Tetrahedron

Printable View

January 5th 2010, 05:16 PM
Vicky1997

1 Attachment(s)

Tetrahedron

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point p is selected at random inside the circumscribed sphere. The probability that p lies inside one of the five shperes is closest to

a) 0 b) 0.1 c) 0.2 d) 0.3 e) 0.4

The answer is c.

And I am clueless how i could get any numbers at all.
Could someone take a look at my attached horrible drawing?
I just want to make sure I have the correct idea.

Vicky.
January 6th 2010, 12:02 AM
earboth

1 Attachment(s)

Quote:

Originally Posted by Vicky1997

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point p is selected at random inside the circumscribed sphere. The probability that p lies inside one of the five shperes is closest to

a) 0 b) 0.1 c) 0.2 d) 0.3 e) 0.4

The answer is c.

And I am clueless how i could get any numbers at all.
Could someone take a look at my attached horrible drawing?
I just want to make sure I have the correct idea.

Vicky.

As far as I understand the question your sketch isn't correct. I've made a sketch but it doesn't take into account that we are dealing with a 3D object.

Probably (:D) the probability could be calculated by:

$p(P\ in\ spheres) = \dfrac{\text{sum of the 5 volumes}}{\text{volume of the circumscribing sphere}}$

Let a denote the length of one edge of the tetrahedron, then

$R_c = \frac23 \cdot h = a\cdot \frac2{9} \cdot \sqrt{6}$

$R_i = \frac12 \cdot R_c$

$r = \frac12 \cdot R_i$

That means the radii are proportional like:

$r : R_i : R_c = 1 : 2 : 4$

Can you take it from here?
January 6th 2010, 05:00 PM
Vicky1997

1 Attachment(s)

Quote:

Originally Posted by earboth

As far as I understand the question your sketch isn't correct. I've made a sketch but it doesn't take into account that we are dealing with a 3D object.

Probably (:D) the probability could be calculated by:

$p(P\ in\ spheres) = \dfrac{\text{sum of the 5 volumes}}{\text{volume of the circumscribing sphere}}$

Let a denote the length of one edge of the tetrahedron, then

$R_c = \frac23 \cdot h = a\cdot \frac2{9} \cdot \sqrt{6}$

$R_i = \frac12 \cdot R_c$

$r = \frac12 \cdot R_i$

That means the radii are proportional like:

$r : R_i : R_c = 1 : 2 : 4$

Can you take it from here?

I think I got the answer.

4/3pi 8r cubed + 4 X 4/3pi r cubed over 4/3 pi 64 r cubed = 0.1875

Thanks a lot for your help.

[COLOR=#ff0000]But I still can't totally drop the idea that the spheres should be similar in size, if not equal. I like making geo-gamis and one of my favorite shapes is actually the tetrahedron. I can pretty much visualize what it will look like. I made my own tetrahedron and spheres so could you please take a look at my attachment. It is quite accurate in proportion.

I don't know how to get the proportion of the radii, so I would be grateful if you could try what you did with my idea. And if it doesn't work please tell me what I have wrong.

Vicky.
January 6th 2010, 08:53 PM
earboth

1 Attachment(s)

Quote:

Originally Posted by Vicky1997

[/COLOR]

I think I got the answer.

4/3pi 8r cubed + 4 X 4/3pi r cubed over 4/3 pi 64 r cubed = 0.1875 <<<<<< correct (Clapping)

Thanks a lot for your help.

[COLOR=#ff0000]But I still can't totally drop the idea that the spheres should be similar in size, if not equal. I like making geo-gamis and one of my favorite shapes is actually the tetrahedron. I can pretty much visualize what it will look like. I made my own tetrahedron and spheres so could you please take a look at my attachment. It is quite accurate in proportion. <<<<<<<< unfortunately no. The circles which are part of the circumscribing sphere must pass through 2 vertices and the midpoint of the opposite edge. Your circles are much to large.

I don't know how to get the proportion of the radii, so I would be grateful if you could try what you did with my idea. And if it doesn't work please tell me what I have wrong.

Vicky.

I've modified your foto a little bit. Have a look.
January 7th 2010, 06:21 AM
Vicky1997

Quote:

Originally Posted by earboth

I've modified your foto a little bit. Have a look.

Thanks!!!

Now I understand what was wrong.(Blush)