# Hidden planet area

A solar system contains some number of stationary perfectly-spherical planets of equal radius. Call a point on a planet's surface private if it can't be seen from any other planet. Show that the total private surface area doesn't depend on how many planets there are and how they're arranged.

(This is in 3D, the number of planets is finite and nonzero, and they are disjoint.)

• @bobbee Same for "stationary" :-) – xnor Jan 15 '15 at 7:45
• If it's a 'solar system' then it must have a sun, surely? That would sometimes interrupt the view from one planet to another. Also, if the planets start out motionless then gravity will accelerate them towards the sun (or if no sun, towards each other) until they impact in a big fireball. – A E Jan 15 '15 at 10:55
• This question does not appear to be about creation and solving of puzzles. meta.puzzling.stackexchange.com/a/2788/228 – Peter Taylor Jan 15 '15 at 15:33
• This puzzle could be made much more difficult (although it's fine as it is) if the question was changed to, "As a function of the number of planets, what is the total private surface area in the solar system? Give upper and lower bounds." Then, the puzzle would require the solver to make the realization that the private surface area is constant wrt position and number of planets, instead of that insight being immediately apparent to solvers. – Kevin Jan 15 '15 at 21:40
• The number of planets may be finite and nonzero, but what about the space? The answers assume that it is Euclidean. What if it's a 3D torus or Klein bottle? Or if it's like our own universe, where the gravity of the planets distorts the space and bends the light? – KSmarts Jan 16 '15 at 19:56

Fix any direction and call it "north." Look at the north poles of all planets. A north pole is private iff there are no planets further to the north. Therefore, there is exactly one private north pole: that of the northernmost planet.

Similarly, for any direction $X$, there is exactly one private $X$ pole. I claim that this means there is exactly one planet's worth of private area. Indeed, if we translate all of the planets' private parts onto a single reference planet, then the private parts combine to cover every point on the planet exactly once.

• I love this line of reasoning. Makes me wonder how well one can argue when two planets are on exactly the same level. – No. 7892142 Jan 16 '15 at 9:52
• @No.7892142 We remove points (one for each two planets, minus cases with three or more planets on one line) from surface. Area stays the same. – Cthulhu Jan 16 '15 at 9:59
• @Cthulu What justifies removing points from a practical point of view? – No. 7892142 Jan 16 '15 at 10:02
• @No.7892142 The fact that we can remove them without changing the area. Alternativly we don't remove the points. Simply note that there is a finite number of these points, so they total to 0 area, so they don't affect the total private area. – Taemyr Jan 16 '15 at 10:40
• @Taemyr Fair enough. – No. 7892142 Jan 16 '15 at 10:44

What one planet can see from another is always a circle/half a planet with $$A = 2\pi r^2$$

If you add one planet to the system you will decrease the private surface on other planets by the same amount you add private area on the new planet. I think it is always one planet of private surface but I don't know how to express that mathematically.

• If you can prove that adding a planet always keeps net private area constant you can easily get to the answer of one planet by induction. – Taemyr Jan 16 '15 at 10:38
• @Taemyr Teach me master :D – Avigrail Jan 16 '15 at 15:35
• It's perfectly straightforward. In the case of a single planet you have 1 planet worth of private area. If adding another planet does not change the private area then this will remain 1 no matter how many planets you add. – Taemyr Jan 16 '15 at 15:39