• Draw a line through two known points

• Get knew known points from intersections of such drawn lines

or with a previously drawn (=given) curve (eg. circle).

• Choose any arbitrary point, as long as the final construction doesn't depend on the choice.

Full stop. Any other action is forbidden.

At last we are in an "ordered" plane, that is we can say on a line
if two points are on the same side or the opposite sides of a third point.

That last note is mandatory here, for the constructions with straightedge alone
have the projective plane as "natural" frame, in which this is not the case, ABC and ACB being undistinguishable.

The plane transforms here are the "collineations", that is the general projective homographies.
The distance ratios are not held by these transforms, only ratio of ratios (cross ratio or anharmonic ratio).
The concept of midpoint doesn't exist.

As for the concept of parallels, it is changed into concept of "points at infinity" .
To draw a parallel to a given line is to draw a line to the intersection point with the line at infinity.
That one must then be given (from two points), or the point at infinity given (by two given parallels)
Draw a parallel to just a given line is then impossible.

We use here a projective property of the complete quadrilateral :

Let I the intersection of diagonals AE and BF, M and N the intersections of DI with the sides, (C,I,M,N) is an harmonic range. |

Applying Thales then suffice to prove that I is the midpoint of MN.

Hence the cross ratio (M,N,C,I) = -1, value held by the homography having sent C to infinity.

For D being midpoint of AB, (A,B,∞,D) = -1, is even obvious here.

However, the projection from center H of (M,N,C,I) onto line AB gives (A,B,C,D) = -1 in general case.

Hence the wanted construction :

We choose a pencil of lines HA,HB,HC from any point H

Draw any secant CEF, and the diagonals AE and CF intersect in I

HI intersects AB in the wanted D.

From a given point P, to draw a concurring line (or parallel).

It is really the previous construction (the reciprocal) performed twice :
the projection from center I holds the cross ratio (S,P,M,N) = (S,J,M',N') = -1, hence IPJ are in line.

If I is at infinity, the given lines are in fact parallel but the construction is the same.

Draw through P any two secants AB and CD.

AC and BD intersect in S

Draw through S any secant SA'C', then A'C and C'D intersect in J.

PJ is the searched line

If P is on the angle bisector (or the median parallel), S is at infinity.

We have then to first construct a concurring line d'2 through any P' clearly not on that angle bisector.

Then construct through P the concurring line with d1 and **d'2**.

Even when P is quite near that angle bisector, S itself would be outside the sheet, and the auxilliary P' should be used.

We are here in the case of the line at infinity defined by two of its points Q1 and Q2.

A transformed view through an homography, in which the line at infinity is "viewable" :

The Desargues theorem

AA', BB', CC' concur in I ⇔ Q1 = AB∩A'B', Q2 = AC∩A'C', Q = BC∩B'C' in line here on d_{∞}

allows then to do the projective construction on that picture of any "parallel" to D.

From the initial drawing (in which the line at infinity d_{∞} is "really at infinity"),
the construction is even simpler to understand
(we can even consider it as a proof of Desargues theorem !) :

Let I any point on line AA'.
Let B' the intersection of IB with d'1 and C' the intersection of IC with d'2

The dilation from center I transforming A into A' transforms AB into A'B' and
AC and A'C', hence finally BC into B'C' which are then parallel.

At last the construction of the parallel through P to the two parallel lines D and B'C' is not detailed here :
previously seen.

• Given a segment AB and a parallel line,
to construct the midpoint of AB.

Here also, Thales in a trapezoid is enough, as already seen.

From any point S, draw the secants SCA and SDB

BC and AD intersect in I

SI intersects AB in its midpoint.

We also can duplicate CD into DE (by constructing a third parallel through S).

Constructing q×CD in that way, and p×AB, then with Thales p/q×AB,
we can then construct all points with rational coordinate on line AB,
segment AB considered as unit, and only those.

Then extend to all rational coordinates points
in the affine coordinate system defined by two pairs of parallel lines
(by the "unit" parallelogram) and only those.

• Given a square ABCD and a line d. To construct through P the perpendicular to d.

We suppose the line not already parallel to the sides of the square (obvious)
nor to the diagonals (midpoint of diagonals etc.)

Construct (not detailed but previously seen) the parallels through P to the sides of the square,
intersecting the diagonal in A and B.

The parallel to the "other" sides from A and B intersect d in C and D.

The parallel to the first sides through C and D intersect the diagonal in E and F

At last the parallels to the second sides through E and F intersect in Q.

The right angled triangles HPQ and KCD are congruent (HP = KC and HQ = KD)

The angles in P and C are then equal and HP _|_ KC ⇒ PQ _|_ CD. QED.

Constructible points

In the affine coordinate system defined by the square (even cartesian, but doesn't care)
we can construct only points with rational coordinates.
The point (1/√2, 1/√2) can't then be constructed.

Although we can construct right angles, it is not yet enough : we can't copy distances on any line.

This is a direct application of the "Cauer theorem" :

All points constructible with compass and straightedge are constructible with straightedge alone, given two intersecting circles without their centers. |

But not with just two non intersecting circles.

This can be proved (through the following construction) by reducing it to the Poncelet-Steiner theorem :

All points constructible with compass and straightedge are constructible with straightedge alone, given one circle with its center. |

We can draw with the tumbler any two intersecting circles, with unknown center

(they are really equal, but doesn't matter here) :

• Construct with straightedge alone the centers of two intersecting circles.

Choose on one circle Γ' two points C and D, on the outside arc.

