## Monday Math 141

Let us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths $d_k=\frac{\sin\frac{k\pi}n}{\sin\frac{\pi}n}$, k=2,3,…,n-2, with d1=dn-1=1 the length of the sides.
Now, then, we consider the product of two diagonal lengths, dkdm. From the above, this is
$d_kd_m=\frac{\sin\frac{k\pi}n\sin\frac{m\pi}n}{\sin^2\frac{\pi}n}$.
We found here that the numerator product is
$\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n$.
Thus,
$\begin{array}{rcl}d_kd_m&=&\frac{\sin\frac{k\pi}n\sin\frac{k\pi}n}{\sin^2\frac{\pi}n}\\&=&\frac1{\sin^2\frac{\pi}n}\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\&=&\frac1{\sin\frac{\pi}n}\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\d_kd_m&=&\sum_{i=1}^{m}\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}\end{array}$.
Now, if km, then km+2i-1 is a positive integer for all i=1,2,…,m; similarly, if k+mn, then the largest value of km+2i-1 in the sum, k+m-1, is then an integer less than n, and thus the term $\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}$ is equal to a diagonal (or side) $d_{k-m+2i-1}$ for all i in the sum, so
$d_kd_m=\sum_{i=1}^{m}d_{k-m+2i-1}$, with 2≤mkn-2, k+mn. This is the “diagonal product formula” named by Peter Steinbach, who outlined a proof using the n-gon formed by the complex n-roots of unity (compare this problem, for example).
Now, we can generalize the formula first by exchanging k and m to see that for mk,
$d_kd_m=\sum_{i=1}^{k}d_{m-k+2i-1}$;
we combine to form
$d_kd_m=\sum_{i=1}^{\min(k,m)}d_{|k-m|+2i-1}$,
which gives us the formula for any 2≤kn-2, 2≤mn-2, k+mn.
For k+mn, we see that one or both of k and m must be greater than $\frac{n}2$. However, recall that dk=dnk. Thus, if $d_k>\frac{n}2$, then $d_{n-k}<\frac{n}2$, and similarly for m; thus, we see we can pick the smaller of k and nk; the latter gives the same terms in the sum, just in opposite order, so only the limit matters. Doing the same with m, we thus find the formula for all k and m in the range 2 to n-2:
$d_kd_m=\sum_{i=1}^{\min(k,m,n-k,n-m)}d_{|k-m|+2i-1}$,

Now, let us look at a few examples. The smallest n to give us a diagonal is n=4, the square. For square of unit side, the length of the diagonal is $\sqrt2$, and so the only diagonal product is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt2)^2=2=1+1$
For the pentagon, the diagonals have length $\phi=\frac{1+\sqrt5}2$, the golden ratio, so the only unique diagonal product, since $d_2^2=d_2d_3=d_3^2$, is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$\phi^2=1+\phi$,
a classic equation for the golden ratio.
For the hexagon, we have $d_2=d_4=\sqrt{3}$ and $d_3=2$.
Thus we have three unique products
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt3)^2=3=1+2$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, which is
$2(\sqrt3)=\sqrt{3}+\sqrt{3}$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$2^2=4=1+2+1$.

Further, this can be used to prove some interesting relations of the (non-constructible) diagonal lengths of the heptagon, and thus the sides of the heptagonal triangle. Letting the heptagon have unit side, labelling the shorter and longer diagonal lengths as $a=\frac{\sin\frac{2\pi}7}{\sin\frac{\pi}7}$ and $b=\frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7}$ respectively, then our diagonal product formula tells us (with $d_1=d_6=1$, $d_2=d_5=a$, and $d_3=d_4=b$):
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, and thus
$a^2=1+b$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, and thus
$ab=a+b$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$b^2=1+a+b$.
Via simple algebra on these equations, we can find formula for the quotients of these numbers; dividing $ab=a+b$ by a or b and solving for the quotient left, we see
$\frac{b}{a}=b-1$ and $\frac{a}{b}=a-1$;
and dividing by the product ab gives
$\frac{1}{a}+\frac{1}{b}=1$,
while further algebra with these formulas lets us find the reciprocals of a and b as linear combinations of a, b, and 1:
$\begin{array}{rcl}b^2&=&1+a+b\\b&=&\frac{1}b+\frac{a}{b}+1\\b&=&\frac{1}b+a-1+1\\b&=&\frac{1}b+a\\\frac{1}b&=&b-a\end{array}$
and
$\begin{array}{rcl}a^2&=&1+b\\a&=&\frac{1}a+\frac{b}{a}\\a&=&\frac{1}a+b-1\\\frac{1}a&=&a-b+1\end{array}$.

These relations appear, for example, in the substitution rules for Danzer’s 7-fold quasiperiodic tiling and Maloney’s 7-fold quasiperiodic tiling