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Conic Sections Before Cones

Menaechmus' Constructions

Let us consider how Menaechmus constructed the two mean proportionals. [14, pp. 278-283]. Let the given lengths be a and b. Let a straight line be given to serve as the axis, with a point D on the line to serve as the origin. Construct a parabola with the given axis, with vertex at D, and latus rectum a. This is equivalent to constructing a square on ordinate y equal to the rectangle on the latus rectum with side x. Note that this makes y the geometric mean of the latus rectum and the side x. Consider the diagram below to see why.

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(The height of the rectangle is x, the abscissa.
Click the red X in the lower right of the sketch to clear the trace.
Download the Geometer's Sketchpad 4.0 file.)

From the above sketch you can convince yourself that this does indeed generate a parabola. (If you would like to see how to construct a parabola pointwise, using Euclidean tools, download this Geometer's Sketchpad 4.0 file.)

Next, using the same axis and origin as above, construct a hyperbola such that the rectangle contained by the abscissa x and the ordinate y makes an area equal to the area of the rectangle contained by the lengths a and b.

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(The base of the rectangle at lower left is a; the height is b.
Click the red X in the lower right of the sketch to clear the trace.
Download the Geometer's Sketchpad 4.0 file.)

From the above sketch you can convince yourself that this does indeed generate a hyperbola. (If you would like to see how to construct a hyperbola pointwise, using Euclidean tools, download this Geometer's Sketchpad 4.0 file.)

Now, what about the mean proportionals of a and b? The abscissa x and ordinate y of the point of intersection of the hyperbola and parabola so generated give the mean proportionals of a and b.

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Click the red X in the lower right of the sketch to clear the trace.
Download the Geometer's Sketchpad 4.0 file.)

In this manner, Menaechmus constructed the mean proportionals to a and b. If we let b = 2a, we have found the side x of a cube with double the volume of a cube with side a. It is important to point out here that this does not solve the problem of doubling the cube as the Greek geometers intended. This solution cannot be completed with ruler and compass only, and hence is not a solution.

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