You and a buddy start with a rectangular cake, represented by rectangle $ABCD$ in Figure 1. Then you get to pick any point $E$ between $C$ and $D$, and the cake is sliced with two straight cuts, from $B$ to $E$ and $E$ to $A$. Then it’s your friend’s turn to pick a point $F$ between $B$ and $C$, so that the cake will be sliced from $D$ to $F$ and $F$ to $A$. The four straight slices intersect at points $G$, $H$, and $I$ as shown in Figure 1. You receive piece $AGHI$ of the cake, and the other player receives pieces $DEG$, $CEHF$, and $BFI$.

Determine all points $E$ you can choose on side $CD$ that will prevent your buddy from ending up with more cake (in all) than you do.

This problem originally appeared in the Prisoner’s Dilemma in the 2024 Spring issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

Show solution?

Solution.

Christopher West (PMP) and the Georgia Southern Eagle Problem Solvers got this one. The answer is that $E$ can be any point at all on $CD$, because no matter what points $E$ and $F$ are chosen, the purple and pink areas are equal. To see this, note that by the usual area formula for triangles, each of the triangles $AFD$ and $ABE$ has area half that of rectangle $ABCD$, so the sum of their areas is equal to that of $ABCD$. On the other hand, simply looking at the regions in Figure 1, the area of $ABCD$ is the sum of triangles $AFD$ and $ABE$, plus the regions of the rectangle that they miss (precisely the purple areas), minus the area of their overlap (since that’s counted twice in the sum), which is exactly the pink area. Since the sum of $AFD$ and $ABE$ is already exactly the area of the rectangle, the purple areas and the pink area must be equal.