So a friend of mine posed to me a question:
"How would you divide a square using three straight lines such that the average area of each piece is minimised"
Now the first observation I had was that if we have a square of area $A$, then the average area of the pieces is determined solely by the number of pieces $N$. Such that clearly we want to maximise $N$, minimising $\mu = \frac{A}{N}$. Thinking about straight lines intersecting it is trivially true that $N = 7$ is the maximum number of pieces implying the minimal average area $\mu = \frac{A}{7}$.
After that she adjusted the question to instead ask me to "minimise the maximum area of any of the pieces", i.e. make all of the pieces have the same area. Now I am really struggling to think of a principle to follow for figuring out whether it is possible to divide a square into 7 equal areas using 3 straight lines. And I am curious, is there a method to do this for $x$ pieces and $y$ lines?
Best Answer
This is not possible; in short, this is because the triangular region cut out by the three lines cannot be large enough. Without loss of generality, let our square have side length $1$.
In any configuration of three lines which divide the plane into seven regions, each line has three regions on one side and four regions on the other. If all of these regions are to have intersection of area $1/7$ with the square, then each line must cut the square into two pieces, one of area $3/7$ and one of area $4/7$. Consider the region inside the square which lies on the heavy side of all three lines. This is one of our seven regions, and must have area $1/7$. Furthermore, it must be a triangle; since all seven regions intersect the square, all three intersections between our lines must lie inside the square. Let these intersection points be $X$, $Y$, and $Z$.
We claim that the area $[OXY]$ of $OXY$ is at most $1/28$. Indeed, let $\ell$ be line $XY$, and let $\ell$ intersect the square at points $A$ and $B$, with $A$, $X$, $Y$, and $B$ on $\ell$ in that order. Let $A'$, $B'$, and $\ell'$ be the reflections of $A$, $B$, and $\ell$ about $O$. The quadrilateral $ABA'B'$ is a parallelogram entirely contained within the intersection of the square (since the square is convex) with the strip bounded by lines $\ell$ and $\ell'$. This intersection has area $1/7$, since $\ell$ and $\ell'$ cut the square into three pieces, two of which (by symmetry) have area $3/7$. So, $[ABA'B']\leq 1/7$. As $O$ is the center of the parallelogram, $[OAB]=[ABB'A']/4\leq 1/28$, and so $[OXY]\leq 1/28$ (since it has the same height and a smaller base $XY<AB$), as desired.
Now, $$[XYZ]\leq [OXY]+[OYZ]+[OZX]\leq \frac 1{28}+\frac 1{28}+\frac 1{28}=\frac{3}{28}<\frac 17,$$ a contradiction.
Note that this argument can be generalized to give a bound strictly smaller than $1/7$ for the minimum area of any piece: if all seven pieces have area at least $m$, then the "small side" of each line has area at least $3m$, and we have $$[OXY]\leq [OAB]\leq \frac{[ABA'B']}4\leq \frac{1-6m}{4},$$ which gives $m\leq 3(1-6m)/4$, or $m\leq 3/22$. This shouldn't be tight, because the inequality $[OXY]\leq [OAB]$ is only tight when $X$ and $Y$ are near the boundary of the square, in which case the regions cut out by two of the lines must be very small.