The five sides and five diagonals of a regular pentagon are drawn on a piece of paper. Two people play a game, in which they take turns to colour one of these ten line segments. The first player colours line segments blue, while the second player colours line segments red. A player cannot colour a line segment that has already been coloured. A player wins if they are the first to create a triangle in their own colour, whose three vertices are also vertices of the regular pentagon. The game is declared a draw if all ten line segments have been coloured without a player winning. Determine whether the first player, the second player, or neither player can force a win.
The solution given is as follows:
We will prove that the first player can force a win. In other words, we will prove that no matter how the second player responds, the first player can win. Suppose that the vertices of the pentagon are $P_1, P_2, P_3, P_4, P_5.$ Let the first player start the game by colouring the line segment $P_1P_2.$ Without loss of generality, we may assume that the second player responds by colouring $P_2P_3$ or $P_3P_4.$
• Case 1: The second player responds by colouring $P_2P_3.$ The first player then colours $P_2P_4,$ forcing the second player to colour $P_1P_4$ to avoid losing. The first player then colours $P_2P_5$ and wins the game by colouring either $P_1P_5$ or $P_4P_5$ on their next move.
• Case 2: The second player responds by colouring $P_3P_4$.
The first player then colours $P_1P_5,$ forcing the second player to colour $P_2P_5$ to avoid losing.
The first player then colours $P_1P_3$ and wins the game by colouring either $P_2P_3$ or $P_3P_5$ on their next move.
I want to understand, why are they making the statement:
"Without loss of generality, we may assume that the second player responds by colouring $P_2P_3$ or $P_3P_4.$"
Then they are showing that under these two cases, the 1st player wins and they assert that this suffices their claim's validity. But I don't get why considering only these two cases, suffices. The 2nd player can color any one of the 9 sides remaining and not just these two.
Furthermore, in the case 1, there is a seemingly controversial statement, which says, "The first player then colours $P_2P_4,$". I dont understand why the first player has to color only this side when he/she can color any of the 8 remaining sides?
Finally, I want to ask, is there any strategy to solve these type of problems?
The main thing to realise in this question is that swapping any $2$ points does not change the question.
This is true because the focus is on the lines, not on the names of the lines. It should be clear that the labelling of the pentagon does not really matter. We can relabel the pentagon $P_1P_3P_5P_2P_4$ and the question stays the same. The players still chooses the same lines, just that the names of the lines are now different. Now, let's go over all the cases.
If the 2nd player chooses $P_1P_3$ (basically $1$ of the points on the line is amongst $P_1$, $P_2$), then we relabel the pentagon from $P_1P_2P_3P_4P_5$ to $P_2P_1P_3P_4P_5$ (swapping $P_1$ and $P_2$). $P_1P_3$ now becomes $P_2P_3$. The same reasoning holds true for $P_1P_4$ (swapping $P_1$ and $P_2$, $P_4$ and $P_3$) and $P_1P_5$ (swapping $P_1$ and $P_2$, $P_5$ and $P_3$). The same reasoning also holds true for $P_2P_3$, $P_2P_4$ and $P_2P_5$.
If the 2nd player chooses $P_3P_5$ (basically none of the points on the line is amongst $P_1$, $P_2$), then we relabel the pentagon from $P_1P_2P_3P_4P_5$ to $P_1P_2P_3P_5P_4$ (swapping $P_5$ and $P_4$). $P_3P_5$ now becomes $P_3P_4$. The same reasoning holds true for $P_3P_4$ and $P_4P_5$ (swapping $P_4$ and $P_3$, $P_5$ and $P_4$).
Now apply the same concept to your 2nd question.
For your 3rd question, the key to answering such question is