A Geometry Puzzle

[Nobody]

I'll try my best to keep posting puzzles

Given: A regular tetratrahedron with side length x

Find: 1. Volume of tetrahedron in terms of x
2. Inscribe a sphere in the tetrahedron. Find its volume in terms of x

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I am Nobody
Nobody is perfect
Therefore, I am perfect

[Bicho the Inhaler]

I like these a lot My very first puzzle here was along these lines. It's soooo tempting, but I won't post the answer. edit: Not now, at any rate.

[This message has been edited by Bicho the Inhaler (edited 03-29-2002 09:14 PM).]

[SaberKitty]

I think- the volume one is V=((sqrt2)*(s^3))/12
but i could be wrong... haven't done geometry(unless you count trig-which sucked) since 8th grade...hence to say it's been a while
i'll actually have to figure out the sphere one-if my answer was right

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"You mean you have other words?" cried the bird happily. "Well, by all means, use them. You're certainly not doing very well with the ones you have now."

[Nobody]

Nice to see that you have an interest in geometry. I'd like to see mathgrant give it a shot. I am, only about a year older than he is. I'll post another in a few days. (That is, when im certain that someone has the solution.)

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I am Nobody
Nobody is perfect
Therefore, I am perfect

[SaberKitty]

Am I right? It would be a shame to attempt the sphere one with an incorrect given

[cubestudent]

nice puzzles.

my quick calculation claims that the first one is: x^3(sqrt(2)/12).
this seems a little wonky to me, but what do i know?


edited to fix the rotten invis tags



[This message has been edited by cubestudent (edited 03-29-2002 09:38 PM).]

[Nobody]

part 1 is solve, get to work on 2

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I am Nobody
Nobody is perfect
Therefore, I am perfect

[cubestudent]

i'll go for it:

my quick calculation #2 sez:
(pi * x^3 * 2^(7/2))/ 3^(11/2)
aka:
(8sqrt(2)/243sqrt(3)) * pi(x^3)

[Nobody]

I got something different

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I am Nobody
Nobody is perfect
Therefore, I am perfect

[cubestudent]

my reasoning (spoiler):

i think the sphere would touch the midpoints of each of the four triangular faces.

which i think would make the center of the sphere overlap the center of the tetrahedron.

the center of the tetrahedron should be in the vertical line from the 'top' point to the center of the 'bottom' face, one third the way 'up'. //the 's refer to the tetrahedron being oriented with one of it's faces parallel to the floor.

thus the radius would be 1/3 that height, which i thought was: x * (sqrt(2)/sqrt(3))

fitting this into the formula for a sphere's volume: 4/3 pi*r^3 gives my answer.

but there's a whole lot of 'i think's and 'should's in there. . .

[Griffin]

CubeStudent - (hidden) "thus the radius would be 1/3 that height, which i thought was: x * (sqrt(2)/sqrt(3))"

The middle point of a triangle or tetrahedron is called a centroid. The centroid is 1/3 the height of a triangle(from the side), but 1/4 the height of a tetrahedron. So the radius is x * sqrt(2) / (4*sqrt(3)).

[Nobody]

Would anyone like to tell me how to post invisible? I'm kinda new at this.

[Nobody]

test test test

[Nobody]

Invisible Reply: You are right about the center of the sphere overlapping the center of the tetrahedron, but I really do not think that the center will be located, so convieniently, 1/3 or 1/4 of the way up. Try finding the center a different way.

[Bicho the Inhaler]

I think Griffin is right, actually. I conjecture that the center of mass is 1/4 the way up along any "median" of any tetrahedron, although I don't have a proof handy. But Nobody, the regular tetrahedron case is easy to see because the medians are also altitudes and there's so much symmetry. The volume of the tetrahedron is Bh/3, where B is the area of a base and h is an altitude from vertex to center of opposite base. Let h* be the distance from base to center of tetrahedron. Then, by decomposing the tetrahedron into 4 congruent tetrahedra with base B and height h*, we see the volume is also 4Bh*/3, so

4Bh*/3 = Bh/3, or h* = h/4, QED.

I take it you found the volume some other way; so did I. Maybe we did it the same way...

[This message has been edited by Bicho the Inhaler (edited 03-30-2002 02:25 AM).]

[Nobody]

like most mathematicians, i won't believe it till i see a proof (so stubborn, are we?) I solved it using very basic algebra and geometry, the answers that i got contradict that the radius is 1/4 the distance of the altitude. I would very much like to see your process for solveing this problem though.

This problem is original to me and i solved it on my own. My answers, of course, have a possibility of error.

[SaberKitty]

I get in invisible
the radius =
sqrt(2((1/2 of the altitude of one face)^2))
so when I figure it out, I'll cube it and multiply by ((4/3)*(pi))
edit 1: can't multiply
edit 2: can't use invisible tags



[This message has been edited by SaberKitty (edited 03-30-2002 02:49 AM).]

[Nobody]

I, again, got something different. I would like to see your process for solving it.

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I am Nobody
Nobody is perfect
Therefore, I am perfect

[SaberKitty]

nobody, I found an error in my calculations-Ican'tmultiplyintigers- and so i posted what I should have done or at least ment to before my numbers went ary

[cubestudent]

i guess i'll go again, then:

thanks to Griffin's fixing my assumption about a tetrahedron's centroid:
(i'll fix you, someday )
my answer becomes:
(1/(36sqrt(6))) * pi * x^3
or, more elegantly:
6^(-5/2) * pi * x^3

[Nobody]

When i solved this problem, i made no assumption that the centriod of the tetrahedron lies 1/4 "up" from the base. I solved it differently. As a result, my answers are different. I still won't believe that the centroid lies 1/4 up untill I see a proof.

[cubestudent]

Nobody: Bicho did prove it, albeit not formally.


i'm not ambitious enough to try to draw 3D in ascii, so i'll prove the 2D triangle, and hope it clicks what is intended for the 3D tetrahedron.

code:

---

                .a triangle, cut thusly, leaves three triangles


      /|\           of equal area, each 1/3 the original area


     / | \          A' = (1/3)A


    /  |  \       .the new triangles have the same base length


   /   |   \          B' = B


  /  _/ \_  \       .so:


 / _/     \_ \          (1/2)BH' = (1/3)(1/2)BH


/_/_________\_\        .and:


                           H' = (1/3)H





Thus the centroid is 1/3 the length along a median line.

---

use similar reasoning for the tetrahedral case, except that there are four faces, which yield four smaller tetrahedrons, each with 1/4 the volume.

they've got the same base area, so the heighth must be 1/4 of the original heighth, and the centroid must be located 1/4 of the way along a median line.

i'm assuming equilaterals, of course, just to make things simpler. and because that's what's asked about in the problem.

make sense?

[edited for formatting and typos]



[This message has been edited by cubestudent (edited 03-30-2002 05:59 AM).]

[cubestudent]

just a little addition question i'm not sure i've got the capability to answer (certainly not quickly or easily):

is the centroid of a tetrahedron ALWAYS the center of an inscribed sphere?

[SaberKitty]

i'd assume it would be if all angles are = (and all sides are equal, which goes without saying)

[Bicho the Inhaler]

Nobody- they're right, I did prove it, although I left out many steps. Cubestudent explained it much better than I did. I don't think there's any way that you could correctly determine the radius to be anything else using another method; you might have just made a small mistake (or we all did). I did get px^3/(36Ö6) for the volume of the sphere, using different methods. (aside: want to find out about using special symbols? Click here)

Here's what I did:
You can imbed the tetrahedron (with great symmetry) in a cube. The 6 edges of the tetrahedron become diagonals on the face of the cube. I can't draw this too well in the reply box; maybe I'll get around to making a graphic of it. (The 4 vertices of the tetrahedron also coincide with 4 of the cubes 8 vertices.) Hence, we have it imbedded in a cube whose side length is x/Ö2. Now, you can see that the distance between opposite edges of the tetrahedron is the side length of the square, and hence the distance from an edge to the center of the square (and tetrahedron, by symmetry) is half of this length, making it x/(2Ö2). The distance from an edge to the center of a base triangle is x/(2Ö3) (There are many ways to get this; you can use the centroid/median 1/3 relationship or dissect the triangle into smaller triangles, etc.) From here we see that the radius of the incircle and the distance from edge to center of a base triangle form the legs of a right triangle of which the distance from edge to center of tetrahedron is the hypotenuse, and use the Pythagorean theorem to get radius = x/(2Ö6).

There might be simpler ways, but I think that one's really cool.

cubestudent- I don't think the centroid of any tetrahedron is necessarily the center of the inscribed sphere; in fact, I wouldn't be surprised if the two were never equal unless the tetrahedron has 4 congruent bases. For tetrahedra with 4 congruent bases (equilateral triangles or not), I think every "reasonable" measure of center would end up in the same place. I think this makes intuitive sense...

Also, would the people who got #1 mind showing how they got it? This is very closely related to a puzzle I once posted, so I'm holding off.

[Nobody]

Here's what I did

This needs a bit of visual thinking or a nice diagram, try to follow me through. We know, that the center of the sphere lies on the altitude of the tetrahedron. Because this is a regular tetrahedron, all altitudes are the same. Therefore, the center of the circle will lie at the intersection of two of the altitudes. I first constructed an altitude to the tetrahedron. On the bottom face of the tetrahedron, I constructed a median, which is perpendicular to the altitude. That median commects the midpoint of the base to its opposite vertex of the base. At the vertex, construct another altitude to its opposite face. At the midpoint, construct another median, which should intersect with both altitudes that were constructed. The entire constructed figure lies in one plane and contains two altitudes. Using the lengths of the medians and the altitudes, it is possible to solve for the radius. I showed that certain triangles in the firgure were simlar to one another, and solve for the unknown length.

[Nobody]

Sorry guys, found an error in my calculations (we all make mistakes). I saw 3 methods to solve this problem, they all work. I like them all! Heres the next problem.

