6. ANGLES AND ELEMENTAL TRIGONOMETRY

Transcription

1 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 5 6. NGLES ND ELEMENTL TRIGONOMETRY Hr, th basic intitis of th lmntal trigonomtr ar uc in clos connction with basic gomtric facts, a vr usful point of viw for our pupils. Sum of th angls of a polgon Firstl lt us s th spcial cas of a triangl. For an triangl th following intit hols: Lt α, β an γ b th trior angls btwn th sis an, an, an rspctivl (figur 6.). ppling th finition of gomtric quotint, th moulus of all th sis ar simplifi an onl th ponntials of th argumnts rmain: Figur 6. ( α ) p( β ) p( γ ) ( α β γ ) p p[ ] Th thr angls hav th sam orintation, which w suppos positiv, an ar lssr than π. Hnc, sinc th ponntial is qual to th unit, th aition of th thr angls must b qual to π: α β γ π Th intrior angls, thos form b an, an, an ar supplmntar of α, β, γ (figur 6.). Thrfor th sum of th angls of a triangl is qual to π: ( π α ) ( π β ) ( π γ ) π This rsult is gnralis to an polgon from th following intit:... YZ YZ Z Z Lt α, β, γ... ω b th trior angls form b th sis Z an, an, an D,..., YZ an Z, rspctivl. ftr th simplification of th moulus of all vctors, w hav: [( α β γ... ω ) ] p Lt us suppos that th orintation fin b th vrtics,,... Z is countrclockwis, although th trior angls b not ncssaril all positiv. Translating That is, it is not n that th polgon b conv.

2 54 RMON GONZLEZ LVET thm to a common vrt, ach angl is plac clos ach othr following th orr of th primtr an summing on turn: α β γ... ω π Th intrior angls form b th sis an Z, an, D an,... Z an ZY ar supplmntar of α, β, γ,... ω. Thrfor th sum of th angls of a polgon is: ( π α ) ( π β ) ( π γ )... ( π ω ) n π π ( n )π n bing th numbr of sis of th polgon. Th uction for th clockwis orintation of th polgon is analogous with th onl iffrnc that th rsult is ngativ. Dfinition of trigonomtric functions an funamntal intitis Lt us consir a circl with raius r (figur 6.). Th trm of th raius is a point on th circumfrnc with coorinats (, ). Th arc btwn th positiv X smiais an this point (, ) has an orint lngth s, positiv if countrclockwis an othrwis ngativ. lso, th X-ais, th arc of circumfrnc an th raius limit a sctor with an orint ara. n orint angl Figur 6. α is fin as th quotint of th arc lngth ivi b th raius : α s r Sinc th ara of th sctor is proportional to th arc lngth an th ara of th circl is πr, it follows that: α r α r Th trigonomtric functions, sin, cosin an tangnt of th angl α, ar rspctivl fin as th ratios: sin α cos α tg α r r an th coscant, scant an cotangnt as thir invrs fractions: With this finition it is sai that th angl is givn in raians, although an angl is a quotint of lngths an thrfor a numbr without imnsions. lso call circular functions u to obvious rasons, to b istinguish from th hprbolic functions.

3 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 55 r cscα sin α r scα cosα cotα tgα bing r rlat with an b th Pthagoran thorm: r Thn th raius vctor v is: v r ( cosα sinα ) From ths finitions th funamntal intitis follow: sin α tgα sin α cos α tg α sc α cos α cos α If w tak th opposit angl α insta of α, th sign of is chang whil an r ar prsrv, so w obtain th parit rlations: ( α ) α cos( α ) cosα tg( α ) tgα sin sin Th sin an tangnt ar o functions whil th cosin is an vn function. Look at th figur 6.: β π/α is th complmntar angl of α. If α is highr than π/, th angl β bcoms ngativ. On th othr han, if th angl α is ngativ thn β is highr than π/. Th trigonomtric ratios for β giv th complmntar angl intitis: sin β cosα r cos β sin α r tg β tgα ngl inscrib in a circl an oubl angl intitis Lt us raw an iamtr PQ an an raius O (figur 6.) in a circl with cntr O. Sinc OP is also a raius, th triangl PO is isoscls an th angls OP an PO, which will b not as α, ar qual. caus th aition of th thr angls is qual to π, th angl OP is π α. Th angl QO is supplmntar of OP, an thrfor is qual to α, th oubl of th angl PQ. Lt us raw from a sgmnt prpnicular to th iamtr an touching it at th point N. th finition of sin w hav: Figur 6.

4 56 RMON GONZLEZ LVET sin α N O N P P O Th first quotint is sinα for th triangl PN. If M is th mipoint of th sgmnt P, thn P M an th scon quotint is qual to cosα for th triangl MO: M sin α sin α sin α cosα O Through an analogous wa w obtain cos α : cos α ON O ( PN PO ) P P O cos α cos α sin α lso this rsult ma b obtain from th scon funamntal intit. In orr to obtain th tangnt of th oubl angl w mak us of th first funamntal intit: tg sin α sin α cosα tgα cosα cos α sin α tg α α Finall, lt us raw an othr sgmnt P (figur 6.). Th angl PQ will b not as β. th sam argumnts as abov th angl OQ is β. Whil th angl P is α β, th angl O is α β : an angl inscrib in a circumfrnc is qual to th half of th cntral angl (angl whos vrt is th cntr of th circumfrnc) which intrcpts th sam arc of circl. onsquntl, all th angls inscrib in th sam circl an intrcpting th sam arc ar qual inpnntl of th position of th vrt on th circl. ition of vctors an sum of trigonomtric functions Lt us consir two unitar vctors forming th angls α an β with th irction (figur 6.4): u cosα sinα v cos β sin β ( α cos β ) ( sin α sin β ) u v cos Th aition of both vctors, u v, is th iagonal of th rhombus which th form, whnc it follows that th long iagonal is th bisctor of th angl α β btwn both vctors. Th short iagonal cuts th long iagonal prpnicularl forming four right triangls. Thn th moulus of u v is qual to th oubl of th cosin of th half of this angl: Figur 6.4

5 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 57 u v α β cos Morovr, th aition vctor forms an angl (α β) / with th irction: u v α β α β α cos cos sin β intifing this prssion for u v with that obtain abov, w arriv at two intitis, on for ach componnt: α β α β cosα cos β cos cos α β α β sin α sinβ sin cos In a similar mannr, but using a subtraction of unitar vctors, th othr pair of intitis ar obtain: α β α β cosα cos β sin sin α β α β sin α sin β cos sin Th aition an subtraction of tangnts ar obtain through th common nominator: ( α ± β ) sin α sinβ sin α cos β ± cosα sin β sin tgα ± tg β ± cosα cosβ cosα cos β cosα cosβ Prouct of vctors an aition intitis Lt us s at th figur 6.4, but now w calculat th prouct of both vctors: v ( β sin β ) ( cosα sin ) u cos α ( sin α cos β cosα sin β ) cosα cos β sin α sin β Sinc u an v ar unitar vctors, thir prouct is a compl numbr with unitar moulus an argumnt qual to th orint angl btwn thm: v ( α β ) ( α β ) u cos sin Th intification of both quations givs th trigonomtric functions of th angls

