I have a Circle separated into 3 sectors. At start each sector has the same central angle, 120°. Therefore each sector should be taking up the same area. I want to be able to move the center point denoted by the Blue Triangle in my poorly drawn pic, while the Red Star points remained fixed. How can I calculate the areas of the resulting Sectors after the Movable point stops?

I'm pretty sure I have to calculate the Angle of each Sector, but my trig (geometry?) knowledge is pitiful. And all of the solutions for finding Area of a sector have a fixed Central point.

Any help would be appreciated!

Circle with evenly sized Sectors

  • 1
    $\begingroup$ I think you need to look at a different problem. Instead of the circle consider the triangle with the red points as corners. The area only differs by a constant you can easily calculate and the problem for the triangle looks easier. (However if you move the point to far, you need to consider negative areas) $\endgroup$
    – mlk
    Jun 6, 2014 at 16:16
  • $\begingroup$ Yes I believe a triangle would work, and I could constrain movement to inside the triangle, so I wouldn't have to worry about negative areas. Although I'm not grasping how the Constant would be calculated and how I would use that to get the 3 different areas. $\endgroup$
    – hoobajue
    Jun 6, 2014 at 16:44

1 Answer 1


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For simplicity, we let (1) the center of the circle be at $O(0, 0)$; (2) the radius be $2$.

We can further let the $3$ fixed points be $A(0, -2)$, $B(\sqrt{3}, 1)$ and $C(–\sqrt{3}, 1)$. These points (together with $O(0, 0)$ divides the circle into sectors $M$, $N$, and $R$.

The following results are also basics:-

(1) Equation of $OB$ is $y = \frac{1}{\sqrt{3}}x$ and that of $OC$ is $y = –\frac{1}{\sqrt{3}}x$

Suppose that the center is moved to $P(h, k)$; where, of course, $h$ and $k$ are known quantities.

The following computations are based on the situation that $P$ is in the yellow-shaded region. (The logic can be equally applied if $P$ is at some other locations.)

Equation of $AP$ can be found by “two-point form” and therefore assumed to be known.

$D(p, q)$ is the point of intersection of $OB$ and $AP$. Thus, $p$ and $q$ of $D(p, q)$ are also known by solving the two equations.

Notation:- [figure] = area of that figure.

$[\triangle OAD]$, $[\triangle OPD]$, $[\triangle PDB]$ and $[\triangle OPC]$ can all be found according to the “Shoe-lace” theorem.

Because of the center is moved to P, sector M has gained $[\triangle OPD]$ and $[\triangle PDB]$ but loses $[\triangle OAD]$.

The gain/lose of the other two regions can be calculated similarly.


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