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Sunday Times Teaser 2767 – Cutting Corners

by Nick MacKinnon

Published: 4 October 2015 (link)

To make an unusual paperweight a craftsman started with a cuboidal block of marble whose sides were whole numbers of centimetres, the smallest sides being 5cm and 10cm long. From this block he cut off a corner to create a triangular face; in fact each side of this triangle was the diagonal of a face of the original block. The area of the triangle was a whole number of square centimetres.

What was the length of the longest side of the original block?

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  1. Brian Gladman permalink

    This problem can be recast as a quadratic diophantine equation as follows. With \(x\), \(y\) and \(z\) as the sides of the rectangular cuboid, Heron’s formula for the area of a triangle can be used to show that:\[2A=\sqrt{(x y)^2+(y z)^2+(z x)^2}\]If we know \(x\) and \(y\) we can recast this as a quadratic diophantine equation in \(A\) and \(z\) as:\[(2A)^2 – (x^2+y^2)z^2=(xy)^2\]and hence \[(2A/xy)^2-(x^2+y^2)(z/xy)^2=1\]Substituting the known values \(x=5\) and \(y=10\) now gives: \[(A/25)^2 – 5(z/10)^2=1\]which is Pell’s equation:\[p^2-5q^2=1\]which has an infinite sequence of solutions \((p_k, q_k)\) where \(p_k\) and \(q_k\) are given by:\[p_k+\sqrt{5}q_k=(9+4\sqrt{5})^k\]Solutions can be generated using the recurrence relations:\[\begin{array}{lrcr}p_{k+1}=&9p_k&+&10q_k\\q_{k+1}=&4p_k&+&9q_k\end{array}\]starting with \(p_0=1\) and \(q_0=0\). In terms of \(A\) and \(z\) these relations become:\[\begin{array}{lrcr}A_{k+1}=&9A_k&+&50z_k\\z_{k+1}=&(8/5)A_k&+&9z_k\end{array}\] starting with \(A_0=25\) and \(z_0=0\) and giving an infinite sequence of solutions for \((z,A)\) pairs as:

    We can also derive an expression for \(z_k\) alone as:\[\begin{array}{llr}z_k&=&\sqrt{5}[(9+4\sqrt{5})^k-(9-4\sqrt{5})^k]\\&=&\sinh{[k\log{(9+4\sqrt{5})}]}\end{array}\] This form can be tested using Python with mpmath:

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