6 58 RMON GONZLEZ LVET iffrnc: sin ( α β ) sinα cos β cosα sin β ( α β ) cosα cos β sinα sin β cos Taking into account that th sin an th cosin ar o an vn functions rspctivl, on obtains th intitis for th angls aition: ( α β ) sinα cos β cosα sin β sin cos ( α β ) cosα cos β sinα sin β Rotations an D Moivr s intit If v' is th rsult of turning th vctor v ovr an angl α thn: ( α sinα ) v' v cos To rpat a rotation of angl α b n tims is th sam thing as to turn ovr an angl nα : v'' v n ( α sinα ) v ( cos nα sin nα ) cos nα sin nα ( cosα sinα ) n cos This is th D Moivr s intit 4, which allows to calculat th trigonomtric functions of multiplis of a crtain angl through th binomial thorm. For ampl, for n w hav: cosα ( cosα α ) sin α sin cos α cosα sin α Splitting th ral an imaginar parts on obtains: ( cos α sin α sin α ) cosα cos α cosα sin α sinα cos α sinα sin α n iviing both intitis on arrivs at: tg tgα tg α tg α α 4 With th Eulr s intit, th D Moivr s intit is: p ( α ) [ ( α )] n n p

7 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 59 Invrs trigonomtric functions Th arcsin, arccosin an arctangnt ar fin as th invrs functions of th sin, cosin an tangnt. Th ar multivalu functions an th principal valus ar takn in th following intrvals: arcsin π π sin an arccos cos an π arctg π π tg an < < arccot cot an < < π From th finitions th parit of th invrs functions follow immiatl: arcsin arcsin( ) arccos arccos( ) arctg arctg ( ) Through th funamntal intitis for th circular functions on obtains th following intitis 5 for th invrs functions: arcsin arccos [ arccos ] arctg [ arcsin ] arctg arctg arcsin arccos whr th brackts inicat that th intit onl hols for positiv valus. From th complmntar angls intitis w hav: π π arcsin arccos arctg arc cot 5 Th onl invrs circular function prfin in th languag asic is th arctangnt TN(X), so ths intitis allows us to program th arcsin an arccosin.

8 6 RMON GONZLEZ LVET Erciss 6. Prov th law of sins, cosins an tangnts: if a, b an c ar th sis of a triangl rspctivl opposit to th angls α, β an γ, thn: a b c (law of sins) sin α sin β sin γ c a b a b cos γ (law of cosins) a a b b α β tg α β tg (law of tangnts) Hint: us th innr an outr proucts of sis. 6. Prov th following trigonomtric intitis: cos sin α β β γ γ α ( α β ) cos( β γ ) cos( γ a ) 4 cos cos cos ( α β ) sin( β γ ) sin( γ α ) α β β γ γ α 4sin sin sin 6. Eprss th trigonomtric functions of 4α as a polnomial of th trigonomtric functions of α. 6.4 Lt P b a point on a circl arc whos trms ar th points an. Prov that th sum of th chors P an P is maimal whn P is th mipoint of th arc. 6.5 Prov th Mollwi s formulas for a triangl: a c b α β cos γ sin a c b α β sin γ cos 6.6 Duc also th projction formulas: a b cosγ c cos β b c cosα a cosγ c a cos β b cosα 6.7 Prov th half angl intitis: α sin ± cosα α cosα cos ± α tg ± cosα cosα sinα cosα sinα cosα

9 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 6 7. SIMILRITIES ND SINGLE RTIO Two gomtric figurs ar similar if th hav th sam shap. If th orintation of both figurs is th sam, th ar sai to b irctl similar. For ampl, th ials of clocks ar irctl similar. On th othr han, two figurs can hav th sam shap but iffrnt orintation. Thn th ar sai to b oppositl similar. For ampl our hans ar oppositl similar. Howvr ths intuitiv concpts ar insufficint an th similarit must b fin with mor prcision. Dirct similarit (similitu) If two vctors u an v on th plan form th sam angl as th angl btwn th vctors w an t, an th hav proportional lngths (figur 7.), thn th ar gomtricall proportional: Figur 7. α ( u, v) α( w, t) u v w t u v w t that is, th gomtric quotint of u an v is qual to th quotint of w an t, which is a compl numbr. This finition of a gomtric quotint is also vali for vctors in th spac provi that th four vctors li on th sam plan. In this cas, th gomtric quotint is a quatrnion, as Hamilton show. Th gomtric proportionalit for vctors allows to fin th similarit of triangls. Two triangls an ''' ar sai to b irctl similar an thir vrtics an sis not with th sam lttrs ar homologous if: '' '' that is, if th gomtric quotint of two sis of th first triangl is qual to th quotint of th homologous sis of th scon triangl. rranging th vctors of this quation on obtains th qualit of th quotints of th homologous sis: '' '' On can prov asil that th thir quotint of homologous sis also coincis with th othr quotints: '' '' '' r Th similarit ratio r is fin as th quotint of vr pair of homologous sis, which is a compl constant. Th moulus of th similarit ratio is th siz ratio an th argumnt is th angl of rotation of th triangl ''' with rspct to th triangl.

10 6 RMON GONZLEZ LVET r ' ' p [ α(, ' ' ) ] Figur 7. Th finition of similarit is gnralis to an pair of polgons in th following wa. Lt th polgons...z an '''...Z' b. Th ar sai to b irctl similar with similarit ratio r an th sis not with th sam lttrs to b homologous if: r '' '' D 'D'... YZ Y'Z' Z Z'' On of ths qualitis pns on th othrs an w o not n to know whthr it is fulfill. Hr also, th moulus of r is th siz ratio of both polgons an th argumnt is th angl of rotation. Th fact that th homologous trior an intrior angls ar qual for irctl similar polgons (figur 7.) is trivial bcaus: '' '' angl ''' angl '' 'D' D angl ''D' angl D tc. Th irct similarit is an quivalnc rlation sinc it has th rfliv, smmtric an transitiv proprtis. This mans that thr ar classs of quivalnc with irctl similar figurs. similitu with r is call a isplacmnt, sinc both polgons hav th sam siz an orintation. Opposit similarit Two triangls an ''' ar oppositl similar an th sis not with th sam lttrs ar homologous if: ( '' '' )* '' '' Th formr qualit cannot b arrang in a quotint of a pair of homologous sis as w hav ma bfor. caus of this, th similarit ratio cannot b fin for th Figur 7. opposit similarit but onl th siz ratio, which is th quotint of th lngths of an two homologous sis. n opposit similarit is alwas th composition of a rflction in an lin an a irct similarit.

