1
SKOLIAD
No. 114
Va lav Linek
Please send your solutions to problems in this Skoliad by August 1, 2009. Solutions should be sent to Lily Yen and Mogens Hansen at the address inside the ba k over. The Skoliad se tion is in transition and, unfortunately, we have lost several of the submitted solutions to past ontests. If you have
opies of solutions that you sent to past ontests, please send them again so that we an mention any orre t solutions we re eive. (This in ludes any
ontest in Skoliad appearing in or after the Mar h 2008 issue of CRUX). Our rst problem set of the year is the Math Kangaroo Contest Pra ti e Set. The Kangaroo Contest is international in s ope and supported in Canada by the Canadian Mathemati al So iety and the Institute of Ele tri al and Ele troni s Engineers (Northern Se tion). Our thanks go to Valeria Pandelieva, the Canadian representative of the Kangaroo Contest, for bringing this ontest to our attention, and for making us aware of the need for ontests and math-parti ipation in the lower years in Canada. For that reason, and also sin e this ontest is straightforward to administer (see www.mathkangaroocanada.com), we are featuring its entire range of questions over all grades. Finally, while it is a multiple hoi e test, we ask our readers to send in
omplete solutions showing all the steps and details so that we an evaluate the solutions and give full redit to the solvers. Math Kangaroo Contest Pra ti e Set
Part A (3 points per question)
. (Grades 3-4) In the addition example, ea h letter represents a digit. Equal digits are represented by the same letter. Di erent digits are represented by di erent letters. Whi h digit does the letter K represent? (A) 0 (B) 1 (C) 2 (D) 8 1
+ W
O K O
K O W
(E) 9
. (Grades 5-6) Ten aterpillars, arranged in a row one behind another, walked in the park. The length of ea h aterpillar was equal to 8 m, and the distan e any two adja ent aterpillars kept for safety reasons was 2 m. What is the total length of their row? (A) 100 m (B) 98 m (C) 82 m (D) 102 m (E) 96 m 2
2 . (Grades 7-8) An ant is running along a ruler ..................................sq...q.r................................................ ... of length 10 m with a onstant speed of 1 m ... . 1 2 3 4 5 6 7 8 9 . per se ond (see the gure). Any time when ....................................................................................... the ant rea hes one of the ends of the ruler, it turns ba k and runs in the opposite dire tion. It takes the ant exa tly 1 se ond to make a turn. The ant starts from the left end of the ruler. Nearest whi h number will it be after 2009 se onds? (A) 1 m (B) 2 m (C) 3 m (D) 4 m (E) 5 m 3
. .... .
4
. .... .
. .... .
.... .. ..... ............... . ..... .... . .
. .... .
. .... .
. .... .
. .... .
. (Grades 9-10) Whi h of the numbers 26 , 35 , 44 , 53 , 62 is the greatest? (A) 26 (B) 35 (C) 44 (D) 53 (E) 62
. (Grades 11-12) A de orator has prepared a mixed paint, in whi h the volumes of red and yellow olours were in the ratio 2 : 3. The resulting
olour seemed too light to him, so he added 2 L of red paint. This way, the ratio of the volumes of the red and yellow olours hanged to 3 : 2. How many litres of paint did the de orator use? (B) 6 L (C) 7 L (D) 8 L (E) 9 L (A) 5 L 5
Part B (4 points per question) 6. (Grades 3-4) Two boys are playing tennis until one of them wins four times. A tennis mat h annot end in a draw. What is the greatest number of games they an play? (B) 7 (C) 6 (D) 5 (E) 9 (A) 8 7. (Grades 5-6) In two years, my son will be twi e as old as he was two years ago. In three years, my daughter will be three times as old as she was three years ago. Whi h of the following best des ribes the ages of the daughter and the son? (A) The son is older; (B) The daughter is older; (C) They are twins; (D) The son is twi e as old as the daughter; (E) The daughter is twi e as old as the son.
. (Grades 7-8) Some points are marked on a straight line so that all distan es
8
1 m, 2 m, 3 m, 4 m, 5 m, 6 m, 7 m, and 9 m are among the distan es
between these points. At least how many points are marked on the line? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8
. (Grades 9-10) Eva, Betty, Linda, and Cathy went to the inema. Sin e it was not possible to buy four seats next to ea h other, they bought ti kets for seats number 7 and 8 in the 10th row and ti kets for seats number 3 and 4 in the 12th row. How many seating arrangements an they hoose from, if Cathy does not want to sit next to Betty? (A) 24 (B) 20 (C) 16 (D) 12 (E) 8 9
3 10. (Grades 11-12) Triangle ABC is isos eles with BC = AC . The segments DE , F G, HI , KL, M N , OP , and XY divide the sides AC and CB into equal parts. Find XY , if AB = 40 m. (A) 38 m (B) 35 m (C) 33 m (D) 30 m (E) 27 m
C ... ... ....
D ..................................... E
F .......................................................... G H ............................................................................. I K .................................................................................................. L M ....................................................................................................................... N O ............................................................................................................................................ P
X ................................................................................................................................................................ Y A ........................................................................................................................................................................... B
Part C (5 points per question)
. (Grades 3-4) Matt and Ni k onstru ted two buildings, shown in the gures, using identi al ubes. Matt's building weighs 200 g, and Ni k's building weighs 600 g. How many ubes from Ni k's building are hidden and annot be seen in the gure? (B) 2 (C) 3 (A) 1 (D) 4 (E) 5 11
.......... ....................... ...................................... .. .. .. ..................... ........................................................... ................................................ .. .. .. .. .. ... ................................................
............ ................................ . . .............................. ......................................... ............................
Ni k's building
Matt's building
12. (Grades 5-6) Consider all four-digit numbers divisible by 6 whose digits are in in reasing order, from left to right. What is the hundreds digit of the largest su h number? (A) 7 (B) 6 (C) 5 (D) 4 (E) 3
. (Grades 7-8) A square of side length 3 is divided by several segments into polygons as shown in the gure. What per ent of the area of the original square is the area of the shaded gure? (A) 30% (B) 33 13 % (C) 35% (E) 50% (D) 40% 13
.........1..........................2................. .. ...................................................................................................................... ..... 1 . ........................................................................................................ . 2 .. ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ .. ................................................................................................................................................................................................................................................................................................................................. ... ......................................................................................................................................................................................................................................................................................................................................................................................................... .. 2 . . .................................................................................................................................................................................................................... .... . 1 .... ........................................................ 2
1
14. (Grades 9-10) A boy always tells the truth on Thursdays and Fridays, always tells lies on Tuesdays, and tells either truth or lies on the rest of the days of the week. Every day he was asked what his name was and six times in a row he gave the following answers: John, Bob, John, Bob, Pit, Bob. What did he answer on the seventh day? (A) John (B) Bob (C) Pit (D) Kate (E) Not enough information to de ide
. (Grades 11-12) An equilateral triangle and a ir le M are ins ribed in a ir le K , as shown in the gure. What is the ratio of the area of K to the area of M ? (A) 8 : 1 (B) 10 : 1 (C) 12 : 1 (D) 14 : 1 (E) 16 : 1
15
.... .................. ......................... ...... K .......... ..... ... ........ .... .... ..... ............... .... . . . ........ ... ... ........ ... .. ... ........ ....M .... . ........ ........ .... ..................... ... ........ ... ...... ...... ......... .. .... ......... . . . ..... . ..... . . . . .. ......................... ........ .... ... . ........ .. ..... ... ....... .. ........ ... ... ........ . . .... .... . ... . . . ... .... ... ............. . . . . ....... ........ ........... ...... ................. ........................ ......
4 Con ours Math Kangaroo Feuille d'entra^ nement
Partie A (3 points par question)
. (Classes 3-4) Dans l'exemple d'addition idessus, haque lettre di erente represente un hi re di erent. Quel hi re la lettre K represente-t-elle ?
O K O
1
(A) 0
(B) 1
(C) 2
+ W
(D) 8
K O W
(E) 9
. (Classes 5-6) Dix henilles se promenaient a la le indienne dans un par . Chaque henille mesurait 8 m et, pour des raisons de se urit e, elles gardaient une distan e de 2 m entre ha une d'elles. Quelle etait la longueur totale de leur ortege ?
2
(A) 100 m
(B) 98 m
(C) 82 m
(D) 102 m
(E) 96 m
. (Classes 7-8) Une fourmi ourt le long ..................................sq...q.r................................................ ... d'une regle de 10 m de longueur, a la vitesse ... . 1 2 3 4 5 6 7 8 9 .
onstante de 1 m a la se onde (voir la gure). ....................................................................................... Chaque fois qu'elle atteint une extremit e, elle
ourt dans la dire tion opposee et elle met exa tement 1 se onde pour hanger de dire tion. La fourmi part de l'extremit e gau he de la regle. Pres de quel hi re sera-telle apres 2009 se ondes ? 3
..... .
(A) 1 m 4
(B) 2 m
(C) 3 m
..... .
(D) 4 m
..... .
... .. ...... .............. .... ..... . .
..... .
..... .
..... .
..... .
(E) 5 m
. (Classes 9-10) Lequel des nombres 26 , 35 , 44 , 53 , 62 est-il le plus grand ? (A) 26
(B) 35
(C) 44
(D) 53
(E) 62
. (Classes 11-12) Un de orateur a prepar e un melange de peinture ou les volumes des ouleurs rouge et jaune etaient dans un rapport de 2 : 3. Trouvant le melange trop lair, il ajouta 2 L de peinture rouge. Le rapport des volumes des ouleurs rouge et jaune devint alors de 3 : 2. Combien de litres de peinture le de orateur a-t-il utilise ? 5
(A) 5 L
(B) 6 L
(C) 7 L
(D) 8 L
(E) 9 L
Partie B (4 points par question)
. (Classes 3-4) Deux gar ons jouent au tennis jusqu'a e que l'un d'eux gagne quatre fois. Un mat h de tennis ne peut nir en un pointage nul. Quel est le plus grand nombre de jeux qu'ils peuvent jouer ? 6
(A) 8
(B) 7
(C) 6
(D) 5
(E) 9
5 . (Classes 5-6) Dans deux ans, mon ls aura deux fois l'^age qu'il avait il y a deux ans. Dans trois ans, ma lle aura trois fois l'^age qu'elle avait il y a trois ans. Quelle reponse de rit-elle le mieux l'^age de la lle et du ls ? (A) Le ls est plus a^ ge ; (B) La lle est plus a^ gee ; (C) Ils sont des jumeaux ; (D) Le ls est deux fois plus a^ ge que la lle ; (E) La lle est deux fois plus a^ gee que le ls. 7
. (Classes 7-8) Sur une droite on marque des points de sorte que toutes les distan es de 1 m, 2 m, 3 m, 4 m, 5 m, 6 m, 7 m et 9 m gurent parmi les distan es entre es points. Combien y a-t-il au minimum de points marques sur ette droite ? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 8
9. (Classes 9-10) Liliane, Ni ole, Katia et Charlotte sont allees au inema. Comme il n'etait pas possible d'a heter quatre pla es ensemble, elles ont et d'autres a hete des billets pour les sieges numero 7 et 8 dans la 10e -rangee pour les sieges numero 3 et 4 dans la 12e -rangee. De ombien de manieres peuvent-elles hoisir de s'asseoir, si Charlotte ne veut pas e^ tre assise a ot ^ e de Ni ole ? (B) 20 (C) 16 (D) 12 (E) 8 (A) 24
. (Classes 11-12) Soit ABC un triangle iso ele ave BC = AC . Les segments DE , F G, HI , KL, M N , OP et XY divisent les ot ^ es AC et CB en parties egales. Trouver XY si AB = 40 m. (B) 35 m (A) 38 m (C) 33 m (D) 30 m (E) 27 m 10
C .. .... ....
D ..................................... E
F ......................................................... G H .............................................................................. I K .................................................................................................. L M ...................................................................................................................... N O ............................................................................................................................................ P
X ................................................................................................................................................................. Y A .......................................................................................................................................................................... B
Partie C (5 points par question)
. (Classes 3-4) En utilisant des ubes identiques, Mathieu et Ni olas ont onstruit deux b^atiments, omme illustres dans les gures. Le b^atiment de Mathieu pese 200 g et elui de Ni olas 600 g. Combien de ubes du b^atiment de Ni olas sont-ils a hes et ne peuvent e^ tre vus dans la gure ? (B) 2 (C) 3 (A) 1 (D) 4 (E) 5 11
.......... ....................... ...................................... .. .. .. ..................... ........................................................... ................................................ .. .. .. .. .. ... ................................................
b^atiment de Ni olas
............ ................................ . .............................. ......................................... ............................
