When I first did this video, I thought the trisecting curve (see video for explanation) was a regular, circular arc. ArjenKiksman (in a separate email) first suggested it would be a hyperbolic curve. 18Brumaire1799 recognized this as well and has done a graphic video of it. Can we verify this curve is correct and that it can be constructed with edge and compass?
@linelites though you can't construct the hyperbolic curve with edge and compass....I have read that, at least with an ellipse, you can find the set of points that define it....do you imagine one could find the point of that hyperbolic curve as it intersects the arc of the subject curve?
I create a video response. You can notice that the two hiperbolas do not depent on the vertex B, only on points A and D. If these points are fixed, your curve is a hyperbola of eccentricity 2.
I believe that is encouraging, since I understand a hyperbolic curve is construcible with edge and compass....Can you tell me how you know it is a hyperbola?
Nice way to investigate the trisection problem. I don't yet understand all of your "trisecting" or "close to trisecting" arcs. Do you have a reference for the "Lost theorem"? I devised a proof for a^2=c^2+bc in less than 10 steps using the zigzagging addition of angles pattern you can find on my blog and would like to compare both proofs.
When I first did this video, I thought the trisecting curve (see video for explanation) was a regular, circular arc. ArjenKiksman (in a separate email) first suggested it would be a hyperbolic curve. 18Brumaire1799 recognized this as well and has done a graphic video of it. Can we verify this curve is correct and that it can be constructed with edge and compass?
linelites 4 months ago
@linelites though you can't construct the hyperbolic curve with edge and compass....I have read that, at least with an ellipse, you can find the set of points that define it....do you imagine one could find the point of that hyperbolic curve as it intersects the arc of the subject curve?
linelites 3 months ago
I create a video response. You can notice that the two hiperbolas do not depent on the vertex B, only on points A and D. If these points are fixed, your curve is a hyperbola of eccentricity 2.
18Brumaire1799 4 months ago
I believe that is encouraging, since I understand a hyperbolic curve is construcible with edge and compass....Can you tell me how you know it is a hyperbola?
linelites 4 months ago
The response to your question is no, one can describe your curve but it is not a circle arc. The correct answer is a hyperbola.
18Brumaire1799 4 months ago
Nice way to investigate the trisection problem. I don't yet understand all of your "trisecting" or "close to trisecting" arcs. Do you have a reference for the "Lost theorem"? I devised a proof for a^2=c^2+bc in less than 10 steps using the zigzagging addition of angles pattern you can find on my blog and would like to compare both proofs.
ArjenDijksman 1 year ago