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# How to Count Squares!

Let me go grab a hamburger real quick...

Ok, I'm back.

How many squares are there in the 6$$\times$$4 grid below?

That's a reaallly good question!

Let's start by counting the smallest 1$$\times$$1 squares, this is just the same as counting the number of unit squares in a 6$$\times$$4 grid, there are $$6\times4=24$$ 1 by 1 squares in the grid.

Now let's move on to the 2$$\times$$2 squares, notice that counting the number of 2$$\times$$2 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 5$$\times$$3 grid. So the number of 2 by 2 squares in the grid is $$5\times3=15$$.

Now the 3$$\times$$3 squares, similarly, counting the number of 3$$\times$$3 squares in a 6$$\times$$4 grid is just the same as counting the number of unit squares in a 4$$\times$$2 grid, which is $$4\times2=8$$.

And again the number of 4$$\times$$4 squares in a 6$$\times$$4 grid is equal to the number of unit squares in a 3$$\times$$1 grid, which is $$3\times1=3$$.

Add up all the number of squares together: $$24+15+8+3=50$$. Tada! We now have our answer! There are 50 squares in a 6$$\times$$4 grid.

Mmm... the hamburger is really good...

Back on topic, in general, what is the total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$?

Again let's start from the 1$$\times$$1 squares, that's trivial, there's $$ab$$ of them.

Now moving on to the 2$$\times$$2 squares, the number of 2$$\times$$2 squares in an $$a\times b$$ grid is equal to the number of unit squares in an $$(a-1)\times(b-1)$$ grid.

Notice the pattern? Counting the number of $$n\times n$$ squares in an $$a\times b$$ grid is the same as counting the number of unit squares in an $$(a-n+1)\times(b-n+1)$$ grid.

The largest square that can contain in an $$a\times b$$ grid given that $$a\geqslant b$$ is $$b\times b$$.

Hence, the total number of squares in an $$a\times b$$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\sum_{i=0}^{b-1}{(a-i)(b-i)}$

This is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit...

\begin{align} \sum_{i=0}^{b-1}{(a-i)(b-i)}&=\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\&=ab^2-\frac{(a+b)b(b-1)}{2}+\frac{b(b-1)(2b-1)}{6} \\&=b\left[ab-\frac{ab-a+b^2-b}{2}+\frac{2b^2-3b+1}{6}\right] \\&=\frac{b}{6}\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\right] \\&=\frac{b}{6}\left[3ab+3a-b^2+1\right] \\&=\frac{b(b+1)(3a-b+1)}{6} \end{align} BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes*

The total number of squares in an $$a\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\geqslant b$$ is $\frac{b(b+1)(3a-b+1)}{6}$

If $$a<b$$, then we just swap the $$a$$ and $$b$$ around.

Special case: If $$a=b$$, the above equation becomes $\frac{a(a+1)(2a+1)}{6}$ which is the formula of the sum of squares from 1 to $$a$$.

Done! Now let me finish my burger...

This is one part of Quadrilatorics.

Note by Tan Kenneth
1 year, 5 months ago

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You should add this to the Brilliant wiki. Great note! · 1 year, 5 months ago

Cool! Thanks for that note bro, you're awesome! · 1 year, 5 months ago

Hope you finished your burger peacefully :P · 1 year, 5 months ago

Thanks, I'm glad you like the note. Well unfortunately, I think my hamburger has become stale. XD · 1 year, 5 months ago

Nice simple way of explaining complex situation. So many thanks. · 1 year, 3 months ago

are you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2) · 1 year, 5 months ago

No, i am 100.1% sure I'm not a robot. · 1 year, 5 months ago

What a note @Tan Kenneth:) · 1 year, 5 months ago

HEY tankenneth, you hyped? · 1 year, 5 months ago

Oh yes I am! $$1+1=3$$ · 1 year, 5 months ago

جميلة · 1 year, 5 months ago