I noticed something curious about intersecting chords in a circle. Suppose two chords have lengths $p$ and $q$ and intersect at right angles at point $O$. The intersection $O$ divides the two chords into four total segments of lengths $a,b,c,d$ (say $a+c=p$ and $b+d=q$).

The radius of the circle turns out magically to be the root-mean-square


I think it's fascinating, because for other pairs of intersecting chords, the radius is just as much an "average" as excenters of triangles are "centers." I came up with a more general formula involving more than just right angles, but it's not satisfying. Does anyone have some good insight into why the right angle case is particularly nice? Maybe there isn't anything deep and it strikes me more only because I am a functional analyst.

My favorite proof so far is saying $a^2+b^2+c^2+d^2=(AB)^2+(CD)^2$, then rearranging the arcs to make another right triangle with the diameter as hypotenuse:

The diameter is the "third diagonal" among the three different cyclic quadrilaterals you can make with those four lengths. Turns out, you can generalize to angles other than $90^\circ$

$$R=\frac{\text{third diag.}}{\sin\theta}$$

It's more symmetry than I would have hoped - you just have to think about all three quadrilaterals at once. But, it doesn't deal much with the root-mean-square bit.


I realize that you aren't looking for proof, but it's hard to gauge why your result is "particularly nice" without knowing compared to what? So, I'll derive the general formula here in hopes of discerning what's going-on in the broader context.

I'll start by calculating, not the circle's radius, but the distance from the circle's center to the chords' point of intersection.

Consider the following ...

enter image description here

In $\bigcirc O$ shown, chords $\overline{AC}$ and $\overline{BD}$ (of lengths $a+c$ and $b+d$, respectively) meet at $P$ and make an angle of $\theta$. Let $A^\prime$ and $B^\prime$ lie on the chords such that $\overline{AA^\prime}\cong \overline{PC}$ and $\overline{BB^\prime}\cong\overline{PD}$. The midpoints of $\overline{PA^\prime}$ and $\overline{PB^\prime}$ are the midpoints of the respective chords, and therefore, the same can be said for the perpendicular bisectors (which, of course, meet at center $O$); consequently, $O$ is the circumcenter of $\triangle PA^\prime B^\prime$. If $p$ is the radius of the new circumcircle, and $q := |\overline{A^\prime B^\prime}|$, then the Law of Sines tells us that ... $$2 p \sin\theta = q \tag{1}$$ so that the Law of Cosines tells us that ...

$$4 p^2 \sin^2\theta = (a-c)^2 + (b-d)^2 - 2(a-c)(b-d)\cos\theta \tag{2}$$

This makes for a satisfyingly-straightforward calculation for distance $p$, but what does it have to do with your radius formula(s)?

Well, recall that the "power" of point $P$ with respect to $\bigcirc O$ is defined by $$k := p^2 - r^2 \tag{3}$$ and is computed (for interior points $P$) by $$k = - |\overline{PX}||\overline{PY}| \quad\text{for any chord}\;\overline{XY}\;\text{through}\;P \qquad\to\qquad k = - a c = - b d \tag{4}$$ so we can re-write $(2)$ as $$4r^2\sin^2\theta = ( a - c)^2 + ( b - d )^2 - 2 ( a - c )( b - d)\cos\theta - 4 k \sin^2\theta \tag{5}$$ Now, notice that if we expand the square terms, we get products $-2ac$ and $-2bd$, each of which conveniently equals $2k$, so that (replacing $\sin^2\theta$ with $1-\cos^2\theta$ on the right) ...

$$4r^2\sin^2\theta = a^2 + b^2 + c^2 + d^2 - 2 ( a - c )( b - d)\cos\theta + 4 k \cos^2\theta \tag{6}$$

(Note: To express everything as explicitly, and symmetrically, as possible in $a$, $b$, $c$, $d$, one can make the substitution $k \to -\sqrt{a b c d}$.) The relation's a little bulky, but I don't know that I'd call it especially unsatisfying. It seems on par with results like Bretschneider's Formula and Brahmagupta's Formula for quadrilateral area.

Evidently, when the chords are perpendicular, $\theta = \pi/2$, and $(6)$ reduces to the equivalent of your root-mean-square formula for the radius. Of course, $(2)$ reduces, as well, and we see that a lot of the complexity of the diagram vanishes:

enter image description here

This may (or may not) help explain why the right-angle case is "particularly nice", but it's the best I can do at the moment.

Incidentally, here's a separate, simple proof of the root-mean-square formula for perpendicular chords:

enter image description here

$$( 2 r )^2 = ( a + c )^2 + ( b - d )^2 = a^2 + b^2 + c^2 + d^2 + 2 ( a c - b d ) = a^2 + b^2 + c^2 + d^2$$

Here, we construct $\overline{C^\prime C}$ parallel to $\overline{BD}$, and invoke Thales' Theorem to conclude that $\overline{AC^\prime}$ is a diameter. Then, the power of a point theorem allows us to cancel $a c - b d$ from the Pythagorean relation. $\square$

(The proof at Cut-The-Knot, to which @dxiv linked in a comment, is close to this, but uses similar triangles to drive the argument.)

  • $\begingroup$ BTW: The root-mean-square relation says that the sum of the areas of the squares on the four sub-segments equals the area of the circle's circumscribed square. By the Wallace-Bolyai-Gerwien Equidecomposability Theorem, it's possible to dissect the smaller squares into finitely-many polygonal pieces that can be arranged to fill the circumsquare. It would be "particularly nice" if there were a standard dissection for this, as there are for demonstrating the Pythagorean Theorem. $\endgroup$ – Blue Oct 30 '16 at 5:53
  • $\begingroup$ Yes, I think would be in the direction I want to go. I'm really not sure. I've just thought about it for a couple weeks and thought I would throw it up. It seems to be the invariance and the Euclidean look that capture me. Speaking of Euclidean, maybe there's something about spheres in $\mathbb{R}^4$... like if you look at all permutations of the coordinates in the point $(\pm a,\pm b,\pm c,\pm d)$, you will get 24*16 points on a sphere. I think it would be interesting to find some corresponding regularity among those. For example, if they all sat on certain cylinders through the origin. $\endgroup$ – Aaron Goldsmith Oct 31 '16 at 13:14

Hint: keep a chord fixed, and move the other chord to make it diameter and calculate the distance it had to be moved to make it a diameter.

  • $\begingroup$ Thanks, but I have multiple proofs already. Another proof is not necessarily what I'm looking for. In fact, I would generally like to steer away slightly from the emphasis on proof. Perhaps my phrasing was ambiguous. I would like to know if this proposition can be generalized or seen in a broader context. As of yet, I am still surprised that it comes out to the root-mean-square even though I am quite convinced of the proof. $\endgroup$ – Aaron Goldsmith Oct 29 '16 at 13:00

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