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jamesshuang
Icarian Member

 Posted: Wed Apr 24, 2002 8:21 pm    Post subject: 1 Here's a really hard 3D geometry problem. You have a cube with 1 cm sides, each side has one diagonal, such that a tetrahedron is formed. A tetrahedron is a figure made with 4 equilateral triangles. What is the altitude of this tetrahedron? In other words, what is the shortest length of a line connecting one tip of the pyramid figure to the triangle opposite of the tip?
kartelite
Daedalian Member

 Posted: Wed Apr 24, 2002 9:20 pm    Post subject: 2 just an initial guess, is it about 1.17? maybe it's more complicated than i gave it credit for if not...
Quailman
His Postmajesty

 Posted: Wed Apr 24, 2002 9:42 pm    Post subject: 3 I get 1.15470054, but I'm no mathematician. The Tetrahedron is SQRT(2) on each edge. The height(?) of each face is SQRT(1.5). The answer you're looking for is the height of an isosceles triangle with base and one side of SQRT(1.5) and the other side SQRT(2).
Bicho the Inhaler
Daedalian Member

 Posted: Wed Apr 24, 2002 10:42 pm    Post subject: 4 The answer is sqrt(3) - 1/sqrt(3), which rounds off to Quailman's answer. You're probably just not thinking inside the box (sorry, you knew I was going to get that in somewhere): The diagonal of the cube is sqrt(3) (the longest one), right? The altitude of the regular tetrahedron is concurrent with this, except it stops short of one endpoint. So the altitude is sqrt(3) - x, where x is the distance from the vertex of the cube to the face of the regular tetrahedron. What is x? x is itself the altitude of a secondary tetrahedron, the base of which is a face of the regular tetrahedron. The volume of this secondary tetrahedron is thus Bx/3, where B is the area of a face of the regular tetrahedron. The edge of the regular tetrahedron is sqrt(2) (diagonal of face of cube), so the area of an equilateral base is just (sqrt(2)/2)^2 * sqrt(3) = sqrt(3)/2 = B (there are many ways to derive this formula and it's pretty well known). But the volume of the secondary tetrahedron is really very easy to find: it has a base that's half of a face of the cube (area = 1/2) and a height equal to the edge of the cube (1), so its area is 1/6. So now we know: Bx/3 = 1/6 Bx = 1/2 sqrt(3)/2 * x = 1/2, so x = 1/sqrt(3). This gives us a final answer of sqrt(3) - 1/sqrt(3).
Bicho the Inhaler
Daedalian Member

 Posted: Wed Apr 24, 2002 10:59 pm    Post subject: 5 Maybe I should add: Newbies, you can read "invisible" text in Quailman's post and mine by highlighting it with the mouse. It's customary to put spoilers in invisible for people who might still want to work on the problem and don't want to see the solution immediately. To see how to use "invisible" text, you can click the "edit/delete message" icon above my last post. I'm saying this because this information isn't in the FAQ or UBB information (yet???), and I think there are newbies here.
jamesshuang
Icarian Member

 Posted: Wed May 08, 2002 1:34 am    Post subject: 6 1.15470054 is the correct rounded answer. The actual answer is 2/3*sqrt(3).
Da5id
Daedalian Member

 Posted: Wed May 08, 2002 12:11 pm    Post subject: 7 Does this qualify as a chestnut yet?
Bicho the Inhaler
Daedalian Member

 Posted: Wed May 08, 2002 3:01 pm    Post subject: 8 As far as I can remember, I have never seen this particular puzzle posted before. ------------------ All your ISE are belong to us!
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