TWO GEOMETRICAL PASSAGES ffci THE MENO 301
the equation of which referred to AB, AG as axes of x, y is
xy = b 2 , where b 2 is equal to the given area. For a real
solution it is necessary that h 2 should be not greater than the
equilateral triangle inscribed in the circle, i. e. not greater than
3 Vs . a 2 / 4, where a is the radius of the circle. If h 2 is equal
to this area, there is only one solution (the hyperbola in that
case touching the circle); if h 2 is less than this area, there are
two solutions corresponding to two points E, E' in which the
hyperbola cuts the circle. If AD = x, we have 0D = x — a,
DE=V{2ax — x 2 ), and the problem is the equivalent of
solving the equation
x V(2ax — x 2 ) — h 2 ,
or x 2 (2ax — x 2 ) = 6 4 .
This is an equation of the fourth degree which can be solved
by means of conics, but not by means of the straight line
and circle. The solution is given by the points of intersec
tion of the hyperbola xy — b 2 and the circle y 2 = 2 ax — x 2 or
x 2 + y 2 = 2 ax. In this respect therefore the problem is like
that of finding the two mean proportionals, which was likewise
solved, though not till later, by means of conics (Menaechmus).
I am tempted to believe that we have here an allusion to
another actual problem, requiring more than the straight
line and circle for its solution,
which had exercised the minds
of geometers by the time of
Plato, the problem, namely, of
inscribing in a circle a triangle
equal to a given area, a problem
which was 'still awaiting a
solution, although it had been
reduced to the problem of
applying a rectangle satisfying the condition described by
Plato, just as the duplication of the cube had been reduced
to the problem of finding two mean proportionals. Our
problem can, like the latter problem, easily be solved by the
‘ mechanical ’ use of a ruler. Suppose that the given rectangle
is placed so that the side AD lies along the diameter AB of
the circle. Let E be the angle of the rectangle ADEC opposite
to A. Place a ruler so that it passes through E and turn
