Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 4 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 $\phi^{\frac{d(d-1)}{2}-2}$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.