Date: Sun, 12 Jul 1998 02:54:05 -0400 (EDT) From: George Baloglou To: Dave Rusin Subject: Re: Area of a tetrahedron inside a sphere Professor Rusin, I believe I have proved the following: of all the cones having as their base a fixed circle on the unit sphere and their vertex on the unit sphere, the cone with the maximum surface is the one whose vertex lies at the "North Pole" -- that is, the point *on the unit sphere* whose distance from the base is maximum (equivalently, the point where a plane parallel to the base is tangent to the sphere). I conjecture that the same result holds if one replaces the cone's circular base with *any* fixed polygon inscribed in it (hence in the unit sphere as well) and the cone with a pyramid having as base that fixed polygon. In the case the polygon is a triangle, my conjecture reduces to what I stated on sci.math ... and its validity implies at once that the maximum area tetrahedron inscribed in the unit sphere is the regular one. I am thinking of submitting this as a problem to the Monthly or some other journal -- provided that no one settles the pyramid part and the cone part stands its ground. Best regards, George Baloglou On Sat, 4 Jul 1998, George Baloglou wrote: > > Yes, by "inscribed in a sphere" I mean that the tetrahedron's all four > vertices must lie on the sphere's surface. Now as that fourth vertex > moves to a higher lattitude some triangular sides' area could decrease, > and that's where the difficulties begin ... > > George > > ... What a brilliant way to celebrate July 4th, by the way :-)) > > > On Sat, 4 Jul 1998, Dave Rusin wrote: > > > In article <6n7ci4$9nm@panix3.panix.com> you write: > > >What is the maximum surface area of a tetrahedron inscribed in a > > >sphere of radius 1? (Old-fashioned Calculus works, with extra care, > > >in case the tetrahedron is a regular triangular pyramid (i.e., with an > > >equilateral triangle for base and its top vertex at the "north pole"); > > >the general case would follow if, for example, one could show that > > >for *every* triangular base the maximum area occurs when the fourth > > >vertex lies at the maximum possible distance from that base.) > > > > You're not suggesting that you think the optimal tetrahedron might > > fail to touch the sphere at all four points, are you? If the base is > > fixed and we consider moving the apex along a line perpendicular to > > the base, then the area of _each_ side increases (*) as the distance of > > the apex to the base increases; in particular the total surface area > > will increase until we must cease raising the apex when we encounter > > the sphere. > > > > (*)area = (1/2)(base of triangle, fixed)(height of triangle) > > but (height of triangle)^2=(height of apex off base of tetrahedron)^2+ > > (dist from triangle-base-line to apex-motion-line)^2 > > > > dave > ============================================================================== From: Dave Rusin Date: Sun, 12 Jul 1998 11:31:18 -0500 (CDT) To: baloglou@Oswego.EDU Subject: Re: Area of a tetrahedron inside a sphere I must be missing something here. Yes, I'm sure the right circular cone has the largest surface area of all those with a fixed base and vertex constrained to a fixed sphere containing the base; by symmetry this is a one-variable maximization problem which, simple or not, is straightforward. I'm also confident that this result is true for _regular_ polygons inscribed in the circle, replacing the cone with a pyramid. Proving this would be considerably more difficult, however. Finally, I would be hard-pressed to disbelieve your original conjecture in sci.math, that the regular tetrahedron maximizes the surface area of all tetrahedra in the sphere, but frankly I would am loathe to begin considering this because the most elementary technique, while sure to succeed eventually, would invoke at least half a dozen variables. Some more geometric shortcuts to symmetry would have to be discovered, and I don't see any offhand. BUT -- I'm confused by your latest conjecture. If you now _fix_ an arbitrary polygon in the circle and draw the tetrahedron of maximum area, the top vertex does NOT necessarily lie above the center of the circle. I worked out just one example, more or less at random, and found a different vertex yielding a larger area. I thought I sent it to you already (if not, let me know). So for a _fixed_, _not-necessarily-regular_ polygon, I don't think you've picked the right conjecture. (I suppose, if there is to be any justice, that the optimal vertex should be related in _some_ predictable way from the triangle. My example shows you can't go straight above the center of the circumscribed triangle. Very well; is it perhaps straight above the center of the inscribed triangle? Or one of the other known "centers" of a triangle? One thing mitigating against this proposal is that it seems quite reasonable that the position of the optimal vertex will _not_ travel in a vertical line at all as the radius of the sphere changes, so that the location of the optimal vertex will not be easily related just to some classic points in the triangle. ... Or maybe there is an elegant answer and I just don't see it! ) Naturally, if you do find an beautiful geometric approach it would be perfect for the Monthly. dave ============================================================================== Date: Sun, 12 Jul 1998 15:07:27 -0400 (EDT) From: George Baloglou To: Dave Rusin Subject: Re: Area of a tetrahedron inside a sphere On Sun, 12 Jul 1998, Dave Rusin wrote: > I must be missing something here. Yes, I'm sure the right circular cone has > the largest surface area of all those with a fixed base and vertex constrained > to a fixed sphere containing the base; by symmetry this is a one-variable > maximization problem which, simple or not, is straightforward. Well, I had to show a certain family of integrals to be increasing by showing the integrands to be increasing -- the integrals could not be evaluated; it was fun, tricky and combined "all" parts of Calculus :-) > I'm also confident that this result is true for _regular_ polygons inscribed > in the circle, replacing the cone with a pyramid. Proving this would be > considerably more difficult, however. Why do we need the polygon to be *regular*? [More on this below.] > Finally, I would be hard-pressed to disbelieve your original conjecture in > sci.math, that the regular tetrahedron maximizes the surface area of all > tetrahedra in the sphere, but frankly I would am loathe to begin > considering this because the most elementary technique, while sure to > succeed eventually, would invoke at least half a dozen variables. Some > more geometric shortcuts to symmetry would have to be discovered, and I > don't see any offhand. Exactly, this is the way I feel about the problem. > BUT -- I'm confused by your latest conjecture. If you now _fix_ an arbitrary > polygon in the circle and draw the tetrahedron of maximum area, the top ^^^^^^^^^^^ > vertex does NOT necessarily lie above the center of the circle. I worked > out just one example, more or less at random, and found a different vertex > yielding a larger area. I thought I sent it to you already (if not, let me > know). So for a _fixed_, _not-necessarily-regular_ polygon, I don't > think you've picked the right conjecture. Why *tetrahedron*? If, say, the "base" is a fixed (not necessarily regular) hexagon, then, by connecting various points on the sphere to each of the six vertices, we obtain a family of "hexagonal pyramids" of common base; and I conjecture(d) that the one with maximum surface area is the one with its top vertex at (what I call) the "north pole". [Am I missing something? Please elaborate and send me your example if you still think that is relevant -- and what you write below suggests that it is indeed relevant!] > (I suppose, if there is to be any justice, that the optimal vertex should be > related in _some_ predictable way from the triangle. My example shows you > can't go straight above the center of the circumscribed triangle. Very well; > is it perhaps straight above the center of the inscribed triangle? Or one > of the other known "centers" of a triangle? One thing mitigating against > this proposal is that it seems quite reasonable that the position of the > optimal vertex will _not_ travel in a vertical line at all as the radius of > the sphere changes, so that the location of the optimal vertex will not > be easily related just to some classic points in the triangle. > ... Well, the above paragraph suggests that you have even disproved my original conjecture on sci.math: I guess I need to see your example then :-) > Or maybe there is an elegant answer and I just don't see it! ) It sounds like a very interesting new problem now ... > Naturally, if you do find an beautiful geometric approach it would be > perfect for the Monthly. Indeed. Thanks a lot for your message(s)! George ============================================================================== From: Dave Rusin Date: Sun, 12 Jul 1998 23:08:19 -0500 (CDT) To: baloglou@Oswego.EDU Subject: Re: Area of a tetrahedron inside a sphere Well, you've succeeded in sucking me in to this problem! First let me say that the problem with a circular base is indeed trickier than I thought, not least because (I am embarrassed to say) I confused myself into thinking that the shapes being compared were in fact ordinary cones (the things you get conic sections from) whereas in fact they will all be slices of _elliptic cones_. I'll stick with the word "cone" though, in the topologists' sense: the set of points of the form t*P + (1-t)*Q ( 0 <= t <= 1 ) where P is fixed and Q ranges, in this problem, over a curve of some sort in a plane. Thus even the tetrahedra, etc., are "cones". When you take a cone over circle, you get a region with a fairly complicated surface area. The way I do it I get an integral which at best might be expressed in terms of elliptic integrals; maximizing using ordinary calculus is likely to be a real chore. I tried viewing the surface area by the usual unrolling procedure, as can be done for the cone over any curve. To compute the area, one must determine the shape of the unrolled edge, which appears to be pretty ugly. But it might improve the visualization of the optimization process -- there's probably some way to cast this as a calculus-of-variations problem, for example. Second, let me apologize for this boo-boo >> BUT -- I'm confused by your latest conjecture. If you now _fix_ an arbitrary >> polygon in the circle and draw the tetrahedron of maximum area, the top > ^^^^^^^^^^^ in which I wanted to discuss fixed but arbitrary polygonal bases, but visualized triangular ones. I think there's plenty to do just with triangular bases, but one may of course look at other polygons too. Finally, let me get to what seems to be confusion on your part: >Well, the above paragraph suggests that you have even disproved my original >conjecture on sci.math: I guess I need to see your example then :-) I thought your first conjecture was, if we may select _any_ triangular base in a sphere, and _any_ vertex in the sphere, then the resulting surface area is bounded by that of the regular tetrahedron in the sphere. I still think this conjecture is true. Here you are asking for the maximum of a certain real-valued function on (S^2)^4. I thought your conjecture in yesterday's email was, if we _fix_ any triangular base in a sphere and may _select_ any vertex in the sphere for it, then the surface area is maximized when the vertex is selected to be directly above the center of the circumscribed circle. Here you are asking for the maximum of a family of real-valued functions on (S^2). This is not the same problem. I'm sure the second conjecture is true if the fixed triangle is equilateral. But it's clear that it's not true in general. My previous example was way too complicated. Try this instead: draw a circle at (or near) the equator and pick on it a nearly-degenerate triangle consisting of three tiny line segments way off to the left. Now matter where the vertex, the triangles you draw for the surface area will have bases roughly w, w, and 2w, and will have height roughly the distance from the vertex to the "West Pole" (-1,0,0). So the surface area is maximized roughly by moving the vertex to the "East Pole" (1,0,0) rather than the North Pole (0,0,1). Capish? By the way, this example shows that as the size of the sphere increases, the location of the optimal vertex changes in a fundamental way: to embed that same triangle in a horizontal circle within a larger sphere, the circle will have to drop down from the equator towards the South Pole. As it falls, the optimal vertex location (always roughly straight across from the tiny triangle) will be directly above ... what? Above points in the plane of the triangle which move from the right edge of the circle (when the circle is at the equator) to the middle of the circle (when the circle is near the South Pole of the sphere). So if there is a "clean" conjecture to make, valid for each triangle in the sphere, it's going to have to say the optimal location for the vertex is directly _opposite_ (something), rather than, as I suggested yesterday, being directly _above_ (something). I don't _know_ that any such conjecture is really valid, of course. dave ============================================================================== Date: Mon, 13 Jul 1998 02:34:12 -0400 (EDT) From: George Baloglou To: Dave Rusin Subject: Re: Area of a tetrahedron inside a sphere On Sun, 12 Jul 1998, Dave Rusin wrote: > Well, you've succeeded in sucking me in to this problem! :-) > First