13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS


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1 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in the negtive) the following geometric problems posed by the Greeks: I (Doubling the Cube) Is it possible using only strightedge nd compss to construct cube with precisely twice the volume of given cube? II (Trisecting n Angle) Is it possible using only strightedge nd compss to trisect ny given ngle θ? III (Squring the Circle) Is it possible using only strightedge nd compss to construct squre whose re is precisely the re of given circle? To nswer these questions we must trnslte the construction of lengths by compss nd strightedge into lgebric terms Let denote fixed given unit distnce Then ny distnce is determined by its length R, which llows us to view geometric distnces s elements of the rel numbers R Using the given unit distnce to define the scle on the xes, we cn then construct the usul Crtesin plne R nd view ll of our constructions s occurring in R A point (x, y) R is then constructible strting with the given distnce if nd only if its coordintes x nd y re constructible elements of R The problems bove then mount to determining whether prticulr lengths in R cn be obtined by compss nd strightedge constructions from fixed unit distnce The collection of such rel numbers together with their negtives will be clled the constructible elements of R, nd we shll not distinguish between the lengths tht re constructible nd the rel numbers tht re constructible Ech strightedge nd compss construction consists of series of opertions of the following four types: () connecting two given points by stright line, () finding point of intersection of two stright lines, (3) drwing circle with given rdius nd center, nd (4) finding the point(s) of intersection of stright line nd circle or the intersection of two circles It is n elementry fct from geometry tht if two lengths nd b re given one my construct using strightedge nd compss the lengths ± b, b nd /b (the first two re cler nd the ltter two re given by the construction of prllel lines (Figure )) b b Fig b b It is lso n elementry geometry construction to construct if is given: construct the circle with dimeter + nd erect the perpendiculr to the dimeter s indicted in Figure Then is the length of this perpendiculr Sec 33 Clssicl Strightedge nd Compss Constructions
2 Fig It follows tht strightedge nd compss constructions give ll the lgebric opertions of ddition, subtrction, multipliction nd division (by nonzero elements) in the rels so the collection of constructible elements is subfield of R One cn lso tke squre roots of constructible elements We shll now see tht these re essentilly the only opertions possible From the given length it is possible to construct by these opertions ll the rtionl numbers Q Hence we my construct ll of the points (x, y) R whose coordintes re rtionl We my construct dditionl elements of R by tking squre roots, so the collection of elements constructible from of R form field strictly lrger thn Q The usul formul ( two point form ) for the stright line connecting two points with coordintes in some fieldf gives n eqution for the line of the formx+by c = 0 with, b, c F Solving two such equtions simultneously to determine the point of intersection of two such lines gives solutions lso in F It follows tht if the coordintes of two points lie in the field F then strightedge constructions lone will not produce dditionl points whose coordintes re not lso in F A compss construction (type (3) or (4) bove) defines points obtined by the intersection of circle with either stright line or nother circle A circle with center (h, k) nd rdius r hs eqution (x h) + (y k) = r so when we consider the effect of compss constructions on elements of field F we re considering simultneous solutions of such n eqution with liner eqution x + by c = 0 where, b, c, h, k, r F, or the simultneous solutions of two qudrtic equtions In the cse of liner eqution nd the eqution for the circle, solving for y, sy, in the liner eqution nd substituting gives qudrtic eqution for x (nd y is given linerly in terms of x) Hence the coordintes of the point of intersection re t worst in qudrtic extension of F In the cse of the intersection of two circles, sy (x h) + (y k) = r nd (x h ) + (y k ) = r, subtrction of the second eqution from the first shows tht we hve the sme intersection by considering the two equtions (x h) + (y k) = r nd (h h)x + (k k)y = r h k r + h + k
3 which is the intersection of circle nd stright line (the stright line connecting the two points of intersection, in fct) of the type just considered It follows tht if collection of constructible elements is given, then one cn construct ll the elements in the subfield F of R generted by these elements nd tht ny strightedge nd compss opertion on elements of F produces elements in t worst qudrtic extension of F Since qudrtic extensions hve degree nd extension degrees re multiplictive, it follows tht if α R is obtined from elements in field F by (finite) series of strightedge nd compss opertions then α is n element of n extension K of F of degree power of : [K : F ] = m for some m Since [F(α) : F ] divides this extension degree, it must lso be power of Proposition 3 If the element α R is obtined from field F R by series