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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.

Polygon

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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: $$\bbox[10px, border:1px solid red]{\therefore I=\left(\frac{\sqrt{3}r}{2\left(\sqrt{3}+2\right)},-\frac{1}{2}\left(\sqrt{3}r\right)\right)}$$

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