Let $\mu_1,\mu_2,\ldots,\mu_k \in \Delta^{d-1}$ be $k\ (k\geq 2)$ distinct vectors on the standard simplex, where $$\Delta^{d-1} = \{\mu\in R^{d}:\|\mu\|_1 = 1,\mu_j \geq 0\}.$$ (With a slight abuse of notation, here $\mu_j$ means the $j$-th entry of $\mu \in R^d$.) We define $\Delta^{k-1}$ similarly.

Our goal is to find a $\mu \in \Delta^{d-1}$ that satisfies $\mu \notin \{\mu_1,\ldots,\mu_k\}$, such that for some $w \in \Delta^{k-1}$,
$$ \sum_{i} w_i \mu_i = \mu, $$ and $$ \sum_{i} w_i \mu_i^{\otimes 2} = \mu^{\otimes 2}. $$ Equivalently, one can think of constructing $\mu_1,\mu_2,\ldots,\mu_k$ for a given $\mu$.

Here is the question: Is there a principled way to do this construction? Or does there exist a solution at all?