Let L and N be adjacent cusps of a cycloid.
Let P be a point on the cycloid between L and N.

Construct with straightedge and compass the tangent line to the cycloid at point P.

Since a cycloid is traced out by a point on the rim of a circle rolling on a straight line, it should be possible to reconstruct that circle and hence a tangent, as follows.
Join LN to get the straight line mentioned, and construct the perpendicular bisector of LN crossing LN at M and the cycloid at K. KM is the diameter of the circle.
Construct the perpendicular bisector of KM crossing KM at C. This is the path of the centre of the circle – call it ‘centre path’.
Draw an arc with centre at P and radius equal to KC, to cut the centre-path at O. This is the centre of the circle.
Draw a perpendicular from O to LN, meeting LN at R. This is the instantaneous centre of rotation of the circle, so P will be moving at right angles to RP at this instant.
Draw a line through P, perpendicular to RP. This is the tangent to the cycloid.

(All this depends on having the point K available to provide a radius measurement. Could this still be done if only L and N were available?
KM = LN/pi, but I don’t know a way of dividing a length by pi through construction).

Posted by Harry on 2010-01-11 14:53:43