THE CONTEST CORNER No. 65 John McLoughlin

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1 THE CONTEST CORNER /185 THE CONTEST CORNER No. 65 Joh McLoughli The problems featured i this sectio have appeared i, or have bee ispired by, a mathematics cotest questio at either the high school or the udergraduate level. Readers are ivited to submit solutios, commets ad geeralizatios to ay problem. Please see submissio guidelies iside the back cover or olie. To facilitate their cosideratio, solutios should be received by October 1, The editor thaks Adré Ladouceur, Ottawa, ON, for traslatios of the problems. CC321. Six boxes are umbered 1, 2, 3, 4, 5 ad 6. Suppose that there are N balls distributed amog these six boxes. Fid the least N for which it is guarateed that for at least oe k, box umber k cotais at least k 2 balls. CC322. Suppose that the vertices of a polygo all lie o a rectagular lattice of poits where adjacet poits o the lattice are at distace 1 apart. The the area of the polygo ca be foud usig Pick s Formula: I + B 2 1, where I is the umber of lattice poits iside the polygo, ad B is the umber of lattice poits o the boudary of the polygo. Pat applied Pick s Formula to fid the area of a polygo but mistakely iterchaged the values of I ad B. As a result, Pat s calculatio of the area was too small by 35. Usig the correct values for I ad B, the ratio = I B is a iteger. Fid the greatest possible value of. (Ed.: For more iformatio o Pick s formula, take a look at the article Two Famous Formulas (Part I), Crux 43 (2), p ) CC323. Evaluate log 10 (20 2 ) log 20 (30 2 ) log 30 (40 2 ) log 990 ( ) log 10 (11 2 ) log 11 (12 2 ) log 12 (13 2 ) log 99 (100 2 ) CC324. O the iside of a square with side legth 60, costruct four cogruet isosceles triagles each with base 60 ad height 50, ad each havig oe side coicidig with a differet side of the square. Fid the area of the octagoal regio commo to the iteriors of all four triagles. CC325. Seve people of seve differet ages are attedig a meetig. The seve people leave the meetig oe at a time i radom order. Give that the yougest perso leaves the meetig sometime before the oldest perso leaves the meetig, the probability that the third, fourth, ad fifth people to leave the meetig Copyright c Caadia Mathematical Society, 2018

2 186/ THE CONTEST CORNER do so i order of their ages (yougest to oldest) is m, where m ad are relatively prime positive itegers. Fid m CC321. Six boîtes sot umérotées 1, 2, 3, 4, 5 et 6. O suppose que N balles sot distribuées das ces six boîtes. Détermier la plus petite valeur de N pour laquelle il est certai que pour au mois ue valeur de k, la boîte uméro k cotiet au mois k 2 balles. CC322. O cosidère u quadrillage avec ue distace de 1 uité etre les liges verticales et etre les liges horizotales. U polygoe est tracé sur le quadrillage et ses sommets sot des poits de treillis. O peut alors calculer l aire du polygoe à l aide de la formule de Pick : A = i + b 2 1, i état le ombre de poits de treillis à l itérieur du polygoe et b état le ombre de poits de treillis sur le bord du polygoe. Pat a utilisé la formule de Pick pour calculer l aire d u polygoe, mais il a chagé l ue pour l autre les valeurs de i et de b avec comme résultat que sa répose était 35 de mois que la boe répose. Si o utilise les boes valeurs de i et de b, le rapport = i b est u etier. Détermier la plus grade valeur possible de. (N.D.L.R. Pour e coaître davatage sur la formule de Pick, voir l article Two Famous Formulas (Part I), Crux 43 (2), p ) CC323. Évaluer log 10 (20 2 ) log 20 (30 2 ) log 30 (40 2 ) log 990 ( ) log 10 (11 2 ) log 11 (12 2 ) log 12 (13 2 ) log 99 (100 2 ). CC324. À l itérieur d u carré de 60 de côté, o costruit quatre triagles isocèles, chacu avec ue base de 60 et ue hauteur de 50, chaque triagle ayat u côté qui coïcide avec u côté du carré. Détermier l aire de la régio octogoale commue aux quatre triagles. CC325. Sept persoes d âges différets assistet à ue réuio. Les sept persoes quittet la réuio ue par ue das u ordre aléatoire. Sachat que la persoe la plus jeue quitte avat que la persoe la plus âgée e quitte, la probabilité que les troisième, quatrième et ciquième persoes quittet la réuio das l ordre de leurs âges (de la plus jeue à la plus âgée) est égale à m, m et état des etiers premiers etre eux. Détermier m +. Crux Mathematicorum, Vol. 44(5), May 2018

