How to determine number of edge pieces in a puzzle?

Question

Given the number of puzzle pieces (1000) and the length and width of the puzzle (19.68*29.52 inches), is it possible to determine the number of edge pieces that make up the puzzle perimeter?

Answer 1

You can get close. If you convert your inches to centimeters the puzzle is $50 \times 75$ which might be easier to work with. Assuming all the pieces are the same size, say the puzzles is $a \times b$ pieces. Then

$$\frac{a}{b} = \frac{50}{75} = \frac{2}{3}.$$

But we also know that $ab = 1000$, so we have

$$\frac{a}{b}= \frac{a}{1000/a} = \frac{a^2}{1000} =\frac{2}{3}$$

So

$$a^2 = \frac{2000}{3}$$

or

$$a = \sqrt{666.7} = 25.82$$

which makes $$b = 1000/25.82 = 38.73$$

Then the puzzle is (about) $26$ pieces by $39$ pieces which makes the perimeter $2\cdot 26 + 2\cdot 39 - 4 = 126$ pieces.

Answer 2

Aside from counting them? Not really. The primary reason for this is that the number of pieces given on the box is often only an approximation. Stand Up Maths has an amusing and informative video about this.

There would be some ways we could approximate the number of edge pieces, or find upper and lower bounds on the number of edge pieces by using the dimensions of the full puzzle and taking measurements of non-edge pieces, though. However, the approaches that follow rest on the assumption that all of the pieces are roughly rectangular in overall shapes and roughly equal in their "rectangular dimensions."

For example, if we want an approximation, we could pick an arbitrary edge piece and obtain the measurements between the four pairs of adjacent corners--let's call them $l_1,l_2,w_1,$ and $w_2,$ where $l_2$ (for example) is the distance between the corners opposite those of $l_1$. Then there should be integers $m$ and $n$ such that either $ml_i\approx 19.68$ and $nw_j\approx29.52$ or $ml_i\approx29.52$ and $nw_j\approx19.68$ for some $i,j\in\{1,2\}.$ Moreover, $mn$ should be around $1000.$ In that case, then, the number of edge pieces should be about $mn-(m-2)(n-2)=2m+2n-4.$

If we want bounds, we could find measurements for two different kinds of non-edge pieces, one with $4$ protrusions, and one with $0$. The former would let us find a "height" and "width" (including the protrusions) that should be an upper bound on the rectangle dimensions. Let's call these upper bound measurements $D_1$ and $D_2.$ Measuring the "inner height" and "inner width" of the piece without protrusions will give us lower bounds on those dimensions, say $d_1,d_2.$ Doing something similar to the above will give us lower and upper bounds on the number of rows and columns, by finding $m,n,M,N$ such that $mD_i\approx 19.68\approx Md_j$ and $nD_k\approx 29.52\approx Nd_l$ for some $i,j,k,l\in\{1,2\}$ with $i\neq k$ and $j\neq l.$ Then there should be between $2m+2n-4$ and $2M+2N-4$ edge pieces.