They Differ Equally

Geometry Level 4

Given two grids, with dimensions $2m\times 3n$ and $(3m-1)\times(3n+2)$ , the difference of the number of unit squares and the difference of the number of $m\times(n+1)$ rectangles in these two grids are equal, what is the value of $m$ ?

Details and assumptions:

Unit square is a square with side 1.
Dimensions are expressed as width $\times$ height . Thus an $m\times(n+1)$ rectangle (that is, a rectangle of width $m$ and height $n+1$ ) is considered to be different to an $(n+1)\times m$ (width $n+1$ height $m$ ) rectangle, in this case, you only want the $m\times(n+1)$ rectangles, not the $(n+1)\times m$ ones.

This is one part of Quadrilatorics .

The answer is 1.

1 solution

Kenneth Tan
Jan 28, 2017

From this note , we know that the number of $x\times y$ rectangles in an $a\times b$ grid given that $x\leqslant a$ and $y\leqslant b$ is $(a-x+1)(b-y+1)$

The number of unit squares in these two grids are $6mn$ and $9mn+6m-3n-2$ respectively, while the number of $m\times(n+1)$ rectangles in these two grids are $(2m-m+1)[3n-(n+1)+1]=2mn+2n$ and $[(3m-1)-m+1][(3n+2)-(n+1)+1]=4mn+4m$ respectively.

Thus, $|(9mn+6m-3n-2)-6mn|=|(4mn+4m)-(2mn+2n)| \\|3mn+6m-3n-2|=|2mn+4m-2n|$

Because $m$ and $n$ are both positive integers, $3mn+6m-3n-2=3n(m-1)+6m-2\geqslant6-2=4>0$ $2mn+4m-2n=2n(m-1)+4m\geqslant4>0$ $\therefore 3mn+6m-3n-2=2mn+4m-2n \\ mn+2m-n-2=(m-1)(n+2)=0$

As $n$ is positive, $m=1$ .

They Differ Equally

This is one part of Quadrilatorics .

The answer is 1.

1 solution

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