Pentadecagon

Construction with the rule and the compass

As one can build the equilateral triangle and the regular pentagon, one applies the theorem of Gauss:

3 and 5 being first between them, by multiplying by $\ frac {2 \ pi} {15}$ the relation of Bezout 2 × 3 - 5 = 1, the equality is obtained: $2 \ frac {2 \ pi} {5} - \ frac {2 \ pi} {3} = \ frac {2 \ pi} {15}$

On a circle, starting from a point has, one places a point G such as $(\ overrightarrow {OA}, \ overrightarrow {OG}) = \ frac {4 \ pi} {5}$ , the point B such as $(\ overrightarrow {OG}, \ overrightarrow {OB}) = - \ frac {2 \ pi} {3}$ is the second top of the regular polygon on side AB.

In practice regular pentagon ADGJM (direct direction) is traced.

From the point G one traces equilateral triangle GBL (direction retrogresses).

By deferring 14 times length AB on the circle one obtains regular polygon ABCDEFGHIJKLMNP.

Cross pentadecagons

The number of crossed regular polygons of N east coasts equal to the number of prime numbers with N contained in continuation 2,3…, ( N -1) /2. There are three cross pentadecagons which one obtains by uniting the tops of two into two, four into four or seven in Sept.

Construction with a mediator

To build regular pentagon ADGJM registered in the circle (c) of center O.

To place G' the symmetrical point of G compared to O.

The mediator of cut the circle (c) in two points B and L, tops of the pentadecagon.

Justification

Triangle OBG' is equilateral because OB = OG' like rays and OB = G' B because B is on the mediator of.

The angle $\ widehat {MOA}$ of two rays of the pentagon is of $\ frac {2 \ pi} {5}$ .

$\ widehat {G' OA} = \ frac 1 2 \ widehat {MOA} = \ frac {\ pi} {5}$ . $\ widehat {AOB} = \ widehat {G' OB} - \ widehat {G' OA} = \ frac {\ pi} {3} - \ frac {\ pi} {5} = \ frac {2 \ pi} {15}$ , angle two rays of the pentadecagon.

Construction with the compass

To build regular pentagon ADGJM of center O.

To place a' points, Of, G', I, symmetrical of have Me, D, G, J, M compared to O.

The points of the pentadecagon are the points of intersection of the circle (c) with the circles of A' centers, Of, G', I, passing to Me by the center O.

Justification

G' OB is an equilateral triangle on side equal to the ray R of the circumscribed circle, $\ widehat {G' OB} = \ frac {2 \ pi} 3$ .

As above one a: $\ widehat {G' OA} = \ frac 1 2 \ widehat {MOA} = \ frac {\ pi} {5}$ (angle in the center of the pentagon).

$\ widehat {AOB} = \ widehat {G' OB} - \ widehat {G' OA} = \ frac {\ pi} {3} - \ frac {\ pi} {5} = \ frac {2 \ pi} {15}$ is the angle in the center of the pentadecagon and the point B is well a top.

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