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Guest
Guest
Tetrahedron ABCD is circumscribed on sphere, which has a centerpoint S and radius 1. We know that SA>= SB >= SC.
Proove that SA>sqrt(5)
Proove that SA>sqrt(5)
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Re: nightmare
Well, my sphere is a small marble...about 1/16 inch in diameter...brolly said:Tetrahedron ABCD is circumscribed on sphere, which has a centerpoint S. We know that SA>= SB >= SC.
Proove that SA>sqrt(5)
The question needs review. We do indeed know SA=SB=SC=SD since they are all radii of the sphere and = 1. 1>sqrt(5) is very dificult to prove.Tetrahedron ABCD is circumscribed on sphere, which has a centerpoint S and radius 1. We know that SA>= SB >= SC.
Proove that SA>sqrt(5)
BTW: It is better to post than edit. No one knows that you have responded to a question if you edit.
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Guest
Guest
You're not right. This sphere is inscribed into the tetrahedron, and it's not a regular tetrahedron...
Pipe down, brolly; here's what you posted:
"Tetrahedron ABCD is circumscribed on sphere...."
A tetrahedron is REGULAR unless otherwise stated.
With your problem, it HAS TO BE REGULAR, else it makes no sense.
And all you have to do is draw a circle radius = 1 (diameter = 2)
inscribed in an equilateral triangle: same thing as far as your proof goes;
simply show that the triangle's height is greater than sqrt(5).
"Tetrahedron ABCD is circumscribed on sphere...."
A tetrahedron is REGULAR unless otherwise stated.
With your problem, it HAS TO BE REGULAR, else it makes no sense.
And all you have to do is draw a circle radius = 1 (diameter = 2)
inscribed in an equilateral triangle: same thing as far as your proof goes;
simply show that the triangle's height is greater than sqrt(5).
G
Guest
Guest
Well, sorry, but I think we misunderstood...
I suppose this tetrahedron might not be regular, because
SA>= SB >= SC. Of course, it's regular, when SA=SB=SC. But in this assignment we have inequality...
And forgive me this linguistic mistake... I meant sphere inscribed into the tetrahedron.[/b]
I suppose this tetrahedron might not be regular, because
SA>= SB >= SC. Of course, it's regular, when SA=SB=SC. But in this assignment we have inequality...
And forgive me this linguistic mistake... I meant sphere inscribed into the tetrahedron.[/b]
Got to defend brolly. He's right, I misinterpreted circumscribed. I've been trying to draw a 3D picture of his tetrahedron with a sphere inside so I can write some equations, but...
I don't see how a triangle around a circle helps in the proof. 2D is MUCH easier than 3D for me. My most promising attack is a 2D side projection of it and showing that the 3D edges are longer that the 2D but that isn't satisfing to me. The sphere doesn't touch anything in the projection 'cause the "walls" are slanted. Still thinkin.
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Gene
I don't see how a triangle around a circle helps in the proof. 2D is MUCH easier than 3D for me. My most promising attack is a 2D side projection of it and showing that the 3D edges are longer that the 2D but that isn't satisfing to me. The sphere doesn't touch anything in the projection 'cause the "walls" are slanted. Still thinkin.
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Gene
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Guest
Guest
To be honest I don't enjoy 3D geometry as well...
And that's not the easiest assignment,too.
Nevertheless, thanks for your help. If you have any idea, please write it down there.
It's time-consuming problem, isn't it ? :/
And that's not the easiest assignment,too.
Nevertheless, thanks for your help. If you have any idea, please write it down there.
It's time-consuming problem, isn't it ? :/
Nobody can have a worse "spatial imagination" than me!
Thinking about this some more, and hanging on to my equilateral triangle idea:
a = sides
height = 3 (since diameter of circle = 2, and center is 1/3 of way up)
then a = 2sqrt(3)
I now see that this 2sqrt(3) is really the height of the tetrahedron,
since the sphere tangentcy points are on the height lines of the 4
equilateral triangles making up the tetrahedron: the triangle I'm
using has sides equal to the height of these "face triangles".
So the distance from the vertices to center of sphere is 2sqrt(3) - 1;
and that equals 3.464~ - 1 = 2.464~
And sqrt(5) = 2.236~
Sorry that I can't explain my stuff any better, but I'll bet a buck
that this will be what the official solution will look like...
if the tetrahedron is not regular, then the 2sqrt(3) will simply be larger.
Thinking about this some more, and hanging on to my equilateral triangle idea:
a = sides
height = 3 (since diameter of circle = 2, and center is 1/3 of way up)
then a = 2sqrt(3)
I now see that this 2sqrt(3) is really the height of the tetrahedron,
since the sphere tangentcy points are on the height lines of the 4
equilateral triangles making up the tetrahedron: the triangle I'm
using has sides equal to the height of these "face triangles".
So the distance from the vertices to center of sphere is 2sqrt(3) - 1;
and that equals 3.464~ - 1 = 2.464~
And sqrt(5) = 2.236~
Sorry that I can't explain my stuff any better, but I'll bet a buck
that this will be what the official solution will look like...
if the tetrahedron is not regular, then the 2sqrt(3) will simply be larger.
Ok, I'm persuaded. The projection works better than I thought.
Just one comment. If I move D out to infinity then SA is exactly sqrt(5) but anything short of that leaves it > sqrt(5) so I don't think it invalidates your proof.
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Gene
Just one comment. If I move D out to infinity then SA is exactly sqrt(5) but anything short of that leaves it > sqrt(5) so I don't think it invalidates your proof.
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Gene