11 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 6 v '' '' v v '' v '' ( v '' v ) ( v '' v ) r whr r is th ratio of a irct similarit whos argumnt is not fin but pns on th irction vctor v of th rflction ais. Notwithstaning, this prssion allows to fin th opposit similarit of two polgons. So two polgons...z an '''...Z' ar oppositl similar an th sis not with th sam lttrs ar homologous if for an vctor v th following qualitis ar fulfill: ( v '' v ) ( v '' v )... Z ( v Z'' v ) that is, if aftr a rflction on polgon is irctl similar to th othr. Th opposit similarit is not rfliv nor transitiv: if a figur is oppositl similar to anothr, an this is oppositl similar to a thir figur, thn th first an thir figurs ar irctl similar. Thn thr ar not classs of oppositl similar figurs. n opposit similarit with r is call a rvrsal, sinc both polgons hav th sam siz both opposit orintations. Th thorm of Mnlaus For vr triangl (figur 7.4), thr points D, E an F ling rspctivl on th sis, an or thir prolongations ar align if an onl if: Figur 7.4 F F D D E E Proof Lt us suppos that D, E an F ar align on a crossing straight lin. Lt us not b p, q an r th vctors with origin at th vrtics, an an going prpnicularl to th crossing lin. Thn vr pair of right angl triangls having th hpotnuss on a common si of th triangl ar similar so that w hav: F F q p D D r q E E p r Multipling th thr qualitis on obtains: E E D D F F Taking th compl conjugat prssion an changing th sign of F, D an E, th thorm is prov in this irction: F F D D E E

12 64 RMON GONZLEZ LVET Proof ing F a point on th lin, D a point on th lin an E a point on th lin w hav: F a ( a) D b ( b) E c ( c) with a, b an c ral. Thn: F ( a) F a D ( b) D b E ( c) E c That th prouct of ths sgmnts is qual to implis that: ( a)( b)( c) a b c ( a)( b) c a b Th substitution of c in th prssion for E givs: E ( a)( b) a b a b a b a b a b a b [( b) b ] [ a ( a) ] a b D F a b a b That is, th point E blongs to th lin FD, in othr wors, th point D, E an F ar align, which is th prov: E D ( ) F with a a b Th thorm of va Givn a gnric triangl, w raw thr sgmnts D, E an F from ach vrt to a point of th opposit si (figur 7.5). Th thr sgmnts mt in a uniqu point if an onl if: Figur 7 5 D D E E F F In orr to prov th thorm, w

13 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 65 nlarg th sgmnts E an F to touch th lin which is paralll to th si an passs through. Lt us not th intrsction points b M an N rspctivl. Thn th triangl DO is irctl similar to th triangl MO, an th triangl DO is also irctl similar to th triangl NO. Thrfor: D DO M O DO D O N Th multiplication of both qualitis givs: D D M N nalogousl, th triangls ME an E ar similar as soon as NF an F. Hnc: E E M F F N Th prouct of th thr qualitis ils: D D E E F F M N M N Th sufficinc of this conition is prov in th following wa: lt O b th point of intrsction of E an F, an D' th point of intrsction of th lin O with th si. Lt us suppos that th formr qualit is fulfill. Thn: D' D' D D Sinc both D an D' li on th lin, it follows that D D'. Homotht an singl ratio homotht with cntr O an ratio k is th gomtric transformation which ilats th istanc from O to an point in a factor k: O' O k O is th uniqu invariant point of th homotht. If k is ral numbr, th homotht is sai to b simpl (figur 7.6), othrwis is call composit. If k is a compl numbr, it ma alwas b factoris in a prouct of th moulus an a unitar compl numbr: k k z with z Figur 7.6 Th moulus of k is th ratio of a simpl homotht with cntr O, an z inicats an

14 66 RMON GONZLEZ LVET aitional rotation with th sam cntr. Thn, a composit homotht is quivalnt to a simpl homotht follow b a rotation. Dirct similaritis an homothtis ar iffrnt nams for th sam transformations (rcis 7.5). lso homothtis with k an isplacmnts ar quivalnt. Th singl ratio of thr points,, an is fin as: ( ) Th singl ratio is a ral numbr whn th thr points ar align, an a compl numbr in th othr cas. Unr a homotht, th singl ratio of an thr points rmains invariant. Lt us prov this. caus an vctor is ilat an rotat b a factor k, w hav: '' O' O' O k O k k '' k '' k ( ' ' ' ) k k ( ) If th singl ratio is invariant for a gomtric transformation, thn it transforms triangls into irctl similar triangls an hnc also polgons into similar polgons: ( ) ( ' ' ' ) '' '' '' '' Thrfor, th homotht alwas transforms triangls into irctl similar triangls as shown in th figur 7.6. It follows immiatl that th homotht prsrvs th angls btwn lins. It is th simplst cas of conformal transformations, th gomtric transformations which prsrv th angls btwn lins. Sinc a lin ma also b transform into a curv, an for th gnral cas a curv into a curv, th conformal transformations ar thos which prsrv th angl btwn an pair of curvs, that is, th angl btwn th tangnt lins to both curvs at th intrsction point. transformation is irctl or oppositl Figur 7.7 conformal whthr it prsrvs or changs th sns of th angl btwn two curvs (figur 7.7). This conition is quivalnt to th consrvation of th singl ratio of an thr points at th limit of accumulation: lim ( ) lim ( ' ' ' ) irctl conformal, ', ' ' lim ( ) lim ( ' ' ' )* oppositl conformal, ', ' '

15 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 67 Erciss 7. Lt T b th triangl with vrtics (, ), (, ), (, ) an T' that with vrtics (,), (5, ) an (4, ). Fin which vrtics ar homologous an calculat th similarit ratio. Which is th siz ratio? Which is th angl of rotation of th triangl T' with rspct to T? 7. Th altitu prpnicular to th hpotnus ivis a right triangl in two smallr right triangls. Show that th ar similar an uc th Pthagoran thorm. 7. Evr triangl with not vanishing ara has a circumscrib circl. Th lin tangnt to this circl at th point cuts th lin at th point M. Prov that: M M 7.4 Lt b an quilatral triangl inscrib insi a circl. If P is an point on th arc, show that P P P. 7.5 Givn two irctl similar triangls an ''', show that th cntr O of th homotht that transform on triangl into anothr is qual to: O ' ( '' ) 7.6 Draw an lin passing through a fi point P which cuts a givn circl. Lt Q an Q' b th intrsction points of th lin an th circl. Show that th prouct PQ PQ' is constant for an lin blonging to th pncil of lins of P. 7.7 Lt a triangl hav sis a, b an c. Th bisctor of th angl form b th sis a an b ivis th si c in two parts m an n. If m is ajacnt to a an n to b rspctivl, prov that th following proportion is fulfill: m a n b

16 68 RMON GONZLEZ LVET 8. PROPERTIES OF TRINGLES ra of a triangl Sinc th ara of a paralllogram is obtain as th outr prouct of two conscutiv sis (takn as vctors, of cours), th ara of a triangl is th half of th outr prouct of an two of its thr sis: a PQR PQ PR QR QP RP RQ If th vrtics P, Q an R ar countrclockwis orint, th ara is a positiv imaginar numbr. Othrwis, th ara is a ngativ imaginar numbr. Not that th funamntal concpt in gomtr is th orint ara. Th moulus of th ara Figur 8. ma b usful in th currnt lif but is insufficint for gomtr. From now on I shall onl rgar orint aras. Writing th sgmnts of th formr quation as iffrncs of points w arriv to: a PQR (P Q Q R R P ) which is a smmtric prssion unr cclic prmutation of th vrtics. Th position vctor P gos from an arbitrar origin of coorinats to th point P. Thn, P Q is th oubl of th ara of th triangl OPQ. nalogousl Q R is th oubl of th ara of th triangl OQR an R P is th oubl of th ara of th triangl ORP. Thrfor th formr prssion is qual to (figur 8.): a PQR a OPQ a OQR a ORP For th arrangmnt of points shown in th figur 8. th ara of OPQ is positiv, an th aras of OQR an ORP ar ngativ, that is, th aras of ORQ an OPR ar positiv. Thrfor, on woul intuitivl writ, taking all th aras positiv, that: a PQR a OPQ a ORQ a OPR Whn on consirs orint aras, th qualitis ar wholl gnral an inpnnt of th arrangmnt of th points. Th intgral of a function is also th orint ara of th rgion nclos b th curv an th X- ais. bout this topic, s. M. Lopshitz, álculo las áras figuras orintaas, Rubiños-86 (994).

