8
$\begingroup$

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

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ But where was I wrong? $\endgroup$ – Omer Jun 14 '18 at 19:29
  • 2
    $\begingroup$ @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. $\endgroup$ – Nominal Animal Jun 14 '18 at 19:55
  • $\begingroup$ "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! $\endgroup$ – immibis Jun 15 '18 at 3:55
7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $$ \sum_{x \in \{1,3,5,7,9\}} \sum_{y \in \{1,3,5,7,9\}} (11-x)(11-y) = 900$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.