THE USE OF HAND-HELD TECHNOLOGY IN THE LEARNING AND TEACHING OF SECONDARY SCHOOL MATHEMATICS

Transcription

1 THE USE OF HAND-HELD TECHNOLOGY IN THE LEARNING AND TEACHING OF SECONDARY SCHOOL MATHEMATICS The funcionali of CABRI and DERIVE in a graphic calculaor Ercole CASTAGNOLA School rainer of ADT (Associazione per la Didaica con le Tecnologie) NRD Diparieno di Maeaica Universià Federico II NAPOLI (Ial) ABSTRACT The workshop will focus on he use of hand-held echnolog o iprove eaching and learning in aheaics in secondar school, wih an ephasis on -6 educaion The alk will be illusraed wih pracical eaples using a new graphic calculaor, which conains a version of CABRI and DERIVE In paricular i will be shown how o use he graphic calculaor o each geoeric plane isoeries, boh fro a snheic poin of view and fro an analical one Inroducion In his workshop we wan o show how a eacher in a secondar school can illusrae geoeric plane isoeries using a new graphic calculaor, which conains he funcionali of CABRI and DERIVE The didacic pah will be divided in several unis Eplaining, b using CABRI, he following Theore: Ever isoer of he plane is a produc of a os hree reflecions in lines The geoeric consrucion (given wo congruen riangles ABC and A B C ) is coposed b he following seps: (i) Using he Perpendicular bisecor ool we draw he perpendicular bisecor r of he line segen AA Then, using he Reflecion ool, we ake he riangle ABC o A B C b he reflecion in r (A A ) [Look a Figure] ii) If B B, using he Perpendicular bisecor ool we draw he perpendicular bisecor s of he line segen B B ; s passes hrough poin A ( A ) because A B A B Then, using he Reflecion ool, we ake A B C o A B C b he reflecion in s (considering ha A A A and B B ) (iii) If C C, using he Perpendicular bisecor ool we draw he perpendicular bisecor of he line segen C C Then, using he Reflecion ool, we ake (necessaril) A B C o A B C b reflecion in Consequenl he coposiion of hree reflecions in lines r, s, is he isoer which ake ABC o A B C ) Verifing he congruence of wo riangles whose verices are given and drawing is graph We are given ABC and A B C, whose verices are A(,), B(7,9), C(0,) and A (,), B (0, ), C (6, ) Using he funcion dis ha we have included ino he graphic calculaor we can do he following check ADT is he Ialian version of T-cubed or T

2 9 AB AC BC 9 B A C A C B The congruence of riangles follows To draw he graph we can begin fro he vecor equaion of line r AB : p a (b a) where p is he posiion vecor of a variable poin P r AB and a and b are he posiion vecors of A(, ) and B(, ) and R is he paraeer The previous equali can also be wrien in he following for fro which we ge he paraeric equaions of he line ) ( ) ( In paricular, o represen he line segen AB i is enough o resric he paraeer o he inerval [0,] For he line segens ha for our wo riangles we have he following paraeric equaions AB: BC: CA: A B : B C : C A : 0 6 REMARK When we graph our wo riangles on he graphic calculaor we can observe ha he firs riangle ABC is raversed in aniclockwise sense and he second one A B C in clockwise sense Thus he wo riangles are relaed b an isoer of second kind (or indirec or odd) I is possible o show in an analic wa ha wo riangles are relaed b an isoer of he second kind or, ore generall, if he have he sae orienaion, using he following heore

3 THEOREM Le A(, ), B(, ), C(, ) and A (, ), B (, ), C (, ) be he coordinaes of he verices of he oriened riangles ABC and A B C, respecivel In order ha he riangles have he sae orienaion, i is necessar and sufficien ha he deerinans have he sae sign For our wo riangles we ge and 0 6 So he wo riangles have opposie orienaion REMARK We noe ha he absolue value of he above deerinan is wice he area of he riangle REMARK Le us consider a ransforaion of he plane given b he following equaion a b c a b c We noe ha we can represen such an equaion using onl one objec: a ari In fac i is equivalen o he equaion a b c a b c 0 0 The las row of he ari ha represens he ransforaion is alwas fored b he vecor [0,0,] A poin of he plane is represened b a vecor wih hree coponens, having he hird coponen alwas equal o This represenaion is paricularl convenien fro an algorihic poin of view, because he ransforaion is copleel described b onl an objec ha is eas o ipleen In paricular, for he isoeries we have a b c Isoeries of he firs kind: b a c 0 0 a b c Isoeries of he second kind: b a c 0 0 wih a b ) Deerining analicall, using Theore in ), he isoer (of second kind) ha ake he riangle ABC o A B C Firsl we have o find he perpendicular bisecor of he line segen AA using he funcion perpbis ha we have included ino he graphic calculaor Such a funcion akes as inpu wo poins given as liss of wo eleens

4 where linesp is he funcion ha akes as inpu he slope and a poin of line and idpoin is he funcion ha calculaes he idpoin of a line segen given as inpu is end poins REMARK As we can see, a suden can build a librar of funcions o use in he resoluion of paricular pes of probles Such funcions can be used o build oher funcions or progras in he sae wa we use he funcions buil in he graphic calculaor In our case we have Perpendicular bisecor AA : Thus he slope is and he inercep q In general he reflecion in a line r whose equaion is q can be obained b he ehod of double ranslaion: b eans of a ranslaion we ake he line r o r which is parallel o r and passes hrough he origin O, han we do a reflecion in r and lasl we do a ranslaion b a vecor ha is opposie o he firs one We can choose he poin on r whose coordinaes are (0,q) and herefore we can use he ranslaion b vecor [0,q] Consequenl we have rifr(,q) rasl([0,q]) rifor() rasl([0,q]), where rifor() denoes he reflecion in a line ha passes hrough he origin, whose equaion is We reeber on his subjec ha he reflecion in a line ha passes hrough he origin and akes an angle α wih he posiive half of he -ais can be obained b eans of a roaion (abou he origin) hrough an angle α (ha akes he line r o he -ais), hen a reflecion in he -ais whose ari is 0, 0 and again a roaion hrough an angle α Therefore he ari we are looking for is ro(α) rif ro(α), where ro(α) denoes he ari ha represens a roaion hrough an angle α

5 The resul of his calculaion b a graphic calculaor is ha, if we keep in ind he double-angle ideniies, coincides wih he ari α α α α cos sin sin cos, ha, as we know [see Ipedovo] represens he reflecion in a line ha passes hrough he origin and akes an angle α wih he posiive half of he -ais Since anα, we can use he doubleangle ideniies o epress cosα e sinα b eans of anα, so he above ari becoes, or we can calculae he previous produc wih he condiion a an () and we obain ha is he sae ari wrien above Thus, using again a ari, he reflecion rifor() has he following for rifor() Suing up, we obain for he ari rifr(,q) he following for

6 In paricular, he reflecion in he perpendicular bisecor of AA is Such a ari will be denoed wih iso Le us denoe wih ver0 he ari whose coluns are he vecors [ i, i,], i, where i, i are he coordinaes of verices A, B, C of our firs riangle Tha is The produc iso ver0 give us a ari, ha we call ver, whose coluns conain he coordinaes of he ransfored poins A, B, C We can noe ha A A Now we find he perpendicular bisecor of he line segen B B, wih B (0,) We ge B, and 70 perpendicular bisecor B B : and we can easil verif ha such a line passes hrough A The reflecion in he line s (s: 70 ) is epressed b he ari

7 ha we call iso The produc iso ver give us a ari, ha we call ver, whose coluns conain he coordinaes of he ransfored poins A A, B B, C Lasl we find he perpendicular bisecor of he line segen C C, wih C (6,) We ge 97 C, and 9 9 perpendicular bisecor C C : and we can easil verif ha such a line passes hough A and B The reflecion in he line (: ) is epressed b he ari ha we call iso We can verif ha he produc iso ver gives us a ari whose coluns conain he coordinaes of verices of A B C Therefore he ari ha represens he isoer we are looking for is 0 iso iso iso Looking a he ari we can easil infer ha we can obain our isoer b eans of a reflecion in he line and a ranslaion b vecor [,6] (firs he reflecion and hen he ranslaion)

8 ) Deerining he glide reflecion ha generaes our isoer To find such glide reflecion we firs observe ha is ais us pass hrough he idpoins of line segens AA, BB e CC If we denoe such idpoins wih M AA, M BB e M CC, we have M AA 0, M BB, M CC, The equaion of line r ha passes hrough M AA and M BB is and r is parallel o he line We can verif ha M CC r The reflecion in he line r is epressed b he ari The vecor v of he ranslaion is parallel o he line r and hus us be of he for v [k,k] Bu he reflecion akes he poin A(,) o he poin A*, and he ranslaion b vecor v has o ake A* o A (,) Consequenl k 7 k k Therefore he vecor v is, The ranslaion b vecor v, is The produc of such a ari wih he ari of he reflecion gives us he sae resul we have obained before and we can verif ha his produc is couaive

9 Bibliograph Dedò, M: 996 Trasforazioni geoeriche Decibel-Zanichelli Eves, H: 99 Fundaenals of Modern Eleenar Geoer Jones and Barle Ipedovo, M: 999 Maeaica: insegnaeno e copuer algebra Springer-Verlag Ialia Marin, GE: 9 Transforaion Geoer An Inroducion o Ser Springer-Verlag Schuann, H & Green, D: 99 Discovering Geoer wih a Copuer - using Cabri Géoère Charwell-Bra