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## #1 2013-01-04 23:59:28

rajinikanth0602
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### challenging problem for all

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

## #2 2013-01-05 00:15:39

bob bundy
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### Re: challenging problem for all

hi rajinikanth0602

Is this a problem you want help with?  Or an exercise for the rest of us?

And is it exactly five rays or at least five?

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

rajinikanth
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only five

## #4 2013-01-05 01:54:46

anonimnystefy
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### Re: challenging problem for all

I would say 72°.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #5 2013-01-05 11:56:49

scientia
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### Re: challenging problem for all

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.

## #6 2013-01-05 12:43:17

anonimnystefy
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### Re: challenging problem for all

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #7 2013-01-05 13:53:07

Agnishom
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### Re: challenging problem for all

#### 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?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

## #8 2013-01-05 19:49:44

bob bundy
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### Re: challenging problem for all

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.

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?

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #9 2013-01-05 22:58:48

anonimnystefy
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### Re: challenging problem for all

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.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #10 2013-01-06 00:52:48

bob bundy
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### Re: challenging problem for all

hi Stefy,

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

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 ??

bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #11 2013-01-06 01:17:07

Agnishom
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### Re: challenging problem for all

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

## #12 2013-01-06 01:21:24

bob bundy
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### Re: challenging problem for all

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.

Bob

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You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #13 2013-01-06 02:27:48

Agnishom
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### Re: challenging problem for all

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

## #14 2013-01-06 02:50:20

anonimnystefy
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### Re: challenging problem for all

Hi Bob

Yes, the 4 rays solutions is ok, sorry.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #15 2013-01-06 02:54:35

Agnishom
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### Re: challenging problem for all

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

## #16 2013-01-06 06:25:11

bob bundy
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### Re: challenging problem for all

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

Bob

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You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #17 2013-01-06 08:51:35

anonimnystefy
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### Re: challenging problem for all

Hi Bob

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #18 2013-01-06 20:25:00

bob bundy
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### Re: challenging problem for all

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.

LATER EDIT:

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.

EVEN LATER EDIT:

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.

Bob

Bob

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You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #19 2013-01-07 01:36:56

Agnishom
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### Re: challenging problem for all

Thanks for the proof

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

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