17 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 69 Mians an cntroi Th mians ar th sgmnts going from ach vrt to th mipoint of th opposit si. Lt us prov that th thr mians mt in a uniqu point G call th cntroi (figur 8.). Sinc G is a point on th mian passing through P an (Q R)/: ( ) R Q k P k G k ral G lis also on th mian passing through Q an (P R) / : ( ) R P m mq G m ral Equating both prssions w fin: ( ) ( ) R P m mq R Q k P k m k R m k Q m k P linar combination of inpnnt points can vanish onl if vr cofficints ar null, a conition which las to th following sstm of quations: m k m k m k Th solution of this sstm of quations is k / an m /, inicating that th intrsction of both mians ar locat at / istanc from th mipoints. Th substitution into th prssion of G givs: R Q P G This prssion for th cntroi is smmtric unr prmutation of th vrtics. Thrfor th thr mians mt at th sam point G, th cntroi. Th mians ar a spcial cas of cvian lins (lins passing through a vrt an not coinciing with th sis) an this statmnt is also prov b mans of va s thorm. Figur 8.

18 7 RMON GONZLEZ LVET Prpnicular bisctors an circumcntr Th thr prpnicular bisctors of th sis of a triangl mt in a uniqu point call th circumcntr, th cntr of th circumscrib circl. Evr point on th prpnicular bisctor of PQ is quiistant from P an Q. nalogousl vr point on th prpnicular bisctor of PR is quiistant from P an R. Th intrsction O of both prpnicular bisctors is simultanousl quiistant from P, Q an R. Thrfor O also blongs to th prpnicular bisctor of QR an th thr bisctors mt at a uniqu point. Sinc O is quall istant from th thr vrtics, it is th cntr of th circumscrib circl. Lt us us this conition in orr to calculat th quation of th circumcntr: Figur 8. OP OQ OR whr is th raius of th circumscrib circl. Using th position vctors of ach point w hav: (P O) (Q O) (R O) Th first qualit ils: P P O O Q Q O O simplifing an arranging th trms containing O at th lft han si, w hav: ( Q P ) O Q P PQ O Q P From th scon qualit w fin an analogous rsult: QR O R Q Now w introuc th gomtric prouct insta of th innr prouct in ths quations: PQ O O PQ Q P QR O O QR R Q subtraction of th scon quation multipli on th right b PQ minus th first quation multipli on th lft b QR, w obtain: PQ QR O O PQ QR PQ R PQ Q Q QR P QR using th prmutativ proprt on th lft han si an simplifing th right han si, w hav:

19 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 7 PQ QR O QR PQ O P QR Q RP R PQ ( PQ QR ) O P QR Q RP R PQ Finall, th multiplication b th invrs of th outr prouct on th lft givs: O ( PQ QR ) ( P QR Q RP R PQ ) ( P QR Q RP R PQ ) ( PQ QR ) a formula abl to calculat th coorinats of th circumcntr. For ampl, lt us calculat th cntr of th circl passing through th points: P (, ) Q (, ) R ( 4, ) P 8 Q R QR R Q RP P R 4 PQ Q P PQ QR 4 O ( 8 ( ) ( 4 ) ( ) ) 4 (, ) In orr to uc th raius of th circl, w tak th vctor OP: OP P O P ( P QR Q RP R PQ ) ( PQ QR ) an tract th invrs of th ara as a common factor: OP ( P PQ QR P QR Q RP R PQ ) ( PQ QR ) [ P ( P Q Q R R P ) P QR Q RP R PQ ] ( PQ QR ) [ P ( P Q Q P Q R R Q R P P R ) P ( R Q ) Q ( P R ) R ( Q P ) ] ( PQ QR ) Th simplification givs: OP ( P Q R P R Q P R P P Q P Q P Q R R Q R P ) ( PQ QR ) (Q P) (R Q) (P R) ( PQ QR ) PQ QR RP ( PQ QR )

20 7 RMON GONZLEZ LVET nalogousl: OQ QR RP PQ ( PQ QR ) OR RP PQ QR ( PQ QR ) Th raius of th circumscrib circl is th lngth of an of ths vctors: OP PQ QR RP PQ QR PQ QR RP sinqrp sin RPQ sin PQR whr w fin th law of sins. ngl bisctors an incntr Th thr bisctor lins of th angls of a triangl mt in a uniqu point call th incntr. Evr point on th bisctor of th angl with vrt P is quiistant from th sis PQ an PR (figur 8.4). lso vr point on th angl bisctor of Q is quiistant from th sis QR an QP. Hnc its intrsction I is simultanousl quiistant from th thr sis, that is, I is uniqu an is th cntr of th circl inscrib into th triangl. In orr to calculat th quation of th angl bisctor passing through P, w tak th sum of th unitar vctors of both ajacnt sis: Figur 8.4 PQ u PQ PR PR QP v QP QR QR Th incntr I is th intrsction of th angl bisctor passing through P, whos irction vctor is u, an that passing through Q, with irction vctor v: I P k u Q m v k, m ral rranging trms w fin PQ as a linar combination of u an v: k u m v Q P PQ Th cofficint k is: PQ v k u v PQ QR RP PQ RP PQ QR QR RP PQ RP PQ QR Sinc all outr proucts ar qual bcaus th ar th oubl of triangl ara, this prssion is simplifi:

21 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 7 k PQ RP RP PQ QR Thn, th cntr of th circumscrib circl is: I P k u P PQ PQ RP QR RP PQ PQ PR PR taking common nominator an simplifing, w arriv at: P I QR Q RP R PQ QR RP PQ For ampl, lt us calculat th cntr of th circl inscrib insi th triangl with vrtics: P (, ) Q (, ) R (4, ) PQ QR 5 RP 4 5 I (,) 4 (,) ( 4,) (,) 5 4 (,) In orr to fin th raius, firstl w must obtain th sgmnt IP: QP RP RP IP QR RP PQ PQ Th raius of th inscrib circl is th istanc from I to th si PQ: ( I, PQ) IP PQ PQ PQ RP PQ QR RP whnc th ratio of raius follows: raius of circumscrib circl raius of inscrib circl PQ PQ QR QR RP RP ltitus an orthocntr Th altitu of a si is th sgmnt prpnicular to this si (also call bas) which passs through th opposit vrt. Th thr altitus of a triangl intrsct on a uniqu point call th orthocntr. Lt us prov this statmnt calculating th

