Given triangle $ABC$ (of sides $a, b, c$) and three circles $O_A, O_B, O_C$, with their centres in the triangle's vertices $A,B,C$, respectively. They all have the same radius lengt $d$. What's more, $d\ge max(a,b,c)$. Determine the common area of circles $O_A, O_B, O_C$.
2026-05-02 00:14:17.1777680857
Common area size
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2 examples of ABC triangles
Since points A, B, and C are given, and sides a,b, and c are known, it is straight forward to obtain the circles radii, which you have said are all the same. What is sought is the area of the union of the 3 circles. For most triangles, A,B,C, this union is made of 3 intersecting arcs. Obtaining the circle intersections gets 6 points, 3 of which bound the desired union. The easiest way to determine which 3 points to use is to graph the 3 circles and color in the union. If you are careful and use a computer to solve for the circle intersections, it appears that the same 3 points will always fall out, but I am not sure about that. Once you have identified the correct 3 intersections, then it is just drudge work to connect them, get the area of their triangle, and the area between each chord and its circle and then add it all up.
However, their are some triangles A,B,C that fit your restrictions but for which the union is not bounded by 3 arcs. Take for example the points $$A=(4,-1.5)$$ $$B=(3.5,0)$$ $$C=(3,1)$$
It is a skinny little triangle with $max(side)=2.69258$ which becomes the radius of all 3 circles. But, the union of the 3 circles is bounded by only two arcs at $$P=(1.33484,-1.11603)$$ $$Q=(5.66506,0.61603)$$
Getting this area is actually a couple of steps easier than the first case, since only the area between each circle and its chord is needed and then they are summed.
What is a little difficult is to be assured, without using a graph, that the correct intersection points have been selected for calculating area of the union. Since this wasn't requested, I haven't really tried to do it, but from a computer gaming perspective, it might be the more interesting problem.