let me say that the problem with a circular base is indeed > trickier than I thought, not least because (I am embarrassed to say) I > confused myself into thinking that the shapes being compared were in > fact ordinary cones (the things you get conic sections from) whereas > in fact they will all be slices of _elliptic cones_. I'll stick with the > word "cone" though, in the topologists' sense: the set of points of the > form t*P + (1-t)*Q ( 0 <= t <= 1 ) where P is fixed and Q ranges, > in this problem, over a curve of some sort in a plane. Thus even the > tetrahedra, etc., are "cones". > > When you take a cone over circle, you get a region with a fairly > complicated surface area. The way I do it I get an integral which at > best might be expressed in terms of elliptic integrals; maximizing using > ordinary calculus is likely to be a real chore. Well, not quite, provided one is a born optimist, of course! In my solution there is an one-parameter, one-variable integrand: what I did was to show that for any fixed value of the variable the integrand is a decreasing function of the parameter (which is no other than the angle phi one encounters in spherical coordinates); to do this I had to show a certain derivative to be negative, and that reduced to a two-variable inequality that was fun indeed :-) > I tried viewing the surface area by the usual unrolling procedure, as > can be done for the cone over any curve. To compute the area, one must > determine the shape of the unrolled edge, which appears to be pretty > ugly. But it might improve the visualization of the optimization process -- > there's probably some way to cast this as a calculus-of-variations problem, > for example. Yes, I thought of that, too, looking for a "clever" approach. > Finally, let me get to what seems to be confusion on your part: Not really confusion, just lack of clarity I would say ... let's see: > >Well, the above paragraph suggests that you have even disproved my original > >conjecture on sci.math: I guess I need to see your example then :-) > > I thought your first conjecture was, if we may select _any_ triangular base in > a sphere, and _any_ vertex in the sphere, then the resulting surface area > is bounded by that of the regular tetrahedron in the sphere. I still think > this conjecture is true. Here you are asking for the maximum of a certain > real-valued function on (S^2)^4. Indeed, that was the original problem, and that was my conjecture. > I thought your conjecture in yesterday's email was, if we _fix_ any > triangular base in a sphere and may _select_ any vertex in the sphere for it, > then the surface area is maximized when the vertex is selected to be > directly above the center of the circumscribed circle. Here you are asking > for the maximum of a family of real-valued functions on (S^2). This is > not the same problem. This conjecture was also stated in my sci.math post, with the indication that, *if true*, it then implies the first one! (Indeed the maximum area tetrahedron would then be forced to have all three edges meeting at each of the four vertices equal to each other, hence it would have to be the regular one!) > I'm sure the second conjecture is true if the fixed triangle is equilateral. > But it's clear that it's not true in general. My previous example was way > too complicated. Try this instead: draw a circle at (or near) the equator > and pick on it a nearly-degenerate triangle consisting of three tiny > line segments way off to the left. Now matter where the vertex, the triangles > you draw for the surface area will have bases roughly w, w, and 2w, and > will have height roughly the distance from the vertex to the "West Pole" > (-1,0,0). So the surface area is maximized roughly by moving the vertex > to the "East Pole" (1,0,0) rather than the North Pole (0,0,1). Capish? Neat! It ruins the strategy described in my previous paragraph, though :-( > By the way, this example shows that as the size of the sphere > increases, the location of the optimal vertex changes in a fundamental > way: to embed that same triangle in a horizontal circle within a > larger sphere, the circle will have to drop down from the equator > towards the South Pole. As it falls, the optimal vertex location > (always roughly straight across from the tiny triangle) will be > directly above ... what? Above points in the plane of the triangle > which move from the right edge of the circle (when the circle is at > the equator) to the middle of the circle (when the circle is near the > South Pole of the sphere). I had all