3 THE CONTEST CORNER /187 CONTEST CORNER SOLUTIONS Statemets of the problems i this sectio origially appear i 2017: 43(5), p CC271. Warre s lampshade has a iterestig desig. Withi a regular hexago (six sides) are two itersectig equilateral triagles, ad withi them is a circle which just touches the sides of the triagles. (See the diagram.) The poits of the triagles are at the midpoits of the sides of the hexagos. If each side of the hexago is 20 cm log, fid: a) the area of the hexago; b) the area of each large equilateral triagle; c) the area of the circle. Origially Questio 5 from the 2011 Uiversity of Otago Juior Mathematics Competitio. We received two submissios to this problem, both of which were correct. preset the solutio by David Maes, modified by the editor. a) The formula for the area A H of a regular hexago is A H = s2, where s is the side legth. Sice s = 20 cm, we have A H = (20)2 = cm 2. b) Note that the two large equilateral triagles are cogruet, the six smaller equilateral triagles are all cogruet ad the six rhombi are also cogruet. Sice the vertices of the equilateral triagles occur at the midpoits of the sides of the hexago, it follows that the side legth of each rhombus ad each smaller equilateral triagle is 10 cm. Thus the side legth for the two larger equilateral triagles is a = 30 cm. Therefore, the area A T for each of We Copyright c Caadia Mathematical Society, 2018

4 188/ THE CONTEST CORNER the larger equilateral triagles is 3 3 A T = 4 a2 = 4 (30)2 = cm 2. c) The circle is the iscribed circle for each of the two equilateral triagles with 3 side legth a = 30 cm. Therefore, its radius r is give by r = 6 a = 5 3 cm. Hece, the area A C of the circle is A C = πr 2 = π(5 3) 2 = 75π cm 2. CC272. A sum-palidrome umber (SPN) is a umber that, whe there are a eve umber of digits, the first half of the digits sums to the same total as the secod half of the digits, ad whe odd, the digits to the left of the cetral digit sum to the same total as the digits to the right of the cetral digit. A productpalidrome umber (PPN) is like a sum-palidrome, except the products of the digits are ivolved, ot the sums. a) How may three-digit SPNs are there? b) The two SPNs 1203 ad 4022 sum to 5225, which is itself a SPN. Is it true that, for ay two four-digit SPNs less tha 5000, their sum is also a SPN? c) How may four-digit o-zero PPNs are there? Origially Questio 2 from the 2011 Uiversity of Otago Juior Mathematics Competitio. We received 2 solutios, oe of which was correct ad complete. We preset the solutio by Ivko Dimitrić. a) A three-digit SPN is of the form aba where b, the digit of tes, ca be ay digit 0 to 9 (thus te choices) ad digit a of hudreds caot be zero but ca be ay umber 1 to 9, thus 9 choices. The there are 10 9 = 90 ways to choose a ad b idepedetly, ad hece there are 90 three-digit SPNs. b) The statemet is ot true, i particular i some situatios that ivolve carryig over of uits to a higher-value place, such as i examples = 7860, = 4186, = 6841, = c) First, we determie the umber of four-digit PPNs i which 0 does ot appear amog the digits. We do the coutig accordig to the umber of distict digits used i the represetatio of a PPN. (1) There are ie PPNs if all the digits are equal. (2) If exactly two distict digits a ad b are used, the the same pair is used i the first half of the umber as i the secod half for the followig four Crux Mathematicorum, Vol. 44(5), May 2018

