Let $P = \langle v_1, \ldots, v_m \rangle$ be the longest path of $G$. Since $P$ is the longest path, $N(v_1)$ and $N(v_m)$ are subsets of $\{v_1,\ldots,v_m\}$. (Otherwise $P$ is not the longest path.)
If there exist $i \in \{2,\ldots ,m-2 \} $, such that $v_i \in N(v_m)$ and $v_{i+1} \in N(v_1)$, then the vertices of $P$ forms a cycle in $G$. (Why?) Since $G$ is connected, $P$ must connect to other vertices (If there's no other vertices, then $m = 100$.) This contradicts to the maximality of the length of $P$.
Hence if $i \in \{2,\ldots ,m-2 \}$ with $v_i \in N(v_m)$, we get $v_{i+1} \notin N(v_1)$. $\quad (*)$
Our goal is to show $m \geq 21.$ Suppose $m \leq 20$, then $N(v_1) \subseteq \{v_2,v_3,\ldots,v_{m-1}\}$. Note that $|\{v_2,v_3,\ldots,v_{m-1}\}| = (m-1)-2+1 = 18$. But $(*)$ restricts $9$ vertices in $\{v_3,v_4,\ldots,v_{m-1}\}$ can't adjacent to $v_1.$ This implies $|N(v_1)| \leq 18-9 = 9,$ which leads to a contradiction. This is our desired result.
Extra Exercise: Use this method to show the general result:
Every connected graph $G \neq K_2$ contains a path or a cycle of length at least $\min\{2 \delta(G), |G|\}$.
Let G be a connected simple graph where every vertex has degree 2. Let P be a path of maximal length in G, and let u and v be the origin and terminus of that path. Let u' be the neighbor of u that is not the "next" vertex in the path, and let v' be the neighbor of v that is not the "next-to-last" vertex in the path. Since P cannot be extended in either direction, it must be that u' and v' are already vertices in P. But the neighbors of each internal vertex of the path already have both of their neighbors used. So it must be that u'=v and v'=u. Hence, connecting the ends of the path lead to a cycle.
Best Answer
Vertices and edges of truncated tetrahedron.
Similarly for the truncated dodecahedron. Every vertex has degree 3. There are many cycles length greater than 4, but none length 4.