# 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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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: