We have a regular polygon $P$ with 2019 vertices, and in each vertex there is a coin. Two players Azul and Rojo take turns alternately, beginning with Azul, in the following way: first, Azul chooses a triangle with vertices in $P$ and colors its interior with blue, then Rojo selects a triangle with vertices in $P$ and colors its interior with red, so that the triangles formed in each move don't intersect internally the previous colored triangles. They continue playing until it's not possible to choose another triangle to be colored. Then, a player wins the coin of a vertex if he colored the greater quantity of triangles incident to that vertex (if the quantities of triangles colored with blue or red incident to the vertex are the same, then no one wins that coin and the coin is deleted). The player with the greater quantity of coins wins the game. Find a winning strategy for one of the players. Note. Two triangles can share vertices or sides.
Problem
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Tags: combinatorics, game strategy
20.06.2019 09:48
Cool Problem. Claim: Azul player wins!!! Azul starts by picking triangle $A_1A_{1010}A_{1011}$, the points are split into two sets $S=\{A_2,A_3,\dots A_{1009}\}$ and $\overline{S}=\{A_{1012},A_{1013},\dots A_{2019}\}$. It is clear that any other triangle must have all its vertices in one of these two sets. On the successive rounds Azul just mimic the previous triangle chosen by Rojo in the complementary set. It is straightforward to show that at the end : 1- Azul and Rojo are even on $S\cup \overline{S}$. 2- Rojo can get at most one of ${A_{1010},A_{1011}}$. 3- $A_1$ is for Azul. So our claim is true.
28.06.2019 01:30
This same strategy works for all odd $n$, but if $n=2m$ and Azul picks $A_1A_{n-1}A_{n+1}$ as works in this very similar problem, Rojo can instead force a tie. For $n=4$ the game ends in a tie regardless. I wonder what happens for even $n>4$...
08.10.2020 04:42
Reminds me of 2019 ELMO SL C1 and USAMO 2008/4. We claim that Azul always wins. Label polygon $P$ with verticies $A_1, A_2, \dots A_{2019}$. First, have Azul pick triangle $A_1A_{1010}A_{1011}$, so the points are split into two separate polygons. Now, each triangle must have all of its vertices on one of the two polygons. Now, have Azul do strategy stealing - each time Rojo does something, Azul just has to do the same thing on the opposite polygon. At the end, Azul and Rojo are have equal amounts of triangles on $A_2, A_3, \dots A_{2019}$, but Azul wins $A_1$, so Azul wins.