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## #1 2006-02-22 22:39:51

rimi
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### fun with logic

Can you place four glasses on a table so that the distances of each glass from the other is exactly the same?

## #2 2006-02-23 01:38:37

ganesh
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### Re: fun with logic

The only possible way it can be done is
(i) Place three glasses on a plane at the three vertices A,B, and C of an equilateral triangle.
(ii) Place the fourth glass at a point D in a different plane, such that AB=BC=AC=AD=BD=CD

Character is who you are when no one is looking.

## #3 2006-02-23 07:52:37

MathsIsFun
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### Re: fun with logic

As in a Tetrahedron ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #4 2006-02-23 08:33:21

Ricky
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### Re: fun with logic

How do you measure the distance?

Assuming that you measure distance by the center of the glass, and you are only measuring in a 2d plane...

Take one glass on the table, and put the other three classes inside of it (stacked).  Then the distance of each glass is.... 0.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #5 2006-02-23 15:43:28

rimi
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### Re: fun with logic

The glasses should be on a plane and therefore the idea of a tetrahedron by Maths Is Fun rules out.
Ganesh,you are trying to draw four equilateral triangles...but then does the angles agree with the theorem that they should be each 60 degrees or the totality of all the angles in a triangle should be 180 degrees?
Ricky ,I didnot quite get your point . May be the fact that I forgot to mention that the glasses should be identical can clear your confusion about the question.

## #6 2006-02-23 16:20:49

Ricky
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### Re: fun with logic

Just like stacking cups, you put one glass inside another.

It's a cheat, but that's what I'm best at.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #7 2006-02-23 16:44:01

rimi
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### Re: fun with logic

No thats not quite right I guess.Anyway the glsses cannot be stacked and should be placed seperately.

## #8 2006-02-23 17:24:27

darthradius
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### Re: fun with logic

After thinking this over at 12:30 in the morning, I have decided that the best solution would be to smash them...

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
-Bertrand Russell

## #9 2006-02-24 03:46:02

mathsyperson
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### Re: fun with logic

I too think that it is impossible to do on just one plane.

If you build a tetrahedron from dry spaghetti and marshmallows, magnetic sticks and balls, or whatever your preferred choice is, then it is rigid, which suggests that there are no other ways to get this to work.

A more mathematical approach would be to place the first glass down.
As the other glasses must all be an equal distance from this glass, so you can draw a circle around this first glass to reflect this. The arrangement is currently infinitely symmetrical, so it doesn't matter where you put the next glass down, as long as it is on the circle.

Drawing a second circle of the same size around this circle shows that there are only two points where the next glass can go, because there are only two points that both circles go through. Again, putting the glass on either of the points will just show a reflection of what would happen if you put it on the other one, so it doesn't matter which you choose.

Putting the third glass down and drawing a circle around it shows that there are no places left where a glass can go so that it will be the same distance from each glass, so it cannot be done.

Why did the vector cross the road?
It wanted to be normal.

## #10 2006-02-24 08:01:11

MathsIsFun
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### Re: fun with logic

Very well reasoned, mathsy! Using two different methods, too.

So any answer to this puzzle is bound to be tricky. Like smash three of them up and place the remains in the fourth or something.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #11 2006-02-24 10:28:13

ryos
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### Re: fun with logic

This is kind of lame, but bear with me.

Lay them on their sides so that they form a square. Two glasses on opposite ends of the square are one glass-length apart.

"Wait," you say, "what about the glasses that are touching? They're much closer to each other than those that aren't!"

Aha, but with logic, I can distort reality. You see, a part of those glasses (the opposite end from the one that's touching) is one glass-length away, so it can be said that all of the glasses are equidistant.

I know, I know...lame. But it's all I've got.

El que pega primero pega dos veces.

## #12 2006-02-24 10:35:54

irspow
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### Re: fun with logic

I have found one possible solution although there would be many others along the same lines.

First for the conditions that I used.  I arbitrarily chose a distance of 4r to exist between any two glasses' geometric centers.  Then I allow the height of the glasses to vary to make this true.

Place two glasses standing up 4r apart.  Next, lay a glass on its side touching the two standing ones.  Repeat second step again on the opposite side of the standing glasses.

No proof is needed to show that the standing glasses are 4r apart.  Also the glasses lying on their sides are 4r apart because they are spaced only by the diameter of the standing glasses and their radii.

Now we let the height vary to accomodate this distance.

Overhead it can be seen that centers along the plane of the table from one lying glass to one standing glass form a triangle two equal sides of 2r.  Thus the base of our vertical triangle will be r√8.

The side of our vertical triangle will be (h/2 - r).  So the hypotenuse of this triangle will be;

d = distance between geometric centers of lying to standing glass

d = √(8r² + h²/4 + r² - hr);

We want d = 4r;

4r = √(9r² + h²/4 - hr);

4r = (1/2)√(36r² + h² - 4hr);

8r = √(36r² + h² - 4hr);

64r² = 36r² + h² - 4hr;

-h² + 4hr + 28r² = 0;

h = [-4r ± √(16r² + 112 r²)] / -2;

h =  [-4r ± √(128r²)] / -2;

h = [-4r ± 4r√8] / -2;

Since h must be positive;

h = (-4r - 4r√8) / -2;

h = [-4r( 1+√8)] / -2;

h = 2r(1 + √8);

So this is just one particular case.  If the dimensions of the glasses and the space wanted were known then a solution could be calculated in a similar manner.

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