First let's remember a definition: A geometric construction is a drawing carried out with only two tools :
• a ruler to draw a line going through two points,
• and a pair of compasses to report a distance or/and draw a (arc of) circle whose center and either the radius or one of its points are known.
The intersections of two lines or (arcs of) circles define then new points which allow to advance the construction.

constructions of regular polygons

The constructions of the equilateral triangle, the square, the regular hexagon and octagon are well known. Caution! constructing a side and then making successive transfers is not the best way to proceed (you rarely fall back on the starting point).
If you need a circle with a diameter, it is more accurate to draw the line first.
 Remark: we can obviously construct a regular octagon starting from a circle with two perpendicular diameters and their bisectors; the construction starting from a square proposed here, is less known (exercise: justify it!). The construction of the regular pentagon is not much more difficult. In a circle with two perpendicular diameters OA and OA', construct the circle with diameter OA'; its center is the midpoint M of OA'. The line AM intersects this little circle at P and Q. The circles with center A and going through P and Q cross the great circle on four vertices of the pentagon; the fifth vertex is diametrically opposite to A. Remark: We can avoid the construction of the midpoint M using the perpendicular bisector of [OA'] by changing the drawing's order of the first two circles: start by drawing a straight line, choose a point M on it, and plot a circle of center M and diameter (OA '); then draw the circle of radius [OA']. The construction of a perpendicular bisector is however useful (but not necessary!) to obtain the diameter (OA) perpendicular to (OA '). I find the proof of this construction nice; reasoning by analysis and synthesis deserves a pause. Let us also take the opportunity to remind the theorems about the angle of a sector whose sides intersect a circle. To be concise, we will use the "classic" vocabulary (inscribed, central, inside, outside angle, when the vertex of the sector is on, at the center, inside, outside the circle), although it is incorrect to speak about the vertex or the sides of an angle. The angle is at the sector of what length is at the segment. Nobody would think of talking about the ends of a length!  •  An inscribed angle is half the angle at the center that intercepts the same arc (the inscribed angle has same measure as the arc it intercepts).  •  An outside angle is equal to the half-difference of the angles at the center intercepting the same arcs.  •  An inside angle is equal to the half-sum of the angles at the center intercepting the same arcs. These results remain valid when one side of the sector (or both!) is tangent to the circle; a tangent is indeed the limit position of a secant. •  Let us suppose that the problem is solved and let us analyze the figure on the right on which AB and AD are the sides of the two regular decagons: the convex ABCDEFGHIJ and the star ADGJCFIBEH. With AB=x, AD=y and r radius of the circle, we have:     r=y-x because ABN and DON are isosceles in A and D (theorems above), so OD=ND=AD-AN=AD-AB,     r²=xy because AON and ADO are similar (they have the same angles), so AO/AD=AN/AO=AB/OA. With y=r+x the second relation leads to the equation r²=x²+rx, which has only one positive solution.  •  The construction proposed above provides precisely the only pair (x, y) satisfying these two properties: x=AP and y=AQ (radii of the two arcs of circles). Indeed in the right-angled triangle AMO we have AM²=OA²+OM² (Pythagoras' theorem), thus (x+r/2)²=r²+(r/2)², hence x²+rx=r²; we have also y=AP+PQ=x+r. And we are done!  Remarks: Knowing that φ²=1+1/φ where φ=(1+√5)/2 is the golden ratio, we compute x=r/φ and y=rφ. OAB, ABN, DNO, ANO and OAD are all golden triangles. The right-angled triangles ABF and ADF give (Pythagoras' theorem) the sides BF and DF of the regular star and convex pentagons, and we verify that the value of their ratio BF/DF is φ. In a convex regular pentagon the ratio of the radius of the circumscribed circle to the side is about 0.85.

Caution! Not all regular polygons can be constructed with ruler and compass; the heptagon is the first example.
You may have a look at all the different kinds of regular polygons (order 3 to 14).

Remark: if one simply needs a drawing, a method of successive approximations makes it possible to obtain any polygon of order n by transferring n times the side along a circle (in three or four steps one arrives at a satisfactory result):
•  on a given circle, from an approximate side c0: we transfer n times the length c0 and we see an error e1; we start again with c1 = c0 ± e1/n (we correct by adding/subtracting one nth of the error), then again with c2 = c1 ± e2/n, and so on...
•  with a given side c, in a circle of approximate radius r0: we transfer n times the length c, and we start again with a circle of radius r1 after correction of the error, and so on ...

Reminder: if we have several exact constructions we must favor those that avoid successive transfers of the side; the result is often disappointing (the slightest drawing error spreads and grows during the transfers). an

small interesting constructions

cutting of a segment in three

 The theorem of equidistant parallels seems to be the right tool, but here is a construction using only midpoints, based on a property of the parallelogram: two segments which join the ends of a diagonal and the midpoints of two opposite sides divide the other diagonal in three. exercise: prove this property You may also use the center of gravity of a triangle, which lays on the third of each median: choose a segment MN with midpoint A and construct the center of gravity of triangle MNB (with I midpoint of NB).
A nice exercise to split a segment AB in n segments of the same length: we have only a non-graduated ruler with parallel edges (of any width l). We are looking for an construction which is always realizable, whatever the values ​​of n, AB and l.
There are many solutions, some are possible only for certain values ​​of n, others are conditioned by the value of the ratio AB / l; all are interesting and we must take the time to explore them before hoping to discover a "general" solution. Personally I know two such solutions, each illustrated by a drawing that does not require additional explanations.

two outstanding rhombi

The faces of the two semiregular rhombic polyhedra are the Varignon's rhombi of two outstanding rectangles:
the format A rectangle and the golden rectangle.
 face of the rhombic dodecahedron: SR2 rhombus (the ratio of the diagonals is √2 = 1.414...) face of the rhombic triacontahedron: golden rhombus (the ratio of the diagonals is the golden ratio 1.618...)

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