equilateral triangles in space, and cyclohexane

David Madore · science forum beginner Joined: 23 Feb 2005 Posts: 24

Here is a question of elementary space geometry which I find very
elegant but which I only know how to solve in an algebraic
(computational) manner. I hope someone can find a geometric proof. I
will first state the problem, then give the algebraic solution, and
make a few comments.

*** Statement of the problem ***

Let a and b be two lengths.

Consider an equilateral triangle P0 P2 P4 in (three-dimensional
Euclidean) space, having side length 2a. Call C1, C3 and C5 the
circles whose axes[#] are respectively (P0 P2), (P2 P4) and (P4 P0),
whose centers are the respective midpoints of the corresponding
segments (I mean, [P0 P2], [P2 P4] and [P4 P0]) and all having radius
b.

[#] I'm not sure how clear/standard that terminology is. In case it
isn't, the "axis" of a circle in space is the line perpendicular to
the circle's plane and passing through its center (i.e., the line such
that any rotation around that axis leaves the circle invariant).

To put it differently, C1 is the intersection of the spheres with
centers P0 and P2 and both having radius c where a^2+b^2 = c^2.
Similarly for C3 and C5.

Now assume P1 is a point on C1 such that there exist two distinct
points on C3 (hence also on C5) at distance 2a from it. Call P3 one
of these points, on C3, and P5 "the other" on C5. By "the other", I
mean that P3 and P5 *are not* symmetric with respect to the plane
containing C1.

Show that P1 P3 P5 is equilateral.

*** Remarks ***

The point is that the "hexagon" P0 P1 P2 P3 P4 P5 (which is supposed
to generalize a cyclohexane molecule) has fixed side lengths (all
lengths P0 P1, P1 P2... through P5 P0 are equal to c) and fixed angles
(all are equal to 2arccos(a/c) or some such thing) and still has a
degree of freedom (since P1 can vary on C1 even though P0, P2 and P4
are fixed).

Or, if you will, we have a fixed equilateral triangle P0 P2 P4 and
another fixed triangle P1 P3 P5 with the same side, and all lengths P0
P1, P1 P2... through P5 P0 are fixed, and still we can move the
triangles one with respect to the other. I made a video showing this:
<URL: http://www.madore.org/~david/images/cyclohexane.avi > (and I
find it quite fascinating to watch).

The reason this is surprising is that the six points P0 P1 P2 P3 P4 P5
with the twelve fixed distances (six being 2a and six being c) are
combinatorily equivalent to an octahedron, and if it were convex then
it would be rigid (=no degree of freedom) by a rigidity theorem of
Cauchy. Except that it is not (convex) and it is not (rigid): there
is an unexpected redundancy of constraints. But some other
configurations are rigid.

*** Algebraic proof ***

Take a system of Euclidean coordinates where P0 has coordinates
(a.sqrt(2),0,0), P2 is (0,a.sqrt(2),0) and P4 is (0,0,a.sqrt(2)).

Take a rational parameterization of the circle C1 with parameter u:
the point P1 has coordinates

((a+b(1-u^2)/(1+u^2))/sqrt(2), (a+b(1-u^2)/(1+u^2))/sqrt(2), b.2u/(1+u^2))

Similarly, rationally parameterize C3 with parameter v and C5 with
parameter w: P3 and P5 have coordinates

(b.2v/(1+v^2), (a+b(1-v^2)/(1+v^2))/sqrt(2), (a+b(1-v^2)/(1+v^2))/sqrt(2))

((a+b(1-w^2)/(1+w^2))/sqrt(2), b.2w/(1+w^2), (a+b(1-w^2)/(1+w^2))/sqrt(2))

Now write the fact that the squared distance from P1 to P3 is 4a^2: by
chasing away the denominator (1+u^2)(1+v^2), it gives us a condition
P(u,v)=0 where P is the polynomial

P(s,t) = -3a^2(1+s^2)(1+t^2) - 2ab(1+st)(-1+sqrt(2)(s+t)+st)
+ b^2(1+2sqrt(2)(s^2 t+s t^2-s-t)+3(s^2+t^2)+s^2 t^2)

Of course, the constraint on the distance from P1 to P5 is the same,
i.e. P(u,w)=0. So we are trying to show the following: that if v and
w are the two distinct roots, with u fixed, of P(u,t)=0, with
indeterminate t, then P(v,w)=0 (that is, the distance from P3 to P5 is
also 2a).

Now the polynomial P(s,t) is symmetric in its two variables s and t:
so if we let q=s+t and p=st, we can write

P(s,t) = -3a^2(1-2p+p^2+q^2) - 2ab(1+p)(1-p+sqrt(2)q)
+ b^2(1+2sqrt(2)(p-1)q-6p+p^2+3q^2)

Call Q(q,p) the polynomial in question (-3a^2(1-2p+p^2+q^2) -
2ab(1+p)(1-p+sqrt(2)q) + b^2(1+2sqrt(2)(p-1)q-6p+p^2+3q^2)). Now call
c0(u), c1(u) and c2(u) the coefficients of P(u,t) as a polynomial in
t: i.e., P(u,t) = c0(u) + c1(u) t + c2(u) t^2. In other words, c0(u)
= -3a^2(1+s^2) - 2ab(-1+sqrt(2)s) + b^2(1-2sqrt(2)s+3s^2), etc. Using
the relations between coefficients and roots of an equation, we know
that if v and w are the two roots of P(u,t)=0 (in t) then v+w =
-c1(u)/c2(u) and vw = c0(u)/c2(u). So all we have to do is check that
Q(-c1(u)/c2(u), c0(u)/c2(u)) = 0. And this is the case (a tedious but
trivial verification best left to a formal computation software
package).

So we have shown what we wanted: if v and w are the two roots, for u
fixed, of P(u,t)=0, in other words if P3 and P5 are the two
"different" points, for P1 fixed, on C3 and C5 at distance 2a from P1,
then P(v,w)=0 also, that is the distance P3 P5 is also 2a, which shows
that P1 P3 P5 is equilateral.

*** Further remarks ***

In the above proof, the curve P(s,t)=0 is (or, better, can be
completed to) an elliptic curve (after an appropriate choice of
origin). The crucial property that "if P(u,v)=0 and P(u,w)=0 (with v
different from w) then automatically P(v,w)=0" corresponds more or
less to the fact that the points (u,v), (v,w) and (w,u) on the curve
are separated by a third-of-period (the difference between two of them
is a 3-torsion point). Or something of the sort.

[PS: I basically translated this post from the French, <URL:
http://groups.google.com/group/fr.sci.maths/browse_thread/thread/d0f7f5fecc1533d/

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