Saudi Arabian IMO Booklet

Let a minor intersection be a point in $S$ where exactly three rectangles meet and let the number of minor intersections be $j$. Let side segments be segments corresponding to a side of a rectangle in the partition and let proper segments be segments into which intersection points cut up maximal segments. Let the number of side segments be $s$ and the number of proper segments be $p$. If we start from maximal segments, we note that each addition of an intersection point forms two new segments. Ultimately, when all the intersection points are included, only proper segments remain and therefore $p=m+2i$. We now multiply all proper segments by 2 to account for both sides of a proper segment and subtract 4 to account for the fact that the sides of the rectangle $R$ (which are by definition proper segments) are counted only once. Thereafter, each addition of a minor intersection increases the number of segments by 1 until we get only side segments and therefore $s=2p+j-4$. Combining the two equations, we obtain $$s=2m+4i+j-4.$$ Now, counting rectangles by side segments we obtain $s=4r$ and counting rectangles by their angles we obtain $4r=4i+2j+4$, the final term accounting for the four corners of $R$. We transform the equation into $2r-6=2i+j-4$. Combining all the obtained equations we get $4r=s=2m+2i+2r-6$ which gives us $2r+6=2m+2i$, i.e. $m+i=r+3$. $\square$