The University of Melbourne Department of Mathematics and Statistics School Mathematics Competition, 2013 JUNIOR DIVISION Time allowed: Two hours

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1 The Universiy of Melbourne Deparmen of Mahemaics and Saisics School Mahemaics Compeiion, 203 JUNIOR DIVISION Time allowed: Two hours These quesions are designed o es your abiliy o analyse a problem and o express yourself clearly and accuraely. The following suggesions are made for your guidance: () Considerable weigh will be aached by he examiners o he mehod of presenaion of a soluion. Candidaes should sae as clearly as hey can he reasoning by which hey arrived a heir resuls. In addiion, more credi will be given for an elegan han for a clumsy soluion. (2) The six quesions are no of equal lengh or difficuly. Generally, he laer quesions are more difficul han he earlier quesions. (3) I may be necessary o spend considerable ime on a problem before any real progress is made. (4) You may need o do considerable rough work bu you should hen wrie ou your final soluion nealy, saing your argumens carefully. (5) Credi will be given for parial soluions; however a good answer o one quesion will normally gain you more credi han skechy aemps a several quesions. Texbooks, elecronic calculaors and compuers are NOT allowed. Oherwise normal examinaion condiions apply.

2 . Averaging. Fill he 3 3 square wih he numbers o 9 such ha for each horizonal, verical and diagonal riple of numbers, he number in he middle is he average of he numbers on eiher side. Soluion. We have 5 odd numbers:,3,5,7,9 and 4 even numbers: 2,4,6,8. The firs and las number in a riple mus have he same pariy: hey are eiher boh even (as in 246) or boh odd (as in 59 or 23). The number, being he smalles number, canno arise as he average of hree numbers. Hence i mus occur in one of he corner squares. The same reasoning applies o he larges number, 9. Up o roaion and/or reflecion his leaves only wo possible scenarios: The righ-mos scenario is impossible because by our earlier pariy consideraions his would lead o a filling of he form 5 9 o o o which uses 6 odd numbers, where we have only 5. Coninuing wih he lef-mos opion, we now know ha he remaining wo odd numbers, 3 and 7, mus also occur in a corner. By reflecion boh choices are equivalen, and we may choose Filling in he missing squares his finally gives Regular polygons. You are given an equilaeral riangle of side lengh 2 and a regular hexagon of side lengh : Find he raio of heir respecive areas. 2 Soluion. By los of hard work you can show ha he respecive areas are = 3 and A = so ha A = 2 3 or, if you prefer, A = 3 2. The idea is no o work harder han necessary, and his is how you can ge he same answer wihou compuing he individual areas. Each of he angles of an equilaeral riangle is 60 and ha of a regular hexagon 20. (More generally, he angles of a regular n-gon are α = /n degrees: α 360/n α = /n.) This means we can ile he regular hexagon wih 6 equilaeral riangles as follows:

3 Hence A = 6 if boh he hexagon and riangle have side lengh. If you scale any shape by a facor of 2 is area scales by a facor of 2 2 = 4. Hence he raio of a regular hexagon of side lengh and an equilaeral riangle of side lengh 2 is 6/4 = 3/2. 3. Year of he snake. Julia Gillard and Tony Abbo aend a pary hosed by heir muual friend Kevin Rudd. Boh have oo much o drink and decide o walk home o clear heir heads. They sruggle o keep on course, however, and Julia snakes home like his: 4 whereas Tony snakes home like his: where R denoes Rudd s place R Julia and Tony each ake one sep of one mere per second, going from R o wih heir firs sep, from o 2 in heir nex sep, and so on. Tony leaves Rudd s house wo seconds afer Julia bu bumps ino her afer jus one sep. Boh being ipsy hey view each oher wih fresh eyes; Tony is suddenly hrilled by Julia s fiery red locks and Julia really likes he look of Tony s powerful ears. As a resul hey share a forbidden kiss, and hen coninue on heir respecive walks. Soon, however, hey mee again and anoher guily kiss follows. Their passion coninues ill hey boh reach home afer exacly 203 seps. Soluion. If Julia and Tony had se ou a exacly he same ime hey would obviously have me along he main diagonal (he line hrough 2, 6, 2, 20, 30, 42, 56,..., n 2 + n,... ). Because hey lef Rudd s place 2 seconds apar hey can now only mee in he wo diagonals ha are one above and one below he main diagonal. Bu only half of hese poins are acual meeing poins, namely hose poins ha Tony can ge o quicker han Julia. Those are he poins Julia reaches immediaely afer she has been on he main diagonal (because Tony ges o hese immediaely before he reaches he main diagonal, hence wo seconds quicker han Julia): R These are he poins of he previous sequence plus : 3, 7, 3, 2, 3, 43, 57,..., n 2 +n+,.... Since = 98 < 203 bu = 207 > 203, Tony and Julia share 44 kisses. Noe ha no oo many guesses are required o find he numbers 44 and 45: he easy squares 40 and 50 ell you you are looking for numbers roughly in beween hese wo: = = 64 and = = 255. Alernaively, you can find all of he poins where kisses are shared and hen coun hese: 3, 7 = 3 + 4, 3 = 7 + 6, 2 = 3 + 8, 3 = 2 + 0, 43, 57, 73, 9,, 33, 57, 83, 2, 24, 273, 307, 343, 38, 42, 463, 507, 553, 60, 65, 703, 757, 83, 87, 93, 993, 057, 23, 9, 26, 333, 407, 483, 56, 64, 723, 807, 893, 98 = Colour coding. You are going on a school camp supervised by he mahs eaching saff. Because mahs eachers are incredibly good a recognising paerns bu hopeless wih names, your principal has come up wih he following plan. Every suden has o wear shoes, a pair of socks, shors, a T-shir and a ha, such ha each of hese 5 iems of clohing is in one of he hree school colours. Moreover, every suden has o wear a leas one iem of each colour, and no wo sudens are allowed o wear exacly he same clohing-colour combinaion. Wha is he maximal number of sudens ha can go on camp? Soluion. Le s say he colours are red, yellow and blue. If your principal had no insised on each suden wearing all hree colours here would have been 3 5 = = 243 differen colour combinaions possible. So wha we mus do is rule ou monochromaic and bichromaic combinaions. There are exacly 3 monochromaic ways o dress yourself: you wear only red, only yellow or only blue. To find he number of bichromaic ways o dress requires a bi more work. Assume firs ha you wear he wo colours red and yellow. You can do his by wearing (a) 4 red and yellow iem, (b) 3 red and 2 yellow iems, (c) 2 red and 3 yellow iems, (d) red and 4 yellow iems. Now (a) can be done in 5 ways, (b) in 0 ways, (c) in 0 ways and (d) in 5 ways. Hence here is a oal of 30 differen combinaions of wearing red and yellow. The same applies o red and blue and o yellow and blue so ha here are a oal of 90 bichromaic colour combinaions. The oal number of richromaic dress combinaions is hus = 50.