5 THE CONTEST CORNER /189 possibilities for each choice of a ad b : abab, abba, baab, baba. Sice a pair of distict digits ca be chose from the set of ie i ( 9 2) = 36 ways, each choice producig four PPNs, the total umber produced this way is 36 4 = 144. (3) If exactly three digits a, b ad c are used, for such a umber to be a PPN the product of two of them equals the square of the third, e. g. a b = c 2, whereas a pair a, b appears i oe half of the umber ad digit c, repeated twice, i the other half. This happes for the followig four products of pairs 1 4 = 2 2, 1 9 = 3 3, 2 8 = 4 4, 4 9 = 6 6, each of the cases producig four PPNs, abcc, bacc, ccab, ccba, therefore there are sixtee PPNs obtaied this way. (4) If four distict digits are used for a PPN with two differet sets of pairs a, b ad c, d i each half, we must have ab = cd. Sice all four digits are o-zero ad differet, it is easily see that ac bd ad ad bc, which meas that the umbers from differet pairs caot be swapped ad combied i the same half of the umber ad would have to stay withi the origial pair. This situatio happes for the followig five products of pairs: 1 6 = 2 3, 1 8 = 2 4, 2 6 = 3 4, 2 9 = 3 6, 3 8 = 4 6. Each of them gives rise to eight PPNs: abcd, abdc, bacd, badc, cdab, cdba, dcab, dcba. Thus, there are 8 5 = 40 PPNs obtaied this way. Addig up the umbers for all the possibilities we get = 209 PPNs i which the product of digits i each half of the umber is the same ad 0 is ot oe of the digits. CC273. Kakuro is the ame of a umber puzzle where you place umbers from 1 to 9 ito empty boxes. There are three rules i a Kakuro puzzle: oly umbers from 1 to 9 may be used, o umber is allowed i ay lie (across or dow) more tha oce, the umbers must add up to the totals show at the top ad the left. The left figure i the diagram shows a small fiished Kakuro puzzle. Solve the Kakuro puzzle o the right i the diagram. Is your solutio uique? Origially Questio 1 from the 2008 Uiversity of Otago Juior Mathematics Competitio. Copyright c Caadia Mathematical Society, 2018

6 190/ THE CONTEST CORNER We received oe submissio. We preset the solutio by Ivko Dimitrić. The oly way (up to the order) to write 16 as a sum of two distict positive itegers which are less tha te is 16 = = Thus, these are the choices for the Sum-16 row ad Sum-16 colum. Assume the Sum-16 row (the first row) is [ 9 7 ]. The Sum-23 colum is the [ 7 a b ] T, where subscript T deotes traspose, i.e. the triple of umbers should be see i a vertical colum. The a + b = 16, where a, b are distict itegers from 1 to 9, so oe of them must be 7 agai, which caot happe, sice we already have oe 7 i the top box of that colum. Hece, our assumptio caot stad ad the first row is the [ 7 9 ], whereupo the Sum-23 colum takes the form [ 9 a b ] T, with a + b = 14. Let the Sum-24 colum be [ 7 c d ] T where c+d = 17 so that oe of the umbers c, d is 9 the other oe 8, i particular c is either 8 or 9. Sice 9 already appears i the Sum-23 colum, accordig to the rules, a + b = 14 is possible oly if it is the sum of 8 ad 6. Assume first that the Sum-23 colum is [ ] T. We show that this is ot possible. If the Sum-16 colum (the last colum) is [ 7 9 ] T, the the Sum-29 row has the form [ f d 6 9 ], with f + d = 14 = = 8 + 6, so oe term i the sum would have to be 6 or 9, clashig with the same umber already i the last row. If the Sum-16 colum is [ 9 7 ] T, the the Sum-30 row would appear as [ e c 8 9 ], so the umber c (which, from the above, is either 8 or 9) would clash with oe of the last two umbers. Thus, both possibilities for the last colum cotradict the choice of the Sum-23 colum as [ ] T, which meas that that colum would have to be chaged to [ ] T istead. The, sice c + d = 17 = i the Sum-24 colum, we have c = 8 ad d = 9 ad that colum is completed as [ ] T so that the last colum could be oly [ 9 7 ] T, to avoid havig two 9s i the last row. The the remaiig cell i the Sum-30 row is filled with e = 7 ad the remaiig cell i the Sum-29 row is filled with f = 5 to make the required sums, icludig the remaiig Sum-12 colum. Sice every choice at every step is forced upo by previous choices or cotradictios resultig from alterative possibilities, every umber is uiquely determied, i. e. there is oly oe solutio to the puzzle, amely Crux Mathematicorum, Vol. 44(5), May 2018

