Berrov s.l.

In a convex $n$-gon, several diagonals are drawn. A drawn diagonal is called good if it intersects (at interior points) with exactly one of the other drawn diagonals. Find the maximum possible number of good diagonals.
We will prove by induction on $n$ that the number of good diagonals does not exceed $n-2$ if $n$ is even, and $n-3$ if $n$ is odd. We will consider a segment as a 2-gon without diagonals. The base cases ($n=2,3$) are obvious.

Inductive step. Let $n \geq 4$; denote our polygon by $P=A_{1} A_{1} \ldots A_{n}$. If no two good diagonals intersect, then their number does not exceed $n-3$ (see problem 35139).

Now suppose there exist two intersecting good diagonals $A_{i} A_{k}$ and $A_{j} A_{l} (i<j<k<l)$. Then each of them does not intersect with any other drawn diagonals. Remove $A_{i} A_{k}$ and $A_{j} A_{l}$ from consideration. Each remaining drawn diagonal $d$ is a diagonal or side of exactly one of the polygons $Q_{1} = A_{i} \ldots A_{j}, Q_{2} = A_{j} \ldots A_{k}, Q_{3} = A_{k} \ldots A_{l}$, or $Q_{4} = A_{l} \ldots A_{n} A_{1} \ldots A_{i}$ (see the left figure). If $d$ is a side of one of them, then it cannot intersect with other diagonals (and therefore is not good).
![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-39.jpg?height=484&width=1310&top_left_y=-2&top_left_x=378)

Let $n$ be even. By the inductive hypothesis, among all diagonals that fall into some polygon $Q_{s}$, the number of good ones does not exceed the number of vertices in it, decreased by 2. Therefore, the total number of good diagonals in $P$ does not exceed

$2 + (j-i-1) + (k-j-1) + (l-k-1) + (n-l+i-1) = n-2. \quad(*)$

For odd $n$, the sum of the number of vertices of polygons $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ is the odd number $n+4$; hence, the number of vertices in one of them is odd. The corresponding term in the sum (*) will decrease by 1, and the total number of good diagonals does not exceed $n-3$.

It remains to provide examples showing the sharpness of the estimate. For even $n$, it is sufficient to draw in the polygon $A_{1} \ldots A_{n}$ the diagonals $A_{i} A_{n-i}$ and $A_{i+1} A_{n-i+1}$ for all $i=1,2, \ldots, n / 2-1$ (see the right figure); they are divided into pairs of intersecting good diagonals. For odd $n$, it is sufficient to draw such diagonals in the even-sided polygon $A_{1} \ldots A_{n-1}$.

## Answer

$n-2$ for even $n$, $n-3$ for odd $n$.