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#1 2006-05-11 01:37:18

doraeyee_u_v
Member
Registered: 2006-05-02
Posts: 13

a chair problem

Show that a chair with four equal legs can be positioned so that all its legs are in contact with the floor.

this question is not an easy question, you cannot just say without four equal legs, the chait will fall. but instead, you have to do this questions in a mathematical way.

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#2 2006-05-11 03:00:18

ganesh
Moderator
Registered: 2005-06-28
Posts: 15,128

Re: a chair problem

doraeyee_u_v,
I think the question can be reduced to showing four points in a three dimensional space are coplanar.
I think this can be done mathematically by forming the equation of the plane with the help of the coordinates of three points and showing that the fourth point lies in the plane.


Character is who you are when no one is looking.

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#3 2006-05-11 07:23:48

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,564

Re: a chair problem

If the seat is flat and the legs are the same length then there exists room to make a perfect chair and an imperfect chair.
There are many scenarios with the legs at various angles that would still work, but perfectly vertical legs would be something to prove.    Make vertical the y-axis and flat across the front of the chair the x-axis.   We know the bottom of the seat where the legs come out is perfectly flat, so we say the the y-values are all the same, hence the slope or rise over run on the seat is zero.
Zero = 0 vertical change / some horizontal distance.
Now the legs are vertical, so the x values do not change for the calculations of the bottom of the legs, only the y values.
The y values all are reduced by the same value, the leg length, so all of the y values stay the same for the bottom of the legs.
Therefore if you do slope calculations between any of the six pairs of legs choosable, they will all have a slope of zero again.


igloo myrtilles fourmis

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#4 2006-05-11 14:01:49

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: a chair problem

This should be a geometry problem, but I've forgotten all the theorems.:(


X'(y-Xβ)=0

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#5 2006-05-12 03:20:34

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: a chair problem

The following are assumed:

The flat part of the chair (where your bottom goes) is contained in a single plane.
The four legs are of the same length.
The four legs are all at 90 degree angles to the chair.

Let the flat part of the chair lie in the plane:

And let all four legs be of length L.

The normal to the plane is (A, B, C).  Also, let the position where each leg intersects the plane be:




The unit normal of the plane is:

As so the normal of length L (the lenght of the leg is:

So the position of the feet of the chair will be at:

And it's the same idea for L2, L3, and L4.  All of these points can be placed into the plane:

So the bottom of all four legs are in that plane, which we call the "floor".


"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..."

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#6 2006-05-12 04:02:30

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,564

Re: a chair problem

What's a unit normal?  Just wondering.


igloo myrtilles fourmis

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#7 2006-05-12 04:13:12

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: a chair problem

A unit normal is a normal with a lenght of 1.


"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..."

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#8 2006-05-13 12:24:10

doraeyee_u_v
Member
Registered: 2006-05-02
Posts: 13

Re: a chair problem

thank you very much ricky, this is very clear and i understand it

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