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Easton Defense reverse engineer


FeruchemicalAbhorsen

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My family and I spent several hours trying to reconstruct the triangle for the Easton. As far as we can tell, this triangle doesn't exist. Does anyone know what it is? ( I'd love to be proven wrong). Another possibility could be that that the drawings in the book are not entirely accurate, because a slightly squished equilateral triangle almost works.

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Sorry, Im not clear on what you are after.  What do you mean by "reconstruct" exactly?

 

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Easton Defense
A defense that is suited for multiple opponents. It has circular Lines of Warding at each of its bindpoints and and Lines of Forbiddance that form a nine-sided figure with three lines missing which act as support for the mine circle. Drawbacks to the defense are the difficulty of nine-point circles and the restriction created by the Lines of Forbiddance. There are a number of variations on this defense, such as adding defensive chalklings to the outer circles.[27] A more advanced iteration of this defense adds a Mark's Cross to each of the outer circles and decreases the internal Lines of Forbiddance from six to three. Defensive chalklings are also bound to a number of the outer circles' bindpoints.[28]

 

 

 
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@TypeIVBioChromaticEntity As far as I'm aware, all 9 point defenses are based around the same triangle. Here's an image of the Easton defense and one of the triangles used to construct 9-points for comparison.

Edit: My mistake. Any non-obtuse triangle works. Since the exact angles don’t matter, its not necessary to look for the exact triangle that leads to the configuration seen in the book.

rithmatist_diagram_ch-17_webres.jpg

rithmatist_diagram_ch-0prologue_webres.j

Edited by ILuvHats
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  • 2 weeks later...

I believe you do the following:

1. Draw any acute triangle.

2. Draw the altitudes.  (These are the line segments from an interior angle of the triangle to the opposite side, that are perpendicular to the opposite side.)

3. The three points where the altitudes hit the opposite sides are called the "foots of the altitudes".  These give three points.

4. The midpoint of each side of the triangle gives another three points.

5. The three altitudes intersect in a single point, called the orthocenter.

6. The points halfway from each vertex to the orthocenter give another three points.

7. It is an old theorem that the nine points above all lie on a unique circle, the nine-point circle.  See: nine-point circle link

Every acute triangle has such a 9-point circle.

Going from a circle to its 9 points, you need to have in mind a triangle somewhere.  There are many different choices for the 9 points.  But not every choice of 9 points will work.

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  • 1 year later...

Everyone here is saying that it's not necessary to figure out the exact triangle in the book, and for purely dueling purposes that's true. But it's still worth questioning whether that diagram is any good from a standpoint of plain curiosity.

I'm a couple of years late, but I think I've figured it out (attached image)

I know Lightning kind of already said this, but the points on a nine-point circle are:

  • The feet of the altitudes (type I points)
  • The midpoints of the segments connecting the orthocenter to the corners of the triangles (type II points)
  • The midpoints of the sides of the triangle (type III points)

These types aren't actual geometric terms, but I'm going to use them to explain my methodology.

It turns out, there's actually a much easier way to construct a nine-point circle without the triangle.

  • Place three points randomly along a circle. These are the type I points.
  • Bisect the minor arc between each pair of type I points. These are the type II points.
  • Find the points diametrically opposed to each type II point. These are the type III points.

With this newfound knowledge, we can jump into our investigation.

Let's consider the attached labeling schema. We can pretty much assume that the point towards which all the grey segments are drawn is the orthocenter of the triangle. The chances that four different pairs of points - AF, BG, DH, and EI - are all colinear with a point chosen inside the circle seems far too unlikely to be purely coincidental. However, if that chosen point is the orthocenter, three pairs of points necessarily have this property, as each type I point must have a corresponding type III point along the same altitude.

Consequently, we can assume that three of AF, BG, DH, and EI must be altitudes, and one is a fake altitude. Meanwhile, C doesn't have an opposing point, so it must not be along an altitude. Therefore, it is a type III point.

Since C is a type III point, the point diametrically opposed to it - G - is a type II point. Thus, the point altitudinally opposed to G - B - must be a type I point.

Also since G is a type II point, we know that it is at the center of a minor arc between two type I points. Arcs HF and IE appear to be the only viable candidates.

Therefore, either HFB or IEB are the sets of type I points. Out of the two, I thought HFB looked more likely.

EastonLabeled.PNG

EastonQuadrangle.jpg

Edited by CrypticSpren
Wrote Lighting instead of Lightning
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Great, thank you!

Just one question - what does it mean to "Bisect the minor arc" to make the type II points? Is it just halfway along the arc between the two points?

If that is true, I don't think IEB is valid, because it doesn't look like there are halfway points for any pair except IE, and even IE is a bit if a stretch.

Thanks again!

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Yup, that's exactly what it means to bisect the minor arc. I say minor because technically there's also a major arc, which is more than 180 degrees, around in the other direction.

Also, you're right about IEB looking pretty bad, which is why I think we can be fairly sure about HFB being the correct type I points. I'm glad someone appreciated this!

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