Quadratrix of Hippias

Definition

Consider a square ABCD with an inscribed quarter circle centered in A, such that the side of the square is the circle's radius. Let E be a point that travels with a constant angular velocity on the quarter circle arc from D to B. In addition the point F travels with a constant velocity from D to A on the line segment AD, in such a way that E and F start at the same time at D and arrive at the same time in B and A. Now the quadratrix is defined as the locus of the intersection of the parallel to AB through F and the line segment AE.

If one places such a square ABCD with side length a in a (cartesian) coordinate system with the side AB on the x-axis and vertex A in the origin, then the quadratix is described by a planar curve γ : ( 0 , π 2 ] → R 2 with:

This description can also be used to give an analytical rather than a geometric definition of the quadratrix and to extend it beyond the ( 0 , π 2 ] interval. It does however remain undefined at the singularities of cot ⁡ ( t ) except for the case of t = 0 , where due to lim t → 0 t cot ⁡ ( t ) = 1 the singularity is removable and hence yields a continuous planar curve on the interval ( − π , π )

To describe the quadratrix as simple function rather than planar curve, it is advantageous to switch the y-axis and the x-axis, that is to place the side AB on y-axis rather than on the x-axis. Then the quadratrix is given by the following function f(x):

Angle trisection

The trisection of an arbitrary angle using only ruler and compasses is impossible. However if the quadratrix is allowed as an additional tool, it is possible to divide an arbitrary angle into n equal segments and hence a trisection (n = 3) becomes possible. In practical terms the quadratrix can be drawn with the help of a template or a quadratrix compass (see drawing).

Since by the definition of the quadratrix the traversed angle is proportional to the traversed segment of the associated squares' side dividing that segment on the side into n equal parts yields a partition of the associated angle as well. Dividing the line segment into n equal parts with ruler and compass is possible due to the intercept theorem.

For a given angle BAE ( ≤ 90°) construct a square ABCD over its leg AB. The other leg of the angle intersects the quadratrix of the square in a point G and the parallel line to the leg AB through G intersects the side AD of the square in F. Now the segment AF corresponds to the angle BAE and due to the definition of the quadratrix any division of the segment AF in n equidistant parts yields a corresponding division of the angle BAE into n parts of equal size. To divide the segment AF into n equidistant parts proceed as follows. Draw a ray a with origin in A and then draw n equidistant segments (of arbitrary length) on it. Connect the endpoint O of the last segment with F and draw lines parallel to OF through all the endpoints of the remaining n − 1 segments on AO, these parallel lines divide the segment AF on AD into n equidistant segments. Now draw parallel lines to AB through the endpoints of those segments on AF, these parallel lines will intersects the trisectrix. Connecting those points of intersection with A yields a partition of angle BAE into n parts of equal size.

Since not all points of the trisectrix can be constructed with circle and compass alone, it is really required as an additional tool next to compass and circle. However it is possible to construct a dense subset of the trisectrix by circle and compass, so while you cannot assure an exact division of an angle into n parts without a given trisectrix, you can construct an arbitrarily close approximation by circle and compass alone.

Squaring of the circle

Squaring the circle with ruler and compass alone is impossible. However if one allows the quadratrix of Hippias as an additional construction tool, the squaring of the circle becomes possible due to Dinostratus' theorem. It allows to turn a quarter circle into square of the same area, hence a square with twice the side length has the same area as the full circle.

According to Dinostratus' theorem the quadratrix divides one of the sides of the associated square in a ratio of 2 π . For a given quarter circle with radius r one constructs the associated square ABCD with side length r. The quadratrix intersect the side AB in J with | A J ¯ | = 2 π r . Now one constructs a line segment JK of length r being perpendicular to AB. Then the line through A and K intersects the extension of the side BC in L and from the intercept theorem follows | B L ¯ | = π 2 r . Extending AB to the right by a new line segment | B O ¯ | = r 2 yields the rectangle BLNO with sides BL und BO the area of which matches the area of the quarter circle. This rectangle can be transformed into a square of the same area with the help of Euclid's geometric mean theorem. One extends the side ON by a line segment | O Q ¯ | = | B O ¯ | = r 2 and draws a half circle to right of NQ, which has NQ as its diameter. The extension of BO meets the half circle in R and due to Thales' theorem the line segment OR is the altitude of the right angle triangle QNR. Hence the geometric mean theorem can be applied, which means that OR forms the side of a square OUSR with the same area as the rectangle BLNO and hence as the quarter circle.

Note that the point J, where the quadratrix meets the side AB of the associated square, is one of the points of the quadratrix that cannot be constructed with ruler and compass alone and not even with the help of the quadratrix compass based on the original geometric definition (see drawing). This is due to the fact, that the 2 uniformly moving lines coincide and hence there exists no unique intersection point. However relying on the generalized definition of the quadratrix as a function or planar curve allows for J being a point on the quadratrix.