Thread: Constructing an ellipse with a triangle

This is a strange question, by I "feel" that the math behind it can be quite simple. It has to with constructing an ellipse (I think)!

A, B and C are angles (or points) of a triangle.
a, b and c are the respective opposing sides.

Side c and its opposing angle C are both held constant.

Now all possible triangles are formed given these restrictions. I.e. all combinations of A, B, a and b's are choosen which allow ABC to remain a triangle.

Given that the location of side c is held constant, the possible point C's will form av curve around that base c. This shape, according to my experiments, seems to be an ellips, although maybe not well defined very close to the base c.

I would like to know if there is a simple relationship between the parameters of the ellips (its two axes and its center point) and the side c and angle C of the triangle.

Code:

      .C .  .
   .   |\      .
 .     | \       .
.      |  \       .
.      |   \      .
  .    |    \   .
     . |_____\.
       A  c   B

The points above is the ellips. On each such point, angle C of triangle ABC will be constant towards opposing side c. (Though this might not be correct for the points very close to base c.)

How do I determine the parameters of such an ellipse given values of C and c?

Hmmm...

Is it a CIRCLE, not an ellipse???

Originally Posted by Optiminimal

Hmmm...

Is it a CIRCLE, not an ellipse???

Hello,

you are right: It is a circle. You find the proof here: PlanetMath: circumferential angle is half the corresponding central angle

Great, thanks!
Your short reference saved me much time and headache.

I knew that there were some relationship like this if one just put the problem in a clear way.

The reason that I first thought that this circle/angle relationship does not work when C is close to A or B, seems to be that I unknowingly calculated the angle OCA instead of BCA when OCA>BCA, due to me using home made formulas where the simplicity of the problem wasn't obvious.

Now, all I need to do in order to construct this circle is to find the circle center O given ABC. This is quite simple:
I consider the midpoint D between A and B.
ODA = ODB is a right angled triangle where AD=BD is known.
Angle AOD = ACB ("half of double" the circumface angle ACB) and known.
With three known in this triangle where circle centre O is one of the corners, things are under total control.