Problem 1293

Answer. No, it cannot.

Solution. We will prove the following simple lemma.

Lemma. If a p-gon is divided by non-intersecting diagonals into $(d+2)$-gons, the number of which is $t$, then $n=td+2$.

Proof. Induction on $t$; the base case for $t=1$ is obvious.

For the inductive step. Assuming that the statement is true for the number of $(d+2)$-gons equal to $1,2, \ldots, t-1$, consider the dissection of an $n$-gon $P$ into $t$ such polygons. Take any diagonal in the dissection. It divides our $n$-gon into two polygons $P_{1}$ and $P_{2}$, one of which is dissected into $s$, and the other into $r$ polygons, where $s+r=t$, and $s>0$. By the inductive hypothesis, the number of vertices of $P_{1}$ is $s d+2$, and the number of vertices of $P_{2}$ is $r d+2$. Therefore, the number of vertices of $P$ is $(s d+2) + (r d+2) - 2 = (s+r)d + 2 = td + 2$. This completes the inductive step.

Now, let's consider a polygon $Q$ with $n$ vertices, and suppose it is dissected by non-intersecting diagonals into $(d+2)$-gons. Let $A_{0}, A_{1}, \ldots, A_{d+1}$ be vertices of $Q$ such that the arcs $A_{0} A_{1}, A_{1} A_{2}, \ldots, A_{d} A_{d+1}$ are equal. Suppose there exists a $k \in \{1,2, \ldots, d-1\}$ such that the arc $A_{0} A_{k}$ is a side of one of the $(d+2)$-gons in the dissection. Let $B_{1}, B_{2}, \ldots, B_{m}$ be the vertices of $Q$ on the arc $A_{0} A_{k}$ (excluding the endpoints $A_{0}$ and $A_{k}$). If $m>0$, cut out the $(m+2)$-gon $A_{0} B_{1} B_{2} \ldots B_{m} A_{1}$ from the polygon $Q$. We see that this $(m+2)$-gon is dissected by non-intersecting diagonals into $(d+2)$-gons. According to the lemma, $m$ is divisible by $d$ (this is true even when $m=0$).

Similarly, we prove that the number of vertices of the polygon $Q$ inside each of the arcs (excluding the endpoints) $A_{1} A_{2}, \ldots, A_{k-1} A_{k}$ is divisible by $d$. Then the number of vertices of the polygon $Q$ inside (the smaller) arc $A_{0} A_{k}$ (excluding the endpoints $A_{0}$ and $A_{k}$) is $t d+k-1$ for some integer $t$. A similar count of the number of vertices of the polygon $Q$ lying inside (the smaller) arc $A_{k+1} A_{d+1}$ gives $s d+d-k-1$. Equating these, we get $t d+k-1=s d+d-k-1$, from which $2 k$ is divisible by $d$. Due to the oddness of $d$, we get that $k$ is divisible by $d$; this contradicts the condition $k \in\{1,2, \ldots, d-1\}$.

Remark. In the given solution, the contradiction arises from counting the number of vertices on the equal arcs $A_{0} A_{k}$ and $A_{k+1} A_{d+1}$ and equating these quantities.

This reasoning has variations; for example, a double count of the sum of angles in (equal) polygons cut off from $Q$ by segments $A_{0} A_{k}$ and $A_{k+1} A_{d+1}$ can be performed.

Comment. Only the lemma or an equivalent statement is formulated and proved - 0 points.

There is progress in the form of the idea of equating the number of vertices on the equal arcs $A_{0} A_{k}$ and $A_{k+1} A_{d+1}$ or equating the sums of angles in (equal) polygons cut off from $Q$ by segments $A_{0} A_{k}$ and $A_{k+1} A_{d+1}$ - 1 point.

In an otherwise correct solution, the proof of the lemma from the solution above is missing or incorrect (but the statement is present) - 1 point is deducted.