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Analyzing Pentagrams Analyzing Pentagrams Paul Elliott March 2014 License This work is licensed under the Creative Commons Attribution 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/. http://www.free.blackpatchpanel.com/pme/geometry/pent.pdf https://github.com/pelliott80/analyzing-pentagrams This work was made with GNU/Linux LYX LATEX beamer carmetal inkscape git Analyzing Pentagrams Intellectual Machinery In order to analyze a pentagram, we must rst create some intellectual machinery rst. Golden Rectangle What is a golden Rectangle? Cut o a square When you cut o a square, what remains has the same proportions. Figure: Equation of Golden Rectangle l s = s l − s Golden Equation l s = s l − s l (l − s) = s2 l 2 − ls = s2 l 2 − ls s2 s2 = s2 l 2 l 1 s − s = l Now let φ (phi) be s the ratio of the long to the short sides. Then this equation is: φ2 − φ = 1 φ2 − φ − 1 = 0 Golden Equation roots. φ2 − φ − 1 = 0 If we apply the quadratic formula to this equation, we nd: p p 1 ± 1 + 4 1 ± 5 φ = 2 = 2 The negative root we can lay aside for the moment, as a distance or a ratio of distances can not be negative. The positive root is: p 1 5 + 1 618033989 2 = : ::: This is the Golden Ratio, which often crops up uncannily in mathematics, nature, art, and the occult. Golden ratio conjugate p 1− 5 Let us consider the other root 2 . This is a negative number, and the ancient Greeks did to believe in negative numbers, so we will consider its p 1− 5 negative, Φ = − 2 . It is called the Golden ratio conjugate. Consider φ Φ: p p 1 5 1 5 + 1 − φ Φ = 2 (− ) 2 1 5 1 − = (− ) 4 = 1 So φ and Φ are reciprocals. To divide by one, multiply by the other. Analyzing the Golden Triangle Denition The golden triangle, sometimes also called the sublime triangle, is an isosceles triangle such that the ratio of the adjacent sides to base is equal to the golden ratio. Dissecting the Golden Triangle l s = φ. Mark o a segment of length s on segment AC at point D, creating segment AD with length AD = s. Then l DC = l − s. s satises the golden equation, so: l 2 l 1 0 s − s − = l s = s l − s AC = BC = φ CB DC Golden Triangle Split Consider the triangles 4ACB and 4BCD they share the angle \ACB = \DCB and the adjacent sides are in proportion, therefore by the SAS Similarity Theorem, the two triangles are similar. 4ACB v 4BCD. It follows that 4BCD is isosceles triangle and that BD = BC = s. It also follows that \BAC = \CBD. AC BC CB = DC Golden Triangle Split Take the triangle on the left, and rotate it so the small triangle is upright and magnify by φ, to get the triangle on the left. Superimpose Take the triangle on the left slide right to superimpose the triangles! The other triangle We have already seen that \CBD ' \BAC. So \ABD ' \CBD ' \BAC, and thus ACB = ABC = 2 BAC. ] −!] ] The ray BD bisects the angle \ABC. Because BD = AD, triangle 4ADB is isosceles. So \ABD ' \BAD = \BAC. More about this important triangle later. The angles of the Golden Triangle But the sum of the three angles of a triangle is two straight angles. So: ◦ ]BAC + ]ABC + ]ACB = 180 ◦ ]BAC + 2 ]BAC + 2 ]BAC = 180 ◦ 5 ]BAC = 180 ◦ ]BAC = 36 ◦ ]ABC = ]ACB = 2 ]BAC = 72 So all golden triangles are 36◦, 72◦, 72◦ triangles. Any other 36◦, 72◦, 72◦ triangle would be similar to this one, and hence a golden triangle. Remember the other triangle? It is a Golden Gnomon Denition The Golden Gnomon, is an It is an isosceles triangle with isosceles triangle such that the base angles 36◦ degrees. The ratio of the base to the vertex angle must therefore be adjacent sides is equal to the 108◦ degrees. The ratio of the Golden Ratio. length of the base to the The golden gnomon has angles length of a leg, l is . s φ 36◦, 36◦, 108◦ as we have seen. The Golden Gnomon can also be dissected The Golden Gnomon can also be dissected, just as the Golden Triangle can. It breaks into a Golden Triangle and another Golden Gnomon, just as the Golden Triangle breaks into a Golden Gnomon and a Golden Triangle. This is important in Penrose tilings. Penrose Tilings Figure: Inscribed Angle Theorem Inscribed Angle Theorem Theorem An angle Θ inscribed in a circle The Greek Philosopher Thales is half the central angle 2 Θ was the rst to prove a special that subtends the same arc on case of this theorem, when the the circle. central angle is 180◦ degrees. He sacriced a