HMMT-Feb 2021 guts, Problem 24

Let $P$ be a point selected uniformly at random in the cube $[0,1]^{3}$. The plane parallel to $x+y+z=0$ passing through $P$ intersects the cube in a two-dimensional region $\mathcal{R}$. Let $t$ be the expected value of the perimeter of $\mathcal{R}$. If $t^{2}$ can be written as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers, compute $100 a+b$.

Solution: Solution: We can divide the cube into 3 regions based on the value of $x+y+z$ which defines the plane: $x+y+z<1,1 \leq x+y+z \leq 2$, and $x+y+z>2$. The two regions on the ends create tetrahedra, each of which has volume $1 / 6$. The middle region is a triangular antiprism with volume $2 / 3$. If our point $P$ lies in the middle region, we can see that we will always get the same value $3 \sqrt{2}$ for the perimeter of $\mathcal{R}$. Now let us compute the expected perimeter given that we pick a point $P$ in the first region $x+y+z<1$. If $x+y+z=a$, then the perimeter of $\mathcal{R}$ will just be $3 \sqrt{2} a$, so it is sufficient to find the expected value of $a$. $a$ is bounded between 0 and 1 , and forms a continuous probability distribution with value proportional to $a^{2}$, so we can see with a bit of calculus that its expected value is $3 / 4$. The region $x+y+z>2$ is identical to the region $x+y+z<1$, so we get the same expected perimeter. Thus we have a $2 / 3$ of a guaranteed $3 \sqrt{2}$ perimeter, and a $1 / 3$ of having an expected $\frac{9}{4} \sqrt{2}$ perimeter, which gives an expected perimeter of $\frac{2}{3} \cdot 3 \sqrt{2}+\frac{1}{3} \cdot \frac{9}{4} \sqrt{2}=\frac{11 \sqrt{2}}{4}$. The square of this is $\frac{121}{8}$, giving an extraction of 12108 . $\fbox{12108}$.

Answer: 12108