#### Solution

Solution 1

We denote the vertices of the 25-gon by $A_1,A_2, … A_{24},A_{25}$. We call the index of the diagonal $A_i,A_j$ $($i $<$ j$)$ the number min {j – i, 25 – j + i}. The index of the diagonal $A_i,A_j$ passing through the point at which nine diagonals intersect can assume the values 9, 10, 11 and 12, since there are 8 diagonal ends between the points $A_i$ and $A_j$ on the contour of the 25-gon.

It is clear that the diagonals of one index are tangent to one circle.

Let nine diagonals of a regular 25-gon intersect at one point. Then among them there are at least three single indexes. This means that from a certain point three tangents to one circle can be drawn. And so, we have a contradiction.

Solution 2

Suppose that the n diagonals $A_1B_1, …, A_nB_n$ intersect at the point M. We can assume that the points $A_1, …, A_n, B_1, …, B_n$ are arranged in this order “clockwise”. Let us describe a circle around a 25-gon. Then the following sum of arcs

$(A_1A_2 + B_1B_2) + (A_2A_3 + B_2B_3) + … + (A_{n-1}A_n + B_{n-1}B_n) + (A_nB_1 + B_nA_1)$ is equal to 2φ. The value of each arc is a multiple of φ = 2π / 25, therefore each of the sums in parentheses equals at least 2φ.

Suppose that one of these sums $(for example, A_1A_2 + B_1B_2)$ is equal to $2φ.$ Then the chords $A_1A_2$ and $B_1B_2$ are equal, hence the quadrilateral $A_1A_2B_1B_2$ is an isosceles trapezoid. The perpendicular dropped from the centre $O$ of the circle to the base $A_1B_2$ of this trapezoid passes through the middle of its bases, and hence through the point M of the intersection of its diagonals. Moreover, the straight line l perpendicular to $OM$ and passing through the point $M$ is parallel to the bases of the trapezium and, therefore, intersects its lateral sides, and hence also the arcs $A_1A_2$ and $B_1B_2$. If one of the sums is equal to $2φ,$ then the same line l intersects two more pairs of the arcs that are under consideration, which is impossible. Consequently, all the sums in parentheses, except perhaps by one, are no less than $3φ.$

And so, we obtain the inequality $(n – 1) \times 3φ + 2φ \leq 2φ = 25φ,$ and hence n $\leq$ 26/3 <9.$