They are not, as the other answers point out. The simplest way (I think) to see it is the following:

1. The discs move on circles. We can think of the cicrcles as the eqautor and a meridian of a sphere. Call the white-disc-circle the equator.
2. At the moment the discs cross the holes, they are at opposite points of the sphere, and their distance is thus maximal.
3. When the black disc is at its highest point ("north pole"), the white one is still on the equator. Hence, they are not at opposite points of the sphere, and their distance is clearly less than the maximum value. This is actually the point of closest approach.

The (spatial) distance oscillates between twice the radius ($2 r$) and $\sqrt{2} \,r$.

Let's say that the sphere has radius 1 centered at the origin in $\Bbb R^3$ and the disks are moving with speed 1. And let's also say that in the picture the $z$-axis points upward, and the $y$ axis points normal to the plane of motion of the black thing. Let's also use the convention that at time zero the black disk has center at $(0,0,1)$ and the white one at $(0,1,0)$ Then You can literally parametrize the two curves which describe the center of the disks in a simple way:

$\gamma_{black}(t) = (\sin t, 0, \cos t)$

$\gamma_{white}(t) = (-\sin t, \cos t, 0)$

So if you compute distances (by subtracting and taking the norm squared), it is clear that they are not always equidistant.

So why does it seem like they are always equidistant in the picture?

Well, if you take a good look, the motion of the two disks is not along an exact sphere but more like some kind of ellipsoid.

So instead of using the spherical model $x^2+y^2+z^2=1$, try the ellipsoidal model $x^2+y^2/2+z^2/2=1$. In other words, let's consider the image of this rigid motion under the map $(x,y,z) \mapsto (x, \sqrt 2 y, \sqrt 2 z)$. So the new parametrizations will be

$\gamma_{black}(t) = (\sin t, 0, \sqrt 2 \cos t)$

$\gamma_{white}(t) = (-\sin t, \sqrt 2 \cos t, 0)$

And in this model we see that they are always equidistant! (Well those are my two cents at least. Might be total BS. Really it depends on how you want to interpret this two dimensional projection of a three dimensional motion - for all we know the planes of motion may not even be orthogonal, which would make the formulas more complicated).

The two points are revolving on circles in the $xy$ and $yz$ planes, in a synchronized way, on the trajectories

$$x=\sin t,y=\cos t,z=0$$ and $$x=0,y=-\cos t,z=\sin t.$$

The phases are such that the points are on opposite locations on the $y$ axis at $t=0$.

The distance is thus

$$\sqrt{\sin^2t+4\cos^2t+\sin^2t}=\sqrt{2(1+\cos^2t)}.$$

It is maximum on every half-turn (from the position at $t=0$), forming a straight line with the center, and minimum every quarter turn later, forming a right angle.

The ratio of the long distance over the short one is $\sqrt2$.

Clearly, the two disks and the center of the sphere alternate bewteen alignment (distance $2r$) and a right angle configuration (distance $\sqrt2r$).

At time $t$ the black disc is at $(\cos t, \sin t, 0)$ and the white at $(-\cos t, 0, -\sin t)$ (up to orientation: Black spins along the unit circle in the $x-y$ plane where $z = 0$. White spins along a unit circle in the $x-z$ plane where $y=0$. White is so oriented that it starts at $z = 0 = -\sin t$, $x = -1, = -\cos 0$ and so forth.)

The distance is $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 } =$

$ \sqrt {(2\cos t )^2 + \sin^2 t + \sin^2 t} = \sqrt {4\cos^2 t + 2 \sin^2 t}=\sqrt{2 + 2\cos^2 t}$ which is not constant.

There's a certain amount of interpretation involved but here is mine.

If this is a bit discursive, OK, I am thinking out loud at the keyboard. I'm letting you in on the thought process.

I see the white comma shape, the yin, being in a horizontal plane, say the equatorial plane of a larger sphere, rotating at a constant rate; and the black comma shape, the yang, in a vertical plane, say the 0/180 longitude great circle of that same sphere, rotating at the same constant rate. You will find it helpful to sketch the large sphere and the two planes. The small circles pass through the holes when the larger shapes are at zero / $2\pi$ and $\pi$. Suppose the yin and yang are rotatng at s rotations per minute. Then the smaller circles, which pass through two times per rotation, might seem to be rotating at 2 rotations per minute, but no, when you watch carefully you see that the small circles are actually on the same planes -- the black dot rotates with the black yin and the white dot rotates with the white yang and they pass once at 0 and once at $/pi$, or two passes per rotation. In fact if you drew the traditional yin-yang with the 2D shapes, just put a pin in the middle and rotate it and you can see the black shapes here. The apparent complexity is hiding simplicity.

So take a look at the black dot. Its center is at a radius r = 1/2 R where R = radius of large sphere. That is unimportant to the question at hand but I just like to set things up properly.

If we put the black yin in the yz plane and the white yang on the xy plane that will work. The center of the black disc follows the circular path $y = -r \cdot cos t, z = r \cdot sin t$ (minus, going clockwise) and the center of the white dot follows the path $x = r \cdot cos t, y = r \cdot sin t$ Hope I got that right, some fudging may be required for directions but the foundation is there. Rotation speed s = 1 rotation per $2\pi $ seconds not that it matters.

So W = $(0,-r \cdot cos t, r \cdot sin t)$, $B = (r \cdot cos t, r \cdot sin t, 0)$

distance $|BW| = \sqrt{r^2cos^2t + (r^2 cos^2 t - r^2 sin^2 t) + r^2 sin^2 t} = r\sqrt{1 + cos(2t)}$ so no it is not constant.

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