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 , *k*=2,3,…,*n*-2, with *d*_{1}=*d*_{n-1}=1 the length of the sides.

Now, then, we consider the product of two diagonal lengths, *d _{k}d_{m}*. From the above, this is

.

We found here that the numerator product is

.

Thus,

.

Now, if

*k*≥

*m*, then

*k*–

*m*+2

*i*-1 is a positive integer for all

*i*=1,2,…,

*m*; similarly, if

*k*+

*m*≤

*n*, then the largest value of

*k*–

*m*+2

*i*-1 in the sum,

*k*+

*m*-1, is then an integer less than

*n*, and thus the term is equal to a diagonal (or side) for all

*i*in the sum, so

, with 2≤

*m*≤

*k*≤

*n*-2,

*k*+

*m*≤

*n*. 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

*m*≥

*k*,

;

we combine to form

,

which gives us the formula for any 2≤

*k*≤

*n*-2, 2≤

*m*≤

*n*-2,

*k*+

*m*≤

*n*.

For

*k*+

*m*≥

*n*, we see that one or both of

*k*and

*m*must be greater than . However, recall that

*d*=

_{k}*d*

_{n–k}. Thus, if , then , and similarly for

*m*; thus, we see we can pick the smaller of

*k*and

*n*–

*k*; 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 , and so the only diagonal product is

, which is

For the pentagon, the diagonals have length , the golden ratio, so the only unique diagonal product, since , is

, which is

,

a classic equation for the golden ratio.

For the hexagon, we have and .

Thus we have three unique products

, which is

;

, which is

;

and

, 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 and respectively, then our diagonal product formula tells us (with , , and ):

, and thus

;

, and thus

;

and

, which is

.

Via simple algebra on these equations, we can find formula for the quotients of these numbers; dividing by

*a*or

*b*and solving for the quotient left, we see

and ;

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:

and

.

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

Tags: Diagonal Product Formula, Geometry, Heptagon, Heptagonal Triangle, Math, Monday Math, Quasiperiodic Tiling, Regular Polygon, Substitution Tiling, Trigonometry

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