22 74 RMON GONZLEZ LVET intrsction H of two altitus. Sinc H blongs to th altitu passing through th vrt P an prpnicular to th bas QR (figur 8.5), its quation is: H P z QR z imaginar bcaus th prouct b an imaginar numbr turns th vctor QR ovr π/, that is, z PQ has th irction of th altitu. H also blongs to th altitu passing through Q an prpnicular to th bas RP. Thn its quation is: Figur 8.5 H Q t RP t imaginar quating both prssions: P z QR Q t RP w arriv at a vctor writtn as a linar combination of two vctors but with imaginar cofficints: z QR t RP PQ In this quation w must rsolv PQ into componnts with irctions prpnicular to th vctors QR an RP. Th algbraic rsolution follows th sam wa as for th cas of ral linar combination. Lt us multipl on th right b RP: an on th lft: z QR RP t RP PQ RP RP z QR RP t RP RP PQ Th imaginar numbrs anticommut with vctors: z RP QR t RP RP PQ aing both qualitis w arriv at: z ( QR RP RP QR ) PQ RP RP PQ PQ RP z QR RP RP PQ PQ RP RP QR QR RP nalogousl on fins t: PQ QR t QR RP

23 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 75 oth prssions iffr from th cofficints of ral linar combination in th fact that th numrator is an innr prouct. Th substitution in an of th first quations givs: H P z QR P ( PQ RP RP PQ ) ( QR RP RP QR ) QR Th vctor QR anticommuts with th outr prouct, an imaginar numbr: H P ( PQ RP RP PQ ) QR ( QR RP RP QR ) Etracting th ara as common factor, w obtain: H [ P ( QR RP RP QR ) ( PQ RP RP PQ ) QR ] ( QR RP RP QR ) using th fact that PQ Q P, tc., w arriv at: H ( P QR RP Q RP QR P PQ QR R PQ QR ) ( QR RP RP QR ) ( P QR Q RP R PQ P QR P Q RP Q R PQ R ) ( QR RP RP QR ) ( P P QR Q Q RP R R PQ ) ( QR RP ) This formula is invariant unr cclic prmutation of th vrtics. Thrfor all th altitus intrsct on a uniqu point, th orthocntr. Th quation of th orthocntr rsmbls that of th circumcntr. In orr to s th rlationship btwn both, lt us raw a lin passing through P an paralll to th opposit si QR, anothr on passing through Q an paralll to RP, an a thir lin passing through R an paralll to PQ (figur 8.6). Lt b th intrsction of th lin passing through Q an that passing through R, b th intrsction of th lins passing through R an P rspctivl, an b th intrsction of th lins passing through P an Q rspctivl. Th triangl is irctl similar to th triangl PQR with ratio : PQ QR RP an P, Q an R ar th mipoints of th sis of th triangl : Figur 8.6 P Q R Thrfor th altitus of th triangl PQR ar th prpnicular bisctors of th sis of th triangl, an th orthocntr of th triangl PQR is th circumcntr of

24 76 RMON GONZLEZ LVET. In orr to prov with algbra this obvious gomtric fact, w must onl uc from th formr rlations th following qualitis an substitut thm into th orthocntr quation: P P QR 8 Q Q RP 8 R R PQ 8 aing th thr trms, th cubic powrs vanish. On th othr han, th ara of th triangl PQR is four tims smallr than th ara of th triangl : QR RP 4 H ( ) ( ) ( ) ( ) This is just th quation of th circumcntr of th triangl. Eulr's lin Th cntroi G, th circumcntr O an th orthocntr H of a triangl ar alwas align on th Eulr's lin. To prov this fact, obsrv that th circumcntr an orthocntr quations hav th triangl ara as a "nominator", whil th cntroi quation os not, but w can introuc it: P Q R G ( P Q R ) QR RP ( QR RP ) Introucing th qualit: QR RP P Q Q R R P P Q Q P Q R R Q R P P R th cntroi bcoms: G ( P Q R ) ( P Q Q P Q R R Q R P P R ) ( 6 QR RP ) ( P P Q P Q P P R P P P R Q P Q Q Q P Q Q R Q R Q R Q R R R Q R R P R P R ) ( 6 QR RP ) ( P QR P QR P Q RP Q RP Q R PQ R PQ R ) ( 6 QR RP )

25 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 77 from whr th rlation btwn th thr points G, H an O follows immiatl: H O G Hnc th cntroi is locat btwn th orthocntr an th circumcntr, an its istanc from th orthocntr is oubl of its istanc from th circumcntr. Th Frmat's thorm Th gomtric algbra allows to prov through a vr as an intuitiv wa th Frmat's thorm. Ovr vr si of a triangl w raw an quilatral triangl (figur 8.7). Lt T, U an S b th vrtics of th quilatral triangls rspctivl opposit to, an. Thn th sgmnts T, U an S hav th sam lngth, form angls of π/ an intrsct on a uniqu point F, call th Frmat's point. Morovr, th aition of th thr istancs from an point P to ach vrt is minimal whn P is th Frmat's point, provi that an of th intrior angls of th triangl b highr than π/. Firstl, w must monstrat that U is obtain from T b mans of a rotation of π/, which will b rprsnt b th compl numbr t: π t cos π sin T t ( T ) t t T t construction, th vctor turn ovr π/ is th vctor U, an T turn ovr π/ is, so: T t U U Figur 8.7 nalogousl, on fins U t an T S t. That is, th vctors S, U an T hav th sam lngth an ach on is obtain from ach othr b succssiv rotations of π/. Lt us s that th sum of th istancs from P to th thr vrtics, an is minimal whn P is th Frmat s point. W must prov firstl that th vctorial sum of P turn 4π/, P turn π/ an P is constant inpnntl of th point P. (figur 8.8). That is, for an two points P an P it is alwas tru that:

26 78 RMON GONZLEZ LVET P t P t P P' t P' t P' fact which is prov b arranging all th trms in on si of th quation: PP' ( t t ) This prouct is alwas zro sinc t t. Hnc, thr is a uniqu point Q such that: P t P t P Q For an point P, th thr sgmnts form a brokn lin as that shown in figur 8.8. Thrfor, b th triangular inqualit w hav: Figur 8.8 P P P Q Whn P is th Frmat s point F, ths sgmnts form a straight lin. Thn, th aition of th istancs from F to th thr vrtics is minimal provi that no angl of th triangl b highr than π/: F F F Q P P P In th othr cas, somon of th vctors F t, F t or F has a sns opposit to th othrs, so that its lngth is subtract from th othrs an thir sum is not minimal. Erciss 8. Th Napolon s thorm. Ovr ach si of a gnric triangl raw an quilatral triangl. Prov that th cntrs of th thr quilatral triangls also form an quilatral triangl. 8. Prov th Libniz s thorm. Lt P b an point on th plan an G th cntroi of a triangl. Thn th following qualit hols: PG P P P ( ) 8. Lt, an b thr givn points on th plan. Thn an point G on th plan can b prss as a linar combination of ths thr points (G is also consir as th cntr of masss locat at, an with wights a, b an c 4 ). 4 S ugust Frinan Möbius, Dr arcntrisch alcul, Lipzig, 87, p. 7 (facsimil ition of Gorg Olms Vrlag, Hilshim,976).

