Straightedge alone

Ah ! the naughty boys ! As they rag with their compass, and it is dangerous, I have taken all their compass away. They have to do the job with just the straightedge. And also, because they use to knock the (hard) head of their buddies with it, one side is bumpy, and they can use only one side. And the bumpy side is the one where are the marks ! No more marks.
Indeed they can't do much with just that but exactly what can they still construct ?
• Midpoint of a segment ? solution
• Given just one line, to construct a parallel ? solution

• Given three inline points A,B,C, to construct the harmonic conjugate of C par rapport Ó AB ?
   (that is DA/DB = -CA/CB). Hint

• Given two lines d1 and d2 intersecting outside the paper sheet (or even parallel).
   To construct a concurring line (or parallel) through a given point P.Hint

As we just saw, drawing parallels requires we already have some !

• Given two pairs of parallel lines, and any line D, to construct a parallel to D through a given point P. Hint

• Given a segment AB and a parallel line, to construct the midpoint of AB. Hint
And conversedly : given a segment AB and its midpoint, to construct a parallel in P to AB.

Let's go with perpendiculars...
We require of course some right angle somewhere !
• Given a square ABCD and a line d. To construct through P the perpendicular to d.

With this square, it seems we can construct a lot of things, but...
• Show a point which can't be constructed, but can with the compass and straightedge.


Prove that, with the circular tumbler they use for painting courses, they can construct with the straightedge alone (only one side) and the tumbler (that is a few circles without their center) all the points they could construct with compass and straightedge.

• Specifically to construct with straightedge alone the center of the tumbler.

And then :
• Copy a given segment on any line.
• Given two points A and B and a straight line D, to construct the intersection points of D with the circle (undrawn) centered in A with radius AB.
• Given points A,B and C,D to construct the intersection points of the (undrawn) circles centered in A with radius AB and centered in C with radius CD.

We have then proved that all points constructible with compass and straightedge are also constructible with just the straightedge and the tumbler.
I had right to take the compasses away !



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