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Three spherical planets of radius r are on orbits that keep them within viewing distances
of one another. At any instant, each planet has a region that cannot be seen from anywhere
on the other two planets. What is the total area of the three unseen regions?
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Let start considering TWO spherical planets. Let the points A and B be their centers.
Will call the planets A and B, using the same letters as their centers.
Consider the straight line AB. Then it is easy to understand what are unseen parts
(regions) of each planet, that can not be seen from anythere of the other planet.
This region for planet B is the "rear" hemi-sphere of the planet B, which is cut by the plane
in 3D space, perpendicular to line AB.
Again, the unseen part of the planet B from anythere of planet A is the hemi-sphere
of B, cut from B by the great circle at B, which is in the perpendicular plane to line AB.
Same is true for the planet B: the unseen part of A from anythere on the B-surface is
the "rear" hemi-sphere of A, in relation to the straigh line AB.
It is clear that these simple configurations of the unseen parts are due to the fact
that the planets have the same radius.
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| We completed consideration of two planets and |
| now we switch our focus to three planets. |
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Let the planets be A, B and C, named according to their centers.
Whichever disposition of the centers A, B, C would be, their centers A, B and C form
some plane in 3D space, and we will consider this plane, assuming that the three planets
do not lie in the same straight line. Let "a", "b" and "c" be the interior angles
of this triangle, opposite to the sides BC, AC, and AB, respectively.
The part of the planet C, unseen from both A and B, is the "rear" part of the C surface between
two planes: one plane is perpendicular to line AC and the other plane is perpendicular to line BC.
Since the angle between the lines AC and BC is "c", the angle between these two planes is
(they are perpendicular to the lines AC and BC).
Therefore, the area of the C surface, which is unseen from A and B is = .
Similarly, the area of the B surface, which is unseen from A and C is =
and the area of the A surface, which is unseen from B and C is = .
To answer the problem question, we should add these three unseen areas, and we will get then
total area of the three unseen regions = = = .
Notice that it is EXACTLY the surface area of one such sphere of the radius "r".
ANSWER. Total area of the three unseen regions is .
CHECK. You can easily check the answer, by considering three spheres located in the vertices of the EQUILATERAL triangle.
Then your common sense will tell you that the combined unseen area of the three spheres is exactly the surface area
of one such a sphere.
Solved.
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It is of a rare beauty geometry problem.