Intersection of perpendicular chords: is this true for a sphere?

Question

It is true that in a circle with radius $R$, if the intersection of any two perpendicular chords divides one chord into lengths $a$ and $b$ and divides the other chord into lengths $c$ and $d$, then $$a^2 + b^2 + c^2 + d^2 = 4R^2.$$

Does this equation hold for a sphere with radius $R$, and three mutually perpendicular chords: $$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6R^2?$$

Answer 1

For circles:

From the circle center draw perpendicular lines to the chords. Let the intersection of the lines with the chords be $X$ and $Y$, respectively. Let $|OX|=x$, $|OY|=y$.

We have (see figure):

$$\begin{align} a&=\sqrt{R^2-x^2}+y,\\ b&=\sqrt{R^2-x^2}-y,\\ c&=\sqrt{R^2-y^2}-x,\\ d&=\sqrt{R^2-y^2}+x, \end{align}$$ so that $$ a^2+b^2+c^2+d^2=4R^2, $$ as claimed.

Try to apply the same construction for the sphere case.

Drawing from the center of the sphere the planes perpendicular to the chords and denoting the intersection points $X,Y,Z$ you will obtain: $$a^2+b^2+c^2+d^2+e^2+f^2=6R^2-2(x^2+y^2+z^2),$$ so that the claim does not hold for a sphere.

Answer 2

HINTS:

For a circle we have constraints

$$ a b = c d = e ( 2 R- e ), ...,....; $$

where e is minimum distance to circle periphery.

Similarly for a sphere

$$ ab=cd = x ( 2p-x), ....,...; $$ plus two more such pair wise relations where $p,q,r $ could be radii of three mutually orthogonal circles contained in such perpendicular planes and $ x,y,z $ could be their corresponding minimum distances.