Having an angle $\alpha$ between the $y$ axis and a line intersecting the origin, natural number $n$ being the number of sides of regular polygon, radius $R$ and assuming the bottom side of the polygon is parallel to the $x$ axis, how can I calculate coordinates of the point $I$?

Angle $\alpha$ being $15°$ in the picture is just an example.


  • $\begingroup$ Lwt the $y$ coordinate of the point where $R$ intersects the circle be $a$. Find the coordinates of the point where the line intersects $y=-a$. $\endgroup$ – Mohammad Zuhair Khan Aug 8 '18 at 15:40
  • $\begingroup$ $a=R \sin{30^\circ}$ and now I believe you can continue yourself. $\endgroup$ – Mohammad Zuhair Khan Aug 8 '18 at 15:42

With this from wikipedia

The circumradius R from the center of a regular polygon to one of the vertices is related to the side length s or to the apothem a by

$$ { R={\frac {s}{2\sin \left({\frac {\pi }{n}}\right)}}={\frac {a}{\cos \left({\frac {\pi }{n}}\right)}}} $$

you can find the length $a$ from the origin to the side of your polygon. That gives you the $y$ coordinate of $I$. For the $x$ coordinate use $\tan \alpha$.

Note that if $n$ is odd the "bottom edge" of the polygon won't be horizontal if one vertex lies on the $x$ axis.

  • $\begingroup$ Thanks! In honesty I needed more general case (where point $I$ may lie on other sides as well, not just the base) for a program, but I think I now know how to deal with it as well - calculate apothem, multiply it by side normal vector, and then use origin, angle and calculated point to calculate $I$ with trigonometry. $\endgroup$ – mrpyo Aug 8 '18 at 16:39
  • $\begingroup$ Yes, that will work to answer the general question. The original question asked just about the special case, and did not indicate how much trig would be acceptable. $\endgroup$ – Ethan Bolker Aug 8 '18 at 16:43

Since a regular hexagon has internal angles of $120^\circ$, the sides, except those parallel to the $x$-axis, will be defined as follows (with $r$ being the radius of the circle): $$y_{1,2,3,4}=\tan\left(\frac{\pm2\pi}{3}\right)(x\pm r)\tag{1}$$

Consider the lone side $y=\tan\left(\frac{2\pi}{3}\right)(x+r)$ below:

enter image description here

The line intersects the circle at $\left(-\frac{r}{2},-\frac{1}{2} \left(\sqrt{3} r\right)\right)$, which tells us that from the original diagram, $I$ lies on the line: $$y=-\frac{1}{2} \left(\sqrt{3} r\right)\tag{2}$$

The line through the origin is then defined as: $$y=(\cot\alpha )x\tag{3}$$ where $\alpha$ is your angle. Just then solve for the intersection of $(2)$ and $(3)$.

In your case, for $\alpha=-\frac{\pi}{12},$ the intersection should be: $$\bbox[10px, border:1px solid red]{\therefore I=\left(\frac{\sqrt{3}r}{2\left(\sqrt{3}+2\right)},-\frac{1}{2}\left(\sqrt{3}r\right)\right)}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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