The exact formulation is: suppose three randomly-placed non-concentric circles expand at the same speed on a plane, will they always intersect at the same point at least once? If not, are there any condition guarantees that they will?

This question looks challenging in its original formulation, but is actually quite simple when looked at a different way. Scroll down for the answer.

... ... ... ... ... ... ... ... Answer: We can look at this question as “is there a point on a plane that's equidistant to the three points”. And the answer is intuitive at that point – circumscribe the triangle (which can always be done for any triangle), and take the center. By definition, that center is equidistant from all the points. Done.

However, since co-linear and non-concentric points cannot be circumscribed, they have no such point.

What about the three-spheres case in 3D space?

Before I thought to look at this question the simple way, the solution was really complicated. But now it's simple – form a plane with the three center-points, take the center of the circumscribing circle on that plane, and any point on the the line perpendicular to that plane passing through the center point will be equidistant to the three spherical centers.

I suspect that this means four non-coplanar spheres expanding at the same speed also have a single meeting point, and pretty sure that a solution always exists even if each circle/sphere was expanding at a different rate – when finding the modified-speed triangle, just imagine a slower circle/sphere's center point is further away from the center of the current triangle, and vice versa.

— Categorized under: #mathematics