Lemma 10

If X has a ramification point i, the full subgraph X - {i} is the union of three chains, and if p - 1, q - 1, r - 1 are the lengths of these chains, the triple {p, q, r} is equal, up to a permutation, to one of the triples {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 1, m} (for some m $\geq 1$).

The vertex i belongs to 3 edges (Lemma 2), and there is no other ramification point (Lemma 7), so the full subgraph X - {i} is the sum of 3 chains X_1, X_2, X_3, each of which has a terminal vertex linked to i in X. Let {i_1, i_2}, {i_2, i_3}, ..., {i_{p-1}, i_p} be the edges of X_1, {j_1, j_2}, {j_2, j_3}, ..., {j_{q-1}, j_q} those of X_2, and {k_1, k_2}, {k_2, k_3}, ..., {k_{r-1}, k_r} those of X_3, with i_1, j_1, k_1 linked to i in X. We can assume that $p \geq q \geq r \geq 1$. Put

Since all the edges of X are of order 3 (Lemma 7), Lemma 8 gives $\| x \| ^2 = \frac{1}{2} p(p+1)$, $\| y \| ^2 = \frac{1}{2} q(q+1)$, $\| z \| ^2 = \frac{1}{2} r(r+1)$. On the other hand, $e_i$ is orthogonal to $e_{i_2}, e_{i_3}, \ldots, e_{i_p}$, so $(e_i | x) = p(e_i | e_{i_1}) = -\frac{1}{2} p$; similarly $(e_i | y) = -\frac{1}{2} q$, $(e_i | z) = -\frac{1}{2} r$. The unit vectors $\| x \|^{-1} x, \| y \|^{-1} y, \| z \|^{-1} z$ are mutually orthogonal, and $e_i$ does not belong to the subspace F that they generate; the square of the distance from $e_i$ to F is

Since this quantity is > 0, we have

(4)

Hence $3(r+1)^{-1} > 1$, so $r < 2$ and finally $r = 1$. Then (4) gives

(5)

hence $2(q+1)^{-1} > \frac{1}{2}$, so $q \leq 2$. Finally, if $q = 2$, (5) gives