[proofplan] Enumerate the $n+2$ points and lift each point $x_i \in \mathbb{R}^n$ to the vector $(x_i,1) \in \mathbb{R}^{n+1}$. Since there are more lifted vectors than the dimension of the ambient space, they satisfy a non-trivial linear relation; separating the last coordinate gives an affine dependence relation among the original points. The positive and negative coefficients in this affine dependence determine the two parts of the partition, and after normalising both sides by the common total coefficient, the same point is written as a convex combination of each part. [/proofplan] [step:Construct a non-trivial affine dependence among the points] Write \begin{align*} S = \{x_1,\dots,x_{n+2}\}, \end{align*} where each $x_i \in \mathbb{R}^n$ and the points are pairwise distinct. For each index $i \in \{1,\dots,n+2\}$, define the lifted vector \begin{align*} y_i := (x_i,1) \in \mathbb{R}^{n+1}. \end{align*} Define the [linear map](/page/Linear%20Map) \begin{align*} T: \mathbb{R}^{n+2} &\to \mathbb{R}^{n+1} \\ (\alpha_1,\dots,\alpha_{n+2}) &\mapsto \sum_{i=1}^{n+2} \alpha_i y_i. \end{align*} Since $\dim \mathbb{R}^{n+2} = n+2$ and $\dim \mathbb{R}^{n+1} = n+1$, the kernel of $T$ contains a non-zero vector. Choose \begin{align*} \alpha := (\alpha_1,\dots,\alpha_{n+2}) \in \ker T \setminus \{0\}. \end{align*} Then \begin{align*} 0 = T(\alpha) = \sum_{i=1}^{n+2} \alpha_i (x_i,1) = \left(\sum_{i=1}^{n+2} \alpha_i x_i,\ \sum_{i=1}^{n+2} \alpha_i\right). \end{align*} Therefore \begin{align*} \sum_{i=1}^{n+2} \alpha_i x_i = 0 \quad\text{and}\quad \sum_{i=1}^{n+2} \alpha_i = 0. \end{align*} [guided] We want an affine dependence, meaning a relation among the points whose coefficients add to zero. To force the condition on the coefficient sum, we add one extra coordinate to every point. For each $i \in \{1,\dots,n+2\}$, define \begin{align*} y_i := (x_i,1) \in \mathbb{R}^{n+1}. \end{align*} Now define the linear map \begin{align*} T: \mathbb{R}^{n+2} &\to \mathbb{R}^{n+1} \\ (\alpha_1,\dots,\alpha_{n+2}) &\mapsto \sum_{i=1}^{n+2} \alpha_i y_i. \end{align*} The domain has dimension $n+2$, while the codomain has dimension $n+1$. Hence $T$ has a non-zero kernel vector. Choose \begin{align*} \alpha := (\alpha_1,\dots,\alpha_{n+2}) \in \ker T \setminus \{0\}. \end{align*} Since $T(\alpha)=0$, we have \begin{align*} 0 = \sum_{i=1}^{n+2} \alpha_i (x_i,1) = \left(\sum_{i=1}^{n+2} \alpha_i x_i,\ \sum_{i=1}^{n+2} \alpha_i\right). \end{align*} Equality in $\mathbb{R}^{n+1}$ means equality in the first $n$ coordinates and in the final coordinate. Thus \begin{align*} \sum_{i=1}^{n+2} \alpha_i x_i = 0 \quad\text{and}\quad \sum_{i=1}^{n+2} \alpha_i = 0. \end{align*} The final coordinate is exactly what turns ordinary linear dependence of the lifted vectors into affine dependence of the original points. [/guided] [/step] [step:Separate the positive and negative coefficients] Define the index sets \begin{align*} I_+ &:= \{i \in \{1,\dots,n+2\} : \alpha_i > 0\},\\ I_- &:= \{i \in \{1,\dots,n+2\} : \alpha_i < 0\},\\ I_0 &:= \{i \in \{1,\dots,n+2\} : \alpha_i = 0\}. \end{align*} Since $\alpha \ne 0$, at least one coefficient is non-zero. Because \begin{align*} \sum_{i=1}^{n+2} \alpha_i = 0, \end{align*} there must be at least one positive coefficient and at least one negative coefficient; otherwise all non-zero coefficients would have the same sign and their sum could not vanish. Hence $I_+$ and $I_-$ are both non-empty. Define \begin{align*} P := \sum_{i \in I_+} \alpha_i. \end{align*} Then $P > 0$. Since the total coefficient sum is zero, \begin{align*} P = -\sum_{i \in I_-} \alpha_i. \end{align*} Using the vector relation $\sum_{i=1}^{n+2} \alpha_i x_i = 0$, and noting that the indices in $I_0$ contribute zero, we obtain \begin{align*} \sum_{i \in I_+} \alpha_i x_i = \sum_{i \in I_-} (-\alpha_i)x_i. \end{align*} [/step] [step:Normalize both sides into convex combinations] Define the point $z \in \mathbb{R}^n$ by \begin{align*} z := \sum_{i \in I_+} \frac{\alpha_i}{P} x_i. \end{align*} For each $i \in I_+$, the coefficient $\alpha_i/P$ is positive, and \begin{align*} \sum_{i \in I_+} \frac{\alpha_i}{P} = 1. \end{align*} Therefore $z \in \operatorname{conv}(\{x_i : i \in I_+\})$. From \begin{align*} \sum_{i \in I_+} \alpha_i x_i = \sum_{i \in I_-} (-\alpha_i)x_i \end{align*} and $P = -\sum_{i \in I_-} \alpha_i$, division by $P$ gives \begin{align*} z = \sum_{i \in I_-} \frac{-\alpha_i}{P} x_i. \end{align*} For each $i \in I_-$, the coefficient $(-\alpha_i)/P$ is positive, and \begin{align*} \sum_{i \in I_-} \frac{-\alpha_i}{P} = 1. \end{align*} Thus $z \in \operatorname{conv}(\{x_i : i \in I_-\})$. [guided] The relation \begin{align*} \sum_{i \in I_+} \alpha_i x_i = \sum_{i \in I_-} (-\alpha_i)x_i \end{align*} already says that the positive side and the negative side determine the same vector after weighting. To make this common vector into a point in a convex hull, the weights must sum to $1$. This is why we divide by the common positive number \begin{align*} P := \sum_{i \in I_+} \alpha_i = -\sum_{i \in I_-} \alpha_i > 0. \end{align*} Define \begin{align*} z := \sum_{i \in I_+} \frac{\alpha_i}{P} x_i. \end{align*} For every $i \in I_+$, $\alpha_i/P > 0$, and the coefficients on the positive side satisfy \begin{align*} \sum_{i \in I_+} \frac{\alpha_i}{P} = \frac{1}{P}\sum_{i \in I_+}\alpha_i = 1. \end{align*} So $z$ is a convex combination of the points indexed by $I_+$. Now divide the equality of the two weighted sums by $P$. This gives \begin{align*} z = \sum_{i \in I_-} \frac{-\alpha_i}{P} x_i. \end{align*} For every $i \in I_-$, $-\alpha_i > 0$, so $(-\alpha_i)/P > 0$. The coefficients on the negative side also sum to $1$: \begin{align*} \sum_{i \in I_-} \frac{-\alpha_i}{P} = \frac{1}{P}\sum_{i \in I_-}(-\alpha_i) = 1. \end{align*} Thus the same point $z$ is a convex combination of the points indexed by $I_-$. This is the core mechanism of Radon's theorem: affine dependence produces two different convex representations of one point. [/guided] [/step] [step:Form the required partition and identify the common point] Define \begin{align*} A := \{x_i : i \in I_+\} \quad\text{and}\quad B := S \setminus A. \end{align*} Since $I_+$ is non-empty, $A$ is non-empty. Since $I_-$ is non-empty and $I_- \subseteq \{1,\dots,n+2\} \setminus I_+$, the set $B$ is non-empty. By construction, $A$ and $B$ are disjoint and $S = A \cup B$. The previous step gives \begin{align*} z \in \operatorname{conv}(A) \quad\text{and}\quad z \in \operatorname{conv}(\{x_i : i \in I_-\}). \end{align*} Because $\{x_i : i \in I_-\} \subseteq B$, monotonicity of convex hull gives \begin{align*} \operatorname{conv}(\{x_i : i \in I_-\}) \subseteq \operatorname{conv}(B). \end{align*} Therefore $z \in \operatorname{conv}(A) \cap \operatorname{conv}(B)$, so \begin{align*} \operatorname{conv}(A) \cap \operatorname{conv}(B) \ne \varnothing. \end{align*} This completes the proof. [/step]