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...