A Puzzle Puzzle

In an $x \times y$ jigsaw, there are $(x-2)\times(y-2)$ pieces without an edge (just a smaller, interior rectangle). We want these to make up $\frac45$ of the total; ie $(x-2)(y-2)=\frac45 xy$ .

Expanding and rearranging, this becomes $xy-10x-10y+20=0$ . Factorising, we get $(x-10)(y-10)=80$ .

It's simple now to check the factor pairs of $80$ to find that the total number of pieces, $xy$ , is minimised when $x=18$ and $y=20$ , or vice-versa, giving $xy=\boxed{360}$ .

This is a $p$ -piece jigsaw, with side lengths $a$ and $p/a$ .

There are $2(a+p/a-2)$ edge pieces.

$p/2(a+p/a-2=5$

$(a-10)p=10a^2-20$

$p=(10a^2-20a)/(a-10)=10a+80+800/(a-10)$

Let $b=a-10$

$p=10b+180+800/b$

Differentiate to find the stationary points.

$10-800/b^2=0$ therefore b minimises at $SQRT(80)$

so we are looking for integer $b$ approx 9 and integer $a$ approx 19.

9 is not a factor of 800, but trying $b=8$ and $b=10$ yields the same $p=360$

Therefore $a=18$ and $a=20$ yield the same $p=360$

Of course they yield the same $p$ because 18 and 20 are the two side lengths of the same jigsaw!

$\boxed{360}$

$\text{Let the Jigsaw puzzle be } x \times y$

$\text{Number of pieces with at least one edge} = (x + x + y + y) - 4 \quad \color{#3D99F6}{[\text{Subtract 4 to remove double counting}]}$

$\begin{aligned} \frac{2(x+y)-4}{xy} &= \frac{1}{5} \\ \\ 10(x+y) &= xy + 20 \\ \\ \frac{x+y}{2} &\geq \sqrt{xy} \quad \color{#3D99F6}{[\text{AM.GM. inequality}]} \\ \\ 20 \cdot \frac{x+y}{2} &\geq 20 \cdot \sqrt{xy} \\ \\ 10(x+y) &\geq 20 \cdot \sqrt{xy} \\ \\ xy + 20 &\geq 20 \cdot \sqrt{xy} \\ \\ t + 20 &\geq 20 \cdot \sqrt{t} \quad \color{#3D99F6}{[\text{Let } xy = t]} \\ \\ t \in [0, 180 - 80\sqrt{5}] &\cup [180 + 80\sqrt{5}, \infty] \\ \\ t &\geq 359 \\ \\ 359 \in \text{Prime} &\implies \boxed{t = 360} \quad \color{#3D99F6}{[x = 18, y = 20]} \end{aligned}$