Consider two nonparallel planes $X,Y$ that go through to the center of a sphere of radius one. This cuts the sphere into 4 parts $A,B,C,D$. The two parts $A,B$ are on the same side of $X$, such that the volume of $A$ is greater than the volume of $B$. If the expected value of the volume of $A$ can be expressed as $\frac{a}b\cdot\pi$, where $\gcd(a,b)=1$ find $a+b$.
I found this problem in a math competition archive and I was curious about how to solve this. I attempted to find the range of the possible values of $A$ but was unable to.
By symmetry, we can suppose that the line along which $X$ and $Y$ intersect is vertical, and fix $X$ to be in a certain direction, so that we only consider the possible angles of $Y$ as it rotates around our polar axis.
Looking at the sphere from above, it should be clear that the volumes of the pieces only depend on the angles between them, and scale linearly with the central angle between the two pieces. In the following picture, we are looking at the sphere from above, with $X$ the red horizontal line and $Y$ the blue line (which could be rotated at any angle),
If the larger of the two angles between $X$ and $Y$ is $\theta$, then the volume of $A$ will just be the volume of the sphere times the fraction of $360^\circ$ that $\theta$ take up, i.e. $\frac{4\pi}{3}\cdot \frac\theta{2\pi}=\frac{2\theta}{3}$.
Since $\theta$ is uniform on $[\pi/2,\pi]$ (because the "true" angle between $X$ and $Y$ uis uniform mon $[0,\pi]$ but we always choose the larger half), this means that the expected value of $\theta$ is $\frac{3\pi}4$, and so the expected volume of $A$ is $\frac{\pi}{2}$. Hence, $a=1, b=2, a+b=3$.