Subfield: Combinatorics


Question: Let $n \geqslant 3$ be a positive integer. Find the maximum number of diagonals of a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other.

Solution:
We consider two cases according to the parity of $n$.

- Case 1. $n$ is odd.

We first claim that no pair of diagonals is perpendicular. Suppose $A, B, C, D$ are vertices where $A B$ and $C D$ are perpendicular, and let $E$ be the vertex lying on the perpendicular bisector of $A B$. Let $E^{\prime}$ be the opposite point of $E$ on the circumcircle of the regular polygon. Since $E C=E^{\prime} D$ and $C, D, E$ are vertices of the regular polygon, $E^{\prime}$ should also belong to the polygon. This contradicts the fact that a regular polygon with an odd number of vertices does not contain opposite points on the circumcircle.
![](https://cdn.mathpix.com/cropped/2023_12_21_9ee4f50247737d9408edg-1.jpg?height=504&width=1248&top_left_y=1062&top_left_x=491)

Therefore in the odd case we can only select diagonals which do not intersect. In the maximal case these diagonals should divide the regular $n$-gon into $n-2$ triangles, so we can select at most $n-3$ diagonals. This can be done, for example, by selecting all diagonals emanated from a particular vertex.

- Case 2. $n$ is even.

If there is no intersection, then the proof in the odd case works. Suppose there are two perpendicular diagonals selected. We consider the set $S$ of all selected diagonals parallel to one of them which intersect with some selected diagonals. Suppose $S$ contains $k$ diagonals and the number of distinct endpoints of the $k$ diagonals is $l$.

Firstly, consider the longest diagonal in one of the two directions in $S$. No other diagonal in $S$ can start from either endpoint of that diagonal, since otherwise it has to meet another longer diagonal in $S$. The same holds true for the other direction. Ignoring these two longest diagonals and their four endpoints, the remaining $k-2$ diagonals share $l-4$ endpoints where each endpoint can belong to at most two diagonals. This gives $2(l-4) \geqslant 2(k-2)$, so that $k \leqslant l-2$.



![](https://cdn.mathpix.com/cropped/2023_12_21_76b69bb4e7ba5ecafb62g-1.jpg?height=420&width=1432&top_left_y=294&top_left_x=298)

Consider a group of consecutive vertices of the regular $n$-gon so that each of the two outermost vertices is an endpoint of a diagonal in $S$, while the interior points are not. There are $l$ such groups. We label these groups $P_{1}, P_{2}, \ldots, P_{l}$ in this order. We claim that each selected diagonal outside $S$ must connect vertices of the same group $P_{i}$. Consider any diagonal $d$ joining vertices from distinct groups $P_{i}$ and $P_{j}$. Let $d_{1}$ and $d_{2}$ be two diagonals in $S$ each having one of the outermost points of $P_{i}$ as endpoint. Then $d$ must meet either $d_{1}, d_{2}$ or a diagonal in $S$ which is perpendicular to both $d_{1}$ and $d_{2}$. In any case $d$ should belong to $S$ by definition, which is a contradiction.

Within the same group $P_{i}$, there are no perpendicular diagonals since the vertices belong to the same side of a diameter of the circumcircle. Hence there can be at most $\left|P_{i}\right|-2$ selected diagonals within $P_{i}$, including the one joining the two outermost points of $P_{i}$ when $\left|P_{i}\right|>2$. Therefore, the maximum number of diagonals selected is

$$
\sum_{i=1}^{l}\left(\left|P_{i}\right|-2\right)+k=\sum_{i=1}^{l}\left|P_{i}\right|-2 l+k=(n+l)-2 l+k=n-l+k \leqslant n-2
$$

This upper bound can be attained as follows. We take any vertex $A$ and let $A^{\prime}$ be the vertex for which $A A^{\prime}$ is a diameter of the circumcircle. If we select all diagonals emanated from $A$ together with the diagonal $d^{\prime}$ joining the two neighbouring vertices of $A^{\prime}$, then the only pair of diagonals that meet each other is $A A^{\prime}$ and $d^{\prime}$, which are perpendicular to each other. In total we can take $n-2$ diagonals.

![](https://cdn.mathpix.com/cropped/2023_12_21_76b69bb4e7ba5ecafb62g-1.jpg?height=447&width=420&top_left_y=1804&top_left_x=798)
Instead of dealing separately with the case where $n$ is even, we shall prove by induction more generally that we can select at most $n-2$ diagonals for any cyclic $n$-gon with circumcircle $\Gamma$.



The base case $n=3$ is trivial since there is no diagonal at all. Suppose the upper bound holds for any cyclic polygon with fewer than $n$ sides. For a cyclic $n$-gon, if there is a selected diagonal which does not intersect any other selected diagonal, then this diagonal divides the $n$-gon into an $m$-gon and an $l$-gon (with $m+l=n+2$ ) so that each selected diagonal belongs to one of them. Without loss of generality, we may assume the $m$-gon lies on the same side of a diameter of $\Gamma$. Then no two selected diagonals of the $m$-gon can intersect, and hence we can select at most $m-3$ diagonals. Also, we can apply the inductive hypothesis to the $l$-gon. This shows the maximum number of selected diagonals is $(m-3)+(l-2)+1=n-2$.

It remains to consider the case when all selected diagonals meet at least one other selected diagonal. Consider a pair of selected perpendicular diagonals $d_{1}, d_{2}$. They divide the circumference of $\Gamma$ into four arcs, each of which lies on the same side of a diameter of $\Gamma$. If there are two selected diagonals intersecting each other and neither is parallel to $d_{1}$ or $d_{2}$, then their endpoints must belong to the same arc determined by $d_{1}, d_{2}$, and hence they cannot be perpendicular. This violates the condition, and hence all selected diagonals must have the same direction as one of $d_{1}, d_{2}$.
![](https://cdn.mathpix.com/cropped/2023_12_21_02deb1202c246c7b11b9g-1.jpg?height=422&width=1220&top_left_y=1046&top_left_x=496)

Take the longest selected diagonal in one of the two directions. We argue that its endpoints do not belong to any other selected diagonal. The same holds true for the longest diagonal in the other direction. Apart from these four endpoints, each of the remaining $n-4$ vertices can belong to at most two selected diagonals. Thus we can select at most $\frac{1}{2}(2(n-4)+4)=n-2$ diagonals. Then the proof follows by induction.

Final answer: $n-2$ if $n$ is even and $n-3$ if $n$ is odd