27 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 79 G a b c with a b c Prov that: a) a, b, c ar th fractions of th ara of th triangl occupi b th triangls G, G an G rspctivl. b) th gomtric locus of th points P on th plan such that: a P b P c P k is a circl with cntr G (pollonius lost thorm). 8.4 Ovr vr si of a conv quarilatral D, w raw th quilatral triangls P, Q, DR an DS. Prov that: a) If D thn PR is prpnicular to QS. b) If PR QS thn is prpnicular to D. 8.5 Show that th lngth of a mian of a triangl passing through can b calculat from th sis as: m In a triangl w raw th bisctors of th angls an. Through th vrt w raw th lins paralll to ach angl bisctor. Lt us not th intrsctions of ach paralll lin with th othr bisctor b D an E. Prov that if th lin DE is paralll to th si thn th triangl is isoscls. 8.7 Th bisction of a triangl (propos b ristóbal Sánchz Rubio). Lt P an Q b two points on iffrnt sis of a triangl such that th sgmnt PQ ivis th triangl in two parts with th sam ara. alculat th sgmnt PQ in th following cass: a) PQ is prpnicular to a givn irction. b) PQ has minimum lngth. c) PQ passs through a givn point insi th triangl.

28 8 RMON GONZLEZ LVET 9. IRLES lgbraic an artsian quations circl with raius r an cntr F is th locus of th points locat at a istanc r from F, so its quation is: (F, P) FP r FP r which can b writtn as: (P F) r P P F F r introucing th coorinats of P (, ) an F (a, b) on obtains th analtic quation of th circl: a b a b r For ampl, th circl whos quation is: has th cntr F (, ) an raius. Thr alwas ists a uniqu circl passing through an thr non align points. In orr to obtain this circl, on ma substitut th coorinats of th points on th artsian quation of th circl, arriving at a sstm of thr linar quations of thr unknown quantitis a, b an c: a a a b b b c c c This is th classic wa to fin th quation of th circl, an from hr th cntr of th circl. Howvr, a mor irct wa is th calculation of th circumcntr of th triangl whos vrtics ar th thr givn points (s th prvious chaptr). Thn th raius is th istanc from th cntr to an vrt. Intrsctions of a lin with a circl Th coorinats of th intrsction of a lin with a circl ar usuall calculat b substitution of th lin quation into th circl quation, which las to a scon gr quation. Lt us now follow an algbraic wa. Lt F b th cntr of th circl with raius r, an R an R th intrsctions of th lin passing through two givn points P an Q (figur 9.):

29 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 8 FR r R P k PQ Figur 9. From th first qualit w obtain: (R F ) R R F F r Th substitution of R givn b th scon qualit ils: ( P k PQ ) ( P k PQ ) F F r P k P PQ k PQ P F k PQ F F r arranging th powrs of k on obtains: k PQ k ( P PQ F PQ ) P F P F r Taking into account that PF F P on fins: k PQ k PF PQ PF r This quation has solution whnvr th iscriminant b positiv: 4 ( PF PQ ) 4 PQ ( PF r ) Introucing th intit (PF PQ ) (PF PQ ) PQ PF th iscriminant bcoms: 4 (PF PQ ) 4 PQ r Th solution of th scon gr quation for k is: k ( PF ) PF PQ ± r PQ PQ Whn r PF PQ / PQ, th lin is tangnt to th circl. In this cas th hight of th paralllogram form b th vctors PF an PQ is qual to th raius of th circl. If u not th unitar irction vctor of th lin PQ (figur 9.): u PQ PQ thn both intrsction points ar writtn as:

30 8 RMON GONZLEZ LVET R P kpq P u PF R' P k' PQ P u PF ( PF ) u r u ( PF ) u r u Powr of a point with rspct to a circl oth intrsction points R an R' hav th following proprt: th prouct of th vctors going from a givn point P to th intrsctions R an R' of an lin passing through P with a givn circl is constant: PR PR' ( PF u ) r (PF u ) Figur 9. PF r PT whr T is th contact point of th tangnt lin passing through P (figur 9.). Th prouct PR PR, which is inpnnt of th irction u of th lin an onl pns on th point P, is call th powr of th point P with rspct to th givn circl. Th powr of a point ma b calculat through th substitution of its coorinats into th circl quation. If its cntr is F (a, b) an P (, ), thn: FP r ( a ) ( b ) r a b a b r Th powr is qual to zro whn P blongs to th circl, positiv whn P lis outsi th circl, an ngativ whn P lis insi th circl. Polar quation W wish to scrib th istanc from a point P to th points R an R' on th circl as a function of th angl α btwn th lin PR an th iamtr (figur 9.). Th vctors PR an PR' ar: PR u PF PR' u PF u r PF u u r PF u Figur 9. W arriv at th polar quation b taking th moulus of PR an PR', which is th istanc to th circl:

31 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 8 PR PF cosα r PF sin α Figur 9.4 PR' PF cosα r PF sin α If P is a point on th circl thn PF r (figur 9.4) an th quation is simplifi: PR r cos α Invrsion with rspct to a circl Th point P' is th invrs point of P with rspct to a circl of cntr F an raius r if th vctor FP is th invrs vctor of FP with raius k: FP' k FP FP FP' k Obviousl, th prouct of th vctors FP an FP' is onl ral an Figur 9.5 positiv whn th points P an P ar align with th cntr F an ar locat at th sam si of F. Th circl of cntr F an raius k is call th invrsion circl bcaus its points rmain invariant unr this invrsion. Th invrsion transforms points locat insi th invrsion circl into outsi points an rciprocall. To obtain gomtricall th invrs of an insi point P, raw firstl (figur 9.5) th iamtr passing through P; aftr this raw th prpnicular to this iamtr from P which will cut th circl in th point T; finall raw th tangnt with contact point T. Th intrsction P' of this tangnt with th prolongation of th iamtr is th invrs point of P. Not that th right triangls FPT an FTP' ar oppositl similar: FP FT FP' FT FP' FP FT k To obtain th invrs of an outsi point, mak th sam construction but in th opposit sns. Draw th tangnt to th circl passing through th point (P' in th figur 9.5) an thn raw th prpnicular to th iamtr FP' passing through th contact point T of th tangnt. Th intrsction P of th prpnicular with th iamtr is th sarch invrs point of P'. Evr lin not containing th cntr of invrsion is transform into a circl passing through th cntr (figur 9.6) an rciprocall. To prov this statmnt lt us tak

32 84 RMON GONZLEZ LVET th polar quation of a straight lin: FP cosα Figur 9.6 Thn th invrs of P with cntr F has th quation: FP' k FP k cosα which is th polar quation of a circl passing through F an with iamtr k /. α is th angl btwn FP an FD, th irction prpnicular to th lin; thrfor, th cntr O of th circl is locat on FD. Lt us prov that an invrsion transforms a circl not passing through its cntr into anothr circl also not passing through its cntr. onsir a circl of cntr F an raius r which will b transform unr an invrsion of cntr P an raius k. Thn an point R on th circl is mapp into anothr point S accoring to: PS k PR PF cosα ± k r PF sin α whr th polar quation of th circl is us. Th multiplication b th conjugat of th nominator givs: PS k ( PF cosα m r PF sin α ) PF r which is th polar quation of a circl with cntr G an raius s: PS PG cosα m s PG sin α k PF k PF whr PG PF r PT k r k r s PF r PT PF r PT is th powr of th point P with rspct to th circl of cntr F. Thn w s that th istanc from th cntr of invrsion to th cntr of th circl an th raius of this circl changs in th ratio of th squar of th raius of invrsion ivi b th powr of th cntr of invrsion with rspct to th givn circl. What is th powr of th cntr of invrsion P with rspct to th nw circl?: ( PF r ) 4 PU PG s 4 k PT 4 k PT That is, th prouct of th powrs of th cntr of invrsion with rspct to an circl an

