Random splitting of a the unit square Consider the unit square $S=[0,1]^2$. Let us choose randomly a point $(x,y)$ in $S$ (uniformly over $S$) and consider the four triangles whose two vertices are a pair of consecutive vertices of $S$ and the third one is $(x,y)$.
What is the expected area of the largest triangle?
 A: I claim that this question is equivalent to: if $X, Y$ are independent random variables uniformly distributed on $[0, 1]$, what is the expected value of $\frac M 2$, where 
$M = \max \{X, 1-X, Y, 1-Y\}$? To see this, draw the picture and observe that the triangle areas are $\frac 1 2 \cdot 1 \cdot U$, where $U \in \{X, 1-X, Y, 1-Y\}$.
From here, we compute the distribution directly; for $k \in [1/2, 1]$, note that
\begin{align*}
 \max\{X, 1-X, Y, 1-Y\} \leq k 
\iff  1-k \leq X \leq k \text{ and } 1-k \leq Y \leq k
\end{align*}
and thus, the CDF for $M$ is given by
\begin{align*}
  \mathbb P (M \leq k) &= \mathbb P(1-k \leq X \leq k) \cdot \mathbb P(1-k \leq Y \leq k) \\
&= (2k-1)^2.
\end{align*}
We differentiate to get the density function:
$$f_M(k) = 4 \cdot (2k-1), \qquad 1/2 \leq k \leq 1$$
then we integrate to get the expected value:
$$\mathbb E[M] = \int_{1/2}^1 k \cdot f_M(k) \, \textrm{d} k = 5/6$$
hence, the expected largest area is $5/12$.
Fun question! Hopefully I didn't make any mistakes.