bullock for the discovery of his theorem. Geometers are tricky Geometers are tricky. The prove the inscribed angle theorem by rst proving a special case in which one of the legs of the inscribed angle is a diameter. They then bootstrap this result, to prove the whole theorem! Proof of the special case Proof of the special case Proof. ! ! Create line CE parallel to line AD. Then ECF is the corresponding angle ! \ for \DAC for the transversal AF . Therefore ]ECF = ]DAC. M ACD is isosceles, two of its sides being radii, thus CAD = ADC. But ADC ] ] ! \ and \DCE are alternate interior angles for the transversal DC. Thus ]ADC = ]DCE. Thus ]CAD = ]ADC = ]DCE. So \DCF is made up by two angles, \DCE and \ECF both of which have the same measure as \CAD. So ]DCF = 2 ]CAD: The General Case The General Case Proof. ! Draw line AF splitting our angles which are not subject to the special case, into 2 angles which have one side as a diameter, and to which our special case does apply. we can write: ]ECF = 2 ]EAF ]FCG = 2 ]FAG ]ECF + ]FCG = 2 ]EAF + 2]FAG ]ECF + ]FCG = 2 (]EAF + ]FAG) ]ECG = 2 ]EAG We now begin on the pentagram There is a pentagon here two There is an inscribed regular pentagon here too! And that pentagon can be sliced into pie slices from the center. Central angle pie slices are all 72◦ Theorem The central angle pie slices of a regular pentagon are all 72◦ Proof. The exterior sides are all of equal length because the pentagon is regular. The slice cuts from the center are all equal because they are all radii. Thus by SSS, the slices are all congruent and the central angles are all equal. 360◦ ◦ 5 = 72 The regular pentagram vertex angles are 36◦ The regular pentagram vertex angles are 36◦ Theorem The regular pentagram vertex angles are 36◦ Proof. The vertex angles span the same arcs as the 72◦ central angles. Inscribed angle theorem. The angles between the pentagon sides and the pentagram sides are also 36◦ The angles between the pentagon sides and the pentagram sides are also 36◦ Theorem The angles between the pentagon sides and the pentagram sides are also 36◦ Proof. The angles between the pentagon sides and the pentagram sides span the same arcs as the 72◦ central angles. Inscribed angle theorem. The interior pentagon angles are 108◦ Proof. 180◦ − 36◦ − 36◦ = 108◦ The secant angles where pentagram sides cross are 72◦ The secant angles where pentagram sides cross are 72◦ Theorem The secant angles where pentagram sides cross are 72◦ Proof. These angles are supplemental to 108◦ angles. The pentagon angles are 108◦ The pentagon angles are 108◦ Theorem The angles of the exterior regular pentagon are 108◦ Proof. 3 · 36◦ = 108◦ We have now gured out all the angles of the inscribed pentagram and pentagon. Let us take a rest and notice how beautiful the pentagram is. I like them. Golden Triangles in Pentagrams The big wedges are 36◦, 72◦, 72◦ and so are Golden Triangles. There are ve of these big triangles in the pentagram. The triangle are isosceles, so all the strokes of the pentagram are the same length. The ratio of the length of a stroke of the pentagram to the length of the side of the regular pentagon is φ. Big Gnomon There are 5 of these big Gnomon in the pentagram. It is a gnomon because the angles are 36◦, 36◦, 72◦. The short sides are sides that start at a vertex cross one secant angle and stop at the other secant angle. There are 10 of these one secant angle crossers in the pentagram. They are all the same length. The ratio of the length of a pentagon stroke to the length of a one secant angle crossers is φ. outside Golden Triangle There are 10 of these Golden Triangles at the edge of the pentagram. They show that the length of a one secant angle crossers is equal to length of a pentagon side. Gnomon outside the pentagram There are ve of these Golden Gnomon. They are isosceles triangles, they show that the lengths from a vertex to the a secant angle is always the same. Anyone want to join Star Fleet? Star Fleet invokes the pentagram! Inner and outer pentagon Triangles M CBA, and M ADB are Golden triangles. AB AC φ = = BD AB φ BD = AB φ AB = AC Inner and outer pentagon φ φ BD = AC φ2 BD = AC AC φ2 = BD Inner and outer pentagon So the ratio of size of the outer pentagon to the size of the inner pentagon is φ2. And all the inner pentagon sides are equal in length.
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