If $P$ is a polytope that is both simplicial and simple, then $P$ is a simplex or an $n$-gon Question: Show that polytopes which are both simple and simplicial are either simplicies or two-dimensional $n$-gons.


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*A polytope $P$ is simple if every vertex is contained in exactly $d$ facets.

*A polytope $P$ is simplicial if every face is a simplex, that is, each facet of $P$ contains exactly $d$ vertices.


Attempt: The case for $d=2$ is immediate, so assume that $d\geq 3$. It suffices to show that if $P$ is not a simplex, then $P$ is an $n$-gon (i.e. $d$=2). Since $P$ is not a simplex, it contains at least $d+2$ vertices. At this point I want to use some sort of Pigeonhole principle argument to assert that at least one of the facets of $P$ contains at least $d+1$ vertices, contradicting that $P$ is simplicial. However, I am stuck on the precise argument to be used.
Would appreciate any help!

Found your question while searching for the answer myself and was able to solve it, so I thought sharing the proof might be useful, even though the question was asked quite a while ago.
I spent quite a lot of time thinking about your Pigeonhole-principle approach, which was rather confusing and brought me to a solution, bot not a really nice one. I think the following solution, which is due to Ziegler, is more intuitive.
Consider the following reasoning:
Suppose $P \subseteq \mathbb{R}^d$ is some $d$-dimensional polytope, which is both simple and simplicial. Suppose $d \geq 3$ and hence, $P$ is not a polygon. Let us look a some vertex $v$ of $P$. As $P$ is simple, we know that $v$ has exactly $d$ neighbors $v_1, \ldots, v_d$, which is equivalent to $v$ lying in exactly $d$ facets. $P$ is simplicial, and hence every facet is a $d-1$ simplex, containing exactly $d$ vertices. Note that the vertex-edge-graph of a simplex is the complete graph, every vertex is connected to every other. $v$ and every selection of $d-1$ of its neighbors form one of the $d$ many facets containing $v$ and hence, all vertices in the facets containing $v$ are connected to each other (using $d \geq 3$). In particular, all neighbors $v_i$ of $v$ have $v, v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_d$ as their $d$ neighbors.
But this means that there cannot be any other vertex $w$ and we conclude $vertP = \{v, v_1, \ldots, v_d\}$. Hence, $P$ is a simplex.