<)CAD = CBD (inscribed angles)

AC intersects again the other circle Γ in A', AD intersects it in A", BC in B' and BD in B"

Angles A"AA' and B"BB' are then equal.

Inscribed angles A'B'A" and B'A'B"
are then equal and therefore A'B" and A"B' are parallel.

A'A"B'B" is then a trapezoid (isosceles as inscribed), the intersect points I of the non parallel sides
A'A" and B'B" , and J of the diagonals A'B' and A"B"
define its symmetry axis IJ, hence a diameter of the circle.

Do it again with another choice for (C, D),
and get another diameter.

The intersect point of these diameters is then the center of circle.

This construction fails if the two circles are orthogonal :
the trapezoid is then a rectangle* and I is at infinity.

* Proof : to prove that angle in I depends only on angle OAO'

But it is then easier :
the center of circle is just the intersection of the rectangle diagonals !

However we should far avoid nearly orthogonal circles when drawing the two circles with the tumbler !

(otherwise I is too far)

But even in this case (I too far), we may as well draw a third parallel from the two first,
building then another trapezoid.

Intersecting the two diagonals always works and gives a point J', and the diameter JJ'.

Having then a circle with its center, we can draw as many parallels and perpendiculars we want :

• Parallels : the previous construction already gives two sets of parallel lines.

Otherwise (starting from circle with center), any two diameters give a rectangle,
and then two pairs of parallel lines, and that's it.

• Perpendicular to d through P.

Draw OP intersecting the circle in A and B, then construct the parallel to d through B,
intersecting the circle in C.

Then the parallel to AC through P is the searched perpendicular
(construction of parallels not detailed here).

The construction fails if OP is parallel to d.

Constructing the perpendicular in P to OP may be obtained from the polar of the foot of the polar of P :

construct the polar IJ of P, intersecting OP in H, the polar of H is the searched perpendicular.

Just one point Q on this polar suffice, we already know P !

• Copy a given segmentAB on any line d from a given M.

Draw the parallels to OA in B and to AB in O to complete the parallelogram OABC

Let D the intersection of line OC with the circle.

Draw the parallel to d in O, intersecting the circle in E E'

The parallel to DE from C intersects EE' in F

The parallel to OM from F intersects d in N, constructing then MN = AB on d,
from the given point M.

Proof : parallelograms and similar isosceles triangles ODE and OCF.

• Given two points A and B and a line d, to construct the intersection points of d with the circle (undrawn) centered in A with radius AB.

The key is here the dilation with center O and ratio AB/R.

Construct the parallelogram OABC, and let D the intersection of AC with the given circle Γ.

Construct the perpendicular AH to d from A.

Construct the parallelogram OAHK.

The parallel to CK in D intersects OK in E

The parallel to d in E intersects Γ in F

The parallel to d in K intersects OF in P

Complete the parallelogram HKPM

• Given points A,B and C,D, to construct the intersect points of circles (not drawn) centered in A with radius AB and centered in C with radius CD.

The key is here to construct the radical axis of the two circles.

We are then reduced to the previous construction of intersect points of this line with one of the circles.

Theorem

Let I a dilation center of the two circles.
The polars of I with respect to the two circles are symmetric through the radical axis. |

Copy the radii on the center line AC to get the diameters B'B" and D'D"

Copy one radius parallel to the other in CE to get a dilation center I.

Choosing the negative dilation center ensures that I is not at infinity
(if the two circles were equal) or too far.

Construct the harmonic conjugates of I with respect to B'B" and to D'D" :

As previously seen, draw a pencil of lines (SB',SB",SI) from any S.
Choose any point J on SI.

B'J gives U and B"J gives V, last UV intersects B'B" in M, conjugate of I with respect to B'B",
hence the foot of the polar of I relative to circle centered in A.

Then similarily for N, foot of the polar of I relative to circle centered in C.

The foot of the radical axis is then the midpoint H of MN, and the radical axis the perpendicular
to line center AC in H.

We have then proved that all points constructible with compass and straightedge are
constructible with straightedge alone, given a circle and its center, hence with just straightedge and tumbler.

The constructions are impracticable, because of the many hidden constructions of parallels.

We shall now give a few

We deal here with classical constructions with straightedge alone and previously drawn circles.

The first ones don't use the center of the circle, which can then be unknown.

Recall the definition of the polar of P : locus of all harmonic conjugates of P with respect to the intersection points of any secant through P.

This locus is a straight line, and we have just to find two points.

The properties of the complete quadrilateral let construct the conjugates M and N for two secants : the polar is then directly line IJ

In reverse (construct the pole of d) we construct the polars of two points of d.

The pole of d is the intersection of these polars.

(the variable secant becomes the tangent when A B M coincide in T)

But it is more speedy to construct the tangent in P with the Pascal theorem :

The opposite sides of an hexagon PPABCD intersect in 3 points in line. ("side" PP is the tangent in P).

For understanding, let's distinguish P and P' in hexagon PP'ABCD.

The opposite sides are then P'A and CD, intersecting in I

AB and PD intersecting in J

PP' and BC intersecting in K

as PP' is unknown, we get here K as intersection of IJ and BC

PP' is then line PK

Here, the hexagon PPABCD is crossing to get I,J,K not too far.

We have of course to know the center of circle to construct the intersection of the polar with line OP.

Another construction (already seen) is to construct the harmonic conjugate of P with respect to the diameter going through P.

we must be given something more.

Christian Gram has then proved (1956) that it is possible to construct their centers given just one point A on the center line.

This construction is very complex. It is given in Appendix (includes applets)