Counting squares... in rectangles.

Geometry Level 1

For an $a \times a$ grid of unit squares, the total number of squares of different sizes is given by:

$N_{a\times a} = \sum_{n=1}^a n^2$

For example, for $a=5$ , $N_{5\times 5} = 1+4+9+16+25 = 55$ . Now rather than looking at a perfect square grid, lets look at rectangular grids.

Find $N_{a\times b}$ , the total number of squares in a $a \times b$ grid, where $a > b$ .

\sum _{ n=1 }^{ b-1 }{ { (a+n })(b+n) }

\sum _{n=0}^{b-1}{ { (a-n })(b-n) }

\sum _{ n=1 }^{ b-1 }{ { (b-n) }^{ 2 } }

\sum _{ n=1 }^{ b-1 }{ { (a-n) }^{ 2 } }

2 solutions

Martin Taylor
Sep 2, 2020

Lets express the squares as the top left corner, for example for a 3 by 3 square it can be expressed as this. this helps as it simplifies the problem ... Showing where each square goes turns the problem from a more complex puzzle into a simple geometry question.

using this information we can work out that the formula is $\boxed{\sum_{n=1}^b-1(a -n)(b-n)}$

@Martin Taylor , the answer should be $\displaystyle \sum_\red{n=0}^\red{b-1}(a-n)(b-n)$ . See my solution. In your problem statement $1+4+9+16+25 = 55$ and not $\red{39}$ . Why use uppercase $\red A$ and $\red B$ in diagram and question but lowercase $a$ and $b$ in answer. I have done all the changes for you.

Chew-Seong Cheong - 9 months, 1 week ago

oh thanks for pointing that out, will fix

Martin Taylor - 9 months, 1 week ago

Chew-Seong Cheong
Sep 3, 2020

For an $a \times b$ grid, where $a > b$ , we note that the number of $1\times 1$ squares in the grid $n_1 = ab$ , the number of $2 \times 2$ squares in the grid $n_2 = (a-1)(b-1)$ , $n_3 = (a-2)(b-2)$ , ... $n_b = (a-b+1)(b-b+1)$ , Therefore

$\begin{aligned} N_{a \times b} & = n_1 + n_2 + n_3 + \cdots + n_b \\ & = ab + (a-1)(b-2) + (a-2)(b-2) + \cdots + (a-b+1)(b-b+1) \\ & = \boxed{\sum_{n=0}^{b-1} (a-n)(b-n)} \end{aligned}$

Counting squares... in rectangles.

2 solutions

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