7 THE CONTEST CORNER /191 CC274. I his office, Shaquille had ie pig pog balls which he used for therapeutic recovery by throwig them ito the waste basket at slack times. Each time he threw the ie balls, some of them would lad i the basket, with the rest of them ladig o the floor. a) If the balls are idetical, how may differet results could there be? b) Suppose ow that the balls are umbered 1 to 9. How may differet results could there be ow? (For example oe possible result is for balls 1 to 4 to lad i the basket, with 5 to 9 o the floor.) c) Suppose istead that the balls are ot umbered, but five are coloured yellow ad four blue. Now how may differet results could there be? (For example oe possible result is for two yellow balls ad three blue balls to lad i the basket, ad the rest to lad o the floor.) d) Oe day aother basket appeared i the office. So ow Shaquille had a choice of baskets to aim at. How did this chage the aswers to (a), (b), ad (c)? e) Now suppose that every time he threw the balls at the two baskets, each basket received at least two balls. How would this chage the aswers to (a), (b), ad (c)? Origially Questio 5 from the 2008 Uiversity of Otago Juior Mathematics Competitio. We received oe solutio to this problem. We preset the solutio by Ivko Dimitrić, modified by the editor. I each case, the umber ad the selectio of the balls that lad o the floor is uiquely determied by the umber ad the selectio of the balls that lad i the basket(s), so it suffices to cout the umber of balls that lad i the basket(s) to get the umber of differet results. a) Ay umber k = 0, 1,..., 9 of balls ca lad i the basket. Therefore there are 10 differet results. b) A umbered ball is either i or out of the basket. Therefore each umbered ball has 2 possible states. As there are 9 balls, there are 2 9 = 512 differet results. c) Sice ay umber y = 0, 1,... 5 of yellow balls ad ay umber b = 0, 1,... 4 of blue balls ca idepedetly lad i the basket we have 6 5 = 30 differet results. d) Let B L ad B R refer to the left basket ad right basket, respectively. To aswer a), for ay umber k = 0, 1,..., 9 of balls that lad i B L there are 10 k choices of balls that may lad i B R. Therefore there are 9 k=0 (10 k) = 55 differet results. To aswer b), a umbered ball is either i B L, i B R, or ot i either. Therefore each umbered ball has 3 possible states. As there are 9 balls, there are 3 9 = 19,683 differet results. Copyright c Caadia Mathematical Society, 2018

8 192/ THE CONTEST CORNER To aswer c), for ay umber y = 0, 1,..., 5 of yellow balls that lad i B L there are 6 y choices of yellow balls that may lad i B R. Therefore there are 5 y=0 (6 y) = 21 ways to distribute the 5 yellow balls. Likewise, for ay umber b = 0, 1,..., 4 of blue balls that lad i B L there are 5 b choices of blue balls that may lad i B R. Therefore there are 4 b=0(5 b) = 15 ways to distribute the 4 blue balls. Therefore there are = 315 differet results. e) To aswer a) for ay umber k = 2, 3,..., 7 of balls that lad i B L there are 8 k choices of balls that may lad i B R. Therefore there are 7 k=2(8 k) = 21 differet results. To aswer (b), choose 4 balls out of ie i ( 9 ) ways to be distributed i baskets B L ad B R. Give the chose balls, k lad i basket B L ad k lad i basket B R. Therefore 2 k 2. Thus, the umber of results is 9 2 =4 k=2 Ç 9 å = k = = k k=2 9 9 [ k k=0 9 9 [2 2 2]. =4 =4 =4 0 Ç 1 1 å ] We complete the sum so that it starts from = 0 by addig ad subtractig terms correspodig to = 0, 1, 2, 3, which are equal to 1, 18, 72, 0, respectively. The above sum becomes 9 =0 9 [2 2 2] + 91 = 9 = Therefore there are 14,142 differet results. 9 = =0 = (2 + 1) 9 2(1 + 1) = = 14, To aswer c), we cout the umber of results i which either or both of the baskets ed up with less tha two balls ad subtract that umber from 315 (the umber of results we foud i part (d) without ay restrictios). Let the sets (L Y, L B ) ad (R Y,R B ) correspod to the umber of yellow ad blue balls that lad i B L ad B R, respectively. If B L eds up with less tha 2 balls the the set (L Y, L B ) {(0, 0), (1, 0), (0, 1)}. We cosider these three cases: 1. If (L Y, L B ) = (0, 0), the there are 6 choices of yellow balls ad 5 choices of blue balls ladig i B R. Totallig 6 5 = 30 differet results. 2. If (L y, L b ) = (1, 0), the there are 5 choices of yellow balls ad 5 choices of blue balls ladig i B r. Totallig 5 5 = 25 differet results. Crux Mathematicorum, Vol. 44(5), May 2018

