I Want Me Less Unit Squares

Let the two sides of the grid (rectangle) be $a$ and $b$ , then the semiperimeter $s=a+b$ and the number of unit squares $n=ab$ . Without loss of generality, let us assume $b \ge a$ and their difference $b-a=d$ , so that $d \ge 0$ . Then, we have:

$\begin{aligned} s & = a+b = a+a+d = 2a + d\\ \implies a & = \frac {s-d}2 \\ \implies b & = \frac {s+d}2 \\ \implies n & = ab = \frac {s-d}2 \times \frac {s+d}2 = \frac {s^2-d^2}4 = \frac {s^2-(b-a)^2}4 \end{aligned}$

This means that for a particular $s$ the smaller the difference $d =b-a$ , the largest the $n$ . The largest $d$ occurs when $a$ is smallest, that is $a=1$ . Since $s=a+b \implies b = s-1$ and $d = b-a = s-2$ when $n$ is minimum. Therefore, $n_{min} = u(s) = \dfrac {s^2 - (s-2)^2}4 = s-1$ .

Therefore, $u(1153) = 1153-1 = \boxed{1152}$ .

In general, suppose the width of the grid is $a$ , the height of the grid is $b$ , and the semiperimeter of the grid is $s=a+b$ . Then the number of unit squares in this $a\times b$ grid is $ab$ . Without loss of generality, let's assume that $a>b$ .

We have $ab=\frac{(a+b)^2-(a-b)^2}{4}=\frac{s^2-(a-b)^2}{4}$

So, to make $ab$ as small as possible, $(a-b)^2$ has to be as small as possible, which means $a-b$ has to be as small as possible, which in turn means that $a$ has to be as large as possible and $b$ to be as small as possible. This can be achieved with $a=s-1$ , $b=1$ , making $ab=\frac{s-(s-2)^2}{4}=s-1$ .

Therefore,

$u(s)=s-1$

Now, evaluating $u(1153)$ , we get $1152$ .