I had a student ask if I could draw a Venn diagram in which each region was of equal area. I have played around with this a little but have not landed on an answer I'm satisfied with.

Sample work.

I was able to find the area of the overlapping region by subtracting the area of the triangle from the sector. I set the area of the overlapped area equal to the area of half of my circle. I am trying to get this down to two variables so that I can plug in a set radius and find an angle or something to that effect.

So far I have been able to simplify this down to radius, triangle height, and angle. Any help would be greatly appreciated.


Start with the unit circle. It has area $\pi$. So you want your areas to be $\frac12\pi$ each. Which means that half the lens-shaped intersection will have area $\frac14\pi$. You could integrate from a given position $x$ to the right to get that:

$$ \int_x^1 2\sqrt{1-t^2} \mathrm dt= \frac12\pi - x\sqrt{1-x^2} - \arcsin x=\frac14\pi \\ x\sqrt{1-x^2} + \arcsin x=\frac14\pi \\ x\approx0.403972753299517209318961740066 $$

I doubt that equation can be solved except numerically, it seems pretty transcendent to me. In any case, now you can compute

\begin{align*} \theta=2\arccos x&\approx2.30988146001005726088663377931\\ &\approx132.346458834092919049353117798° \end{align*}


  • $\begingroup$ Reminds me of a question I asked myself in the past. $\endgroup$
    – MvG
    Apr 25 '14 at 22:35

I was just doing this as well. Came out with a numeric approximation like MvG, not very satisfying. My working involves just trig, no calculus.

$A_{cir}$: Area of a circle = $\pi r^2$
$A_{mid}$: Area of the central region of the Venn diagram
$A_{seg}$: Area of a segment of a circle $= \frac{1}{2}r^2(\theta-sin\theta)$
$$A_{mid}=2*A_{seg}$$ $$A_{mid}=r^2(\theta-sin\theta)$$ In order that the areas of each part of the Venn diagram be equal,
$$A_{mid}=A_{cir}-A_{mid}$$ $$2*A_{mid}=A_{cir}$$ $$2r^2(\theta-sin\theta)=\pi r^2$$ $$2(\theta-sin\theta)=\pi $$ $$\theta-sin\theta=\frac{\pi}{2}$$ I can't think of a way to solve for $\theta$ here. Approximation:
$$\theta\approx2.31$$ Now, to find the correct distance between the centres, a right triangle with a corner on the centre of a circle and on one of the intersection points:


$$\frac{\theta}{2}\approx1.155$$ $$cos(1.155)=\frac{d}{r}$$ $$rcos(1.155)=d$$ So, to get 3 equal areas, the separation distance between the centres of the circles should be: $$2d=2rcos(1.155)$$

Here's a demonstration.


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