[FOM] Alleged impossibility of "direct" proof of Steiner-Lehmus theorem
Timothy Y. Chow
tchow at alum.mit.edu
Mon Aug 9 10:28:15 EDT 2004
As an offshoot of the recent discussion here about elementary geometry,
Charles Silver drew my attention to another notorious result known as
the "Steiner-Lehmus theorem." To state this theorem, recall that by an
"angle bisector" of a triangle is meant the finite line segment from a
vertex of a triangle to the opposite side of the triangle (of course,
that bisects the angle at the vertex). The Steiner-Lehmus theorem says
that if two angle bisectors of a triangle are equal in length, then the
triangle is isosceles.
This innocuous-looking theorem is surprisingly difficult to prove,
especially if one requires a "geometric" rather than an "algebraic"
proof. If one further asks for a "direct" rather than an "indirect"
proof, then one discovers that there has been a long controversy over
whether a "direct" proof is even possible. Sylvester argued that
direct proofs are impossible; I haven't read his paper, but evidently
the "impossibility proof" wasn't sufficiently rigorous to deter other
people from claiming to have found, and publishing, direct proofs.
But these proofs haven't convinced everyone either; some have been
attacked on the grounds that they rely on indirectly proven lemmas.
John Conway has given an intriguing argument that there is no direct
proof---or more precisely, that there is no "equality-chasing" proof
---and although his argument seems compelling, I confess that I don't
fully understand it. (It can be found at
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus
but since URL's tend to be ephemeral and FOM messages are archived
indefinitely, I've included the full text below.) Specifically, I don't
think I could produce a precise definition of an "equality-chasing proof"
and prove that it would work for "all values of the variables."
So my question is, can one formulate and prove in a completely precise
manner that "The Steiner-Lehmus theorem has no direct proof"? Perhaps by
using Conway's ideas, or by using concepts from intuitionistic logic?
Tim
[Conway's article follows]
I've often responded to questions about why this theorem is so hard to
prove by saying that it isn't true if you allow the various lengths
involved to be in more general fields, like the complex numbers. After
someone asked for details about this the other day, I took the trouble
to work everything out, and found to my surprise that it isn't even
true when you allow them to be negative reals, which makes the
situation much easier to understand.
Here's what's going on. The lengths of the A and C angle
bisectors turn out to be
root{bc(a+b+c)(b+c-a)} root{ab(a+b+c)(a+b-c)}
---------------------- and ---------------------- ,
b+c a+b
so the equality we're given is
bc(b+c-a).(a+b)^2 = ab(a+b-c).(b+c)^2
or
(c-a){ ac^2 + (a^2+3ab+b^2)c^2 + b^2(a+b) } = 0.
Now algebraically, this does not imply that c=a, because
it might be the second factor that vanishes.
However, if a,b,c are positive, then so is this second factor,
and so with this assumption we CAN deduce c = a, but only by a proof
that somehow distinguishes positive numbers from negative ones. Now
the most common type of proof by a chain of equalities, such as:
"This length equals that length, therefore those two triangles
are congruent, so this angle equals that angle, whence ... whence
ultimately AB = BC."
cannot possibly do this, because it would work for all values of the
variables, and it's just not possible to deduce that c=a from
(c-a)(ac^2 + (a^2+3ab+b^2)c + b^2(a+b)} = 0.
The 60 proofs you mention avoid this by using order properties
of the reals, typically:
"Suppose AB > BC. Then this angle exceeds that angle, whence ...
whence ultimately the first angle-bisector is greater than the
second. Now suppose instead that AB < BC : this implies similarly
that the second is greater than the first. But we are given that
they are equal, so the only possibility is that AB = BC."
We can actually smoothly vary the parameters of a triangle in such
a way that one of its "edgelengths" changes sign and get a
"counterexample" to the Steiner-Lehmus theorem. Here's how:
C
/ \
E D
/ I \
/ \
A--> F <--B
Fix C, but let A and B move freely on the x-axis. Then if
they move so as to pass right through each other:
C
/ \
/ \
/ \
/ \
<--B F A-->
I
D E
it's natural to say that c has changed sign. As they moved, the
incenter I moved smoothly down, and crossed the x-axis as they
passed through each other, so that it became an excenter of the
final inside-out triangle.
The angle-bisector segments AD,BE,CF of the original triangle
also vary smoothly during this process, and I've drawn the final
positions that they get into in the above picture. Now take a = b = 1;
then the above equation for c becomes
c^2 + 5c + 2 = 0,
which is satisfied by c = (-5 + root17)/2, which is about -.44 ;
then with this value of c ALL THREE of these segments are equal,
but the triangle is only isosceles, not equilateral!
If there were a proof of the Steiner-Lehmus theorem of the above
equality-chasing type, it would continue to work when we varied the
triangle smoothly in this way, and so would prove this triangle
equilateral, which it isn't. So there's no such proof!
John Conway
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