## Problem I-2

Let $n \geq 3$ be an integer. We say that a vertex $A_{i}(1 \leq i \leq n)$ of a convex polygon $A_{1} A_{2} \ldots A_{n}$ is Bohemian if its reflection with respect to the midpoint of the segment $A_{i-1} A_{i+1}$ (with $A_{0}=A_{n}$ and $A_{n+1}=A_{1}$ ) lies inside or on the boundary of the polygon $A_{1} A_{2} \ldots A_{n}$. Determine the smallest possible number of Bohemian vertices a convex $n$-gon can have (depending on $n$ ).

(A convex polygon $A_{1} A_{2} \ldots A_{n}$ has $n$ vertices with all inner angles smaller than $180^{\circ}$.)

(proposed by Dominik Burek, Poland)

Answer. $n-3$.

In the following we write for short 'reflection of $A$ in $P$ ' instead of 'reflection of the vertex $A$ with respect to the midpoint of the segment connecting the two neigbouring vertices of $A$ in the polygon $P$ ?


## Solution.

Lemma. If $A B C D$ is a convex quadrilateral with $\angle B A D+\angle C B A \geq \pi$ and $\angle B A D+\angle A D C \geq$ $\pi$ then $A$ is a Bohemian vertex of $A B C D$.

Proof. Let $E$ be the reflection of $A$ in $A B C D$. It is clearly seen that $E$ belongs to the halfplanes containing $C$ determined by lines $A B$ and $A D$. Since $\angle B A D+\angle C B A \geq \pi$ and $\angle B A D+$ $\angle E B A=\pi$, point $E$ belongs to the (closed) halfplane containing points $A, D$ determined by the line $B C$. Analogously, using the assumption $\angle B A D+\angle A D C$ we infer that $E$ belongs to the closed halfplane containing points $A, B$ determined by the line $C D$.

![](https://cdn.mathpix.com/cropped/2024_06_05_7e99b1556acc0291915eg-3.jpg?height=499&width=713&top_left_y=2189&top_left_x=680)

Therefore $E$ lies inside or on the boundary of $A B C D$. Thus $A$ is Bohemian.

Consider a convex $n$-gon $A_{1} A_{2} \ldots A_{n}$. Choose any four vertices $A_{i}, A_{j}, A_{k}, A_{l}$ with $i<j<k<l$ as in the picture below. Consider quadrilateral $A_{i} A_{j} A_{k} A_{l}$. It is clear that one of the points $A_{i}, A_{j}, A_{k}, A_{l}$ satisfies assumption of the lemma, let's say this point is $A_{i}$. We claim that $A_{i}$ satisfies the assumption of the lemma in quadrilateral $A_{i-1} A_{i} A_{i+1} A_{k}$. Observe that the point $X:=A_{k} A_{i+1} \cap A_{i} A_{i-1}$ lies in the triangle bounded by lines $A_{k} A_{j}, A_{j} A_{i}$ and $A_{i} A_{l}$. So

$$
\angle A_{k} A_{i+1} A_{i}+\angle A_{i+1} A_{i} A_{i-1}=\pi+\angle A_{k} X A_{i} \geq \pi
$$

(Note: it may happen that $X$ does not exist. It happens iff $j=i+1, l=i-1$ and $A_{k} A_{j} \| A_{l} A_{i}$. In that case $\angle A_{k} A_{i+1} A_{i}+\angle A_{i+1} A_{i} A_{i-1}=\pi$.)

Analogously $\angle A_{i+1} A_{i} A_{i-1}+\angle A_{i} A_{i-1} A_{k} \geq \pi$. Using lemma we conclude that $A_{i}$ is a Bohemian vertex of quadrilateral $A_{i-1} A_{i} A_{i+1} A_{k}$. This implies that $A_{i}$ is a Bohemian vertex of $A_{1} A_{2} \ldots A_{n}$ since the quadrilateral $A_{i-1} A_{i} A_{i+1} A_{k}$ is a subset of the $n$-gon and the reflexion point is the same.

![](https://cdn.mathpix.com/cropped/2024_06_05_7e99b1556acc0291915eg-4.jpg?height=786&width=916&top_left_y=1069&top_left_x=570)

Therefore, amongst any four vertices of a convex $n$-gon there exists a Bohemian vertex. So, every $n$-gon has at least $n-3$ Bohemian vertices.

An example of a convex $n$-gon with exactly $n-3$ Bohemian vertices is the following: take any kite $A_{1} A_{2} A_{3} A_{4}$ with $A_{4} A_{1}=A_{1} A_{2}<A_{2} A_{3}=A_{3} A_{4}$ and place points $A_{5}, \ldots, A_{n}$ very close to $A_{1}$. Then $A_{2}, A_{3}, A_{4}$ are not Bohemian vertices of $A_{1} A_{2} \ldots A_{n}$.
