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Sunday Times Teaser 2955 – Go Forth and Multiply

by Nick MacKinnon

Published May 12 2019 (link)

Adam and Eve have convex hexagonal gardens whose twelve sides are all the same whole number length in yards. Both gardens have at least two right-angled corners and the maximum possible area this allows. Each garden has a path from corner to corner down an axis of symmetry. Adam multiplies the sum of the path lengths by the difference of the path lengths (both in yards) and Eve squares Adam’s answer, getting a perfect fifth power with no repeated digit.

What was Eve’s answer?

2 Comments Leave one →
  1. Brian Gladman permalink

    There are three equal sided convex hexagons with two or more right angles and an axis of symmetry:

    Finding the maximum areas and path lengths for the gardens on the left and right is straightforward, giving path lengths \(p = (1 +\sqrt{2})\;a\) and \(r=(1+\sqrt{3})\;a/\sqrt{2}\) respectively.  For the centre garden, the maximum area occurs when the area of triangle EFG (and its symmetric partner) is as large as possible and, since the sides EF and FG are fixed in length, this maximum occurs when they are perpendicular, giving the length of the path DE as \(q=\sqrt{3}\;a\).

    Now forming the product of the sum and difference of the path lengths \(p\) and \(q\) gives:\[(p-q)(p+q)=p^2-q^2= 2\sqrt{2}\:a^2\]which gives Eve’s answer as \(8a^4\), where we must choose \(a\) to make this a perfect fifth power.  If we let \(a=2^e\) we obtain the equation:\[2^{4e+3}= (2^k)^5\] for some integer \(k\) from which we can deduce that \(e\) is equal to 3 mod 5,  giving Eve’s value as\[(8.16^k)^5\]for any integer \(k\), leading to the sequence:\[32768, \;34359738368, \;36028797018963968\; …\]only the first of which is valid since all the others must contain duplicate digits.

    Performing the the same procedure for the other two pairs of garden shapes fails because the results for Eve are irrational.

  2. GeoffR permalink

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