Solution
The Gaussian Triangle
Show the problem again
Three points are chosen independently in the plane, each with independent standard normal x- and y-coordinates. What is the probability that the triangle they form is obtuse? Give a reduced fraction.
Worked solution
The answer is 3/4. At most one angle is obtuse, so compute the probability that the angle at A is obtuse and multiply by 3. That angle is obtuse iff U·V < 0, where U = B − A and V = C − A. Use the polarisation identity U·V = (|U + V|² − |U − V|²)/4. Per coordinate, U and V each have variance 2 and covariance 1 (from the shared A). Then U + V has variance 6 per coordinate, U − V has variance 2, and Cov(U + V, U − V) = Var U − Var V = 0, so the two are independent Gaussian vectors. Thus |U + V|² = 6S and |U − V|² = 2T, where S and T are independent exponentials with the same mean (chi-square with 2 degrees of freedom). P(U·V < 0) = P(6S < 2T) = P(T > 3S) = 1/(1 + 3) = 1/4. So P(obtuse) = 3·(1/4) = 3/4, exactly the same as for three random points on a circle.
Source: Known result on triangles with independent Gaussian vertices, in wide circulation; no traceable original source. Statement written for AxiomIQ.