Consider a square ${\displaystyle ABCD}$ , and an inscribed quarter circle arc centered at ${\displaystyle A}$  with radius equal to the side of the square. Let ${\displaystyle E}$  be a point that travels with a constant angular velocity along the arc from ${\displaystyle D}$  to ${\displaystyle B}$ , and let ${\displaystyle F}$  be a point that travels simultaneously with a constant velocity from ${\displaystyle D}$  to ${\displaystyle A}$  along line segment ${\displaystyle {\overline {AD}}}$ , so that ${\displaystyle E}$  and ${\displaystyle F}$  start at the same time at ${\displaystyle D}$  and arrive at the same time at ${\displaystyle B}$  and ${\displaystyle A}$ . Then the quadratrix is defined as the locus of the intersection of line segment ${\displaystyle {\overline {AE}}}$  with the parallel line to ${\displaystyle {\overline {AB}}}$  through ${\displaystyle F}$ .^[1][2]

If one places such a square ${\displaystyle ABCD}$  with side length ${\displaystyle a}$  in a (Cartesian) coordinate system with the side ${\displaystyle {\overline {AB}}}$  on the ${\displaystyle x}$ -axis and with vertex ${\displaystyle A}$  at the origin, then the quadratix is described by a planar curve ${\displaystyle \gamma :(0,{\tfrac {\pi }{2}}]\rightarrow \mathbb {R} ^{2}}$  with

To describe the quadratrix as simple function rather than planar curve, it is advantageous to swap the ${\displaystyle y}$ -axis and the ${\displaystyle x}$ -axis, that is to place the side ${\displaystyle {\overline {AB}}}$  on the ${\displaystyle y}$ -axis rather than on the ${\displaystyle x}$ -axis. Then the quadratrix forms the graph of the function^[5][6]

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 ${\displaystyle n}$  equal segments and hence a trisection (${\displaystyle n=3}$ ) becomes possible. In practical terms the quadratrix can be drawn with the help of a template or a quadratrix compass (see drawing).^[1][2]

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 ${\displaystyle n}$  equal parts yields a partition of the associated angle as well. Dividing the line segment into ${\displaystyle n}$  equal parts with ruler and compass is possible due to the intercept theorem.

For a given angle ${\displaystyle \angle BAE}$  (at most 90°) construct a square ${\displaystyle ABCD}$  over its leg ${\displaystyle {\overline {AB}}}$ . The other leg of the angle intersects the quadratrix of the square in a point ${\displaystyle G}$  and the parallel line to the leg ${\displaystyle {\overline {AB}}}$  through ${\displaystyle G}$  intersects the side ${\displaystyle {\overline {AD}}}$  of the square in ${\displaystyle F}$ . Now the segment ${\displaystyle {\overline {AF}}}$  corresponds to the angle ${\displaystyle \angle BAE}$  and due to the definition of the quadratrix any division of the segment ${\displaystyle {\overline {AF}}}$  into ${\displaystyle n}$  equal segments yields a corresponding division of the angle ${\displaystyle \angle BAE}$  into ${\displaystyle n}$  equal angles. To divide the segment ${\displaystyle {\overline {AF}}}$  into ${\displaystyle n}$  equal segments, draw any ray starting at ${\displaystyle A}$  with ${\displaystyle n}$  equal segments (of arbitrary length) on it. Connect the endpoint ${\displaystyle O}$  of the last segment to ${\displaystyle F}$  and draw lines parallel to ${\displaystyle {\overline {OF}}}$  through all the endpoints of the remaining ${\displaystyle n-1}$  segments on ${\displaystyle {\overline {AO}}}$ . These parallel lines divide the segment ${\displaystyle {\overline {AF}}}$  into ${\displaystyle n}$  equal segments. Now draw parallel lines to ${\displaystyle {\overline {AB}}}$  through the endpoints of those segments on ${\displaystyle {\overline {AF}}}$ , intersecting the trisectrix. Connecting their points of intersection to ${\displaystyle A}$  yields a partition of angle ${\displaystyle \angle BAE}$  into ${\displaystyle n}$  equal angles.^[5]

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 one cannot assure an exact division of an angle into ${\displaystyle n}$  parts without a given trisectrix, one can construct an arbitrarily close approximation by circle and compass alone.^[2][3]

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 lets one 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 ${\displaystyle {\tfrac {2}{\pi }}}$ .^[1] 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 ${\displaystyle \left|{\overline {AJ}}\right|={\tfrac {2}{\pi }}r}$ . 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 ${\displaystyle \left|{\overline {BL}}\right|={\tfrac {\pi }{2}}r}$ . Extending AB to the right by a new line segment ${\displaystyle \left|{\overline {BO}}\right|={\tfrac {r}{2}}}$  yields the rectangle BLNO with sides BL and 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 ${\displaystyle \left|{\overline {OQ}}\right|=\left|{\overline {BO}}\right|={\tfrac {r}{2}}}$  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-angled 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.^[7]

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 two 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.^[8][9]

The quadratrix is mentioned in the works of Proclus (412–485), Pappus of Alexandria (3rd and 4th centuries) and Iamblichus (c. 240 – c. 325). Proclus names Hippias as the inventor of a curve called quadratrix and describes somewhere else how Hippias has applied the curve on the trisection problem. Pappus only mentions how a curve named quadratrix was used by Dinostratus, Nicomedes and others to square the circle. He neither mentions Hippias nor attributes the invention of the quadratrix to a particular person. Iamblichus just writes in a single line, that a curve called a quadratrix was used by Nicomedes to square the circle.^[10][11][12]

Although based on Proclus' name for the curve it is conceivable that Hippias himself used it for squaring the circle or some other curvilinear figure, most historians of mathematics assume that Hippias invented the curve, but used it only for the trisection of angles. Its use for squaring the circle only occurred decades later and was due to mathematicians like Dinostratus and Nicomedes. This interpretation of the historical sources goes back to the German mathematician and historian Moritz Cantor.^[11][12]

• Greek mathematics

• Claudi Alsina, Roger B. Nelsen: Charming Proofs: A Journey Into Elegant Mathematics. MAA 2010, ISBN 9780883853481, pp. 146–147 (excerpt, p. 146, at Google Books)
• Felix Klein: Famous Problems of Elementary Geometry. Cosimo 2007 (Nachdruck), ISBN 9781602064171, pp. 57–58 (excerpt, p. 57, at Google Books) (complete online copy at archive.org)

• Michael D. Huberty, Ko Hayashi, Chia Vang: Hippias' Quadratrix
• Weisstein, Eric W., "Quadratrix of Hippias", MathWorld
• O'Connor, John J.; Robertson, Edmund F., "Quadratrix of Hippias", MacTutor History of Mathematics Archive, University of St Andrews, p. cur