I'm working through Irme Barany's convexity lectures (http://wiki-math.univ-mlv.fr/gemecod/lib/exe/fetch.php/barany_lecture_2.pdf) and I am struggling to prove an exercise the author left to the reader. I'll include the context and my thoughts on it below, though you are also welcome to just read the proof of Theorem 5 in the linked document.
Having just introduced the Colorful Caratheodory's Theorem, Barany presents the following theorem as an application:
Let $G(V,A)$ be a directed graph with $|V|=n$ and let $C_1, \ldots, C_n$ be directed cycles in $G$. Then there exist arcs $a_i \in C_i$ such that the set $\{a_1, \ldots, a_n\}$ contains a directed cycle.
The proof goes as follows (paraphrased):
Let $V = \{v_1, \ldots, v_n\}$ and for an arc $a\in A$ from $v_i$ to $v_j$ consider the point $p_a\in \mathbb R^n$ given by $p_a = e_j -e_i$, where $e_1, \ldots, e_n$ is the standard basis. Note that $p_a \in H = \{x\in \mathbb R^n: 1^Tx = 0\}$ for all $a\in A$. This implies that the points $p_a$ lie in an affine space of dimension $(n-1)$.
Additionally, note that for any directed cycle $C$ in $G$ we have that $\sum_{a\in C} p_a =0$. Let $K_i = \text{Conv}\{p_a : a\in C_i\}$, and note that $0 \in K_i$ for every $i\in [n]$ since, $$ 0 = \sum_{a\in C} p_a = \sum_{a\in C} \frac{1}{|C|} p_a. $$ It follows from the Colorful Caratheodory Theorem that there exist points $p_{a_1}, \ldots, p_{a_n}$ such that $a_i \in C_i$ for each $i$ and $0 \in \text{Conv}\{p_{a_1}, \ldots, p_{a_n}\}$. WLOG let $k$ be the number of non-zero terms, so there exist constants $\alpha_i >0$ such that $$ 0 = \sum_{i=1}^k \alpha_i p_{a_i}, \text{ and } \sum_{i=1}^k \alpha_i =1. $$
It is at this point that I get stuck on an assertion made in the document: it follows from the above that the non-zero constants are all equal, i.e. that $\alpha_1 = \alpha_2 = \ldots = \alpha_k =\frac{1}{k}$.
From there it is easy to show that the corresponding arc set $\{a_1 , \ldots, a_k\}$ has a directed cycle. I did this by a proof by contradiction. Start at the first arc, $a_1 = v_i \to v_j$. By the results above, there must be an another arc in our set with start node $v_j$ to cancel the contribution of the first arc. Proceeding in this way we show that eventually we must return to an arc we already visited, giving us the directed cycle.
Any help understanding the assertion in bold would be appreciated. I figure there must be some counting argument, since among the points $p_a$ we have exactly $k$ 1's and $k$ -1's, but I am not seeing it.
First note, be very careful in how you restate the proof. Saying "$k$ is the number of non-zero terms" is very different from letting $k$ be minimal so that $\mathbf{0} \in \text{conv}(\{p_{a_i}\}_{i=1}^k)$. After all, there are no zero terms!
Every vector $p_{a_j}$ is the same distance from $\mathbf{0}$. Indeed, for each $i$ we have $d(\mathbf{0},p_{a_i}) = 2^{1/n}$. It follows that the unique expression of $\mathbf{0}$ as a convex combination of the $p_{a_i}$ has some constant scalar whose $k$-fold sum is $1$. The only candidate for such a scalar is $1/k$.