**Page 9**

In general, the perimeter of a \(a\times b\) grid is \(2a+2b\), the semiperimeter of a \(a\times b\) grid is \(a+b\).

For example, the perimeter of a \(3\times 5\) grid is 16, its semiperimeter is 8.

If two grids have the same perimeter, then these 2 grids are "perimeter-like grids".

For example, \(5\times 2\) grid and \(3\times 4\) grid are perimeter-like grids.

Suppose there's a grid, \(a+b=10\), if the number of unit squares in the grid reaches its maximum, then what is the value of \(a\) and \(b\)?

Let's write out the possibilities: \[1\times9,~~~~2\times8,~~~~3\times7,~~~~4\times6,~~~~5\times5,~~~~6\times4,~~~~7\times3,~~~~8\times2,~~~~9\times1\] The corresponding numbers of unit squares are: \[9,~~~~~~16,~~~~~~21,~~~~~~24,~~~~~~25,~~~~~~24,~~~~~~21,~~~~~~16,~~~~~~9\] We see that when \(a=b=5\), then the number of unit squares reaches its maximum.

We say that the maximum number of unit squares it can form when \(a+b=10\) is 25.

What if \(a+b\) is an odd number? Let's try this out...

Suppose \(a+b=9\), the possibilities are: \[1\times8,~~~~2\times7,~~~~3\times6,~~~~4\times5,~~~~5\times4,~~~~6\times3,~~~~7\times2,~~~~8\times1\] The corresponding numbers of unit squares are: \[8,~~~~~~14,~~~~~~18,~~~~~~20,~~~~~~20,~~~~~~18,~~~~~~14,~~~~~~8\] We see that when \(a=b\pm1\), then the number of unit squares reaches its maximum.

We say that the maximum number of unit squares it can form when \(a+b=9\) is 20.

If \(a+b\) is a fixed even number, then when \(a=b\), the number of unit squares it can form in the grid will reach its maximum.

Another way of thinking about this is among all the perimeter-like grids which \(a+b\) is even, then the grids which has the property of \(a=b\) will have the most number of unit squares.

If \(a+b\) is a fixed odd number, then when \(a=b\pm1\), the number of unit squares it can form in the grid will reach its maximum.

Another way of thinking about this is among all the perimeter-like grids which \(a+b\) is odd, then the grids which has the property of \(a=b\pm1\) will have the most number of unit squares.

Proof:

Consider \(a+b\) is even. Suppose \(a+b=2k\), \(a=k+x\), \(b=k-x\) (\(k\) is a positive integer while \(x\) is an integer). The number of unit squares in the grid is \[\begin{align} a\times b&=(k+x)(k-x) \\&=k^2-x^2 \end{align}\] Since \(x^2\geqslant 0\), to maximize the value of \(k^2-x^2\), \(x\) must be equal to 0, thus \(a=b=k\).

Consider \(a+b\) is odd. Suppose \(a+b=2k+1\), \(a=k+x\), \(b=k+1-x\). The number of unit squares in the grid is \[\begin{align} a\times b&=(k+x+1)(k-x) \\&=k^2+k-x^2-x \\&=k^2+k-x(x-1) \end{align}\] Since \(x(x-1)\geqslant 0\), to maximize the value of \(k^2+k-x(x-1)\), \(x\) must be equal to 0 or 1, thus \(a=k\), \(b=k+1\) or \(a=k+1\), \(b=k\), \(a=b\pm1\).

**Generalization:**

If \(a+b\) is a fixed even number, then when \(a=b=k\), the number of unit squares in the grid will reach its maximum, which is \(ab=k^2=\left (\frac{a+b}{2} \right )^2\).

\(\bullet\) **Hence if \(a+b\) is even, the maximum number of unit squares it can form is \(\left (\frac{a+b}{2} \right )^2\).**

If \(a+b\) is a fixed odd number, then when \(a=k+1\), \(b=k\) or when \(a=k\), \(b=k+1\), the number of unit squares in the grid will reach its maximum, which is \(ab=k(k+1)=\left (\frac{2k}{2} \right )\left (\frac{2k+2}{2} \right )=\frac{(a+b-1)(a+b+1)}{4}\).

\(\bullet\) **Hence if \(a+b\) is odd, the maximum number of unit squares it can form is \(\frac{(a+b-1)(a+b+1)}{4}\).**

If \(a+b=14\), what is the maximum number of unit squares it can form?

**Solution:**

Since 14 is even, according to the greatest rule, the maximum number of unit squares it can form is \(\left( \frac{a+b}{2} \right)^2=7^2=49\).

If \(a+b=19\), what is the maximum number of unit squares it can form?

**Solution:**

Since 19 is odd, according to the greatest rule, the maximum number of unit squares it can form is \(\frac{(a+b-1)(a+b+1)}{4}=\frac{18\times 20}{4}=90\).

This is one part of Grids and Quadrilaterals.

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