$(X,Y)$ is uniform over the triangular region with vertices at $(0,0)$,$(\theta, 0)$, and $(0,\theta)$, $\theta$ is unknown. Let $(X_i,Y_i)$ be iid as $(X,Y)$. Find a one dimensional sufficient statistic $T$ for $\theta$, and prove it's sufficient. Find an unbiased estimate of $\hat{\theta}$ which is a function of $T$.
Attempt
$f_{X,Y}(x,y) = \frac{2}{\theta^2}$
$L(\theta| \mathbf{X},\mathbf{Y}) = (\frac{\theta^2}{2})^{-n} \prod I_{x_i+y_i < \theta}$
Sufficient statistic: $T = max_i (X_i+Y_i)$
$\hat{\theta} = ?$
Next you need to find the CDF and pdf of $T$ in order to get $\mathbb E[T]$.
Find CDF of $X_i+Y_i$: $$ F_{X_i+Y_i}(t)=\mathbb P(X_i+Y_i \leq t) = \mathbb P(Y_i \leq t-X_i) $$ This is the probability that a point chosen at random in the triangular region with vertices at $(0,0)$, $(\theta, 0)$, and $(0,\theta)$ appears in the triangular region with vertices at $(0,0)$, $(t, 0)$, and $(0,t)$. For $0 \leq t\leq \theta$ this probability equals to ratio of squares of the triangles: $$ F_{X_i+Y_i}(t)=\frac{t^2/2}{\theta^2/2}=\frac{t^2}{\theta^2}. $$
So, the CDF of $T$ for $0 \leq t\leq \theta$ is $$ F_T(t)=\mathbb P(\max_i (X_i+Y_i)\leq t)=\mathbb P(X_1+Y_1\leq T,\ldots,X_n+Y_n\leq T)=\left(\frac{t^2}{\theta^2}\right)^n=\frac{t^{2n}}{\theta^{2n}} $$ Find pdf $f_T(t)$: $$ f_T(t)=2n\frac{t^{2n-1}}{\theta^{2n}}, \quad 0 \leq t\leq \theta. $$ Find expectation of $T$ and get $$ \mathbb E[T]=\frac{2n}{2n+1}\theta. $$ Finally, $\hat\theta=\frac{2n+1}{2n}T$ is unbiased.