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Is there an obvious way to slice two spheres (no necessarily equal) simultaneously such that the sections share common areas? I mean, I cannot see another way (easier) different from the way provided by the Praying Eyes theorem and related theorems.

You may want to take a look of this first:

http://geometriadominicana.blogspot.com/2014/03/praying-eyes-theorem.htmlenter image description here

Edit: just wanted to say that the construction is possible for three spheres too. For the construction in 2D take a look of this page http://www.cut-the-knot.org/pythagoras/LightingTheBall.shtml

enter image description here

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    $\begingroup$ You might have wanted to include this link, Emmanuel; mathoverflow.net/questions/152192/… $\endgroup$ Dec 29, 2014 at 18:14
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    $\begingroup$ @EmmanuelGarcía Yes, sure, I know "praying mantis", although I still don't understand what that has to do with anything. $\endgroup$
    – Todd Trimble
    Dec 29, 2014 at 18:48
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    $\begingroup$ The name is due to Alexander. This is what he wrote at the Acknowledgment part of the page: "I named it the "Praying Eyes" theorem because of the association (which I shall explore on a separate page) with the Eyeball Theorem".cut-the-knot.org/m/Geometry/PrayingEyes.shtml $\endgroup$ Dec 29, 2014 at 18:55
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    $\begingroup$ What do you do about concentric spheres of unequal size? $\endgroup$
    – S. Carnahan
    Dec 29, 2014 at 23:52
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    $\begingroup$ If there are tangents to the spheres that do not intersect the other, then by continuity there is some plane with the same area intersection with both. If one sphere is inside the other (not just concentric) then there are no such tangents or sections. $\endgroup$ Dec 30, 2014 at 1:09

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EDIT: the construction far below suffices when the center P is outside the circle with center Q, including when the circles do not intersect at all. For the missing case, I'm thinking a segment through an intersection point of the two circles, the only requirement being same chord length. I'll think about it.

EDIT TOO: if the circles intersect, we can simply take their intersection as the collinear chords of equal length:

enter image description here

ORIGINAL:This is actually a two dimensional construction, no continuity argument required; also, it is sometimes impossible.

Start with a circle with center P of small diameter (red) and a circle of center Q of larger diameter. Draw the red segment as a chord of the second circle, and find the segment midpoint W. Draw another circle around center Q, this time with radius QW.

Next, draw the midpoint of the segment PQ, at M. Draw a semicircle around M with radius MQ. This intersects the QW circle in a point, call this T.

By construction, QTP is a right angle. Since length QT is the same as QW, the intersection of the line PT with the larger circle is the same length as the diameter of the smaller circle, always in red.

In higher dimension, instead of just the line PT, take the codimension one hyperplane... that is, take any 2-plane through both centers and perform this construction, then fill in

enter image description here

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