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Let's first look at a general solution for how to find a rectangle with all vertices of the same color.

Assume that our plane looks like this:

Here I have made the colors stronger than what's said in the problem text, just so that we can see them. The coloring of the plane is also done in a very non-complex way, but the method I will present will work regardless of how complex the coloring is made.

First draw 3 horizontal lines and let them be crossed by 9 vertical lines.

For each of the 9 vertical lines we now have 3 intersection points. Now, notice that there are only 2³ = 8 ways to color 3 points red or blue (red, red, red  or  red, red, blue  or  red, blue, red  etc.), and since we have 9 vertical lines, at least 2 of those lines must have intersection points colored the same way (by the pigeonhole principle).

In the following picture we have marked the intersection points of 2 such vertical lines, notice that both triplets have the color combination (blue, red, red) counted from the top and down.

Now we will use the pigeonhole principle one more time. Since a triplet has 1 more point than the number of colors we use (which is 2, red and blue), at least 2 of the points of the triplet must be of the same color.

In our case 2 of the points are red in each triplet, as we have clarified in the following picture.

This means we have found a rectangle with all 4 vertices of the same color, and we can fill it with black.

The worst case scenario would be that only one of the small rectangles could be filled with black, as exemplified in the following picture.

So here 15/16 of the original big rectangle remain white.

Still, after that, we can treat all the remaining small "white" rectangles as new cases.

And in the same way we can find a way to construct black rectangles in them too, which in the worst case could look something like this.

Here 15/16 remain white of the 15/16 that remained white in the first step.

We can continue like this in as many steps as we like. If we do n steps, the part that remains white in the end will be (15/16)ⁿ, and as n approaches infinity the limit value for this expression is 0.

In other words, we can force the remaining white area as close to zero as we like, or in other words we can make the entire rectangle as black as we like, like this.

So, now the rest is simple. We have shown than any rectangle we choose can be made as black as we like. So we just have to select a suitable pixle size of the photo, and, if we like, we can divide those pixles into tiny parts and fill a percentage of them black, in order to give each of our pixles any shade of gray, from white to black.