Suppose you have a sphere of radius R and you have four planes that are all tangent to the sphere such that they form an arbitrary tetrahedron (it can be irregular). What is the ratio of the surface area of the tetrahedron to its volume?
On the assumption that, as the puzzle implies, the ratio does not depend on the actual configuration of the particular tetrahedron, we can take the ratio for a particular example.
One possibility is to consider the base being an equilateral triangle and the vertex to be somewhere on a line connecting its point of tangency to the center of the sphere. We'll consider four such situations.
First, as that vertex point approaches actually being in contact with the spherical surface, the three faces other than the base approach a total matching the area of the base so the total area would approach two times the area of the base, while the height approaches 2 so that the volume approaches two thirds the area of the base, so the total area would approach three times the volume.
Second, consider the opposite situation, as the distance to the opposite vertex increases without limit (approaches infinity). By projection, the base approaches one where the equilateral triangle is circumscribing a circle of radius 1. Such a triangle has a perimeter of 6*sqrt(3), making the area of the faces other than the base (which becomes more and more negligible for the area part) total approach h*3*sqrt(3). The base area is important for the volume and is 3*sqrt(3), making the volume h*sqrt(3). Note that while the h in one formula is the slant height of a triangular face, and in the other formula is the height of the pyramid, the ratio of the two is approaching 1. So again, the total area approaches three times the volume.
The remaining two are specific concrete cases, rather than limits.
Consider the regular tetrahedron case:
We'll need the dihedral angle for a tetrahedron, which can be found from spherical trigonometry using a sphere about one vertex, where the arc sides of the spherical triangle are the 60-degrees of the faces' angles. Solve cos60=(cos60)^2 - (sin60)^2 * cos d, where d is the dihedral angle, which comes out to arccos(1/3).
Consider a wedge that is a tetrahedron that is 1/6 of the whole tetrahedron: one vertex (call it C) at the center of the sphere (the original sphere of the problem, that is), an edge that coincides with an edge of the original tetrahedron, and another vertex (call it B) that is at the foot of a perpendicular from the center of the sphere to one of the two faces of the original tetrahedron that is adjacent to the coincident edge.
The dihedral angle at that coincident edge that belongs to this sub-tetrahedron is 1/2 that of the whole tetrahedron, so it comes out to (1/2)*arccos(1/3). The distance from vertex B of the little tetrahedron is therefore 1/tan((1/2)*arccos(1/3)). The height of the larger (regular) tetrahedron is then this value times tan(arccos(1/3)). That product is 4. So the volume of the original tetrahedron is 4 times the area of the base divided by 3. The area of the original tetrahedron is 4 times the area of the base, so the area is again 3 times the volume.
Finally, consider the inscribed sphere to be tangent to the xy plane, the yz plane and the xz plane of Cartesian 3-D coordinates, and a plane that is tangent to the sphere and intersects the three axes equally distant from the origin. The equilateral triangle that is cut off on that plane by the coordinate planes, together with the right triangles cut off that meet at the origin, form another tetrahedron that can be tested:
First, at the point of tangency, (x-1)^2+(y-1)^2+(z-1)^2 = 1, with x=y and x=z, so that
3(x-1)^2 = 1
and the point of tangency is at 1+1/sqrt(3), and the plane is defined by x+y+z=3+sqrt(3).
The intersections with the axes then occur at a distance of 3+sqrt(3) from the origin. The area of each of the three right triangles is then (3+sqrt(3))^2 / 2, or (12+6*sqrt(3))/2 = 6+3*sqrt(3). One side of the equilateral triangle is sqrt(2 * (12+6*sqrt(3)), so it's area is (2 * (12+6*sqrt(3)) * sqrt(3)/4, so the total area of the surface is (2 * (12+6*sqrt(3)) * sqrt(3)/4 +3*(6+3*sqrt(3)) =52.9807621135331.
The volume is (3+sqrt(3))^3/6 = 17.6602540378445 and the ratio of area to volume is 3.
Of course these do not prove that the ratio holds for all possible shapes of tetrahedron. But any one of them, together with the assurance of the puzzle poser that the actual configuration need not be known, give the answer as 3. Each of the tetrahedra we tried above has one equilateral triangle that can be used as a base and three equal isosceles triangles to complete the tetrahedron, so we haven't even used a varied set.
Posted by Charlie
on 2004-06-23 15:39:10