{ecc, p, r1, t1} = {0.35, 1., 1.4, 0.8}
EQ = {p/r2 == 1 - ecc Cos[t2], p/r3 == 1 - ecc Cos[t3], 
  r1^2 + r2^2 - 2 r1 r2 Cos[t1 - t2] == 
   r3^2 + r2^2 - 2 r3 r2 Cos[t3 - t2], 
  r3^2 + r1^2 - 2 r3 r1 Cos[t3 - t1] == 
   r3^2 + r2^2 - 2 r3 r2 Cos[t3 - t2]}
NSolve[EQ, {r2, r3, t2, t3}]

First radius vector is given/defined in the first line. Also given are ellipse properties ( $ecc, p =$ eccentricity and semi-latus rectum).

I am trying to inscribe an equilateral triangle in order to find two more vector arms $r2,r3$ along with their polar angles $ {t2,t3}. $

What assumptions or essential changes did I miss that makes this to hang? Is there a version problem?

The question has relevance to this MSE link.Thanks in advance.

  • $\begingroup$ I suspect I know what ecc and your other variables represent, but you might want to indicate what they're supposed to be for reference. See this as well. $\endgroup$ Commented Mar 30, 2018 at 15:21
  • $\begingroup$ Yes sir, classical symbols of Newton planetary motion focal conic used. $\endgroup$
    – Narasimham
    Commented Mar 30, 2018 at 19:51
  • $\begingroup$ Perhaps you did not see the image I just deleted or gone through the program. There are three radial arms. First one has length $1.4.$ Other two $(r2,r3)$ are unknown at start of computation. It would be also evident from the program. Feel free to say if my problem expression is in some way lacking. $\endgroup$
    – Narasimham
    Commented Mar 31, 2018 at 17:46
  • $\begingroup$ Because I chose origin / ellipse focus as reference of vectors starting , not its center of gravity. The link has an image of sorts. $\endgroup$
    – Narasimham
    Commented Mar 31, 2018 at 18:39

1 Answer 1


You can use NMinimize

NMinimize[{1, EQ}, {r2, r3, t2, t3}]
(*{1., {r2 -> 1.343, r3 -> 1.31612, t2 -> 0.75289, t3 -> 0.814458}}*)
  • $\begingroup$ you might try FindInstance as well.. $\endgroup$
    – george2079
    Commented Mar 30, 2018 at 21:27
  • $\begingroup$ Thanks,shall try it $\endgroup$
    – Narasimham
    Commented Mar 31, 2018 at 6:15
  • $\begingroup$ @Ulrich Neumann I wianted to plot a solution triangle. When plotted three points they do not form vertices of an equilateral triangle. Why so? Are inverse functions giving wrong choices? $\endgroup$
    – Narasimham
    Commented Mar 31, 2018 at 14:53
  • $\begingroup$ @Narasimham: I don't think so. Please show your code, how you plot it... $\endgroup$ Commented Apr 1, 2018 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.