kinds of crazy ideas, like -- in the more general case of a plygon rather than just a triangle -- draw all vectors from the center of the circumscribed circle to each of the edges, take their vector sum and then ... try to do something with the tip of the resulting vector! (In the case of a regular polygon, that tip is still the center of the circle; in the case of your triangular example above, that tip moves way outside our sphere, directly *opposite* your optimal vertex ...) > So if there is a "clean" conjecture to make, valid for each triangle > in the sphere, it's going to have to say the optimal location for the > vertex is directly _opposite_ (something), rather than, as I suggested > yesterday, being directly _above_ (something). Yes, I do share that intuition ... > I don't _know_ that any such conjecture is really valid, of course. It all looks wonderfully complicated and mysterious ... kind like an unhappy marriage between two incompatible spouses (the spherical and the planar/triangular/polygonal) :-) Many thanks, George ============================================================================== From: Dave Rusin Date: Mon, 13 Jul 1998 01:54:25 -0500 (CDT) To: baloglou@Oswego.EDU Subject: Re: Area of a tetrahedron inside a sphere >This conjecture was also stated in my sci.math post, with the indication >that, *if true*, it then implies the first one! Oopsie on me. Next time I read more carefully. OK, it sounds like we agree on what the problem is, and it sounds like you have a proof of part of it. Sure, send it to the monthly -- be sure to warn them that it's not as direct as it looks. dave ============================================================================== Question: maximum surface area of a tetrahedron embedded in a sphere? The surface area of a tetrahedron is the sum of the areas of its sides. It is thus a root of the polynomial Prod(t +- area1 +- area2 +- area3 +- area4) which is of degree 8 in t^2. The coefficients of (t^2)^0, ..., (t^2)^8 are symmetric functions of the (area_i)^2, and so may be expressed in terms of S=Sum( (area_i)^2 ), ..., V=Prod( (area_i)^2 ) . These coefficients are, respectively S^8+256*T^4-16*T*S^6+96*T^2*S^4-256*T^3*S^2-128*V*S^4-2048*V*T^2+4096*V^2+1024*V*T*S^2 -8*S^7+6144*V*T*S-1536*V*S^3-8192*V*U+1024*U*T*S^2-128*U*S^4-2048*U*T^2+96*T*S^5+512*T^3*S-384*T^2*S^3 28*S^6-240*T*S^4+576*T^2*S^2-256*T^3-2048*U*T*S+4096*U^2+512*U*S^3+1280*V*S^2-7168*V*T -56*S^5-384*T^2*S+320*T*S^3+1024*U*T-768*U*S^2+2560*S*V -240*T*S^2+96*T^2+512*U*S+70*S^4-2176*V 96*T*S-56*S^3-128*U 28*S^2-16*T -8*S 1 so the polynomial in X is poly:= 1*(S^8+256*T^4-16*T*S^6+96*T^2*S^4-256*T^3*S^2-128*V*S^4-2048*V*T^2+4096*V^2+1024*V*T*S^2) +X*(-8*S^7+6144*V*T*S-1536*V*S^3-8192*V*U+1024*U*T*S^2-128*U*S^4-2048*U*T^2+96*T*S^5+512*T^3*S-384*T^2*S^3) +X^2*(28*S^6-240*T*S^4+576*T^2*S^2-256*T^3-2048*U*T*S+4096*U^2+512*U*S^3+1280*V*S^2-7168*V*T) +X^3*(-56*S^5-384*T^2*S+320*T*S^3+1024*U*T-768*U*S^2+2560*S*V) +X^4*(-240*T*S^2+96*T^2+512*U*S+70*S^4-2176*V) +X^5*(96*T*S-56*S^3-128*U) +X^6*(28*S^2-16*T) +X^7*(-8*S) +X^8*(1) : Now, these area_i^2 are given by Heron's formula in terms of the lengths of the sides x y z u v w by equations ug:={ a2=2*x^2*y^2+2*x^2*v^2+2*y^2*v^2-x^4-y^4-v^4, b2=2*x^2*z^2+2*x^2*u^2+2*z^2*u^2-x^4-z^4-u^4, c2=2*y^2*z^2+2*y^2*w^2+2*z^2*w^2-y^4-z^4-w^4, d2=2*u^2*v^2+2*u^2*w^2+2*v^2*w^2-u^4-v^4-w^4 }: Note: here a2 is SIXTEEN TIMES the area of the corresponding side, etc. Note that S, T, U, V are _not_ in general expected to be symmetric functions of the squares of the lengths of the sides; rather, they can be expanded in terms of the polynomials invariant under the 24 symmetries of the set of sides of the tetrahedron (transitive on 6 sides; stabilizer can swap two pairs of sides or swap within a pair). So we should note the products p1...p4 and sums s1...s4 of the opposite pairs of lengths-squared; get a symmetric function in the pi and si, so it's a function in....um... Magma failed! ============================================================================== > dooy:=subs({x1=3/5,y1=0,z1=-4/5,x2=-9/\ > 25,y2=12/25,z2=-4/5,x3=-3/13,y3=-36/65,z3=-4/5},doox)); Yuk -- just solve numerically with Newton;s method starting near xyz=001, T=175 Optimal top point is [ -.00089635319461962538592677061731510009269099135498175, .013658432738894847204561117673500000316877834818587, .99990631749482790263728680568599324925847487327975] 16(Area)^2 is 174.90347057766992756683275705606768056711547627531 x-coordinate does not seem to be algebraic of degree 12 with small coeffs...