4 No more ha 50 sudens can go on camp. 5. No so popular. Our lovers, Julia and Tony, are equally popular on he firs of May 203. In days ime on he 4h of Sepember elecions will be held, and boh Julia and Tony are campaigning hard o win your voe. Unforunaely, neiher heir policies nor heir personaliies are very appealing and hey boh lose suppor a a consan rae wih Julia losing voers wice as fas as Tony. On elecion day, Julia ges a hird of he number of voes ha Tony ges. On wha dae would Julia have received wo hirds of he number of voes ha Tony would ge? Soluion. Assume ha Julia and Tony boh have N supporers on he firs of May, and ha Tony loses supporers a a rae of r a day. Then he has N r supporers on day. Since Julia loses suppor a double Tony s rae, she has N 2r supporers on day. Afer days we know ha Julia ges one hird of he number of voes Tony ges: N r N 2 r = 3. Clearing he denominaor his gives N r = 3 N 6 r or N = 5 68 r. We now wan o find for which Again clearing he denominaor his gives N r N 2r = N 2r = 3N 6r or N = 4r. Since N = 5 68 r his gives 4r = 5 68r = r so ha = 5 7 = 85. This corresponds o he 25h of July. For hose no keen on algebra, he required equaions may also be derived using he following diagram: voes N T T J J days The respecive raios are expressed as: T J = 3, and T J = 3 2 and he fac ha Julia loses voes wice as fas as Tony means ha he difference in rise over run is a facor of wo: N J = N J = 2 N T = 2 N T Hence N J = 2N 2T = 2N 6J so ha N = 5J and N J = 2N 2T = 2N 3J so ha N = 2J. In oher words, J = N/2 and J = N/5. Subsiuing his ino. gives so ha = 85. N N/5 = N N/2 N J = N J which implies ha 4/5 = /2

5 6. Bused. Channel Nine s 2008 TV series Underbelly gave an accoun of he deadly gangland war, feauring some of Melbourne s mos noorious criminals such as Carl Fa Boy Williams and Tony The Wig Mokbel. Sadly, i never old he rue sory of how Mokbel was caugh imporing illici drugs from Sicily by hiding his merchandise inside one of 2 cricke balls he carried in a suicase. The cusoms officer a Melbourne airpor who checked Mokbel s wares was a former Mahs Comp winner and very smar. She realised ha if one of he balls had been ampered wih i would no weigh he same as he oher balls. By using a se of balance-scales no more han hree imes she quickly idenified he ball conaining drugs. Show ha someone clever can always do his using he scales no more han hree imes. Soluion. For simpliciy le us number he balls from o 2. Pu balls,2,3,4 on one side of he scales and 5,6,7,8 on he oher. Two hings can happen: (i) he scales are balanced (ii) one side weighs more han he oher. In he case of (i) one of 9,0,,2 mus conain he drugs and all of,...,8 are clean. Pu 9 and 0 on he scales. If hey weigh he same, or 2 mus conain he drugs. In ha case you compare wih one of he en clean balls. If hey do no weigh he same, is he one you are afer, oherwise i is 2. If 9 and 0 do no weigh he same one of hem mus conain he drugs, and again, by comparing one of hem wih a clean ball you can figure ou which one. In he case of (ii) one of,2,3,4 or one of 5,6,7,8 mus conain he drugs and 9,...,2 are all clean. Le s assume ha,2,3,4 weighs more han 5,6,7,8. (You can hopefully repea he required reasoning in he case ha 5,6,7,8 weighs more han,2,3,4.) Now weigh,5,6 versus 2,7,2. There are hree possibiliies: (a) They weigh he same, (b),5,6 weighs more han 2,7,2, or (c),5,6 weighs less han 2,7,2. In he case of (a), one of 3,4,8 mus conain he drugs. Weigh 3 agains 4. If hey are he same, 8 conains he drugs. If hey are no he same, he heavies of he wo conains he drugs. In he case of (b), here are again wo possibiliies: is heavier han all oher balls and hus conains he drugs, or 7 is ligher han all oher ball and conains he drugs. Simply weigh one of hem agains a clean ball and you will know which one is he odd one ou. In he case of (c), here are once again wo possibiliies: 2 is heavier han all oher balls or one of 5,6 is ligher han all oher balls. Simply weigh 5 agains 6. If hey weigh he same 2 is a heavier ball conaining drugs and if hey do no weigh he same hen he ligher of he wo conains he drugs.