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.

Thanks for all the Euclidian advice.. I’m hoping that this version, with the offending paragraphs replaced by the bold section, will meet the stringent requirements. More fun than anticipated.

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.
Construct the perpendicular bisector of KM crossing KM at C. This is the path of the centre of the rolling circle.
Draw a ‘central circle’ with centre at C and diameter KM.
Construct a line from P, perpendicular to KM, to first meet this central circle at a point Q.
Draw in the line MQ and construct a line through P perpendicular to MQ. This is the tangent to the cycloid at P.

[N.B. the position of the rolling circle is that of the ‘central circle’ displaced by a distance PQ. MQ is therefore parallel to the former RP where R is the centre of rotation of the rolling circle. RP is no longer needed as part of the construction.]

Incidentally, this plan involves constructing perpendicular lines. I got the feeling that no-one objected to that process in the first offering, but I’d be interested to know how it’s done without lifting the compasses (and therefore losing their setting)?

Posted by Harry on 2010-01-13 16:38:35