[proofplan] We reduce the affine intersection statement in $\mathbb{R}^d$ to a colorful selection problem in the [tensor product](/page/Tensor%20Product) $\mathbb{R}^{r-1} \otimes \mathbb{R}^{d+1}$. A regular simplex $u_1,\dots,u_r$ centered at the origin records which part of the partition each point belongs to, while the lift $(a,1)$ records affine rather than merely linear combinations. The Colorful Carathéodory Theorem supplies one tensor from each color class whose convex hull contains $0$. The unique linear dependence among the simplex vertices then forces the lifted barycentres of all $r$ parts to be equal, and projecting back to $\mathbb{R}^d$ gives a common point in all convex hulls. [/proofplan] [step:Choose simplex vectors with a unique barycentric dependence] Let \begin{align*} N := (r-1)(d+1)+1. \end{align*} Since $A$ is finite with $|A|=N$, enumerate it as \begin{align*} A = \{a_1,\dots,a_N\}. \end{align*} Choose vectors $u_1,\dots,u_r \in \mathbb{R}^{r-1}$ forming the vertices of a regular simplex centered at $0$. Thus \begin{align*} \sum_{j=1}^r u_j = 0. \end{align*} Moreover, because these $r$ vertices are affinely independent in the affine hyperplane they span, the space of linear relations among $u_1,\dots,u_r$ is one-dimensional. Equivalently, whenever $c_1,\dots,c_r \in \mathbb{R}$ satisfy \begin{align*} \sum_{j=1}^r c_j u_j = 0, \end{align*} there exists $\lambda \in \mathbb{R}$ such that $c_j=\lambda$ for every $j \in \{1,\dots,r\}$. [/step] [step:Build color classes in the tensor product and select a colorful convex combination] Define the affine lift map \begin{align*} \iota: \mathbb{R}^d &\to \mathbb{R}^{d+1}, \qquad x \mapsto (x,1). \end{align*} Let \begin{align*} V := \mathbb{R}^{r-1} \otimes \mathbb{R}^{d+1}. \end{align*} We identify $V$ with $\mathbb{R}^{(r-1)(d+1)}$ by the standard tensor basis. For each $i \in \{1,\dots,N\}$, define the finite color class \begin{align*} F_i := \{u_j \otimes \iota(a_i) : j \in \{1,\dots,r\}\} \subset V. \end{align*} Since the simplex is centered at $0$, \begin{align*} \frac{1}{r}\sum_{j=1}^r u_j \otimes \iota(a_i) = \left(\frac{1}{r}\sum_{j=1}^r u_j\right)\otimes \iota(a_i)=0 \end{align*} for every $i$. Hence $0 \in \operatorname{conv}(F_i)$ for every $i$. Now $V$ has dimension \begin{align*} \dim V = (r-1)(d+1)=N-1. \end{align*} By the Colorful Carathéodory Theorem (citing a result not yet in the wiki: Colorful Carathéodory Theorem), applied in the [vector space](/page/Vector%20Space) $V$ to the $N=\dim V+1$ color classes $F_1,\dots,F_N$, there exist indices \begin{align*} j(i) \in \{1,\dots,r\} \quad \text{for } i \in \{1,\dots,N\} \end{align*} and coefficients $\alpha_1,\dots,\alpha_N \in [0,\infty)$ with \begin{align*} \sum_{i=1}^N \alpha_i = 1 \end{align*} such that \begin{align*} \sum_{i=1}^N \alpha_i\, u_{j(i)} \otimes \iota(a_i)=0. \end{align*} [/step] [step:Convert the tensor equation into equal lifted barycentres] For each $j \in \{1,\dots,r\}$, define \begin{align*} A_j := \{a_i \in A : j(i)=j\}. \end{align*} These sets are pairwise disjoint and their union is $A$, because each index $i$ is assigned exactly one value $j(i)$. For each $j \in \{1,\dots,r\}$, define the lifted weighted sum \begin{align*} z_j := \sum_{\{i: j(i)=j\}} \alpha_i\, \iota(a_i) \in \mathbb{R}^{d+1}. \end{align*} Then the colorful tensor equation becomes \begin{align*} \sum_{j=1}^r u_j \otimes z_j = 0. \end{align*} We prove that $z_1=\cdots=z_r$. Let $\ell:\mathbb{R}^{d+1}\to\mathbb{R}$ be an arbitrary linear functional. Applying the [linear map](/page/Linear%20Map) \begin{align*} \operatorname{id}_{\mathbb{R}^{r-1}} \otimes \ell : \mathbb{R}^{r-1}\otimes\mathbb{R}^{d+1} \to \mathbb{R}^{r-1} \end{align*} to the tensor equation gives \begin{align*} \sum_{j=1}^r \ell(z_j) u_j = 0. \end{align*} By the uniqueness of the linear dependence among $u_1,\dots,u_r$, the [real numbers](/page/Real%20Numbers) $\ell(z_1),\dots,\ell(z_r)$ are all equal. Since this holds for every linear functional $\ell:\mathbb{R}^{d+1}\to\mathbb{R}$, the vectors $z_1,\dots,z_r$ are equal. [guided] The tensor equation \begin{align*} \sum_{j=1}^r u_j \otimes z_j = 0 \end{align*} contains the essential information of the proof: the vector $u_j$ records the part label $j$, and the vector $z_j$ records the lifted affine combination of the points assigned to that part. We want to prove that the $z_j$ are all the same vector in $\mathbb{R}^{d+1}$. To extract scalar information from the second tensor factor, let $\ell:\mathbb{R}^{d+1}\to\mathbb{R}$ be any linear functional. The map \begin{align*} \operatorname{id}_{\mathbb{R}^{r-1}} \otimes \ell : \mathbb{R}^{r-1}\otimes\mathbb{R}^{d+1} \to \mathbb{R}^{r-1} \end{align*} is the linear map determined on simple tensors by \begin{align*} (\operatorname{id}_{\mathbb{R}^{r-1}} \otimes \ell)(u\otimes v)=\ell(v)u. \end{align*} Applying it to the equation $\sum_{j=1}^r u_j\otimes z_j=0$ gives \begin{align*} \sum_{j=1}^r \ell(z_j)u_j=0. \end{align*} Now we use the defining linear-algebraic feature of the regular simplex. The vectors $u_1,\dots,u_r$ satisfy exactly one linear dependence up to scalar multiple, namely \begin{align*} \sum_{j=1}^r u_j=0. \end{align*} Therefore any relation $\sum_{j=1}^r c_j u_j=0$ must have all coefficients equal. Applying this with $c_j=\ell(z_j)$ shows that \begin{align*} \ell(z_1)=\ell(z_2)=\cdots=\ell(z_r). \end{align*} Because $\ell$ was an arbitrary linear functional on $\mathbb{R}^{d+1}$, the equalities above hold after testing against every linear functional. Linear functionals separate points in finite-dimensional Euclidean space: if $z_p-z_q \ne 0$, one may take $\ell(v)=(v,z_p-z_q)_{\mathbb{R}^{d+1}}$ to get $\ell(z_p-z_q)=|z_p-z_q|^2 \ne 0$. Hence no two of the vectors $z_j$ can differ, and so \begin{align*} z_1=z_2=\cdots=z_r. \end{align*} [/guided] [/step] [step:Project the common lifted barycentre to a common point of all convex hulls] Let $z \in \mathbb{R}^{d+1}$ denote the common value of $z_1,\dots,z_r$. For each $j \in \{1,\dots,r\}$, define \begin{align*} s_j := \sum_{\{i: j(i)=j\}} \alpha_i. \end{align*} The last coordinate of $z_j$ is $s_j$, because the last coordinate of every lifted vector $\iota(a_i)=(a_i,1)$ is $1$. Since $z_1=\cdots=z_r$, all $s_j$ are equal. Also \begin{align*} \sum_{j=1}^r s_j = \sum_{i=1}^N \alpha_i = 1. \end{align*} Therefore \begin{align*} s_j = \frac{1}{r} \end{align*} for every $j \in \{1,\dots,r\}$. In particular, each $s_j$ is positive, so each $A_j$ is nonempty. Let $p \in \mathbb{R}^d$ be the vector formed by the first $d$ coordinates of $r z \in \mathbb{R}^{d+1}$. For each $j$, the equality defining $z_j$ gives \begin{align*} z_j = \sum_{\{i: j(i)=j\}} \alpha_i (a_i,1). \end{align*} Taking the first $d$ coordinates and using $s_j=1/r$ gives \begin{align*} p = \sum_{\{i: j(i)=j\}} r\alpha_i a_i. \end{align*} The coefficients $r\alpha_i$ appearing in this sum are nonnegative, and their sum over the indices with $j(i)=j$ is \begin{align*} \sum_{\{i: j(i)=j\}} r\alpha_i = r s_j = 1. \end{align*} Thus $p$ is a convex combination of the points in $A_j$ for every $j \in \{1,\dots,r\}$. Hence \begin{align*} p \in \operatorname{conv}(A_1) \cap \cdots \cap \operatorname{conv}(A_r). \end{align*} The sets $A_1,\dots,A_r$ are nonempty, pairwise disjoint, and have union $A$, so they form a Tverberg partition of $A$ into $r$ parts. [/step]