9 THE CONTEST CORNER / If (L y, L b ) = (0, 1), the there are 6 choices of yellow balls ad 4 choices of blue balls ladig i B r. Totallig 6 4 = 24 differet results. A total of 79 results. If the situatio is reversed ad the basket B R receives less tha two balls, we will also have 79 results. I their sum the umber of results where both baskets got less tha two balls is couted twice. That umber is 3 3 = 9, sice there are three possibilities for each basket as listed above. The total umber of results i which at least oe basket has less tha two balls is = 149. The umber of results i which both baskets received at least two balls is therefore = 166. CC275. The local sailig club is plaig its aual race. By traditio the boats always start at P, sailig due North for a distace of a km util they reach Q. They the tur ad sail a distace of x km to R (which is due East of P ). Next they tur ad sail a distace of y km to S (which is South of P ) before fially sailig due North for a distace of b km util the fiish lie, which is back at the startig poit P (see the diagram). Berie, the Club Commader, makes four extra rules for this year s race: a + b = 40, x + y = 50, a < b, the four legths (a, b, x, y) must each be a whole umber of kilometres. (Berie does t like decimals.) a) Fid four umbers (a, b, x, y) which satisfy Berie s four rules. b) Are there four differet umbers (a, b, x, y), which also satisfy Berie s four rules apart from the four umbers you foud i part (a)? Explai. Origially Questio 3 from the 2008 Uiversity of Otago Juior Mathematics Competitio. We received four solutios to this problem, all of which were correct. We preset the composite solutio of Kostatie Zelator, Digby Smith, ad Ivko Dimitrić, modified by the editor. We solve a) ad b) by determiig all solutios to Berie s problem. From the right agle triagles P QR ad P RS we have that (P R) 2 = x 2 a 2 = y 2 b 2 Sice a + b = 40 b = 40 a. Sice x + y = 50 y = 50 x. Through substitutio we see that x 2 a 2 = (50 x) 2 (40 a) 2, Copyright c Caadia Mathematical Society, 2018

10 194/ THE CONTEST CORNER which simplifies to x = 9 + 4a 5. Sice x is a whole umber it follows that a is divisible by 5. Note that, sice a < b, a < a+b 2 = 20. Therefore a is a whole umber that is less tha 20 ad divisible by 5. Assumig a > 0 the possible values of a are 5, 10, ad 15. Below we list these 3 cases of a: a = 5 b = 35, x = 13, y = 37. a = 10 b = 30, x = 17, y = 33. a = 15 b = 25, x = 21, y = 29. The three quadruples that solve Berie s problem are therefore (5, 35, 13, 37), (10, 30, 17, 33) ad (15, 25, 21, 29). Note that if a 0 the (0,40,9,41) is also a solutio to Berie s problem. Square Garlads Usig costructio paper, a strig ad some glue, make 5 garlads cosistig of 5 squares each (for a sturdy garlad, glue squares to both sides of the strig). Usig these garlads, ca you completely cover the figure show above? Puzzle by Nikolai Avilov. Crux Mathematicorum, Vol. 44(5), May 2018