Given is a finite set of spherical planets, all of the same radius and no two intersecting. on the surface of each planet consider the set of points not visible from any other planet. prove that the total area of these sets is equal to the surface area of one planet.

Step-by-step explanation:

To answer this question, First, we define a direction in space, which defines

the North Pole of each planet.

We also assume that the direction is not at 90° perpendicular to the axis of any two planet.

Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.

If we refer to the direction of the planet in the diagram, we discover that,

1, A ≥ A

2, If A ≥ B and B≥ A,

Then A = B

3, If A ≥ B, and B ≥ C , then A ≥ C

4, Either A ≥ B or B ≥ A

It should also be noted that the array of order above has a unique maximal element  (M). This is the only  planet whose North Pole is not visible from another.

If we now consider a sphere of the same radius as the planets. Remove

from it all North Poles defined by directions that are perpendicular

to the axis of two of the planets. This is a set of area zero.

For every other point on this sphere, there exists a direction in

space that makes it the North Pole, and for that direction, there

exists a unique North Pole on one of the planets which is not visible

from the others. Hence the total area of invisible points is equal

to the area of this sphere, which in turn is the area of one of the  planets.

answer: it's d

step-by-step explanation: