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 A Geometry Puzzle Goto page Previous  1, 2
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Bicho the Inhaler
Daedalian Member

 Posted: Sun Mar 31, 2002 2:21 am    Post subject: 41 I don't know if I can prove all of it, but I'll do what I can: In the above figure, rectangular solid has a definite center, agreed? This point may be identified as its center of mass, the center of its circumscribed sphere, the intersection of any two of its 4 diagonals, or the intersection of any two perpendiculars from the centers of two adjacent rectangular sides. 1. The center of the circumscribed sphere of the tetrahedron coincides with the center of the rectangular solid. Proof: almost trivial because the two figures are inscribed in the same sphere. 2. The center of the inscribed sphere of the tetrahedron coincides with the center of the rectangular solid. Proof: take the tetrahedron, and at its vertices make 4 similar tetrahedrons of 1/2 the size as the original (reminiscent of the Sierpinski construction). Now, you have some space in the middle of the tetrahedron that isn't part of any of the smaller tetrahedrons. This space comprises a very symmetric octahedron (it is formed by connecting the centers of adjacent rectangular faces of the rectangular solid). This octahedron clearly has an unambiguous center that coincides with the center of the rectangular solid, and because it shares half of its surface with the tetrahedron, the inscribed sphere is also inscribed in it. Its center must be the center of the octahedron and thust the center of the rectangular solid. I was wrong about the "sphere tangent to the edges" because it doesn't actually exist unless the rectangular solid is a cube, e.g., the tetrahedron is regular. I'm sorry I mentioned it. 3, 4, 5, etc. All the other "reasonable" measures of center coincide with the center of the circumscribed sphere, and therefore the center of the rectangular solid. Proof: Suppose we have a "center" C of the tetrahedron that does not coincide with the center of the circumscribed circle. Then there exist 2 vertices A and B such that d(C, A) < d(C, B) (Euclidean distance function). Call the other two vertices G and H. But notice that with one simple 180 degree rotation along one of the principle axes of the rectangular solid, we can exchange the positions of A and B (and we will also exchange the positions of G and H). Other than the names of the vertices, everything is exactly like the original picture; it isn't transformed in any way. For the center C to be "reasonable", I think this means it should be in exactly the same place. This means d(C, A') < d(C, B'), where A' and B' are the vertices A and B were mapped to by our single rotation. But A' = B and B' = A, so d(C, B) < d(C, A), which is a contradiction. So our center C was not reasonable. Actually, that last proof is the only thing we need, but I'll leave the other two in. I think this proves that the center of the tetrahedron will be the same no matter how we measure it.
CrystyB
Misunderstood Guy

 Posted: Sun Mar 31, 2002 3:21 am    Post subject: 42 No matter how we measure it? "Euclidean distance function" is only ONE measure...
cubestudent
3D Member

 Posted: Sun Mar 31, 2002 3:36 am    Post subject: 43 in the same vein as Geo3 (the cut-a-sphere-using-planes-puzzle): Geo5: take a space of infinite dimension. begin cutting it with (infinity minus one)-dimensional spaces. into how many pieces can you cut the space using five cuts? six? what's the general equation? note: the phrasing may be intimidating, but this is actually quite easy.
CrystyB
Misunderstood Guy

 Posted: Sun Mar 31, 2002 3:50 am    Post subject: 44 i'm too tired to look it up. I've seen this solved somewhere, and i even know where! Go figure...
Bicho the Inhaler
Daedalian Member

 Posted: Sun Mar 31, 2002 4:02 am    Post subject: 45 cubestudent, I think you mean 32, 64, and 2^n, where n is the number of cuts. However, I must strenuously object to the notion of an "(infinity minus one)-dimensional space". CrystyB, what exactly do you mean by your comment about Euclidean distance?
cubestudent
3D Member

 Posted: Sun Mar 31, 2002 4:14 am    Post subject: 46 you're correct, obviously. i'd knew you'd get it super-quick, and i almost included a plea that you not answer it and i think the idea of an (infinity minus one)-dimensional space is a fantastigorically stupendiriffic idea! so, OVERRULED!!
Nobody
Icarian Member

 Posted: Sun Mar 31, 2002 7:15 am    Post subject: 47 Heres the next problem Given: 3 congruent circles, which are tangent to one another. The three circles are inscribed within a circle with radius r. Find: the area bounded by the 3 points of tangency, and the arcs in between them in terms of r.
cubestudent
3D Member

 Posted: Sun Mar 31, 2002 11:29 pm    Post subject: 48 well, i've got an answer. it's pretty ugly, which makes me skeptical of its correctness, but here it is: (3r^2(Ö3 - pi/2)) / (7 + 4Ö3) [edited to make it look prettier (and because i forgot the ^2 )] [This message has been edited by cubestudent (edited 03-31-2002 08:52 PM).]
Bicho the Inhaler
Daedalian Member

 Posted: Mon Apr 01, 2002 1:13 am    Post subject: 49 Not what I got: R^2(30pÖ3 - 51.5p - 21Ö3 + 36) Also not very pretty. edit: small typo [This message has been edited by Bicho the Inhaler (edited 03-31-2002 08:14 PM).]
Nobody
Icarian Member

 Posted: Mon Apr 01, 2002 2:38 am    Post subject: 50 I got something different from both of you. My answer is not far off from Bicho's. Of course, the process is more important than the answer. ------------------ I am Nobody Nobody is perfect Therefore, I am perfect
Griffin
Daedalian Member

 Posted: Mon Apr 01, 2002 3:01 am    Post subject: 51 Nobody - I see two interpretaions of this puzzle: I worked in as in situation B and got the same answer as cubestudent.
Nobody
Icarian Member

 Posted: Mon Apr 01, 2002 3:20 am    Post subject: 52 I meant this puzzle to be interpreted as for B, sorry if it was unclear. This is what i got, perhaps it's the same as cubestudent's answer (i simplify my answer as much as possible and never leave radicals in denominators) A= r^2(6pi(sqrt3)-(21/2)pi+21(sqrt3)-36) I would like share different methods of solving this problem with you guys. [This message has been edited by Nobody (edited 03-31-2002 10:28 PM).]
Bicho the Inhaler
Daedalian Member

 Posted: Mon Apr 01, 2002 4:57 am    Post subject: 53 I worked out the "negative" of Griffin's figure A; that is, pR^2 - (red area). I guess there are more than 2 interpretations Now I get exactly what Nobody got. edit: and I can "simplify" (if that's the right answer) cubestudent's answer to this, also. [This message has been edited by Bicho the Inhaler (edited 04-01-2002 12:08 AM).]
Bicho the Inhaler
Daedalian Member

 Posted: Mon Apr 01, 2002 5:01 am    Post subject: 54 Oh, sorry. My method was subtracting half the area of a small circle (three sixths) from the area of the equilateral triangle with vertices at the centers of the 3 circles (side length = diameter of small circle), and then converted to units in terms of big radius.
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 5:15 am    Post subject: 55 i did it the same way. great minds. . .
Nobody
Icarian Member

 Posted: Mon Apr 01, 2002 5:54 am    Post subject: 56 Indeed, i did it the same way too. The hardest part is not finding the area, but the radius of one of the 3 congruent circles in terms of r. Hmm ... running outa puzzles. I'll look for some more, in the mean time, would any of you like to post one up?
Bicho the Inhaler
Daedalian Member

 Posted: Mon Apr 01, 2002 8:47 pm    Post subject: 57 All right, I've managed to prove it, so I'll make it a puzzle: Prove that the centroid of an arbitrary tetrahedron (the intersection of its 4 medians (the segments connecting vertices to centroids of opposite bases)) lies 1/4 the way up (from a base) along any one of its medians. You may find the explanation more difficult than the proof edit: actually, depending on your method, you might very well not (there are several ways to prove it). [This message has been edited by Bicho the Inhaler (edited 04-01-2002 05:27 PM).]
Bicho the Inhaler
Daedalian Member

 Posted: Tue Apr 02, 2002 3:18 am    Post subject: 58 In fact, this is very easy, and it is no harder to prove the general case: The centroid of an n-simplex (for n > 0) is 1/(n + 1) the way up from a base along any one of its medians, a median being the segment connecting the centroid of a base (which is an (n-1)-simplex) to the only non-adjacent vertex. An "n-simplex" is the generalization of a tetrahedron to n-dimensional Euclidean space: 0-simplex: point (the only possibility) 1-simplex: segment (bounded by 2 points) 2-simplex: triangle (bounded by 3 segments) 3-simplex: tetrahedron (bounded by 4 triangular faces) 4-simplex: (bounded by 5 tetrahedral facets) 5-simplex: (bounded by 6 4-simplex facets) etc. The case n = 3 (tetrahedron) is fun to prove geometrically, which is what I had originally. My algebraic proof is far more straightforward and general but not nearly as cool, IMO.
cubestudent
3D Member

 Posted: Tue Apr 02, 2002 3:30 am    Post subject: 59 i think most of my problem is that it's too intuitive for me to grasp how to prove it. i'm sure i could do it with calculus, but i haven't done any real math in six years, so i'm woefully out of practice.
Bicho the Inhaler
Daedalian Member

 Posted: Tue Apr 02, 2002 8:04 pm    Post subject: 60 Looks like this one isn't as fun as I thought it would be...I'll post the solution tomorrow if nobody's solved it. I'll try to come up with a better one next time
Griffin
Daedalian Member

 Posted: Tue Apr 02, 2002 10:42 pm    Post subject: 61 I've been working on it Bicho and think I have something based on your proof for the regular tetrahedron. I liked the problem and would like to see other ways to prove it. Spoiler space instead of invisible (I needed to use some diagrams) . . . . . . . . . . . . . . . . . . . . . . . Take a triangle XYZ. Let C be the centroid and N be the midpoint of XY. The line ZN splits triangle XYZ into two triangles (XZN and NZY) of equal area (because they have the same base length and same height). Similarly, the line CN spilts triangle XCY into two triangles of equal area. By simple subtraction we can see triangles ZCX and ZCY have the same area (and by symmetry triangle XCY is also the same) A similar technique can be used so show that a tetrahedron can be split into four smaller tetrahedrons of the same volume. Here M is the centroid of the of the tetrahedron and C is the centroid of triangle XYZ. Since WCXZ, WCXY, and WCYZ all have the same volume and MCXZ, MCXY, and MCYZ all have the same volume, WMXZ, WMXY, WMYZ and MXYZ must all have the same volume. Now since tetrahedron MXYZ has 1/4 the volume of tetrahedron WXZY and the same base, it must be 1/4 the height. By similar triangles, the line MC must be 1/4 the median WC, which is what we are trying to prove.
Bicho the Inhaler
Daedalian Member

 Posted: Wed Apr 03, 2002 12:22 am    Post subject: 62 Nice. It took me a little while to figure out just what you meant by "similar triangles"; you mean the two right triangles with one leg being the altitude of a tetrahedron and the hypotenuse being the median of one of the two tetrahedra. Very different from mine; like I said, there are several ways. I think I'll still wait until tomorrow before posting my proofs. [This message has been edited by Bicho the Inhaler (edited 04-02-2002 07:22 PM).]
Mrs.princess
Icarian Member

 Posted: Wed Apr 03, 2002 4:19 pm    Post subject: 63 This is the most confusing thing I have ever seen.
impossibleroot
Hi-Keeba!

 Posted: Wed Apr 03, 2002 4:57 pm    Post subject: 64 You should see the IT 3 thread, Fraulein Princess...
Bicho the Inhaler
Daedalian Member

 Posted: Wed Apr 03, 2002 8:34 pm    Post subject: 65 Oh, come now, it ain't so bad. It gets in your blood. Anyway, I think it's time to kill this sucker. My original puzzle was solved by Griffin. Here are my solutions to that problem and the general case: Geometric proof for the arbitrary tetrahedron. For this proof we will assume that the centroid-median 1/3 relationship is true for an arbitrary triangle. This fact itself is easy to prove, using Griffin's method above or a far simpler version of this tetrahedron argument. Pick any arbitrary tetrahedron. Map each vertex of the tetrahedron to the centroid of the opposite face to form another tetrahedron. As you might intuitively expect, this tetrahedron is similar to the original (with ratio 1/3). There are many ways to prove this. For example, pick any face of the small tetrahedron, and notice that it can be formed by connecting the midpoints of a triangle with vertices on edges of the tetrahedron. Elementary triangle facts (I'm skipping some easy steps here because I don't have a picture) tell us that this latter triangle is parallel to a face of the original tetrahedron, and also that it is similar with ratio 2/3 (use the centroid-median relationship). The face of our small tetrahedron is thus similar to the face of the big one with ratio 1/3. Repeat for all the faces, and conclude that the small tetrahedron is similar to the big one with ratio 1/3. It is also easily verified that every median of this small tetrahedron is contained in a median of the big one (skipping steps because I don't have a picture), and thus the two tetrahedra have the same centroid. (If we had any doubts, now would be a perfect time to prove that the 4 medians of the tetrahedron are concurrent, but I digress.) Pick a vertex A of the tetrahedron. Call the centroid of the opposite base B. Call the centroid C. We wish to show that BC = AB/4, or equivalently (by collinearity), that BC = 1/3AC. But this is easy to see because by using similarity, BC on the small tetrahedron is the corresponding part of AC on the big tetrahedron (because we mapped A to the centroid of the opposite base, which happens to be B), and because the ratio of similarity is 1/3, AC/3 = BC, QED. Not as simple as Griffin's proof. But anyway: Algebraic proof for the arbitrary simplex. It's a pretty well known fact that the centroid of a finite "set of points" (every simplex belongs to this category) can be found by averaging the coordinates of the vertices, i.e., the ith coordinate of the centroid is the average of the ith coordinates of all the points. You can stop reading now; the rest is easy algebra. In fact, I think I won't even write it out. To prove it, all you have to do is represent each of your n+1 points as an ordered n-tuple in Euclidean n-space. Pick n of these points to be the base, and let the other be the vertex V. Find formulas for the centroids A of the base and B the whole simplex, and merely compare the vectors AB and AV.
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