# Sunday Times Teaser 748

*by G H Dickson*

Young Bob had been given a carpentry set and with it a rectangular board (whose diagonal measured less than 1 1/2 metres) on which to practise using his saw. With straight cuts of successively greater length and aggregating to an exact number of metres, he first proceeded to saw dissimilar triangular pieces off each corner of the board in turn, the remaining quadrilateral piece being put aside for another day. All the 12 sides of the sawn-off pieces measured exact and different numbers of centimetres.

What were the length and width of the board?

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I solved this teaser today after a couple of days worth of working a spreadsheet pretty hard; no idea how they were supposed to solve it in 1975 with pencil and paper.

have a Merry Christmas

I am interested in you answer as I am seeing multiple solutions.

best wishes for Christmas too!

I have access to the ST archive, so as soon as I’d found a valid solution after about two days of off-and-on work, I looked up the published answer, saw that was it and saw the comment (which I’d been expecting) to the effect that this was an unusually hard puzzle; I was surprised anyone got an answer out at all in 1975. I didn’t go on to check for uniqueness; existence had been hard enough to establish.

Prompted by your reply I fired up the spreadsheet again and thought of some new short cuts (still not short enough) and yes, I can confirm there is more than one valid solution.

A letter (or postcard) of complaint is in order, I think.

Hi Peter,

I have been wondering if there is a Sunday Times archive where these old puzzles can be found. Where is this available?

There is (arguably) a condition that the four cuts have to be in increasing length means that no diagonal corners were cut off consecutively. If this is so, then putting (8, 15, 17), (36, 48, 60), (40, 42, 58) and (16, 63, 65) together as the offcuts of a 56 x 78 rectangle would not yield a valid solution.

No matter; there are multiple valid solutions even with that condition observed.

This gives multiple solutions:

The first of these, which has the minimum area, is the ‘official’ solution. But the puzzle can be read to suggest the cut off triangles are to have increasing hypotenuses in sequence around the rectangle and solutions meeting this criterion are preceded by a ‘*’. Those for which the original rectangle has an integer diagonal are preceded by a ‘!’.

The official title is the “Sunday Times Digital Archive”. Available through major libraries here in Australia and I’m guessing in the UK too – or if you want to pay for it, of course, which the libraries are doing.

I’m still baffled as to how anyone was supposed to find any solution to this in 1975; if only we had the worked answer from back then!

I finally completed my analysis; I found 23 solutions also, and I’ve checked the widths and lengths with your 23 and they match up. With my clumsy methods I won’t claim to have confirmed your result; on the contrary, I’m glad to have my analysis confirmed by yours.

My methodology turned up a couple of potential solutions that were infeasible because the diagonals were longer than 150cm, eg:

92 x 168: (28, 96, 100) (72, 65, 97) (27, 120, 123) (48, 64, 80).

I’ve come to the conclusion that the 1975 reader was supposed to start with a list of Pythagorean triples arranged by side, and start thinking about the shortest side length of the piece of wood. It has to be at least 7 for there to be one way of making it from two sides of Pythagorean triples; but 7 won’t work, because there’s only one way for that (3 and 4) and there has to be at least two. Okay, so let’s try 8. Oh, no, that doesn’t work, so we’ll try 9 … try, try again.

I agree, Peter, the shortest side is short enough to make a manual search among Pythagorean triangles feasible, albeit rather tedious. Thanks for the information on the digital archive.