Triangle greater than (probability) This one is a follow up of my previous question. But a different problem. And this one should have a more interesting answer. I don't really know how to approach this problem nonetheless reach a solution, so again help is appreciated.
Question:
You have a circle with radius $R$. If three points are randomly chosen inside this circle.
What is the probability that the three points form a triangle with an area greater than $\displaystyle \frac{R^2}{5}$?
Edit: Is anyone trying or maybe found an approach that might work? Is there any similar problems you've seen before that could work as a guide towards solving this one? What do you consider being the difficulties? I literally don't have any idea on where to even start.

This isn't an answer, but just a simulation. I get the approximate value
$$P(A\geq \frac{1}{5}) \approx 0.45$$
Here is my Sage-code if someone wants to check it. It agrees with the mean value of mathworld
def randPt():
    r = random()**0.5 #sqrt to make it uniform
    a = random()*2*float(pi)
    return (r*cos(a), r*sin(a))

def simuTriArea():
    a,b,c = [randPt() for _ in range(3)]
    return 0.5*abs(a[0]*b[1] + b[0]*c[1] + c[0]*a[1] - b[0]*a[1] - c[0]*b[1] - a[0]*c[1])

#points([randPt() for _ in range(1000)]).show(aspect_ratio=1)
simuN = 100000
triAreas = [simuTriArea() for _ in range(simuN)]
print ("simulated P(A>0.2): %f" % (sum(1 for a in triAreas if a>0.2) / float(simuN),) )
print ("mean A: %f" %mean(triAreas))
graph = Graphics()
graph += histogram(triAreas, density=True, bins=50)
maxArea = float(3*3**0.5 / 4)
#graph += plot(???, xmin=0, xmax=maxArea)
graph.show()

