Zef Damen
Non ruler-and-compass constructions (1) | ||||||||||||||||||||||||||

Non ruler-and-compass construction of odd-numbered regular polygons | ||||||||||||||||||||||||||

The central part of the Ivinghoe Beacon formation of 26-July-2002 shows a threefold propeller-like pattern. The reconstruction of this part is interesting, since it gives rise to a different way of constructing regular polygons with an odd number of sides. To show this, the reconstruction of that part is shortly repeated here. | ||||||||||||||||||||||||||

Inscribed in a circle 1 is a nonagon 2, a regular 9-sided polygon. Circle 3 has its center at the lower end-point of the vertical centerline and passes through two adjacent angular points of nonagon 2. Its radius, therefore, is equal to one side of the corresponding regular 18-sided polygon (not shown). A circle 4 has been constructed, concentric to circle 1, and tangent to circle 3 at the upper side. Circle 3 has been copied 4 times, one to the center of circle 1, and the other three to the intersections of circle 4 and the lines 2 and upper half of the vertical centerline. These lines are 120° apart. Circle 4 has been copied 3 times, to the intersections of the central circle 3 and the same 120°-lines. An interesting question arises here: why do circles 4 pass through the intersections of circles 3 (e.g. the point marked with an arrow)? Or do they really? | ||||||||||||||||||||||||||

The answer is yes. If the picture is zoomed in, it can be clearly seen that three circles pass through a single point. But this is not yet a proof. In the next sections, I'll show why this is indeed true. And this will give rise to another (although already known) way of constructing regular polygons with an odd number of sides. Not ruler-and-compass constructions, but with "sliding rulers" connected by "rotating points". | ||||||||||||||||||||||||||

The line of reasoning is as follows. I first do the construction a second time, but with an arbitrary circle 3. I then show under what condition circles 4 pass through the intersections of circles 3. And I finally will show, that this condition can only be fulfilled, if the radius of circles 3 is equal to a side of the regular 18-sided polygon corresponding to nonagon 2, as was chosen in the first place. Here is the reconstruction again with an arbitrary circle 3, slightly greater than the original one. | ||||||||||||||||||||||||||

Let's call the radius of circles 3 r, and that of circles 4 R. The line marked x connects the center of the marked circle 4 to the intersection of circles 3 indicated by the arrow. For circles 4 to pass through these intersections, x has to be equal to R.So, we want x = R (1)to be true. | ||||||||||||||||||||||||||

Looking at triangles ABC and BCD (see figure), this means, both triangles must be equal (congruent). That implies, that angle a should be equal to angle b:a = b (2)One of the lines 2 has been extended to the other side of circle 1. Since both lines 2 are 120° apart, this extension and the other line 2 enclose an angle of 60°. Therefore, angle b plus angle c sum up to 60°:b + c = 60° (3)Angle c is an outer angle of triangle BCD, and is therefore equal to the sum of both non-adjacent angles of BCD. Or:c = 2a (4)Together with (2) and (3), it follows, that: a + 2a = 60° (5)Or: a = b = 20° (6)Note, that 20° is ^{1}/_{18}th of a full circle of 360°, and thus covers one sector of a regular 18-sided polygon. | ||||||||||||||||||||||||||

So, if we make angle b indeed 20°, and extend line CB up to circle 1, we get this figure. Now, line FG (length z) is indeed equal to one side of the regular 18-sided polygon. We now have to prove, that z = r.From the figure, we can derive the following: d = 2b = 40° (7)e = 2d - b = 80° - 20° = 60° (8)That implies, triangle AEF is equilateral. Thus: f = 60° (9)and: y = r (10)g = 180° - d - f = 180° - 40° - 60° = 80° (11)CFG is an isosceles triangle (both sides are equal to the radius of circle 1), giving: h + f = i = ^{1}/_{2}(180° - b) = ^{1}/_{2}(160°) = 80° (12)So, i = g (13)which means, that triangle EFG is also an isosceles triangle, and we can finally conclude (from (10)): z = y = r (14) | ||||||||||||||||||||||||||

We now have one sector of an 18-sided polygon, in which 5 lines of length r are layed out, head to tail, exactly covering it. On the next page, I'll show that this can be seen as a way of constructing a regular 9-sided (and 18-sided) polygon, and how this can be generalised to any odd-numbered regular polygon. | ||||||||||||||||||||||||||

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Since 1-February-2005 |