# A puzzle from outer space

Suppose that we mark five arbitrary points, anywhere on Earth.

Is it then always possible to find a point in outer space from which you would see at least 4 of these 5 points?

• I think it's not very well defined. Obviously if the points are next to each other there is an angle but I suspect you are looking for a non trivial solution. – rhsquared Aug 20 '16 at 8:16
• You never said that in the question currently. – haykam Aug 20 '16 at 15:55
• Are you sure you want the [lateral-thinking] tag? It allows a lot of "unconventional" answers that I don't think you intended. – Deusovi Aug 20 '16 at 18:41
• Do we have to account for physickey stuff like space curvature, and light taking a non rectiline path when going near massive bodies? I'm pretty sure the answer can be arbitrarily circumvoluted if we play light billard with all available stars to make sure we can see the 5 points. – Clement C. Aug 20 '16 at 23:25
• This question is ambiguous. What do you mean by "on Earth"? What do you mean by "see"? If I mark points in caves (or under the ocean, or in buildings), they will not be visible from space. A bit more seriously, if I mark a point on the side of a mountain (or cliff), then it will be visible from only about half of the vantage points from which you can see the mountain. So, do you mean "mark five points on the surface of an opaque perfect sphere"? What do you mean by "see"? Can you see an entire hemisphere from space, or is that impossible because it requires viewing from an infinite distance? – Peregrine Rook Aug 21 '16 at 4:50

If we assume that Earth is a perfect mathematical sphere, then

Example:

Put one point into the Northern pole, one point into the Southern pole, and three points onto the equator that form an equilateral triangle.
(1) There is no way to simultaneously see Northern and Southern pole (you would have to move to infinity for this).
(2) There is no way to simultaneously see all three points on the equator. Otherwise you would see the full equator (and you would have to move to infinity for this).
(3) Hence you can see at most one of the poles, and at most two of the three equator points.

• Great. And what about Earth's shape (oblate spheroid )? – ABcDexter Aug 20 '16 at 8:40
• @ABcDexter I believe the technical term is "potato-shaped." – Clement C. Aug 20 '16 at 23:21
• @ABcDexter The exact same argument would apply for an oblate spheroid. In more generality, differentiability of the surface is all that is needed - that implies that the set of points from which a given point $p$ on the surface is visible is contained in a half-space. If you choose any vector $v$ and take two points $p_1$ and $p_2$ respectively minimizing and maximizing the dot product $p_1\cdot v$ and $p_2\cdot v$, you'll find that it is not possible to see both $p_1$ and $p_2$ at once. Choosing $3$ such pairs gives $6$ points such that only $3$ are ever visible at once. – Milo Brandt Aug 21 '16 at 0:13
• It's actually more difficult with a perfect sphere as more points are visible than with any other shape. So any proof for a sphere holds true for a potato etc. – Paul Evans Aug 21 '16 at 23:12

It's:

Not always possible to observe $4$ points. BONUS: Not even for $6$ points!

Because:

For example, place $4$ points on the equator so that they form $2$ pairs of antipodal points.
Now, we can only observe a maximum of $2$ of these points at a time from outer space.
As an aside: we can only observe a maximum of $1$ point from outer space if our line-of-sight is perpendicular to one of the antipodal-pair's connecting diameter.
Add a fifth point point and we can only observe a maximum of $3$ points.
BONUS: Place the fifth point on the south pole and a sixth point on the north pole we still can only observe a maximum of $3$ points.

• nice observation in the bonus section with the 6 points! – elias Aug 20 '16 at 9:28
• Any pair of antipodal points can't be seen at the same time, so you could pick any three arbitrary pairs of antipodes. – f'' Aug 20 '16 at 13:13
• @f'' That's it! Only realised that after answering this puzzle. If $N$ points can be seen from a point in outer space, then you can simply use their antipodal points to have $2\cdot N$ points where only a maximum of $N$ can be seen anywhere from any point in outer space. – Paul Evans Aug 21 '16 at 22:51

Since this puzzle was tagged [lateral-thinking], I am compelled to submit this answer:

Yes, because mirrors. No matter where the points are placed it will always be possible to use (appropriately positioned) mirrors to see all 5 points (or more) simultaneously.

Alternatively:

Yes because we have two eyes. All we need to do is extend the organic connection between them and our brain, enough so that we can see two hemispheres of earth simultaneously. (This would require a whole diameter of earth between our eyes, and some impressive parallax)

Yes, all 5 points should be visible eventually, because the question does not state that they need to be viewed at the same time.

Given that the earth rotates about its axis, most points could be seen in a day from a point approximately on the earth's orbital plane. For the points closest to the poles, the tilt of the earth's axis would reveal them over the course of a year.

• Yeah, my thought too - You can view all the points fro most any spot, eventually, since some will be visible from that spot, and the rest will eventually move into a visibility. – Megha Aug 21 '16 at 1:26
• I believe this is the intended lateral answer. – Glen O Aug 21 '16 at 5:32

It is almost always possible if we consider the Earth a perfect sphere. There are some edge cases which can cause that there is no such angle though.

Reasoning:

If viewed from an ideally far distance, all points on one half of the sphere (not including the bordering orthodrome) can be seen at the same time.
So the puzzle is equivalent to show a half of the sphere which contains at least 4 of the 5 points.

For example if all the 5 points are on a same orthodrome, then this might be impossible - think about 5 points which divides it in 5 equidistant pieces. In this case the problem reduces to finding a half of a circle with 5 marked points, which contains 4 of those points.
Another impossible configuration is like the one given by Gamow.
Or generalizing Paul Evans' idea, even if you have 6 points which are pairwise opposite, you can see at most one from each pair.

However, generally you can find 2 marked points, which are not on opposing poles of the sphere, and that the orthodrome defined by them does not contain any of the other 3 marked points. This orthodrome halves the sphere into two equal parts. By pigeonhole principle, at least 2 of these points will be on the same half of the sphere. These 2 and the 2 on the orthodrome can be seen from a well chosen angle.

NO, because you can choose a point inside of a house or within a cave etc where the view of the point is blocked from viewing from overhead by roof or other physical construct.

Here's a solution similar to Gamow's but with a different arrangement and an illustration.

Place the five points equidistant in a regular pentagon along the equator. The maximum number you can see at any one time is 3 because if you observe from anywhere except along the North/South pole axis 2 or 3 of the points will be behind the earth. 