
We give two excellent and quite different methods of proof, one
from Benjamin Girard, age 18, Lycee Pierre Bourdan, Gueret, France,
and the other from Hyeyoun Chung, age 14, St Paul's Girls School,
London.
BENJAMIN'S EXISTENCE PROOF
Benjamin gives a method for constructing the diagram using only
a ruler and compasses (NB not a protractor).
Let us call the three distinct parallels (1), (2) and (3) and
choose a fixed point $A$ on the middle one. We shall construct the
image $d_1$ of line (3) under the rotation $r$ by angle $\pi /3$
about the centre $A$.
To every point $M$ of (3) this rotation associates one point $N$ of
$d_1$, such that
\begin {eqnarray} AM &=& AN \\ \angle MAN
&=& {\pi \over 3}.\\ \end{eqnarray}
So all such triangles $AMN$ are equilateral and we now need to
identify a particular one, namely triangle $ABC$ with vertex $C$ on
line (1) and vertex $B$ on line (3).
Line $d_1$ cuts (3) so it cuts (1). Let us call this intersection
point $C$. So $C$ is on line (1), A is on line (2) and we know that
a point $B$ on line (3), and only one, exists (because the rotation
is bijective). So given any three parallel lines an equilateral
triangle exists with one vertex on each of the three lines.
RULER AND COMPASS CONSTRUCTION
To construct the triangle by finding the positions of $B$ and
$C$, first draw the line $d_1$. To do this, use compasses and,
using any two radii, draw two arcs $s_1$ and $s_2$ with centre $A$
cutting line (3) at two points $M_1$ and $M_2$. Then with centre
$M_1$ draw an arc of radius $AM_1$ to cut arc $s_1$ at $N_1$ and
with centre $M_2$ draw an arc of radius $AM_2$ to cut arc $s_2$ at
$N_2$. Draw the line $d_1$ through $N_1N_2$. Mark the point $C$
where the line $d_1$ cuts line (1). With centre $A$ draw an arc of
radius $AC$ cutting line (3) at point $B$.
ANOTHER EXISTENCE PROOF
The dynamic geometry 'interactivity' suggested a proof based on
a continuity argument.
The point $A$ is fixed. As $B$ moves along line (3), point $C$
moves on line (1) such that $AB=AC$ making triangle $ABC$
isosceles. The shortest length for $AB$ occurs when $AB$ is
perpendicular to the parallel lines and in this case $\angle BAC$
is obtuse. As $B$ moves the angle $BAC$ changes continuously
getting smaller and there must be one position where $\angle BAC=
{\pi \over 3}$ and triangle $ABC$ is equilateral.
Hyeyoun also uses a continuity argument, but quite a different
one, rotating an equilateral triangle of a fixed size about one of
its vertices so that three parallel lines through the vertices move
continuously.
HYEYOUN'S PROOF
An equilateral triangle is positioned with its bottom edge along a
horizontal line so that the vertices lie on three parallel lines,
one through the uppermost vertex of the triangle, and the other two
horizontal lines on top of each other and passing through the base
of the equilateral triangle.
One vertex on the baseline is kept fixed and if the triangle is
rotated about this point it can be imagined that the parallel lines
through the other two vertices move with them, being translated
upwards but remaining parallel.
If the triangle is rotated through an angle $\pi /3$ the middle of
the three parallel lines leaves the bottom line and eventually
merges with the top line. Therefore the ratio of the perpendicular
distances between the parallel lines varies continuously from
infinity to zero and so takes the required ratio $a:b$ at some
point between. To achieve the actual distances $a$ and $b$ requires
simply apply an enlargement.
Therefore it can be said that for every set of three parallel lines
there is an equilateral triangle with a vertex on each line.