33 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 85 its transform is constant an qual to th fourth powr of th raius of invrsion: PT PU k 4 In th quation of th transform circl th smbol m appars insta of ±. It mans that th sns of th arc from R to R' is opposit to th sns of th arc from S to S'. Th nin-point circl Th Eulr s thorm stats that th mipoints of th sis of an triangl, th ft of th altitus an th mipoints from ach vrt to th orthocntr li on a circl call th nin-point circl. Lt an triangl PQR whos orthocntr is H b. Now w valuat th prssion: ( ) ( ) ( ) ( ) 4 R Q H P R Q H P R Q H P which vanishs bcaus th sgmnt PH ling on th altitu passing through P is prpnicular to th bas QR. nalogousl: ( ) ( ) ( ) ( ) 4 R P H Q R Q H P R Q H P an ( ) ( ) ( ) ( ) 4 P Q H R R Q H P R Q H P Hnc: ( ) ( ) ( ) H R Q P H R Q P H R Q P Sinc th sign ma b opposit insi th squar w hav: ( ) ( ) ( ) H R Q P H R Q P H R Q P ( ) ( ) ( ) H R Q P H R Q P H R Q P Now w introuc th point N (P Q R H )/4 to obtain: H P N R P N Q P N R Q N H Q N H R N That is, N is th cntr of a circl (figur 9.7) passing through th thr mipoints of th sis (P Q)/, (Q R)/ an (R P)/ an th thr mipoints of th vrtics an th orthocntr (P H)/, (Q H)/ an (R H)/. Sinc w can writ:

34 86 RMON GONZLEZ LVET P H Q R N it follows that (PH)/ an (QR)/ ar opposit trms of th sam iamtr. Thn th vctors J(PH)/ an J(QR)/, which ar orthogonal: Figur 9.7 P H Q R J J ar th sis of a right angl which intrcpts a half circumfrnc of th nin point circl. Thrfor, this angl is inscrib an its vrt J also lis on th nin-point circl. lso, w ma go through th algbraic wa to arriv at th sam conclusion. Dvloping th formr innr prouct: J Q R J P H J P H Q R Introucing th cntr N of th nin-point circl, w hav: J Q R J N N Q R an aftr aing an subtracting N w arriv at: Q R ( J N ) N which shows that th lngth of JN is th raius of th nin-point circl. From th formulas for th cntroi an orthocntr on obtains th cntr N of th nin-point circl: G H N QR 4 ( P QR P Q RP Q R PQ R)( 4PQ ) that is, th point N also lis on th Eulr s lin. Th rlativ istancs btwn th circumcntr O, th cntroi G, th nin-point circl cntr N an th orthocntr H ar shown in th figur 9.8. Figur 9.8

35 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 87 clic an circumscrib quarilatrals First, lt us s a statmnt vali for vr quarilatral: th ara of a quarilatral D is th outr prouct of th iagonals an D. Th prov is writtn in on lin. Sinc w can ivi th quarilatral in two triangls w hav: ra D Figur 9.9 ( D ) D D sin(, D) Th quarilatrals inscrib in a circl ar call cclic quarilatrals. Th fulfil th Ptolm s thorm: th prouct of th lngths of both iagonals is qual to th aition of th proucts of th lngths of opposit sis (figur 9.9): D D D bautiful monstration maks us of th cross ratio an it is propos in th first rcis of th nt chaptr. For th cclic quarilatrals, th sum of opposit angls is qual to π, bcaus th ar inscrib in th circl an intrcpt opposit arcs. nothr intrsting proprt is th rahmagupta formula for th ara. If th lngths of th sis ar not b a, b, c an an th smiprimtr b s thn: ra ( s a) ( s b) ( s c) ( s ) a b c s With this notation, th proof is writtn mor brifl. Th law of cosins appli to ach triangl of quarilatral givs: a b a b cos λ c c cos ( π λ) Equating both quations taking into account that th cosins of supplmntar angls hav opposit sign, w hav: a b a b cosλ c c cosλ a b c ( a b c ) cos λ

36 88 RMON GONZLEZ LVET a b c ( a b c ) cos λ Th ara of th quarilatral is th sum of th aras of th triangls an D: ra a b c sin λ sin( π λ) oth sins ar qual; thn w hav: ra 4 ( a b c ) 4 sin ( a b c ) ( a b c ) 6 λ ( ) a b c ( a b c ) 4 4 ( ) a b c ( s a)( s b)( s c)( s ) nothr tp of quarilatrals ar thos whr a circl can b inscrib, also call circumscrib quarilatrals. For ths quarilatrals th bisctors of all th angls mt in th cntr I of th inscrib circl, bcaus it is quiistant from all th sis. Whn tracing th raii going to th tangnc points, th quarilatral is ivi into pairs of opposit triangls (figur 9.), an hnc on ucs that th sums of th lngths of opposit sis ar qual: Figur 9. D z t z t D If a quarilatral has inscrib an circumscrib circl, thn w ma substitut this conition (a c b ) in th rahmagupta formula to fin: ra a b c D D Sinc th sum of th angls of an quarilatral is π an ithr of th four small quarilatrals of a circumscrib quarilatral has two right angls (figur 9.), it follows that th cntral angls ar supplmntar of th vrtics: UT π TIU D WDV π VIW If th quarilatral is also cclic thn th angls an D ar inscrib an intrcpt opposit arcs of th outr circl, so that th sgmnts joining opposit tangnc points ar orthogonal:

37 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 89 D π TIU VIW π TV UW ngl btwn circls Lt us calculat th angl btwn a circl cntr at O with raius r an anothr on cntr at O' with raius r'. If O' O < r r' thn th intrsct. Th points of intrsction P must fulfil th quations for both circls: ( P O) ( P O' ) r r' P P O O P P O' O' r r' ing both quations w obtain: ( O O' ) O O' r P P r' It is trivial that th angl of intrsction is th supplmntar of th angl btwn both raius, its cosins bing qual: cosα ( O O' ) OP O'P P P O O' OP OP' r r' Th substitution of th first qualit into th scon givs: r cosα r' r r' ( O O' ) Th cosin of a right angl is zro; thrfor two circls ar orthogonal if an onl if: r ( O ) r' O' which is th Pthagoran thorm. Raical ais of two circls Th raical ais of two circls is th locus of th points which hav th sam powr with rspct to both circls: OP r O'P r' Lt X b th intrsction of th raical ais with th lin joining both cntrs of th circls. Thn:

38 9 RMON GONZLEZ LVET OX r O'X r' OX O'X r r' ( OX O'X ) ( OX O'X ) r r' O O' X OO' r r' Sinc OX an O'X ar proportional to OO', th innr an gomtric proucts ar quivalnt an w ma isolat X: O O' X ( r r' ) OO' Now lt us sarch which kin of gomtric figur is th raical ais. rranging as bfor th trms, w hav: O O' P OO' r r' Introucing th point X w arriv at: ( P X ) OO' That is, th raical ais is a lin passing through X an prpnicular to th lin joining both cntrs. If th circls intrsct, thn th raical ais is th straight lin passing through th intrsction points. Th raical cntr of thr circls is th intrsction of thir raical ais. Lt O, O' an O'' b th cntr of thr circls with raii r, r' an r''. Lt us not with X (as abov) th intrsctions of th raical ais of th first an scon circls with th lin OO', an with Y th intrsction of th raical ais of th scon an thir circls with th lin joining its cntrs O'O''. Th raical cntr P must fulfil th formr quation for both raical ais: ( P X ) ( P Y ) OO' O'O'' O' P OO' X OO' O'' P O'O'' Y O'O'' O r r' O' r' r'' Now w hav ncountr th cofficints of linar combination of an imaginar composition. This procss is plain in pag 74 for th calculation of th orthocntr. Howvr thr is a iffrnc: w wish not to rsolv th vctor into orthogonal componnts but from th known cofficints w wish to rconstruct th vctor P. Thn: P O'O'' P OO' P OO' O'O'' OO' O'O'' OO' O'O''

39 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 9 Th substitution of th known valus of th cofficints rsults in th following formula for th raical cntr which is invariant unr cclic prmutation of th circls: P ( OO' O'O'' ) ( O'' r'' ) OO' ( O r ) O'O'' ( O' r' ) O''O) Erciss 9. Lt,, b th vrtics of an triangl. Th point M movs accoring to th quation: M M M k Which gomtric locus os th point M scrib as a function of th valus of th ral paramtr k? 9. Prov that th gomtric locus of th points P such that th ratio of istancs from P to two istinct points an is a constant k is a circl. alculat th cntr an th raius of this circl. 9. Prov th Simson s thorm: th ft of th prpniculars from a point D upon th sis of a triangl ar align if an onl if D lis on th circumscrib circl. 9.4 Th rtschnir's thorm: lt a, b, c an b th succssiv sis of a quarilatral, m an n its iagonals an α an γ two opposit angls. Show that th following law of cosins for a quarilatral is fulfill: m n a c b a b c cos ( α γ ) 9.5 Draw thr circls passing through ach vrt of a triangl an th mipoints of th concurrnt sis. Thn join th cntr of ach circl an th mipoint of th opposit si. Show that th thr sgmnts obtain in this wa intrsct in a uniqu point. 9.6 Prov that th invrsion is an opposit conformal transformation, that is, it prsrvs th angls btwn curvs, but changs thir sign.

40 9 RMON GONZLEZ LVET. ROSS RTIOS ND RELTED TRNSFORMTIONS ompl cross ratio Th compl cross ratio of an four points,, an D on th plan is fin as th quotint of th following two singl ratios: ( D ) ( D ) ( D ) D D If th four points ar align, thn th cross ratio is a ral numbr, othrwis it is a compl numbr. Dnoting th angls D an D as α an β rspctivl, th cross ratio is writtn as: ( D ) D D D Figur. D D [( α β ) ] p If th four points ar locat in this orr on a circl (figur.), th cross ratio is a positiv ral numbr, bcaus th inscrib angls α an β intrcpt th sam arc D an hnc ar qual, so: ( D ) D D ut if on of th points or D is locat btwn an (figur.), thn α an β hav istinct arcs D with opposit orintation. Sinc α an β ar th half of ths arcs, it follows that α β π an th cross ratio is a ngativ ral numbr: Figur. ( D ) D D nalogousl, if th four points ar align an or D is locat btwn an, th cross ratio is ngativ. Othr finitions of th cross ratio iffr from this on givn hr onl in th orr in which th vctors ar takn. This qustion is quivalnt to stu how th valu of th cross ratio changs unr prmutations of th points. For four points, thr ar 4 possibl prmutations, but onl si iffrnt valus for th cross ratio. Th calculations ar somwhat laborious an I shall onl show on cas. straight lin is a circl with infinit raius.

41 TRETISE OF PLNE GEOMETRY THROUGH GEOMETRI LGER 9 Th rar ma prov an othr cas as an rcis. I also inicat whthr th prmutations ar vn or o, that is, whthr th ar obtain b an vn or an o numbr of changs of an two points from th initial finition ( D ): () ( D ) ( D ) ( D ) ( D ) r vn () ( D ) ( D ) ( D ) ( D ) r o () ( D ) ( D ) ( D ) ( D ) r o (4) ( D ) ( D ) ( D ) ( D ) r (5) r ( D ) ( D ) ( D ) ( D ) r (6) r ( D ) ( D ) ( D ) ( D ) r vn vn o Lt us prov, for ampl th qualitis (): () ( D ) D D ( ) D ( D ) D D D D D D In th last trm, unr th rflction in th irction, th compl numbr D D bcoms conjugat, that is, ths vctors ar chang: D D D D D D ( D ) D D D D D r r Lt us s that this valu is th sam for all th cross ratios of th cas (): ( D ) D D D D D D ( D ) whr in th thir stp th prmutativ proprt has bn appli. lso b using this proprt on obtains: ( D ) D D D D D D ( D ) ( D ) D D D D D D ( D )

42 94 RMON GONZLEZ LVET Harmonic charactristic an rangs From th cross ratio, on can fin a quantit inpnnt of th smbols of th points an onl pnnt on thir locations. I call this quantit th harmonic charactristic -bcaus of th obvious rasons that follow now- noting it as [ D]. Th simplst wa to calculat it (but not th uniqu) is through th altrnat aition of all th prmutations of th cross ratio, conns in th altrnat sum of th si iffrnt valus: ( D ) ( D ) ( D ) ( D ) ( D ) ( D ) r r ( r) ( r )( r )( r ) r ( r) r r r r r Whn two points ar prmut, this sum still changs th sign. Th squar of this sum is invariant unr vr prmutation of th points, an is th suitabl finition of th harmonic charactristic: [( D) ( D) ( D) ( D) ( D) ( )] [ D] D [ D] ( r ) ( r ) ( r ) r ( r) Th first notabl proprt of th harmonic charactristic is th fact that it vanishs for a harmonic rang of points, that is, whn th cross ratio ( D) taks th valus /, or pnnt on th nominations of th points. Th scon notabl proprt is th fact that th harmonic charactristic of a gnrat rang of points (two or mor coincint points) is infinit. Prhaps, th rar blivs that othr harmonic charactristics ma b obtain in man othr was, that is, b using othr combinations of th prmutations of th cross ratio. Howvr, othr attmpts la whthr to th sam function of r (or a linar pnnc) or to constant valus. For ampl, w can tak th sum of all th squars, which is also invariant unr an prmutation of th points to fin: ( D) ( D) ( D) ( D) ( D) ( D) r r ( r) ( r) ( r ) r r ( r ) [ D ] On th othr han, M. rgr (vol. I, p. 7) has us a function that is also linarl pnnt on th harmonic charactristic: