Without loss of generality assume that one of the vertices of our triangle is the orgin
. Thus we are really examining
, where
are the other verticies. We may assume that
for some
. From here we break this into two cases, namely:
, or
. Clearly both cannot be zero since the origin and
are distinct points.
Case 1,
: It is clear that we assume WLOG that
and
for all other cases are analgous. Thus this triangle will have
and
for some
. Clearly seeing though that the length of
is
(since they are both on the y-axis) we may conclude that the length of
is as well. Realizing though that the lengths of
are given independently by the distance equations
and appealing once again to the equilaterallness of the triangle we may conclude that, in fact,
. Solving this equation yields
. As stated earlier though
and focusing mainly on
we see that
which is clearly irrational for any rational
.
Case 2,
: Now we have that
where neither
or
is zero. Constructing the midpoint
of
gives us that
and that the length of the segment
is
. Now, form
to be the median connecting
and
. If we consider the two segments
to be the restrictions of the lines
respectively we can see that
and since
we may conclue that
(now you see why we needed
). Letting
for some
again and realizing that
both lie on
we may conclude that
where
represents the change in
between the points
and
. Appealing to this we see that
. Remembering though that forming the median
bisected
we see that the new triangle formed,
is a
triangle. In particular, this tells us that
. Equating our two lengths of
gives us
and since
we may conclude that
. Thus, substitution into
gives us that
which is clearly irrational since
.
The conclusion follows.