle, but I was wrong, for there are multiple points converged into one. You're right. I was mistakenly drawing a line from each of the apexes to the midpoint of their associated opposite side rather than to their altitude point (making a right angle with their opposite side). As far as points combining go, I suppose you could say that it is still a nine point circle with a right triangle, but it seems more proper to describe it as a five point circle. Especially so in terms of rithmatics. Else, five and eight point circles would be viable lines of warding since they are the effective depiction of nine-point circles for right and isosceles triangles respectively. That would also suggest that four and six-point circles are actually nine-point circles for right-icosceles and equilateral triangles respectively. A two-point circle would then be the effective depiction of a nine-point circle for a triangle having an apex of zero degrees. Again however, if everything was really just nine-point circles in (many in disguise), then a five-point and 8-point circle should be rithmatically viable. Edited April 3, 2014 by Shardlet 0 Quote Link to comment Share on other sites More sharing options...

johni92 Posted April 21, 2014 Report Share Posted April 21, 2014 It seems like five-point and 8-point circles should be viable. The two-point is the most difficult to explain in terms of a nine-point circle. If you look at the limit as the vertex angle of an isosceles triangle approaches 180 degrees, you end up with a circle of infinite radius, with 5 points converging to lie on a straight line. This can't exist on a circle, but if you also take the limit as the side lengths approach 0, they converge to a single point. The other three points lie an infinite distance away, along three parallel lines, which, again, can converge to a single point for zero side lengths. I suppose this can be seen as the two-point circle, but it is somewhat more abstract than the others. Note: taking the limit as the base angles of an isosceles triangle approach 90 degrees and as the length of the base approaches 0 and the lengths of the other two sides approach infinity gives an identical result. 1 Quote Link to comment Share on other sites More sharing options...

Aidulin41 Posted March 23, 2015 Report Share Posted March 23, 2015 A note on ellipses, if I remember correctly they only have two bindpoints, at the point where they curve the most. It only makes sense to use those two points, but there's another topic(http://www.17thshard.com/forum/topic/22801-hardcore-rithmatic-theory/), first post. The author suggests that more points are possible, but would be weak. So it's not impossible, but not really smart either. 0 Quote Link to comment Share on other sites More sharing options...

WeiryWriter he/him Posted March 23, 2015 Report Share Posted March 23, 2015 It only makes sense to use those two points, but there's another topic(http://www.17thshard.com/forum/topic/22801-hardcore-rithmatic-theory/), first post. The author suggests that more points are possible, but would be weak. So it's not impossible, but not really smart either. True, however Kalyna only asked Brandon about that in January of this year, the post you are quoting is from February 2014. At that point in time I was only going off the book itself. 0 Quote Link to comment Share on other sites More sharing options...

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