Problem:
Given a matrix $A$ with shape $n\times n$ and a scalar value $v$, consider the equation system $$x^TAy = v,\\ \mathbf{1}_n^Tx=1, x\geq \mathbf{0}\\ \mathbf{1}_n^Ty=1, y\geq \mathbf{0}$$ where $x, y$ are in a $n-1$ probability simplex.
Consider a simplest case
Given a matrix $A=\begin{bmatrix} a_{1} & a_{2} \\ a_{3} & a_{4} \\ \end{bmatrix}$ and $v$, the decision variable are $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}$, $y=\begin{bmatrix}y_1\\y_2\end{bmatrix}$.
the problem can be written as $$a_1x_1y_1+a_2x_1y_2+a_3x_2y_1+a_4x_2y_2=v\\ x_1+x_2=1, x\geq \mathbf{0}\\ y_1+y_2=1, y\geq \mathbf{0}$$
Some of my tries
I actually put it to Gurobi directly without any reformulation, and the solver always gives a correct solution, so I think the existence of a solution can be proved.
What is such a problem called? and what method can solve it? Any idea to prove that there always exists at least one pair of $(x, y)$ satisfy the above equation system?
The inequalities $$ \begin{array}{rl} \mathbf{1}_n^Tx =1, &x\geq \mathbf{0}\\ \mathbf{1}_n^Ty =1, &y\geq \mathbf{0} \end{array} $$ implies $0\leq x_{i}\leq 1$, $0\leq y_{j}\leq 1$, $$ x_{1}+\ldots+x_{n}=1 \quad \mbox{ and } \quad y_{1}+\ldots+y_{n}=1. $$ These inequalities make it easy to proof that $$ \min_{ij}A_{ij}\leq x^{T}Ay=\sum_{i=1}\sum_{j=1}x_{i}\cdot A_{ij}\cdot y_{j}\leq \max_{ij}A_{ij} $$ Then the equation $$ x^{T}Ay=\sum_{i=1}\sum_{j=1}x_{i}\cdot A_{ij}\cdot y_{j}=v $$ has solution if, only if, $\min_{ij}A_{ij}\leq v \leq \max_{ij}A_{ij}$. In fact, supose $A_{\alpha,\beta}=\min_{ij}A_{ij}$ and $A_{\mu,\nu}=\max_{ij}A_{ij}$ set $y_{\beta}=t$, $y_{\nu}=1-t$, $x_{\alpha}=t$ and $x_{\mu}=1-t$ for $0\leq t\leq 1$. Then $$ x^{T}Ay = t\cdot A_{\alpha \beta}\cdot t+(1-t)\cdot A_{\mu \nu}\cdot (1-t) = t^{2}\cdot A_{\alpha \beta}+(1-t)^{2}\cdot A_{\mu \nu}. $$ Now just note that the quadratic equation $$ t^{2}\cdot A_{\alpha \beta}+(1-t)^{2}\cdot A_{\mu \nu}=v $$ has solution in the $[0,1]$ interval when $A_{\alpha \beta}\leq v\leq A_{\mu \nu}$.