b^atiment de Mathieu
. (Classes 5-6) On onsidere tous les nombres de quatre hi res, divisibles par 6 et dont les hi res, lus de gau he a droite, vont en ordre roissant. Quel est le hi re des entaines dans le plus grand de es nombres ? (A) 7 (B) 6 (C) 5 (D) 4 (E) 3 12
6 . (Classes 7-8) On divise un arre de ot ^ e 3 en polygones ave plusieurs segments omme indique dans la gure. Quel est le per entage de l'aire de la gure ombree par rapport a elle du arre ? (A) 30% (B) 33 13 % (C) 35% (E) 50% (D) 40% 13
.........1..........................2................ .. ..................................................................................................... ..... 1 . ........................................................................................................................ . 2 .. .............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. .. .............................................................................................................................................................................................................................................................. .. ................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .... 2 ................................................................................................................................. ... 1 ..... ...... ................................................ 2
1
. (Classes 9-10) Un gar on dit toujours la verit e les jeudis et vendredis, ment toujours les mardis et, les autres jours de la semaine, soit il dit la verit e soit il ment. On lui demanda son nom haque jour de la semaine et les six premieres fois, il donna les reponses suivantes : Jean, Bernard, Jean, Bernard, Paul, Bernard. Quelle fut sa reponse le septieme jour ? (A) Jean (B) Bernard (C) Paul (D) Lu (E) Pas possible de de ider 14
. (Classes 11-12) On ins rit un triangle equilat eral et un er le M dans un er le K , omme indique dans la gure. Quel est le rapport de l'aire de K a elle de M? (B) 10 : 1 (C) 12 : 1 (A) 8 : 1 (D) 14 : 1 (E) 16 : 1 15
..................................... .......... ........ ................. .....K ..... ... ............... .... ........ .... ..... ... . . ........ ... ... .. . . . . . ........ .. .... ........ .... .M ........ .... ............................. ........ .. .......... ... ... ... ... . . . . ........ ....... .... ........ .... ......................... . ........ ... .. .. ....... . . . . . ... ... . . . . . . ... .. ........ .... ..... ... ....... . . . . . . . . . . .... ... ......... ............... ..... ......... ...... ...........................................
Next we shall give solutions to the Mathemati s Asso iation of Quebe Contest (Se ondary level) February 9, 2006 [2008 : 67-68℄. We apologize to any readers who sent in solutions to this ontest but whose solutions we have lost. 1. A parti ular magi square. It is well known that a magi square is obtained by putting numbers in a square su h that the sum of ea h row, olumn, and diagonal is the same, as for example, 8 1 3 5 4 9
6 7 2
Imagine now that we de ide to invent a new form of su h squares by repla ing the sum by a produ t. We ask you to nd su h a square by repla ing the asterisks, ∗, by natural numbers, not ne essarily distin t or onse utive, in the following square: ∗ 1 4 ∗ ∗ ∗
∗ ∗ 2
7 Solution by the editor. Suppose that the square A below is a magi square. Then the square B is a magi square for produ ts. For example, by the Law of Exponents, the produ t along the rst row of B is xa xb xc = xa+b+c and the produ t along the rst olumn of B is xa xd xg = xa+d+g and these are the same be ause a + b + c = a + d + g . The same is true for the other rows, olumns, and diagonals of B . A =
a b d e g h
c f i
,
xa xd xg
B =
xb xe xh
xc xf xi
.
Now, if we subtra t 1 from every entry of the rst square given in the question and if we take x = 2, then the square B below is a solution to the problem. A =
7 0 2 4 3 8
5 6 1
,
B =
27 22 23
20 24 28
25 26 21
=
128 4 8
1 16 256
32 64 2
.
2. Clovis' outing. Clovis likes to take an outing in the natural numbers. Ea h day, he starts with a natural number of his hoi e, the biggest possible. Then, during his day, he passes from number to number using the following rules. Suppose that the sequen e of numbers is urrently at n.
