Questions about "Draw From Triangle..." Extension
As the subject says, my questions are about the utility accessed from the menus: Extension > Render > Draw From Triangle... and specifically about whether one of the several 'circle' options is what I want, which is what I am thinking of, which is the following:
Nodes in a path are ordered, right? So an SVG renderer knows how to connect them up. Therefore, the individual lines or Bézier curves (not B-*splines*, as a spline is multiple curves connected together at their ends, and having the same slopes [for Spiro, which is just a subset of B-splines, curvature also] at the points of those connections as their neighbors; polylines are not splines since they are meant not to have the same slope as their neighbors [or else two such connected lines could be replaced with a single line]) can also be ordered: the curve between nodes 1 and 2 is curve 1, n2 & n3 = c2, and so on. If the path is closed, node *last*'s connection to node 1 is the last curve.
So far so good (if all I have said above is technically correct).
I have happily drawn circumcircles and incircles around triangles before in Inkscape. Geometrically speaking, they do not have to be triangles in the literal sense, since:
- they can be closed or open - a quick test just now shows that using nonlinear sides for a three-sided, three-node closed shape produces a circumcircle (for instance, selected just for the test), although it neither encompasses the path nor contains the points where the nodes are at.
I tried, but three lone nodes (or any number of nodes not connected to any other) cannot be created in Inkscape, not sure about SVG itself.
Anyway, now I realised I wanted something else. Given three points, in either a closed or open triangle, I want a circle for which the two sides connected to the 'middle' node (they are ordered, remember) are both tangent to said circle: tangent at the first and last nodes, to be specific. There are infinitely many circles tangent to two true (that is, infinite) lines, see something like http://i.stack.imgur.com/2NFBD.png http://tex.stackexchange.com/questions/146942/inscribing-multiple-incircles-to-a-triangle-in-tikz#http://mathematica.stackexchange.com/questions/42706/counting-and-extracting-hit-circles-triangles-for-randomly-chosen-points but without the intervening lines if you want proof :P but as far as I can tell, only one of those infinitely many circles fits the description I just typed, because either one of the sides connected to the middle node establishes a *distance* from said node, which selects a unique circle.
Is there a mathematical name for this circle in relation to the triangle that generates it that I simply have not learned yet? If so, is it already in 'Draw From Triangle'? If not, could it be? Another feature request in respect to that module could be informative error messages. Often it just fails silently, either producing no objects or something different from the description, especially when multiple desired shapes are requested.
In any case, if that ramble if readable, I would love some thoughts.
-Arlo James Barnes
On 03/20/2016 11:54 PM, Arlo Barnes wrote:
Anyway, now I realised I wanted something else. Given three points ABC, in either a closed or open triangle, I want a circle for which the two sides connected to the 'middle' node (they are ordered, remember) are both tangent(1) to said circle: tangent at the first(2) and(4) last(3) nodes, to be specific.
Hi,
Your triangle will have to be specific (isosceles, probably) for that circle to exist.
(1:tangent to AB and BC lines) <=> center is on angle bisector, radius is k|OB| for some angle-dependent constant k (it's too late for trigonomerty) (2: tangent to AB in A) <=> center is on perpendicular to AB going by A, radius is |OA| (3: tangent to BC in C) <=> center is on perpendicular to CB going by C, radius is |OC| (4: going through A and C) <=> center is on the median of AC, radius is |OA|=|OC|
those 4 lines have no reason to intersect if the triangle is not isosceles, so choose any compatible 2 (1&2,1&3,2&4,3&4)
Hm, perhaps I should play around a bit in GeoGebra, which I have heard can export SVG, and then come back to this. -Arlo James Barnes
participants (2)
-
Arlo Barnes
-
Marc Jeanmougin