General circle identity proof

Question

I very much suspect this is true but I don't have a proof of it.

Take a circle of radius $R$. Pick a point $P$ in the circle, a distance $d$ from the center of the circle. From $P$, extend $N$ ($N>1$) line segments equally-angularly-spaced connecting $A$ to the circle (orientation relative to the circle is irrelevant) (see the figure).

The geometric mean of the lengths of the $N$ segments is always $\sqrt{R^2-d^2}$, irrespective of $N$ or how the segments are oriented relative to the circle.

Does anyone know how to prove this? It seems like such a fundamental and general theorem but I haven't been able to find anything.

Answer 1

As described in the first comment, if the number of segments is even the statement is true:

When $N$ is even, this boils down to the Power of a Point Theorem. (There are $N/2$ full chords, each of which is separated into sub-chords by $P$; the product of each pair of sub-chords gives the same value, the "power" of $P$, which is $R^2−d^2$. In this case, it doesn't even matter that the segments are equally-angularly-spaced.)

As pointed out in another comment, the statement is false when $N$ is odd, as is easily verified in a dynamic geometry package like Geogebra. If $A$ is the set of $N$ segments, we can create another set $B$ of segments consisting of segments $b_i$ that, when paired with segments $a_i$ form full chords. The best we can say is that, if $a$,$b$ are the geometric means of the segments in $A$ and $B$, then $\sqrt {ab} =\sqrt{R^2-d^2}.$ This is hardly earthshaking, because $A\bigcup B$ is the set of segments for $2N$, which is even.

Generally, $a$ and $b$ have only small variance as the orientation of the segments changes as in the OP animation, due to similar mixes of long and short segments in each set.