TITLE: Each $2\times 5$ rectangle contains $1\times 3$ rectangle
QUESTION [5 upvotes]: A $60\times 60$ board is partitioned into rectangles of size $2\times 5$ (or $5\times 2$). Is it true that there always exist another partition into rectangles of size $1\times 3$ (or $3\times 1$) such that any $2\times 5$ (or $5\times 2$) rectangle contains a $1\times 3$ (or $3\times 1)$ rectangle?
For the "simple" partition into $2\times 5$ rectangles, this is certainly true: use the simple partition into $1\times 3$ rectangle.
Another way to partition is to first partition the $60\times 60$ board into $12\times 10$ boards, and then for each such board, put two $2\times 5$ at the top and ten $5\times 2$ below. It is not hard to tile each such $12\times 10$ board with $1\times 3$ (or $3\times 1)$ rectangles so that each $2\times 5$ (or $5\times 2$) contains a $1\times 3$ (or $3\times 1)$ rectangle.
(I'm assuming that an $m\times n$ rectangle has $m$ rows and $n$ columns.)

REPLY [6 votes]: We can even get every 2×5 to contain two 1×3s:
After tiling your board with 2×5 rectangles, divide it into squares of 3×3 each, and then subdivide each of these squares into three 1×3 strips either horizontally or vertically such that if possible one of the strips lies completely within a 2×5 rectangle. (This can't be possible both horizontally and vertically; if neither direction works then just choose one arbitrarily).
Now, consider one of the 2×5 rectangles; let's say wlog that it is lying horizontally. No matter how it is placed with respect to the 3×3 grid, there will be 3 of its 5 columns that make up one column of the 3×3 grid. One or two 3×3 squares in this column will overlap our 2×5 rectangle, and due to the above principle, it or they will have been divided horizontally into strips.