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Define a golden parallelepiped as a $d$-dimensional box with side lengths $(1, \phi, \phi^2, \ldots, \phi^{d-1})$, where $\phi$ is the golden ratio:
          Golden Box
The volume of the parallelepiped is $\phi^{d(d-1)/2}$, e.g., $\phi^3$ for $d=3$. I wonder if there is a natural geometric explanation of the relationship $\phi^n = \phi^{n-1} + \phi^{n-2}$, e.g., $\phi^3 = \phi^2 + \phi$ for $d=3$? I am imagining some type of partition or dissection of the volume of the golden parallelepiped that corresponds to the $\phi$ equation.

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I don't get the "geometric" part of what you are trying to explain. Isn't multiplying the $\phi$-equation $\phi^2=\phi+1$ by $\phi^{n-2}$ good enough as explanation? – Marc van Leeuwen Jun 5 '12 at 14:13
up vote 5 down vote accepted

Since $\phi^2=\phi+1$, we can break up the edges of side length $\phi^2$ into parts of length $\phi$ and $1$, and breaking up the solid along that plane divides it into pieces of volume $\phi\times \phi\times 1=\phi^2$ and $1\times\phi\times1=\phi$, thereby demonstrating $\phi^3=\phi^2+\phi$. The same idea should work for any $d$: given the box with lengths $1$, $\phi$, ..., $\phi^d$, dividing the sides of length $\phi^k$ (for any choice of $k$) into parts of length $\phi^{k-1}$ and $\phi^{k-2}$ and then dividing the solid along that hyperplane gives pieces of volume $\phi^{\frac{d(d-1)}{2}-1}$ and $\phi^{\frac{d(d-1)}{2}-2}$.

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Very clear, Zev! I was missing that idea of partitioning the edge lengths. Thanks! – Joseph O'Rourke Jun 5 '12 at 15:17

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