of compss nd strightedge constructions then [F(α) : F ] = k for some integer k 0 Theorem 4 None of the clssicl Greek problems: (I) Doubling the Cube, (II) Trisecting n Angle, nd (III) Squring the Circle, is possible Proof: (I) Doubling the cube mounts to constructing 3 in the rels strting with the unit Since [Q( 3 ) : Q] = 3 is not power of, this is impossible (II) If n ngle θ cn be constructed, then determining the point t distnce from the origin nd ngle θ from the positive x xis in R shows tht cos θ (the xcoordinte of this point) cn be constructed (so then sin θ cn lso be constructed) Conversely if cos θ, then sin θ, cn be constructed, the point with those coordintes gives the ngle θ The problem of trisecting the ngle θ is then equivlent to the problem: given cos θ construct cos θ/3 To see tht this is not lwys possible (it is certinly occsionlly possible, for exmple for θ = 80 ), consider θ = 60 Then cos θ = By the triple ngle formul for cosines: cos θ = 4cos 3 θ/3 3cos θ/3, substituting θ = 60, we see tht β = cos 0 stisfies the eqution 4β 3 3β / = 0 or 8(β) 3 6β = 0 This cn be written (β) 3 3(β) = 0 Let α = β Then α is rel number between 0 nd stisfying the eqution α 3 3α = 0 But we considered this eqution in the lst section nd determined [Q(α) : Q] = 3, nd s before we see tht α is not constructible (III) Squring the circle is equivlent to determining whether the rel number π = 3459 is constructible As mentioned previously, it is difficult problem even to prove tht this number is not rtionl It is in fct trnscendentl (which we shll ssume without proof), so tht [Q(π) : Q] is not even finite, much less power of, showing the impossibility of squring the circle by strightedge nd compss Sec 33 Clssicl Strightedge nd Compss Constructions 3
4 Remrk: The proof bove shows tht cos 0 nd sin 0 cnnot be constructed The question rises s to which integer ngles (mesured in degrees) re constructible? The ngles nd re not constructible, since otherwise the ddition formule for sines nd cosines would give the constructibility for 0 On the other hnd, elementry geometric constructions (of the regulr 5gon for n ngle of 7 nd the equilterl tringle for n ngle of 60 ) together with the ddition formule nd the hlfngle formule show tht cos 3 nd sin 3 re constructible It follows from this tht the trigonometric functions of n integer degree ngle re constructible precisely when the ngle is multiple of 3 Explicitly, cos 3 = 8 ( 3 + ) ( 6 )( 5 ) sin 3 = 6 ( 6 + )( 5 ) 8 ( 3 ) 5 + 5, showing tht these re obtined from Q by successive extrctions of squre roots nd field opertions After discussing the cyclotomic fields in Section 45 we shll consider nother clssicl geometric question: which regulr ngons cn be constructed by strightedge nd compss? (cf Proposition 49) We hve been creful here to consider constructions using strightedge rther thn ruler, the distinction being tht ruler hs mrks on it If one uses ruler, it is possible to construct mny dditionl lgebric elements For exmple, suppose θ is given ngle nd the unit distnce is mrked on the ruler Drw circle of rdius with centrl ngle θ s shown in Figure 3 nd then slide the ruler until the distnce between points A nd B on the circle is Then some elementry geometry shows tht (cf the exercises) the ngle α indicted is θ/3, ie, this construction (due to Archimedes) trisects θ In prticulr, the second clssicl problem in Theorem 4 (Trisecting n Angle) cn be solved with ruler nd compss α B A θ Fig 3 The first of the clssicl problems in Theorem 4 (Dupliction of the Cube), which mounts to the construction of 3, cn lso be solved with ruler nd compss The following gives construction for k /3 for ny given positive rel k which is less thn This construction ws shown to us by JH Conwy Drwing circle of rdius nd using the point A = (k, 0) s center, construct the point B = (0, k ) Dividing this distnce by 3, construct the point (0, 3 k ) nd drw the line connecting this point with A Slide the ruler with 4
5 mrked unit length so tht it psses through the point B nd so tht the distnce from the intersection point C to the intersection point D with the xxis is of length, s indicted in Figure 4 Then the distnce between A nd D is k /3 nd the distnce between B nd C is k /3 (cf the exercises) B k C 3 k k A D Fig 4 E X E R C I S E S Prove tht it is impossible to construct the regulr 9gon Prove tht Archimedes construction ctully trisects the ngle θ [Note the isosceles tringles in Figure 5 to prove tht β = γ = α] α β θ γ Fig 5 3 Prove tht Conwy s construction indicted in the text ctully constructs k /3 nd k /3 [One method: let (x, y) be the coordintes of the point C, the distnce from B to C nd b the distnce from A to D; use similr tringles to prove () y k = +, (b) x = b + k +, (c) y k x k =, nd lso show tht (d)( k )+(b+k) = (+) ; 3k solve these equtions for nd b] 4 The construction of the regulr 7gon mounts to the constructibility of cos(π/7) We shll see lter (Section 45 nd Exercise of Section 47) tht α = cos(π/7) stisfies the eqution x 3 + x x = 0 Use this to prove tht the regulr 7gon is not constructible by strightedge nd compss 5 Use the fct tht α = cos(π/5) stisfies the eqution x + x = 0 to conclude tht the regulr 5gon is constructible by strightedge nd compss Sec 33 Clssicl Strightedge nd Compss Constructions 5
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