Into how many regions will 5 random planes divide space? Explain. Give a rule for x random planes.

(In random, i mean dividing space into the most regions.)

[mikegoo]

Bicho could you describe a tetrahedron with 4 congruent bases that are not equilateral triangles?...I, for the life of me, can not come up with one.

Could you explain why you don't think the centroid would be the center of an inscribed sphere? (Intuition tells me it should be, but i'm too dang lazy to any actual work on the problem right now).

[cubestudent]

new problem:

i think the answer's 26.

the general form should be y = (1/6)x^3 + (5/6)x + 1

where y is max # of regions from x cuts.

[Bicho the Inhaler]

26<---?

In general, I think it's x^3/6 + 5x/6 + 1.<---?

grrr...less than a minute!

[This message has been edited by Bicho the Inhaler (edited 03-30-2002 05:01 PM).]

[Bicho the Inhaler]

As for the tetrahedron, you can have one with four congruent bases that are any* triangle. You can "unfold" a tetrahedron by taking one of its vertices and pretending to pry the three converging faces apart at that vertex. Now you can lay it on a plane: you have one triangle in the middle and 3 triangles adjacent to it. We're going to reverse this process, and we'll start with 4 congruent triangles. How? Take your arbitrary* triangle and connect the midpoints of its edges. You've dissected your triangle into 4 congruent triangles (1 in the middle and 3 surrounding it) that are similar to the original triangle. Now you can fold up along the edges of the middle triangle and get the tetrahedron. It will work*. Try it with paper and scissors!

I'm thinking about the centroid-- give me a minute

*edit: actually, it has to be acute (see below)

[This message has been edited by Bicho the Inhaler (edited 03-30-2002 05:45 PM).]

[cubestudent]

bicho: i'm pretty sure that folding trick doesn't work.
prove it by using a simple 1 * 1 * Ö2 right triangle.

[This message has been edited by cubestudent (edited 03-30-2002 05:25 PM).]

[Bicho the Inhaler]

All right: pick a tetrahedron that is very thin and tall. For simplicity, give it an equilateral base with edge = 1 and make the other 3 isosceles, say with base = 1 and other edge = 128. Very thin. Sketch in the inscribed circle. It will stay close to the equilateral base, see? But the centroid: remember, centroid = center of mass. It will be skewed toward the vertex and won't be close to the base like the center of the incircle. We can even try making the tetrahedron arbitrarily tall...as we increase the height of the tetrahedron (keeping the same base) without bound, the center of the incircle approaches a limit near the base (specifically, the limit is 1/(2Ö3) units above the center of the base), but the centroid does not approach a limit; it shoots off in the direction of the vertex.

[cubestudent]

agreed.

so i guess my new question is:
(i'll label it Geo4 for formality's sake)

Geo4:
a)Is there a non-equilateral tetrahedron whose centroid is the center of its inscribed sphere?
b) if so, what's it look like?

[answered by Bicho: any tetrahedron with four congruent faces works.]

i think the answer's no, but for questionable reasons. i certainly can't prove it either way. [edit]guess i was wrong. *shrugs* go Bicho![/edit]


[This message has been edited by cubestudent (edited 03-30-2002 10:45 PM).]

[Bicho the Inhaler]

Good catch. I should have said "any acute triangle." You can include or exclude right triangles based on your attitude toward degenerates. Thanks.

[cubestudent]

i think i'm still a little skeptical.

my mental images keep telling me (perhaps erroneously) that the edges don't quite meet up.

i'll have to try it out.

[Bicho the Inhaler]

edit: I forgot to mention that this post refers to "Geo4".

On the contrary; I think the answer is yes. Take a tetrahedron with 4 congruent (acute) bases like I described. Remember the embedding/inscribing in a cube thing? Well, similarly, you can embed a tetrahedron with 4 congruent bases in a rectangular solid (a cube if it's regular). Now, this rectangular solid has an unambiguous center. I claim that this center is also, unambiguously, the center of the tetrahedron embedded in it. Thus the centers of the inscribed sphere, the circumscribed sphere, and the sphere tangent to the edges, as well as the centroid and intersection of altitudes (they should intersect in a point, I think) will be coincident.

[This message has been edited by Bicho the Inhaler (edited 03-30-2002 05:54 PM).]

[Bicho the Inhaler]

I think a mediocre picture says it the best:

[CrystyB]

You're rushing a bit, Bicho. If you want all that to be equal, then i'll bet my last cent on the fact that you'll get a regular. Think 2d for a minute. If any two of the four important 'interior' (they might not be) points of a triangle coincide, the triangle is equilateral.

Ow and i'm sorry i didn't believe your fold will work. It does.

[This message has been edited by CrystyB (edited 03-30-2002 08:17 PM).]

[cubestudent]

Bicho: you have me convinced. The centroid of a tetrahedron with four congruent sides is the center of its inscribed sphere.

which begs: how about tetrahedrons without four congruent sides? what about those with three congruent sides, or two pairs of two? other cases?