[alogo] Equilateral pentagon construction

Given the length of the side of an equilateral pentagon DEFGH and the angles of a side-triangle (build on one side with the prolongations of adjacent sides), the pentagon is uniquely defined. Below we assume that angle alpha is less than 60 degrees, but this is not a severe restriction.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

Triangles EFL, GHM, FGO are equilateral. Quadrilaterals EJHK and FGHJ are rhombi. They define the parallelogram EJHK.

a) Triangles HMK and ELJ are equal, having EL=MH, EJ=KH and the angle(MHK) = (1/2)angle(MGK) =(1/2)(pi/3 - angle(KGH)) = (pi/6)-(angle(A))/2 = fi. A similar calculation shows that this angle is equal to JEL. Notice that all the angles of MKH are known, angle(MKH) being 150 degrees. Thus, given the length of DE=EF=etc. triangle MKH is constructible.

b) Triangle OMJ is equilateral. Indeed MJ=JO since triangles MJH=OFJ, having equal sides and angles at H and F equal to pi/3 - angle(JHG) = pi/3 - angle(JFG)= angle(OFJ). Thus OJM is isosceles. Besides angle(OJG) = angle(MJG) = 150 Deg, being the angle on an arc of central angle pi/3. Thus angle(OJM)=pi/3 and OJM is equilateral.

c) Triangle OKL is equilateral. Since (proved in b) OM=OJ and MK=JL (proved in a), sides OL and OK are equal. Thus, since LEK is also isosceles, EO is the bisector of angle LEK. But EOK = 150 Deg. Thus the angle at O is pi/3, which shows that OKL is equilateral.

d) K, L, Q are on a line. Since angle(OLQ)=angle(OGQ)=(2pi/3) and (shown in c) angle(OLK)=pi/3. M, J, Q are on a line for a similar reason: angle(OJQ)=(2pi/3) and (shown in b) angle(OJM)=pi/3. Besides angle(MQK)=angle(MHK)=(1/2)angle(MGK)= (pi/6)-(angle(A)/2) = x.
A similar reasoning shows that K, O, R are collinear and O, J, P are also collinear. Besides the angle(KOJ)=angle(MOJ)-angle(MOK)= (pi/3)-angle(MHK) = pi/3-x = (pi/6)+(angle(A)/2).
Triangle RKQ is equilateral. F sees RO and LQ under equal angles. JPQ is also equilateral.

Given the length t of DE=EF= etc. and the angle at A, the isosceles triangle EFG is constructible, since its angles and side EF is known. Then the rhombus EFHJ is constructible and the isosceles triangle EHD is determined. This completes the pentagon construction.

Info on animation: Points G and K and X are free movable (switch to the selection tool: ctrl+1). Moving G and K changes only the length of the side t = (DE) and modifies the shape by similarity. Moving X changes the angle beta and modifies the shape.
Y is modifiable through the tool [Select on contour] (ctrl+2). Moving it changes angle alpha and modifies the shape. Look at the file Pentadivision.html for a nice application.

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