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:

  1. Find the intersection points of each pair circles by solving the simultaneous equations
  2. Decide which of the six points are involved in your bumpy triangle by seeing which of them lies inside the other circle
  3. Split your bumpy circle up into four parts: three segments (the bumpy bits), and one triangle between them
  4. 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:

  1. Find the distances between each pair of circle centres – call the distance between $C_1$ and $C_2$ $d_{12}$, and so on.
  2. 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 $\d_{23}$ ((In each of those, the last item in the list is the one opposite the angle you’re trying to find.)) 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.
  3. Find the areas of the segments. The first segment, for instance, is $\frac 12 r_1^2\left( \theta_1 - sin(\theta_1)\right)$
  4. 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.
  5. 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.

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