A Short Proof of a Ptolemy-Like Relation for an Even number of Points on a Circle Discovered by Jane McDougall

Marc CHAMBERLAND and Doron ZEILBERGER

In fond memory of Andrei Zelevinsky (1953-2013) who loved Ptolemy’s theorem

Jane McDougall[M] has discovered (and apparently found a complicated proof using heavy machinery) the following beautiful Ptolemy-style theorem relating the distances amongst any even number of points on a circle.

McDougall’s Theorem: Let n be a positive integer, and let P_{i} (1 ≤ i ≤ 2n) be points on a circle. Let d_{i,j} be the distance between P_{i} and P_{j}, and let

R_{i} := ∏_{{1≤j≤2n \atop j≠i}} d_{i,j} .

Then

∑_{i=1}^{n} \frac{1}{R_{2i}} = ∑_{i=1}^{n} \frac{1}{R_{2i−1}} .

Proof: Without loss of generality the circle is the unit circle. Let

P_{i} = (cos 2t_{i} , sin 2t_{i}) (1 ≤ i ≤ 2n) .

Thanks to trig, d_{i,j} = 2 sin(t_{j} − t_{i}), and thanks to DeMoivre, this equals

−\sqrt{−1} (e^{\sqrt{−1}(t_{j}−t_{i})} − e^{−\sqrt{−1}(t_{j}−t_{i})}) .

Let

u_{i} = e^{\sqrt{−1}t_{i}} .

It follows that

d_{i,j} = −\sqrt{−1} (\frac{u_{j}}{u_{i}} − \frac{u_{i}}{u_{j}} ) = −\sqrt{−1}\frac{u_{j}^{2} − u_{i}^{2}}{u_{i}u_{j}} (i < j) .

Hence

(−\sqrt{−1})^{2n−1}(−1)^{i−1} \frac{1}{R_{i}} = (∏_{j=1}^{2n} u_{j})\frac{u_{i}^{2n−2}}{∏_{j≠i}(u_{j}^{2} − u_{i}^{2})} .

So we have

(\sqrt{−1})^{2n−1} ∑_{i=1}^{2n} \frac{(−1)^{i}}{R_{i}} = (∏_{j=1}^{2n} u_{j}) ∑_{i=1}^{2n} \frac{u_{i}^{2n−2}}{∏_{j≠i}(u_{j}^{2} − u_{i}^{2})} .  (Jane)

Recall the Lagrange Interpolation Formula: If P (z) is a polynomial of degree ≤ N − 1 in z then, for any distinct numbers z_{1}, …, z_{N},

P (z) = ∑_{i=1}^{N} \frac{(z − z_{1}) ⋯ (z − z_{i−1})(z − z_{i+1}) ⋯ (z − z_{N})}{(z_{i} − z_{1}) ⋯ (z_{i} − z_{i−1})(z_{i} − z_{i+1}) ⋯ (z_{i} − z_{N})}P (z_{i}) .  (Joseph)

(Let’s recall the trivial proof: both sides are polynomials of degree ≤ N − 1 that coincide at the N values z = z_{1}, …, z = z_{N}, so they must be identically equal.)

Taking the polynomial to be P (z) = z^{r} (with r < N − 1), and equating the coefficient of z^{N−1} on both sides of (Joseph), gives the identity:

0 = ∑_{i=1}^{N} \frac{z_{i}^{r}}{(z_{i} − z_{1}) ⋯ (z_{i} − z_{i−1})(z_{i} − z_{i+1}) ⋯ (z_{i} − z_{N})} .  (Joseph′)

Now take N = 2n, r = n − 1, and z_{i} = u_{i}^{2} in (Joseph′) and conclude that the right side of (Jane) is indeed zero.

It follows that if our 2n points lie on a line (the case R = ∞), the theorem is true as well. Furthermore, in that case it is even true for an odd number of points. We leave this as an exercise to the dear readers.

Reference

[M] Jane McDougall, private communication:

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/JaneMcDougallMessage.html

.

Marc Chamberland, Department of Mathematics and Statistics, Grinnell College, Grinnell, Iowa

50112, USA. chamberland at grinnell dot edu

Doron Zeilberger, Mathematics Department, Rutgers University (New Brunswick), Piscataway, NJ

08854, USA. zeilberg at math dot rutgers dot edu

April 15, 2013