#

The simplex in the plane

#### (or, the silly guy who can't remember high school
algebra)

Well, I tried. On page 164 of B3D, Prof. Banchoff
mentions that "there is no way to avoid using irrational numbers in
the coordinates of an equilateral triangle so long as it is in the plane."
As if to tease me, Prof. Banchoff cleverly avoided providing any proof
for this would-be axiom. A worthy challenge, I thought.
I started with a hopelessly over-difficult assembly of values and coordinates,
in which I tried to compute the slope of the altitude of the triangle and
use a barrage of pythaogrean constructions to generate irrational numbers.
Fortunately, Prof. Banchoff came to my rescue by suggesting that I begin
by normalizing the triangle such that it has a vertex at the origin, so
that the other two vertices now lie on a circle, as in the diagram. A simple
relationship can then be found such that `a^2 + b^2
= c^2 + d^2`, by pythagoras, if one considers that the hypoteneuse
in each case is the radius of the circle and therefore equal. the dashed
line is also equal in length to the others, as it is an equilateral triangle;
its length is given by the planar geometry distance formula, which, when
squared, is equal to `(a-c)^2 +(b-d)^2`.
So then we know that: `a^2
+ b^2 = c^2 + d^2 = (a-c)^2 +(b-d)^2`. Here
is where I get stuck. What I have is two equations in four unknowns, and
I'm trying to find a relationship between two of the unknowns, so that
I can prove that a relationship exists between the values for the
coordinates such that they cannot all be rational, something like a equals
the square root of three times d. But, if memory serves me, I need three
equations in order to find the relationship between two variables, right?
Another way would be to use trig, e. g. if h is the angle at the origin
of the side in the first quadrant, and k is the angle at the origin of
the side in the second quadrant, then `a/d = sin(120-k)/sin
k`. Unfortunately, I can't recall any useful trig identities.
Oh well, I tried.

Prof. Banchoff's appreciative response.

-david stanke