Given a point $P(x,y)$ in the unit square, two polygons Blue and Green are defined by drawing a 45-degree line through $P$ and creating polygons with the top-left and bottom-right corners, respectively, as shown in the following figure:
Moving P around the square changes the areas of Blue and Green.
My question is: given two numbers $B\in [0,1]$ and $G \in [0,1]$ such that $b+g \in [0,1]$, is it always possible to locate the point $P$ such that the Blue area is $B$ and the Green area is $G$?
One way to prove this is to find explicit expressions for the areas, $b(x,y)$ and $g(x,y)$, then show that the equations $b(x,y)=B$ and $g(x,y)=G$ have a solution for every combination of $B$ and $G$.
Is there a simpler way to prove this, without using the explicit formulas?
The easiest thing that I can think of is to consider the white leftover rectange. Let $C>0$ be the area of this rectange, so $B+G+C=1$. We can think $C$ to be fixed and vary $B$ and $G$. Possible coordinates $(x,y)$ for the white rectangles corner $P$ are now $(x,C/x)$, where $x\in[C,1]$, because the area is $C=xy$. Because $b(x,y)$ is continuous, so is $b(x,C/x)$ as long as $x>0$ (this is the case when $C>0$, as we assumed in the beginning). Now when $x=C$, $b(C,1)=0$ because there is no blue area present as the point $(C,1)$ is at the upper edge of the unit square. Similarly when $x=1$, $b(1,C)=1-C$ as the point $(1,C)$ is at the right edge and blue area is at maximum value $(1-C)$. Because we know the function $b$ to be continuous and we know the values at two points we can use the intermediate value theorem. By the intermediate value theorem for all $B\in[0,1-C]$ there exists $x_0\in [C,1]$ so that $b(x_0,C/x_0)=B$. For the green area this leaves automatically $g(x_0,C/x_0)=1-C-b(x_0,C/x_0)=1-C-B=G$, so there exists always point $P$ on the curve $y=(1-B-G)/x$ that the green area is $G$ and the blue $B$.
This proof works when $B+G<1$. If $B+G=1$ it is fairly easy to see that the solution always exists and the point $P$ lies either on the left or the bottom side of the unit square.