For a set $A\subset\mathbb R^d$ of $d+2$ points, Radon's Lemma states that we find disjoint $B,C\subset A$ with $\mathrm{Conv}(B)\cap\mathrm{Conv}(C)\neq\emptyset$. Elements belonging to such intersections are called Radon points of $A$.
Radon points of $A$ are in general not unique. However, I read that under assumption that $A$ is in general position (that is, no $d+1$ points of $A$ are affinely dependent), then $A$ does have a unique Radon point!
I tried to prove this, but without success. Here is how far I've come.
Consider two Radon points $x$ and $y$ of $A$. Write $A=\{a_1,\dots,a_{d+2}\}$. We can then write
$$\sum_{i\in I_1}\alpha_i a_i=x=\sum_{i\in I_2}-\alpha_ia_i,\qquad \sum_{i\in J_1}\beta_i a_i=y=\sum_{i\in J_2}-\beta_ia_i,$$
where we have:
- $I_1\sqcup I_2=\{1,\dots,d+2\}=J_1\sqcup J_2$ (disjoint unions);
- $\alpha_i\geq 0$ for $i\in I_1$ and $\alpha_i\leq 0$ for $i\in I_2$, as well as $\beta_i\geq 0$ for $i\in J_1$ and $\beta_i\leq 0$ for $i\in J_2$;
- $\sum_{i\in I_1}\alpha_i=1=\sum_{i\in I_2}-\alpha_i$ and $\sum_{i\in J_1}\beta_i=1=\sum_{i\in J_2}-\beta_i$.
Using that $A$ is in general position, we conclude that $\alpha_i>0$ and $\beta_i>0$ for all $1\leq i\leq d+2$. Then we can argue that there is $\lambda>0$ for which $\alpha_i=\lambda\beta_i$ for every $i$.
Further, since $\sum_{i\in I_1}\alpha_i-\sum_{i\in I_2}\alpha_i=2=\sum_{i\in J_1}\beta_i-\sum_{i\in J_2}\beta_i$, we obtain $\lambda=1$.
But still, how can we conclude that $x=y$? After all, it is not guaranteed that $I_1=J_1$ and $I_2=J_2$, is it?