3 points in a unit square

The solution involves clumping the dots together in such a way that they create an "L" shape, but they are arbitrarily close together.

That is, the points are of the form:

• $(x,y)$
• $(x+\delta,y)$
• $(x,y+\delta)$

Where $\delta \rightarrow 0$

Consider this picture:

The three points you choose are shown in red (size and distance between them exaggerated for clarity).The maximum sized rectangles your friend can then draw are depicted in this figure, in blue, green, and orange, one for each point. The brown region is where the orange and blue rectangles overlap.

Note: The optimal solution requires minimizing the area of this overlap and distributing the area equally between the three rectangles.

Here is my stab at a "proof"...

Finding the optimal solution consists of three steps:

• 1) Showing that at least two of the points need to share the same $x$ coordinates and two need to share the same $y$ coordinate.
• 2) Showing that the points must be as close together as possible without touching.
• 3) Finding the coordinate $(x,y)$ of the three clustered points.

So, for (1) WLOG, lets assume that none of the three share the same $x$ coordinate. Then one could draw a line segment at $x=\frac{1}{2}$ . If one point happens to lie on the line segment, then choose a location just to the left or right of that point. Now, if you have three dots on one side, extend the line toward the dots until it contains one point. If you have two dots on one side and one dot on the other side, then draw the rectangle around the one dot. Either way, you have a rectangle of at least $\frac{1}{2}$ which we will show later is greater than our final answer. Therefore, we can assume (1) to be correct (as soon as we show that a lower bound can be achieved).

For (2), lets assume that the points aren't "clustered arbitrarily close together", then the distance between the two that aren't "close together" will only give the second player a degree of freedom that is removed if the two dots are moved arbitrarily close to the single dot (or vice versa)

So, if we assume (1) and (2) to be true, then we are left with a cluster of three dots in an "L" shape and need only to determine x and y in the above equations. Lets assume $\delta \rightarrow 0$ but remains positive in the above equations. Then, your friend has three choices of rectangles:

• Containing the first dot : Just above and left of the "cluster", just containing the first dot in the lower right corner. Area $= xy$ .
• Containing the second dot : Just to the right of dots one and three, extending to the top and bottom and far right of the square. Area $= 1-x$ .
• Containing the third dot : Just below dots one and two, extending to the left, right and bottom of the square. Area $= 1-y$ .

Now the optimal solution is obtained when the three above three areas are equal, since a purturbation from that position would allow your friend to choose a dot which increases the area of the final rectangle.

So,

$y = x = (1-x)^2$

Solving:

$(1-x)^2 = x$

$x^2 - 2x + 1 = x$

$x^2 -3x +1 = 0$

$x = \frac{3 \pm \sqrt{(9-4)}}{2}$

$x = \frac{3 - \sqrt{5}}{2}$ (since $x<1$ )

So, Area $= x = \boxed{0.382}$