# Number of rectangles with odd area.

We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?

I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.

The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.

What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.

• But where was I wrong? – Omer Jun 14 '18 at 19:29
• @Omer: In counting the number of possible starting vertices (121). Not all of them are valid for all of the sub-cases you count. So, you go wrong at the very first step. Compare your thinking to how Hagen von Eitzen counted the cases, to correct the error. – Nominal Animal Jun 14 '18 at 19:55
• "and than assume that each possible odd width and height could be realized with this vertex." - no, s/he just assumed that 5 different odd x pairs and 5 different odd y pairs included the x and y of this vertex. If you start at (0,0) then in the x direction you can go to (1,0), (3,0), (5,0), (7,0), (9,0) but if you start at (5,0) you can go to (0,0), (2,0), (4,0), (6,0), (8,0), (10,0), that's 6 choices! – immibis Jun 15 '18 at 3:55

The reasoning is ok, except for this:

There are 5 possibilities to each

Sadly, no. If the chosen vertex is at $(1,1)$ (grid starting at $(0,0)$) you have 6 possibilities for each direction.

In general, you have 6 possibilities if the coordinate is odd, 5 if it's even.

Fix: Because there $36$ all-even-coordinates vertices, $25$ all-odd, and $121-36-25=60$ mixed vertices, the correct counting is

$$36 \times 5^2+ 25 \times 6^2 + 60 \times5 \times 6=3600$$

Dividing by $4$ you get the $900$ rectangles.

I assume you're looking for rectangles with sides parallel to those of the square and vertices that are lattice points. A rectangle with integer sides has odd area iff the sides are both odd. For each pair of odd integers $x, y$ with $1 \le x,y \le 9$, there are $(11-x)(11-y)$ possible positions for a rectangle of size $x \times y$. Thus the number of rectangles is $$\sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$