How to draw a directed arc with an initial point $(x_i,y_i)$, a terminal point $(x_t,y_t)$, and another given, intermediate point $(x_m,y_m)$ on the arc (say in the middle of the arc)? For instance, we may have $(x_i,y_i)=(0,0)$, $(x_t,y_t)=(2,0)$, and $(x_m,y_m)=(1,1/3)$.

I have noticed the nice answers to a related problem. However, with my very limited proficiency in Mathematica, I do not see a good way to modify those answers to get what I need here. I can certainly draw an arc -- but how to attach an arrow to it nicely?

  • $\begingroup$ The documentation for Arrow gives examples of how to use Bezier curves to describe the path of an arrow, or a list of points. I expect you could create a function to draw a directed arc based on one or other of those. No Mathematica on this machine or I might have experimented a wee bit myself. $\endgroup$ Commented Jan 14, 2022 at 14:49
  • $\begingroup$ @HighPerformanceMark : Thank you very much for your hint. $\endgroup$ Commented Jan 14, 2022 at 14:57
  • $\begingroup$ pts = {{0, 0}, {2, 0}, {1, 1/3}}; and Graphics[Arrow[BezierCurve[pts]]]. $\endgroup$
    – Syed
    Commented Jan 14, 2022 at 14:58
  • $\begingroup$ @Syed : Thank you very much. Can you post this as an answer, so that I could accept it? $\endgroup$ Commented Jan 14, 2022 at 15:04
  • $\begingroup$ It is directly from the docs. If you make the question more interesting, then I would be happy to try to solve it and post an answer. $\endgroup$
    – Syed
    Commented Jan 14, 2022 at 15:07

1 Answer 1


With the hint from HighPerformanceMark, the answer is simple. For instance,



enter image description here

  • $\begingroup$ But this is not a circle arc. Compare with pts = {{0, 0}, {1, 1/3}, {2, 0}}; Graphics[{Circumsphere[pts], Point[pts], Red, {Arrowheads[.06], Arrow[BezierCurve[{{0, 0}, {1, 1}, {2, 0}}]]}}] $\endgroup$
    – cvgmt
    Commented Jan 14, 2022 at 15:08
  • $\begingroup$ @cvgmt : Thank you for your comment. Of course, you are right. However, I am as happy with the Besier version, if not more -- because Besier curves seem more elegant to me than circular arcs. $\endgroup$ Commented Jan 14, 2022 at 15:26

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