(1) If n is divisible by 3 without remainder, then the next number is n/3. (2) If the remainder after dividing 2n + 1.
n
by
3
is 1, then the next number is
(3) If the remainder after dividing 2n − 1.
n
by
3
is 2, then the next number is
(4) If n = 1, then the sequen e stops. Over the years that he has played this game, he noti ed that, whatever the starting number, the sequen e always ended up with the number 1. However, he wonders if there is a sequen e that in reases inde nitely, with larger and larger numbers on average, or su h that it ends up in a loop of numbers that does not ontain 1. Determine if su h a sequen e is possible and give an example, or show that su h a sequen e does not exist by showing that all sequen es using the above rules inevitably end up at the number 1. Here is an example of su h a sequen e: Starting with 55, we get 111, 37, 75, 25, 51, 17, 33, 11, 21, 7, 15, 5, 9, 3 and 1, whi h ends the sequen e. Solution by the editor. Note that Clovis' sequen e starting with 55 has a de reasing subsequen e that goes to 1, given by the underlined numbers: 55, 111, 37, 75, 25,
8 51, 17, 33, 11, 21, 7, 15, 5, 9, 3, 1.
We will show that for any number a > 1 in one of Clovis' sequen es, there is always a number b oming after a in the sequen e su h that a > b. Thus, if Clovis starts with n > 1, then there will be a subsequen e n, m, p, . . . with n > m > p > · · · and this subsequen e must eventually hit the number 1 (be ause all of the terms in it are positive, it annot de rease forever). If a > 1 and a = 3k, k > 1, then by rule (1) the number b = k omes right after a and a > b. If a > 1 and a = 3k + 1, k > 0, then by rule (2) the number 2a + 1 = 2(3k+1)+1 = 6k+3 omes right after the number a, and then by rule (1) the number (6k + 3)/3 = 2k + 1 omes after 2a + 1. Sin e a = 3k + 1 > 2k + 1, we see that the number b = 2k + 1 omes after a and a > b. If a > 1 and a = 3k + 2, k ≥ 0, then by rule (3) the number 2a − 1 = 2(3k+2)+1 = 6k+3 omes right after the number a, and then by rule (1) the number (6k + 3)/3 = 2k + 1 omes after 2a − 1. Sin e a = 3k + 2 > 2k + 1, we see that the number b = 2k + 1 omes after a and a > b. Thus, in all ases where a > 1, there is a number b oming after a in the sequen e su h that a > b, and we are done. 3. Eight balls in two urns. We give you two similar urns, four white balls, and four bla k balls. You must separate the balls amongst the two urns (not ne essarily the same number in ea h urn), after whi h both urns will be made indistinguishable. How should the balls be distributed to maximize the han es that, if you draw a ball randomly from a randomly hosen urn, you will obtain a white ball?
Solution by the editor. Put 1 white ball in one urn and all the other balls in the other urn. The probability of hoosing the urn with 1 white ball and then drawing that white ball from it is 12 ·1 = 12 and the probability of hoosing the other urn and then 3 3 = 14 drawing a white ball from it is 12 · 3+4 . Thus, with this distribution, the overall probability of ultimately obtaining a white ball is p = 12 ·1+ 12 · 37 = 57 . Now let p1 and p2 be the probabilities of drawing white balls from the two urns (p1 = 1 and p2 = 37 above). The overall probability of ultimately p2 obtaining a white ball is then p = 12 p1 + 12 p2 = p1 + , whi h is the average 2 5 of the probabilities p1 and p2 . Therefore, p > 7 implies that p1 > 57 or p2 > 57 . To make an urn with p1 > 57 (say) we must have (w, b) = (1, 0), (2, 0), (3, 0), (3, 1), (4, 0), or (4, 1), where the urn ontains w white balls and b bla k balls. These are the only distributions that ould yield p > 75 . In the ase of (w, b) = (2, 0), (3, 0), or (4, 0) we have p < 57 , as moving all but one white ball to the other urn in reases p2 but leaves p1 = 1. Finally, (w, b) = (3, 1) or (4, 1) yields p = 12 or p = 25 , ea h less than 57 . Our rst distribution maximizes our han e of obtaining a white ball.
