Minimization vs. Maximization of the area of a triangle inscribed in a triangle

Question

We’ve been given a triangle ABC with an area = $1$. Now Marcus gets to choose a point $X$ on the line $BC$, afterwards Ashley gets to choose a point $Y$ on line $CA$ and finally Marcus gets to choose a point $Z$ on line $AB$. They can choose every point on their given line (Marcus: $BC$; Ashley: $CA$; Marcus: $AB$) except of $A$, $B$ or $C$. Marcus tries to maximize the area of the new triangle $XYZ$ while Ashley wants to minimize the area of the new triangle. What is the final area of the triangle $XYZ$ if both people choose in the best possible way?

Answer 1

Your guess in the comment is (almost) correct. The first two points are the midpoints, and the last move makes no difference; the resulting area is $\frac14$.

To prove this, consider Ashley's move. If her move is further away from $AB$ than the first move, Marcus will make the last move arbitrarily close to $A$, whereas if it's closer to $AB$ than the first move, Marcus will make the last move arbitrarily close to $B$. In either case, Ashley would have been better off making her move at the same distance from $AB$ as the first move, so this is what she does. Then the last move makes no difference, and the area is given by

$$ \frac12\lambda h(1-\lambda)c $$

in terms of the length $c$ of $AB$, the height $h$ above $AB$ and the fraction $\lambda\in(0,1)$ at which the first move is placed along $BC$. The maximum is at $\lambda=\frac12$; the corresponding area is $\frac18hc$, and this is $\frac14$ of the area $\frac12hc$ of triangle $ABC$.