# There is a space for a circle in a rectangle filled with squares

I placed 120 unit squares inside a $$20\times 25$$ rectangle. Prove that it is possible to place a circle with an unit diameter (with a diameter witj length 1), such that it doesn’t have a common point with any of the squares!

I love combinatorical problems, but I couldn’t solve it. I don’t know how to start. Maybe coloring? I am sure it uses pigeonhole/principle.

• A circle with unit diameter is contained inside an unit square. Then a $20\times 25$ rectangle could contain up to $500$ unit square then after you've added all the $120$ squares remain certainly space for another square (and then for a circle). – Jihlbert Nov 21 '18 at 22:11
• Are the vertices of the squares required to have integer coordinates? (I suspect that it's true without this requirement, but that it will be quite a bit trickier to prove.) – Rob Arthan Nov 21 '18 at 22:12

First note that the center of the circle needs to be at least 0.5 units away from the edge of the rectangle, so we're working with an area of $$19*24 = 456$$ square units.
If you place the center of the circle anywhere in this region, the center of the circle will be within 0.5 units from some point in the unit square, which means the circle will overlap the square. So we can't put the center of the circle in any of these points. The area of this set of points is $$3+\frac{\pi}{4}\approx3.785$$ units (One unit square, four rectangles, four quarter-circles). We can surround each unit square with a set of disallowed points like this for a total area of $$120 * (3+\frac{\pi}{3}) \approx 454$$ units. But the area we started with was $$456$$ units, so even if we are able to optimally spread out the unit squares so that no two of the forcefields overlap, there will still be a small area left over in which we can put the center of the circle.