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Let $x_1, ..., x_n\in\mathbb{R}^d$. We know that a point $x$ in the convex hull $\text{conv}(x_1, ..., x_n)$ may be expressed as convex combinations $x=\sum_{i=1}^n r_ix_i=\sum_{i=1}^n s_ix_i$ with distinct $(r_1, ..., r_n)$ and $(s_1, ..., s_n)$ with $r_i, s_i\geq 0$, for all $i$ and $\sum_{i=1}^n r_i=1=\sum_{i=1}^n s_i$. With this fact, I have the following question:
For $x\in \text{conv}(x_1, ..., x_n)$, what is the number of distinct ways (i.e. distinct coefficients $(r_1, ..., r_n)$) that $x$ can be expressed as a convex combination of $x_1, ..., x_n$? Is there a simple answer/formula in terms of geometry under appropriate condition (like, in general position) on the points? Any reference of books or papers is appreciated. Thanks

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    $\begingroup$ If $x$ can be expressed as a convex combination in more than one way, then it can be expressed in uncountably many ways. $\endgroup$
    – user44191
    Sep 28, 2019 at 19:05
  • $\begingroup$ @user44191 Thanks, I have figured out a way to construct infinitely many such convex combinations. $\endgroup$
    – Min Wu
    Sep 28, 2019 at 20:25

1 Answer 1

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Let $F$ be the smallest face of the polytope $\text{conv}(x_1, \dots, x_n)$ containing $x$. If $F$ is a simplex intersecting $\{x_1, \dots, x_n\}$ only in its vertices, then there is a unique convex combination giving $x$: the points $x_i$ that do not belong to $F$ must have the coefficient $r_i$ equal to $0$. Otherwise there are continuum many such convex combinations, as user44191 mentioned.

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