59th Ukrainian National Mathematical Olympiad

Question

Suppose there are 2019 points A_1, A_2, , A_2019 in a circle, that form the regular 2019-gon A_1A_2 A_2019. Olesya and Andrew take turns, Olesya is the first to go. According to certain rules, Olesya makes a move to form an obtuse triangle, and Andrew – an acute triangle. The rules for forming triangles are as follows. At the beginning vertices A_1 and A_2 are supposed to be marked. In the first step Olesya marks an unmarked vertex A_i so that A_1A_2A_i is formed and vertex A_i becomes marked. Let in the next step Olesya (Andrew) marked an unmarked earlier vertex A_j and formed an obtuse (acute) A_kA_lA_j. Then in the next step Andrew (Olesya) can mark such an unmarked vertex A_m, for which for an acute (obtuse) triangle A_nA_jA_m is formed, where n=k or n=l. One loses if not able to foll…

Accepted Answer

Let's place the circle so that the diameter through the vertex

A_{1}

is vertical and this vertex itself is located below. Let's renumber all the vertices as shown on Fig. 44:

A_{i} = B_{i - 1}, i = 1, 1010

and

A_{i} = B_{1010 - i}, i = 1011, 2019

.

The strategy of Olesya is as follows: she always marks a vertex, that is symmetrical to the one Andrew marked with respect to the vertical diameter. In her first move she marks

B_{- 1}

. As we see, if vertex

B_{i}

is not marked, then the vertex

B_{- i}

is unmarked too and vice versa.

If Andrew is able to make a move, he marks some point

B_{- k}

and

Δ B_{m} B_{l} B_{- k}

(Fig. 44). Since this triangle is acute, then the center of the circle is inside this triangle. Therefore, we will have a diameter

B_{- k} O

, then vertices

B_{m}

and

B_{l}

lie on different sides of that diameter. One of them is located in one half-plane with a vertex

B_{k}

, let's say,

B_{m}

. Thus Olesya can mark vertex

B_{k}

and

Δ B_{m} B_{- k} B_{k}

, it is obtuse, because it doesn't have the circle center inside.

Thus, Olesya will always be able to answer Andrew's move, even if all the vertices are marked, then Andrew will not be the first to make a move and lose.

Fig. 44