Mathematics in Art and Architecture GEK1518K

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1 Mthemtics in Art nd Architecture GEK1518K Helmer Aslksen Deprtment of Mthemtics Ntionl University of Singpore The Golden Rtio The Golden Rtio Suppose we wnt to divide line AB t point C such tht AC is longer thn BC nd the rtio of AC to BC is equl to the rtio of AB to AC. In other words, whole / long = long / short. We will determine the vlue of this common rtio φ = AC = AB. BC AC If CB =, then AC = φ nd AB = φ 2. Hence, φ 2 = φ + φ 2 = φ + 1 The positive solution is φ = (1 + 5) / This vlue of φ is known s the golden rtio. It is lso known s the golden section or the divine rtio. Conversely, given line segment AB, we my wnt to extend it to line AE such tht AE = φ. We will cll this golden rtion extension. To construct this extension AB geometriclly, strt with line AB of length nd construct squre ABCD. Bisect AB nd let the midpoint be M. With M s center, drw n rc with rdius MC to cut AB produced t E. Let us now check the rtios AE/AB nd AB/BE. To do so, we need to find the lengths MC nd AE. 1

2 By Pythgors Theorem, Hence nd since (MC) 2 = (MB) 2 + (BC) 2 = ( / 2 ) = ( 5 / 4 ) 2 MC = ( 5 / 2 ) AE = AM + ME = AM + MC = ( 1 / 2 ) + ( 5 / 2 ) = (1 + 5)/2 AE = AB ( 1+ 5 / 2 ) = (1 + 5)/2, AB = = (1 + 5)/2, BE ( 5 / 2 1/ 2) AE we see tht = AB = φ. AB BE Observe tht the rectngle AEFD hs length φ nd height. We cll rectngle with dimensions by φ golden rectngle. If the short segment hs length 1, then the long segment hs length φ nd the whole segment hs length φ 2. If the long segment hs length 1, then the short segment hs length 1/ φ nd the whole segment hs length φ. If the whole segment hs length 1, then the long segment hs length 1/ φ nd the short segment hs length 1/ φ 2 A B E /φ Extending line AB to ABE by length of /φ to get golden rtio extension 2

3 The bove construction shows how to get golden rtio extension of AB to ABE. We now wnt to construct golden rtion cut, tht is, strt with AB nd find point C such tht AB/AC=AC/CB = φ. It turns out tht we hve done ll the hrd work lredy. In the bove golden rtion extension, the length of BE is /φ. But this is exctly the length of the long piece in golden rtio cut. Using compss centered t B with rdius BE = / φ, cut the line AB t X. So, BX = BE = / φ. Notice tht AX is the smller prt while BX is the lrger prt of the line segment becuse BX = / φ > ½. Since AB = φ nd BX = BE, BE AB = AB = φ. BX BE X B /φ /φ Cutting given line AB into 2 prts to chieve the golden rtio Now we know how to do both golden rtion extension nd golden rtio cut. Constructing Regulr Pentgon using the Golden Rtio We will use the golden rtio to help us construct regulr pentgon. Observe tht if we divide regulr pentgon ABCDE into 5 isosceles tringles s shown in Figure 5.2.1, then, for ech isosceles tringle, the interior ngle t F is 72 o. A B 72 o E F A regulr pentgon C D We will construct regulr pentgon within circumscribing circle. To do tht, we need wy to construct the ngle of 72 o. Euclid cme up with the following method to get this ngle. 1) Begin with line ABE, which is divided into the golden rtio t B, with AB longer thn BE. 2) Using compsses centered t B nd E nd of rdius AB, drw two rcs of circles to intersect t point C. So AB = BC = CE nd we will prove lter tht AC = AE. 3) Form the tringle CEB. Since BC = CE, so CBE = CEB. Let these two ngles be α. So BCE = 180 o - 2 α. 4) Consider the tringle ACE. Since AC = AE, ACE = CEB = α. So CAB = 180 2α. 3

4 5) Consider the tringle ACB. ACB = ACE BCE = 3 α Since AB = BC, CAB = 3 α ) From (4) nd (5), CAB = 3α 180 = 180 2α. Solving, we get α = 72. Now tht we obtin the ngle α = 72 by the bove construction, construct circle t center E, rdius less thn BE. The circle cuts BE nd CE t X nd Y respectively. X nd Y form two vertices of our regulr pentgon. We cn get the remining three vertices of the pentgon using compsses with rdius XY nd cutting the circle. C Constructing 3α regulr o pentgon 180 o - 2α Y A B α X α E Now, we prove tht AC = AE. Let the length of AB = nd so, BE = / φ. So AE = (1 + 1 / φ ). By Eqution (5.1), AE = φ. So we wnt to prove tht AC = φ. Find the mid-point N of BE nd the line CN is perpendiculr to BE. We pply Pythgors Theorem to get AC. Therefore, AC = φ = AE. (AC) 2 = CN 2 + AN 2 = (CE 2 - NE 2 ) + (AB + BN) 2 = ( 2 ( / 2φ ) 2 ) + ( + ( / 2φ )) 2 = 2 (2 + 1 / φ ) = 2 (1 + φ) by Eqution (5.1) = 2 φ 2 by Eqution (5.2) 4

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