Coloring problem’s

I recently came across Sperner’s Lemma. Again. Sperner’s lemma is a combinatorial result which can prove some pretty strong facts, as Brouwer fixed point theorem. A fast description goes like this: Take a triangle RGB (R red, G green, B blue) labelled counterclockwise, and subdivide it into smaller triangles in whatever way you like (see the picture below). Then label all the new vertices as follows:

• vertices along RG may be labelled either R or G, but not B
• vertices along BC may be labelled either G or B, but not R
• vertices along CA may be labelled either B or R, but not G
• vertices inside triangle ABC may be labelled R or G or B.

Now shade in every small triangle in the subdivision that has three different labels.
Use two different shadings to distinguish the triangles which have been labelled counterclockwise (i.e. in the same sense as triangle RGB) from the triangles which have been labelled clockwise (i.e. in the sense opposite to that of as triangle RGB).

Then there will be exactly one more counterclockwise triangle than clockwise triangles. In particular, the number of shaded triangles will be odd.

For a nice proof and more details take a look here

So two problems I tried that have to do with coloring and combinatorics:

Problem 1: (All Russian Math Olympiad 1980) Some unit squares in an infinite sheet of squared paper are colored red so that every 2 x 3 and 3 x 2 rectangle contains exactly two red squares. How many red squares are there in a 9 x 11 rectangle?

Problem 2: (JBO 2004)  Consider a convex polygon having vertices, We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.