Solution

a) The second player can use a symmetric strategy as follows: every turn draw a line which is centrally symmetric to the line drawn by the first player in the previous turn. Using this strategy, the second player will always be able to make a move and so cannot lose.
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b) In this case the first player can use a symmetric strategy: first draw a line connecting vertices opposite each other, and then for the rest of the turns use the strategy described in b).
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1. In a) the first player couldn’t have acted in this way since vertices opposite each other have different colours.
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2. In b) instead of the central symmetry one can use the axial symmetry (with respect to the first diagonal line drawn).

Answer

a) The second player. b) The first player.

Hint

Solution

Solution 1

We denote the 25-gon by $A_1A_2 … A_{24}A_{25}$. We call the index of the diagonal $A_iA_j$ $($i $<$ j$)$ the number min {j - i, 25 - j + i}. The index of the diagonal $A_iA_j$ 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.

Answer

6 points

Hint

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Solution

We consider all segments with endpoints at the marked points. Since no three of the marked 64 points lie on a single straight line, there will be $64\times63$/2 = 2016 $>$ 1981 such segments. From problem 116573 it follows that among these segments there will be at least one pair of parallel points. A quadrilateral with vertices at the ends of these segments is the desired one $($it cannot be a parallelogram, since there is no parallelogram with vertices at the vertices of the regular 1981-gon$)$.

Hint

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Hint

In a regular pentagon any group of three vertices describes an equilateral triangle.

Solution

Consider a regular pentagon that just fits into the circle. Of its 5 vertices, at least 3 will be the same colour. These three vertices represent the three points that can form an equilateral triangle with vertices of the same colour.

Answer

Comparable to problem 35643.

Hint

In a regular pentagon any group of three vertices describes an equilateral triangle.

Hint

If three vertices of a regular pentagon are painted, then they lie on the vertices of an isosceles triangle.

Solution

Let’s split all of the vertices of the regular 5000-gon into 1000 groups of five vertices, located on 999 consecutive vertices. The points in each group are the vertices of a regular pentagon. There are 2001 = $2 \times 1000$ + 1 painted vertices, therefore, at least three vertices are coloured in a certain group. But every three vertices of a regular pentagon lie at the vertices of an isosceles triangle.

Answer

Compare with problem number 35758.

Hint

If three vertices of a regular pentagon are painted, then they lie on the vertices of an isosceles triangle.

Solution

Split the regular hexagon into 6 equilateral triangles of side 1. Then at least one of the triangles will contain two points. The distance between them will not be greater than the side length of the triangle.

Hint

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Solution

The total number of nodes is $2 \times (6 + 7 + 8 + 9 + 10)$ = 91. Each node, with the exception of the central node, belongs to one of 11 concentric circles centred at the centre of the hexagon $($see the figure$)$.

Suppose that there are no five marked nodes lying on the same circumference. Then the total number of marked nodes is not more than $11 \times 4$ + 1 = 45, that is, not more than half. This is a contradiction.

Hint

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