Help! What is the greatest number of white squares you can place so that each covers part of the blue square without overlapping one another?

Question

You have a large pile of squares that each have a side length of 1 inch. One square is blue, while all the other squares are white. You want to arrange several white squares so they cover part of the blue square but don’t overlap with each other.

What is the greatest number of white squares you can place so that each covers part of the blue square without overlapping one another? (The entire blue square does not have to be covered, while the blue area that each white square covers must be nonzero.)

Answer 1

The correct answer is seven. The first white square is placed on the top of the blue square, so the sides of it will make 45 degree angle and the intersection of the two squares is an octagon.

We place two squares on opposite sides of the white square, so we form a band of three square that will cover central part of the square.

Each of the remaining two angles can be covered by two squares. 2 + 3 + 2 = 7.

Any questions?

Answer 2

Sava already gave a great answer, but I think a picture would help:

https://www.desmos.com/calculator/nc1h8ywvud

This shows that at least seven white squares are possible. I do not yet have a proof that this is the most possible, but I will update this answer if I come up with one.

Answer 3

What is the area of the blue square?

Answer 4

Since the white blocks are squares all four angles are 90°.

360° / 90° = 4

4 white are the most that can be placed