The fastest way is to draw that circle.

Looking at name of this blog entry, my first thought was about fastest in terms of number of operations, not in terms of coding :)

Maybe I am wrong, but it seems that solving geometry without using structures is not the best idea. Yeah, I also like to write pair<pair<int, int> ,pair<int, int> > stuff, but it is a bad habit :)

And what is so hard and boring in intersecting two lines? It is something like:

Point intersect(Line l1, Line l2) 
{
 if (parallel(l1,l2)) return Point(-1e100,-1e100);
 double det=l1.b*l2.a-l1.a*l2.b;
 double x=l1.c*l2.b-l1.b*l2.c;
 double y=l1.a*l2.c-l1.c*l2.a;
 return Point(x/det,y/det);
}

Or you may use vector <Point> instead, depending on which implementation of geometry is more comfortable for you and what is needed for given task (possible solution — one may return 2 points for same lines, 0 for parallel and 1 for intersecting).

Of course, I also would be glad to see some even better solution :)

You can calculate coordinate of point I at complex plane. To make equations easier, put A in 0.

I faced this problem only once in my life — during last CH24 contest.

I have these two long lines from wikipedia in my solution:

xc = ((a.x * a.x + a.y * a.y) * (b.y - c.y) + (b.x * b.x + b.y * b.y) * (c.y - a.y) + (c.x * c.x + c.y * c.y) * (a.y - b.y)) / d;
yc = ((a.x * a.x + a.y * a.y) * (c.x - b.x) + (b.x * b.x + b.y * b.y) * (a.x - c.x) + (c.x * c.x + c.y * c.y) * (b.x - a.x)) / d;

If you want to calculate only R, then there is a compact formula: .

In your original approach, if you are intersecting medians, then you will get centroid, which is not the same, as center of circumscribed circle! Maybe you wanted to say perpendicular bisectors?)

Denote a = |BC|, b = |AC|, c = |AB|, .

Then .

This is true since ray AI divides BC side at a ratio of b·cos(β) to c·cos(γ).

To avoid precision problems, you can use the equation a² + b² - 2ab·cos(γ) = c².