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
We found here that the numerator product is
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,
we combine to form
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:

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
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
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
d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4, which is
d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5, which is

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
d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4, and thus
d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5, which is
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
while further algebra with these formulas lets us find the reciprocals of a and b as linear combinations of a, b, and 1:

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


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