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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2013-01-07 01:36:56

Thanks for the proof

bob bundy
2013-01-06 20:25:00

hi Stefy,

In post 12, I did a similar thing or a tetrahedron.  It's hard to show this in 2D.  I am assuming that it is possible to do this so that the diagram looks the same whichever point (A,B,C,D or E) is "at the top".  OR to prove this cannot be done.

If you can prove what you have said, then it only remains to prove that, for a solution,  all rays must come from a single point.  If this is so, then you have proved impossibilty.


5 rays are drawn from a point in space

I re-read the question.  So all five come from one point.  proving impossibility has suddenly got easier.


Assume it is possible.

Let O be the point the rays come from and OA,OB,OC,OD,OE be the rays.

There is no loss of generality in placing A,B,C,D,E on a sphere, since the rays can always be extended until they cut the sphere without changing the angles between them.

Let A be at the "top" of the sphere.  B,C,D,E must lie in a horizontal plane since these angles are equal: AOB,AOC,AOD,AOE.

I consider three cases: (i) B,C,D,E lie in a plane that cuts O (ii) B,C,D,E lie in a plane below O (iii) B,C,D,E lie in a plane above O.

(i)  angle DOB = 180; angle DOC = 90 =><=

(ii) angle AOB is obtuse.

Let DB and EC cross at F.  angle BFC = 90.

O is the vertex of a square based pyramid OBCDE so BOC < 90  =><=

(iii)  DOA + AOB = DOB  =><=

So it is impossible.



2013-01-06 08:51:35

Hi Bob

No. The OE end wouldn't be at the same angle to OC as OB and OD...

bob bundy
2013-01-06 06:25:11

I'm working on whether the picture is a possible solution or not.  ie.  Can this exist?


2013-01-06 02:54:35

One more thing I must say,
there are infinitely many number of Three ray solutions

2013-01-06 02:50:20

Hi Bob

Yes, the 4 rays solutions is ok, sorry.

2013-01-06 02:27:48

Conjecture: In a plane(or a 2D space), you cannot have more than two rays from a point such that all the angles so formed are equal. Similarly, in a 3D space you cannot have more than four rays with such a property

bob bundy
2013-01-06 01:21:24

That depends on what you say.

If you need to ask the question, then best to say nothing.

Here's a tetrahedron to show what I meant above.  I claim each red line is at the same angle to the other 3.


2013-01-06 01:17:07

Hey Moderators,
Can I say something about the OP? Will it be offending?

bob bundy
2013-01-06 00:52:48

hi Stefy,

6 lines;  Oh yes, I forgot the 180s.  Silly me.  shame

I don't see what is wrong with the tetrahedron though.

I'm going from the middle of the solid out to the 4 vertices.  As it is symmetrical they must all be at the same angle to each other surely ??


2013-01-05 22:58:48

Hi Bob

Those answers for 4 and 6 rays are incorrect. First of all, a tetrahedron has 3 edges from every vertex. Second, if you put the 6 rays along the three axes, you will have three pairs of rays which are at 180 to each other.

bob bundy
2013-01-05 19:49:44

I assumed that these rays are in 3D space with every ray at the same angle to the other 4.

eg. 6 rays can all be at 90 to each other by taking them along the coordinate axes x,y and z.

eg.  4 rays:  Take the start point as the centroid of a regular tetrahedron with each ray being a ray to one vertex.

If O is the centre of a sphere and A,B,C,D, and E are points on the surface of the sphere, I can imagine these points being equally spaced so that every radius is at the same angle to the other 4.  I would think the angle would be obtuse.  But my imagination isn't providing the positions at the moment.  dunno

rajinikanth0602:  You didn't answer my other question.  Is this a puzzle that you can do and have set as a challenge; or one that you want help with?


2013-01-05 13:53:07

rajnikanth wrote:

5 rays are drawn from a point in space such that angle between any two rays is equal then find that angle?

Hi Rajnikanth, Hope you have seen a lot of Tamil Films. Are the angles on the same plane? Or did you mean any adjacent?

2013-01-05 12:43:17

Then it seems it is not possible... I might be wrong though...

2013-01-05 11:56:49

It would be 72 if all the rays were in the same plane.  Judging from the word "challenging" in the thread title, I'd assume the OP wanted the rays to be spread out in 3D space.

I have absolutely no idea. faint

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