Using technology to teach and solve challenging math problems

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1 Using tehnology to teh nd solve hllenging mth prolems Ellin Grigoriev, PhD Astrt Dr. Ellin Grigoriev is Professor of Mthemtis t Tes Womn's University (TWU). he grduted from Mosow tte University first in her lss of 300, defended her PhD in mthemtis nd physis, nd hs pulished over 50 sientifi ppers in the field of differentil equtions, eonomis nd optiml ontrol theory. he is former UR Mth Olympid winner, memer of the UAMO grding ommittee, nd pnelist for AIME, AHME nd AM. For the pst 5 yers, whenever she spotted n espeilly interesting or triky prolem she dded it to her noteooks, long with her originl solutions. he hs umulted thousnds of these prolems, nd she uses them in her everydy tehing. Her students re wre tht her lsses re unlike nything they hve ever enountered. Her prolems enle them to disern reusle ptterns nd tehniques in prolem solving. The uthor will shre her ides with other edutors; she will show how prolems involving numers nd proofs develop the intelletul skills of students. trting from simple emple nd moving to more omplited prolems, you will lern how students in your lulus lss n onstrut nd investigte miroeonomi model nd even predit the est prodution or sles strtegy tht mimizes profit. The workshop will e sed on her own ook Comple mth prolems nd how to solve them nd other soures of mth ontest prolems, inluding Mosow tte University entrne em prolems, UR mth Olympid nd UAMO ontest prolems. Methods will e presented for solving omple prolems tht use tehnology nd reognition of ptterns tht etend from lger. In this tlk, the uthor will teh how to use tehnology nd pttern reognition in order to solve hllenging prolems from different res of mthemtis, suh s lulus, geometry, lger, nd trigonometry.. Trjetories of ystem Let us onsider the following system of two equtions with prmeter ( ost y( sin t (.) Wht is the urve? Cn you rewrite it s y=y()?

2 quring oth sides nd dding the two equtions we otin trjetory of the system (.), irle eqution of rdius, entered t (0,0): ( y ( Ask your students whih urve is desried y the following system. ( 4ost y( 3sin t (. ) ( ost 4 y( sin t 3 (4) y (3) This is n ellipse.

3 Ask students out the trjetory of this system: ( sin t y( 6sin t 5 (. 3) This would give you line y 3 5. You n grph it on lultor nd disuss why the grph looks like segment, ut not line. (Hint: Look t the rnge of ( : [-, ] nd y(: [-, ].) We use the oundedness of the funtion sin t. Equtions with prmeters help us to understnd rel world prolems. For emple, while modeling miroeonomi system of the proess of prodution, storge nd sles of some onsumer good y system of differentil equtions, nd solving it numerilly, we otined the following trjetory desriing ( (the numer of units of the good on the mrke nd ( (the mount of the good sold ut not onsumed in onsumers homes). ([]) Otining grph like the one ove, we do not speifilly see the time, ut this is trjetory desriing the ehvior of our system on the time intervl [0, T]. Every point on this trjetory n e visulized s n ordered pir ( (, ( ), nd the end of the trjetory is the stte t t=t (the end of the given time intervl). Modern tehnology llows us to grph (, ( s funtions of time. Look t these trjetories. They re so different from the phse plne. Compre it with the stright line segment (system (. 3)) nd the grphs of ( nd y( (sin urves).

4 We n lern how to understnd trjetory nlytilly y rewriting prmetri system of equtions in different form. For emple, system of differentil equtions (. 4) n e solved in ny AP lulus lss: y y (.4) We find solutions s t ( C e nd y( C e t, (.5) where C nd C re some onstnts, nd t is n independent vrile (time); so now we n eliminte time in (.5) nd rewrite the solution of the system (.4) s trjetory y=y(): K y, where K C C (.6) Formul (6) tells us tht solutions (trjetories) of system (4) re hyperols, whih, depending on the vlue nd sign of onstnt K, will e loted in ll four qudrnts of the X-Y oordinte plne. On the other hnd, we ould grph trjetories on TI 83 lultor in prmetri MODE, giving different vlues to onstnts C nd C.

5 . Methods of evlution nd estimtion: Non-stndrd prolems nd prolems with prmeters You n deide how neessry this mteril is for you fter tking the quiz elow. I give prolems like these in order to see how deeply my students understnd funtions nd their properties. Here they re, just three prolems. Find ll rel solutions of the following equtions nd inequlity: (.) (.) sin( 3 ) 3 (.3). 5 5 Did you get the solutions? Are you still working? Is it hrd? Where did you get stuk? Hint: Don t try to solve them, think out properties of the funtions, nd try to use ommon sense. None of three prolems hs solution. Mthemtiins sy tht their solutions re the empty set. Now, look t eh prolem gin, like you re solving it from the end. Cn you eplin why, for emple, the first eqution does not hve solution? Look t the left hnd side of the eqution, nd ssume first tht >0. Applying the reltionship etween the rithmeti nd geometri mens of two positive numers nd (.4) nd rememering tht the equlity ppers if nd only if =, we n see tht o the left side of eqution (.) is lwys greter thn or equl to, ut the right side is onstnt,. Beuse >, eqution (.) hs no solutions for ll positive. But you n sy tht mye there is solution when <0, nd tht we n t pply formul (.4) for negtive numers. Yes, when <0 we nnot, ut euse >0, we n rewrite the left side s ( ) nd pply the fmilir formul to the epression within prentheses. For ll <0,, then

