Math competition problem involving ratios of areas. What's a good way to attack similar problems?

Question

9 lines each separately partition a square into two quadrilaterals with areas having the ratio 2: 3. Show that 3 of these lines intersect at the same point.

Any answer or hints is greatly appreciated, as I'm very puzzled by this problem.

Clarification

I have only tried drawing a few drawings and examples of the problem, since when I tried to read it over again, I guess I don't fully understand it either.

I translated it from my main language, but the way I understand it, you have a square, and draw 9 lines from some side to a side. Now there's going to a be lots of different shapes. And there's two quadrilaterls with areas ratio 2:3, and out from that it should be proven, that 3 lines intersect, atleast that's what I think.

Answer 1

Suppose the square has the coordinates $(0,0)$, $(0,10)$, $(10,10)$, $(10,10)$.

Consider the four points $(4,5)$, $(6,5)$, $(5,4)$, $(5,6)$. Prove that each line must go through one of those points. Then apply the pigeonhole principle.

Answer 2

Consider a unit square $ABCD$: to divide it into two quadrilaterals you need a line such as $EF$, connecting one side with the opposite side. A quick computation shows that the area of trapezoid $ABFE$ is $2/3$ of the area of trapezoid $EFCD$ if $AE+BF=4/5$. And you can soon realise that all such lines pass through point $M$ whose distances from $AB$ and $BC$ are $2/5$ and $1/2$.

Of course we could as well impose that the area of trapezoid $EFCD$ is $2/3$ of the area of trapezoid $ABFE$, in which case all lines would pass through point $M'$, symmetric of $M$ with respect to the center of the square. And of course we considered lines cutting $AD$ and $BC$, but they could also cut the other pair of opposite sides, namely $AB$ and $CD$, passing then through other two points $N$ and $N'$.

Now you have $9$ lines which must pass through one among four points: apply the pigeonhole principle to prove that at least three of them pass through the same point.

Answer 3

You ask : "What is a good way to attack such problems" ?

Instead of giving a solution that would not differ much from the solutions given by @Aretino or @Jaap Scherphuis, I will stress a feature that is often useful in these issues. I will call it pompously "the principle of area balance". Take a look at the following picture.

Fig. 1 : Representation of area loss and area gain when quadrilateral $ABFE$ is transformed into $ABF'E'$.

Imagine you have already found a solution, i.e., line $EF$ (such that area($ABFE$)=$\frac12$ area($CDEF$)). Then, it suffices to move upwards $E$ into $E'$ and downwards $F$ into $F'$ from the same distance to have another solution. Why ? If $M$ denotes the midpoint of line segment $[EF]$, triangles $MEE'$ and $MFF'$ are opposite images one of the other, thus have the same area. In this way, there is a perfect balance between area loss and area gain.

Thus, out of a solution, one can generate an infinity of solutions...

Besides, this explains the central (!) rôle of pivoting point $M$ ; then, as this property (dividing the area of the square in a ratio 2:1) is invariant by $\tfrac{\pi}{2}$ rotations around the centre of the square, these rotations generate 3 avatars of $M$...

Connected : https://scholarworks.umt.edu/cgi/viewcontent.cgi?article=1437&context=tme