Kilua and Ndoti play the following game in a square $ABCD$: Kilua chooses one of the sides of the square and draws a point $X$ at this side. Ndoti chooses one of the other three sides and draws a point Y. Kilua chooses another side that hasn't been chosen and draws a point Z. Finally, Ndoti chooses the last side that hasn't been chosen yet and draws a point W. Each one of the players can draw his point at a vertex of $ABCD$, but they have to choose the side of the square that is going to be used to do that. For example, if Kilua chooses $AB$, he can draws $X$ at the point $B$ and it doesn't impede Ndoti of choosing $BC$. A vertex cannot de chosen twice. Kilua wins if the area of the convex quadrilateral formed by $X$, $Y$, $Z$, and $W$ is greater or equal than a half of the area of $ABCD$. Otherwise, Ndoti wins. Which player has a winning strategy? How can he play?
Problem
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Tags: geometry
29.12.2014 15:49
Ndoti wins. If Kilua chooses a side and a vertex of that side as $X$, Ndoti chooses the other side incident with $X$ and $Y\equiv X$. Then $XYZW$ is a triangle inscribed in the square, of area at most half of that of the square. So say Kilua chooses side $AB$ and point $X \in (A,B)$. Now Ndoti chooses side $DA$ and $Y\equiv A$. If Kilua next chooses side $BC$, then Ndoti may choose side $CD$ and $Z\equiv C$, thus the quadrilateral $XYZW$ is contained in the triangle $ABC$. Thus Kilua must choose side $CD$, but then Ndoti may choose side $BC$ and $Z\equiv B$, thus $XYZW$ is again a triangle inscribed in the square.
29.12.2014 16:22
Hey, I didn't mention, but Ndoti and Kilua cannot choose the same vertex as their points. For example, if Kilua chooses the side $AB$ and $X \equiv B$, Ndoti cannot choose the side $BC$ and $Y \equiv B$, but he can choose any other point at this side.
29.12.2014 17:08
Then Kilua trivially wins. First she chooses side $AB$ and $X\equiv A$. If Ndoti chooses side $CD$ or $DA$, next Kilua chooses side $BC$ and $Z\equiv B$, otherwise next Kilua chooses side $DA$ and $Z\equiv D$. Now the quadrilateral $XYZW$ has a common side with the square, a vertex on the opposite side, and is not degenerated to a triangle, hence its area is strictly larger than half of that of the square.