6 And gin there is no solution for negtive euse the left side is less thn or equl to, ut the right side is onstnt. Therefore, eqution (.) hs no rel solutions. I guess you now know in wht diretion to go when looking for n eplntion for the seond prolem. Mye the seond eqution will e esy to understnd. We wnt to ompre the rnges of the funtions on the left nd on the right. If you hve ompletely forgotten trigonometry, let s rell the definitions of sin nd osine funtions. in is the y- oordinte of point on the unit irle (of rdius ) Cosine is the - oordinte of point on the unit irle. Any point on the unit irle n hve oordintes only within intervl [-, ]. We n onlude tht the sin nd osine hve the sme rnge [-, ]; this n e written s sin nd - os. Now, n you eplin why eqution (.) hs no solution for rel? The left side is lwys less or equl to, ut the right side is 3. This eqution hs no solutions. Wht do you think out inequlity (.3)? It hs no solutions s well. Cn we prove it? The left side is positive nd the right is positive. Relling properties of eponents, we notie tht nd re reiprols of eh other. Now we n rewrite the left side s. ine > 5 5, in the inequlity (. 3) hs no solutions. We wnt to mention tht ll of these three prolems hve very understndle grphil interprettions: the funtions on the left hnd side hve no points of intersetion with funtions on the right hnd side. The purpose of this presenttion is to show how to pply properties of funtions nd their oundries to the solution of some non-stndrd prolems, nd prolems with prmeters. Those of you who solved ll three prolems without my help will find mny new ides nd pprohes you might enjoy nd use in your lssroom. Prolem olve the eqution: olution You n notie tht the left side of the eqution is lwys positive nd deresing in. It is less thn or equl to for ny rel. The epression on the right is positive s well nd

7 ( ) ( ) = (We gin used the inequlity etween geometri nd rithmeti mens) The solution n pper only when oth sides equl. Thus, =0 is the solution. Answer: =0 Prolem Let f()= nd g()=. olve the inequlity f ( g( )) 0 for. olution Completing the squre under the rdil of the first funtion nd using properties of n solute vlue, we n rewrite f() s f() ( ) 3 3 ( g()= ) Net, we find the omposition of f() nd g(), ( f g ): f(g()) ( ) 3 () By the ondition of the prolem epression () must e less thn or equl to zero. f ( g( )) 0 ( ) () Let us onsider three ses for ():. If 5 0 (<-5), then we hve no solutions.. If 0 5 ( 5 ), then 5 0 ( 5). 3. If 0 ( ), then 5 ( ) ( 5) Answer: no solutions for <-5; [0,( 5) ] if 5<<; [( ),( 5) ] if. Hving solved this prolem in generl, for ny vlue of prmeter, you n lwys find some prtiulr solution of the prolem, nd nswer question like: If =0, wht will stisfy the given inequlity f(g()) 0? Anlytilly we otined tht if, then [( ),( 5) ]. If this is true, then this must e true for =0 s well, nd the grph of funtion y=f (g()) must go elow the X-is only for [(0),(0 5) ] or 8 5. Let us hek the nswer on TI 83/84 plus grphing lultor..

8 First we mke omposition of two funtions on our grphing lultor. Here Y g( ) nd Y f ( g( )). 4 5 We epet negtive ehvior etween =8 nd =5, then we need to set up n pproprite window for our lultor. Net we use [ nd ] [TRACE] (CALC) uttons to find zeros of f(g()): As we epeted, they re preisely 8 nd ome Geometry Chllenge imilr Tringles: You need to rememer tht two tringles re similr to eh other y two sides, y two ngles (AA), y two sides nd the inluded ngle (A), nd y three sides (). Espeilly importnt is the ft tht in similr tringles the rtio of orresponding sides, medins, heights, nd isetors equls k, the oeffiient of similitude. The rtio of the res of similr tringles equls k, the squre of the oeffiient of similitude. The following prolem will illustrte how I teh plne geometry with the use of tehnology. Tringle ABC is given. Point P is interior with respet to the tringle. Through this point P three lines prllel to eh side of the tringle re drwn, nd form three tringles with res,, nd 3. Find the re,, of tringle ABC. olution In order to even strt thinking out this prolem I reommend drwing piture.

9 B B F 3 D A P C A L M P N E C originl tringle with point P inside lines prllel to sides of the tringle pss through point P Hving piture, we notie tht LE//AC, MD//AB, nd FN//BC, then we n onlude tht LP=AM nd PE=NC. It seems resonle to denote three vriles: =AM, =MN, nd =NC. Besides, eh of the smll tringles is similr to tringle ABC. How n we use this informtion? Of ourse, you rememer the rtio of res of similr tringles equls squre of the oeffiient of similitude, k. However, how n we find these three oeffiients? Agin we know the nswer to this question: the oeffiient of similitude, k, is the rtio of orresponding sides of similr tringles. Now we re redy to solve the prolem. FLP ABC MPN ABC 3 PDE ABC LP AC PE AC MN AC Repling AC= + +, LP=, MN=, nd PE= nd tking the squre root of the left nd the right sides of eh equlity, we hve

10 3 Does it look too fr from the nswer? Try to dd the left nd the right sides. Putting epressions over the ommon denomintor, we otin: 3 tht is equivlent to 3 quring oth sides of the lst eqution gives us the nswer: 3 Answer: 3. Referenes [] Grigoriev E. V. nd E.N. Khilov, 005. An Attinle et of Nonliner Controlled Miroeonomi Model. Journl of Nonliner Dynmis nd Control, Vol., No.,