Written by Colin+ in ask uncle colin, circles, geometry.

Dear Uncle Colin,

I've been challenged to find the area of the intersection of three circles while drawing a Venn diagram. I don't know where to start!

-- Triangle Unpredictably Rounded; I'm No Genius

For a moment, TURING, I thought there wasn't a problem in this problem, but then I realised: of course you have a problem. You're drawing Venn diagrams, which -- after the pie chart -- are probably the worst kind of graph.

But, turning to your question, there's a fairly standard way to sort out this kind of question, which is as follows:

- Find the intersection points of each pair circles by solving the simultaneous equations
- Decide which of the six points are involved in your bumpy triangle by seeing which of them lies inside the other circle
- Split your bumpy circle up into four parts: three
*segments*(the bumpy bits), and one triangle between them - Find the area of each of these shapes: the segments by working out the relevant sector areas and taking off the triangle areas, or using the formula $A_{seg} = \frac 12 r^2(\theta - \sin(\theta))$, working in radians as God intended. You can find the triangle area using Heron's formula or general trigonometry.

But I don't want to give you that, TURING! That faffing about with finding which points to use is, frankly, enough to make Euclid turn in his grave. I have a more reliable way:

- Find the distances between each pair of circle centres -- call the distance between $C_1$ and $C_2$ $d_{12}$, and so on.
- Find the angles for each sector using the cosine rule. To do this for (e.g.) the first circle, we'll need three angles: first, use the side lengths $r_1,\, d_{12}$ and $r_{2}$; then $r_1,\, d_{13}$ and $r_{3}$; lastly, use $d_{12},\, d_{13}$ and $\d_{23}$1 Add the first two angles and take off the third; this gives you $\theta_1$, the sector angle; repeat this for the other two centres.
- Find the areas of the segments. The first segment, for instance, is $\frac 12 r_1^2\left( \theta_1 - sin(\theta_1)\right)$
- Now for the triangle. Start by finding the lengths of its sides: the length of the chord associated with $C_1$ is $x_1 = \sqrt{\frac 12 r_1^2 \left( 1 - \cos(\theta_1)\right)}$ (using the cosine rule), and similarly for the other chords.
- Lastly, use Heron's formula $A_\Delta^2 = S (S-x_1)(S-x_2)(S-x_3)$, where $S = \frac{x_1 + x_2 + x_3}{2}$, and add on the segment areas.
- In each of those, the last item in the list is the one opposite the angle you're trying to find. [↩]

Oof. It's a tricky, ugly problem, and I'm not sure there's enough here to make it clear. Let me know if there's something I should expand on!

-- Uncle Colin