Here follows Heath's presentation:

"After Hippocrates had discovered that the duplication of the cube was equivalent to finding
two mean proportionals in continued proportion between two given straight lines, the problem seems
to have been attacked in the latter form exclusively. [. . .]

**Archytas**

The solution of Archytus is the most remarkable of all, especially when his date is considered
(first half of fourth century B. C.), because it is not a construction in a plane but a bold
construction in three dimensions, determining a certain point as the intersection of three surfaces
of revolution, (1) a right cone, (2) a cylinder, (3) a *tore* or anchor-ring with inner
diameter *nil*. The intersection of the two later surfaces gives (says Archytas) a certain
curve (which is in fact a curve of double curvature), and the point required is found as the point
in which the cone meets this curve.

Suppose that *AC*, *AB* are the two straight lines between which two mean proportionals
are to be found, and let *AC* be made the diameter of a circle and *AB* a chord in it.

Draw a semicircle with *AC* as diameter, but in a plane at right angles to the plane of the
circle *ABC*, and imagine this semicircle to revolve about a straight line through *A*
perpendicular to the plane of *ABC* (thus describing half a *tore* with inner diameter
*nil*).

Next draw a right half-cylinder on the semicircle *ABC* as base; this will cut the surface of
the half-*tore* in a certain curve.

Lastly let *CD*, the tangent to the circle *ABC* at the point *C*, meet *AB*
produced in *D*; and suppose the triangle *ADC* to revolve about *AC* as axis. This
will generate the surface of a right circular cone; the point *B* will describe a semicircle
*BQE* at right angles to the plane of *ABC* and having its diameter *BE* at right
angles to *AC*; and the surface of the cone will meet in some point *P* the curve which
is the intersection of the half-cylinder and the half-*tore*.

[The white triangle is *APC'.*]

Let *APC'* be the corresponding position of the revolving semicircle, and let *AC'*
meet the circumference *ABC* in *M*.

Drawing *PM* perpendicular to the plane of *ABC*, we see that it must meet the
circumference of the circle *ABC* because *P* is on the cylinder which stands on
*ABC* as base.

Let *AP* meet the circumference of the semicircle *BQE* in *Q*, and let
*AC'* meet its diameter in *N*. Join *PC', QM, QN*.

Then, since both semicircles are perpendicular to the plane *ABC*, so is their line of
intersection *QN* (Eucl.XI. 19).

Therefore *QN* is perpendicular to *BE*.

Therefore *QN.QN = BN.NE = AN.NM*, (Eucl. III. 35) so that the angle *AQM* is a
right angle.

But the angle *APC'* is also right; therefore *MQ* is parallel to *C'P*.

It follows, by similar triangles, that

that isC'A:AP = AP:AM = AM:AQ;

AC:AP = AP:AM = AM:AB,

and

In the language of analytical geometry, if

where

From the first two equations we obtain

and from this and (3) we have

or

AC:AP = AP:AM = AM:AB.

Compounding the ratios, we have

therefore the cube of side

In the particular case where

and the cube is doubled."

Quoted with permission from: