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Sunday Times Teaser 3076 – Bee Lines

by Brian Gladman on September 4, 2021

by Nick MacKinnon

Published Sunday September 05 2021 (link)

Three bees are trapped inside three empty cuboidal boxes of different sizes, none of whose faces are squares. The lengths of the edges of each box in centimetres are whole numbers, and the volume of each box is no more than a litre. Starting at a corner, each bee moves only in straight lines, from corner to corner, until it has moved along every edge of its box. The only points a bee visits more than once are corners of its box, and the total distance moved by each bee is a whole number of centimetres. Given the above, the sum of these three distances is as small as it could be.

What is the sum of the distances that the bees moved?

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  1. Brian Gladman permalink

    To traverse all the edges of a cuboid, each only once, requires the traverse of three diagonals and I could not immediately find such a path with all three on the same pair of faces with the same dimensions.  But even if such a path exists, it would cross on one of the faces which the teaser does not allow.

    So, after a lot of sketching of possible paths, I believe that when all the diagonals are face diagonals, two of the three must lie on opposite cuboid faces with the third on another face. When one of the diagonals is a space diagonal, the two face diagonals do not need to be on opposite faces but there is always a path of this kind. (I have written a program which verifies these observations by generating all 1,421,716 different paths).

    This means that two different path arrangements need to be considered:

    In the left hand arrangement, the third diagonal is on another face, which means that the shortest integer length path traversing all edges of a cuboid with edges \((a, b, c)\) has a length of \[4(a+b+c)+2\sqrt{a^2+b^2}+\sqrt{a^2+c^2}\]where \((a,b)\) and \((a, c)\) are the non-hypotenuse sides of Pythagorean triangles.  In the right hand arrangement, the third diagonal is a space diagonal of the cuboid, giving the length as:\[4(a+b+c)+2\sqrt{a^2+b^2}+\sqrt{a^2+b^2+c^2}\].  In this case the third diagonal has to be what is known as a Pythagorean quadruple.

    • Tony Smith permalink

      The lengths of the diagonals in question are integers but not squares.

      But it is also conceivable that the sum of the second and third terms of your expression in (a,b,c) is an integer while the terms themselves are not.

      • Brian Gladman permalink

        Tony, thanks for the input. Although it is possible that there are solutions in
        which the diagonals are not integers, I believe the intention of the teaser is
        that they are. But my current answer is wrong for another reason so the text
        and code above will need to be updated anyway (now done).

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