9 ........................................ 4. The three atta hed barrels. Three big ylindri al ........ ..... .... ...... ... .... . . ... .. barrels, lying parallel to the earth, are atta hed by .. .. ....................... . . . . . . . . .. . . ... . . ........ ... ..... .... .... .......... .... .. a steel able at their onta t points, A and B , su h .................................. ........ ... ........................... .... ..... ... .... .... ... that they stay xed in pla e. Knowing that the two ........... .............. .. .... . ............................ .... .... ............................... . . . smaller ones ea h have a radius of 4 metres and the ............ ........q....A . . . . B ....... ........................................................................q . .. biggest one has a radius of 9 metres, what is the ............ .............................. ................................. ............ ..................... .............................. ..................... length of the steel able? Solution by the editor. Join the entres of the smaller barrels ........................................ .......... ....... ....... ..... ..... .... and drop a perpendi ular to this segment from .... .... . . ... ... . ... . . . .. . the entre of the larger barrel, as in the dia.. ... .. ... ... C . .. . ... . . gram at right. Sin e the barrels rest on the . . . . . . . . . . . . . . . . ... ...... .... ............... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... B........................................ ........A .......... ... ...... . . . . . . . . . . . . . . . . . . . earth, the length of CZ is the di eren e of ...... ...........................q............................................................................................................................................q.................. ......... ................................................................................... ... .. their radii, that is, |CZ| = 9 − 4 = 5. Also ........... X ..............................................................................................Z ... ... ..... ..... Y ......... ........ ...... ..... ..... . .................. ......................... . ............................... . . . . ..................... ....... the length of CY is the sum of the radii, that is, |CY | = 9 + 4 √ = 13. By the Pythagorean Theorem |ZY | = 132 − 52 = 12. Thus, |XY | = 24. Finally, sin e triangle 9 216 CAB is similar to triangle CXY , we have |AB| = |XY | = . 13 13
5. The magi words. An illusionist is sear hing for magi words to a
ompany his many magi tri ks. R He de ides to onstru t his magi words starting D......................... .. ............ ......... .... ......... ... ........ ... ......... . . . . . . . . . . ......... ... ... with the diagram on the right. He takes a path ..... . . . . . . ......... . ... ... . ......... .............. ... ... ............ . ... ... . . . . . . . ......... through the diagram and jots downs the letters he ... ... ...... . . . . . . . . . . . ... ... ......... ..... . . . . . . . ... ... . . . . . . ......... A ... .............. . . . . ......... ..... nds on it. Ea h magi word must have exa tly 11 ... .......... . . . . . . .... ... letters and must start and end with the letter A. C B Two onse utive letters must never be identi al. How many magi words are there? Note: Here are two possible magi words: ABRACADABRA and ARADCABARBA.
Solution by the editor. Let Mk be the number of k-letter words that start with A and end with A and that an be formed by travelling through the bowtie. Let Nk be the number of k-letter words starting with A but not ending with the letter A that an be similarly formed. If k ≥ 2, then we see that Mk = Nk−1 , be ause removing the letter A from a word ending with A leaves a word not ending in A (but still starting with A) and the pro ess an be reversed. Similarly, by deleting the last letter of a word of length k that starts with A but does not end in A, we see that Nk = Nk−1 + 4Mk−1 , be ause any of the four letters di erent from A an be added to a word not ending in A or else there is only way to extend a (k − 1)-letter word not ending in A to one that still does not end in A. Sin e Mk−1 = Nk−2 the last equation be omes Nk = Nk−1 + 4Nk−2 , where k ≥ 1.
10 We now have N1 = 0, N2 = 4, N3 = N2 + 4N1 = 4 + 4 · 0 = 4, and so forth. The results of al ulating the Ni are summarized in the table below: N1 0
N2 4
N3 4
N4 20
N5 36
N6 116
N7 260
N8 724
N9 1764
N10 4660
Finally M11 = N10 , so there are 4660 magi words altogether. 6. All ten digits. Find the smallest positive natural number N su h that, in the de imal notation, N and 2N together use all ten digits: 0, 1, 2, . . . , 9.
Solution by the editor. We have 2(13485) = 26970, and we will prove that if N1 and 2N1 together use all ten digits and N1 ≤ N = 13485, then N1 = N . As N1 has ve digits and N1 ≤ N , then N1 = 1 . . . and 2N1 = 2 . . .. Digits 1 and 2 are now used and 2N1 uses 0 (otherwise N1 uses 0 and 2N1 then uses 0 or 1, a ontradi tion). Thus, N1 = 13 . . .. The smallest available digit for N1 is now a 4 and N1 ≤ N , hen e N1 = 134 . . . and 2N1 = 26 . . .. The number N1 uses 5, be ause 2N1 uses 0. If N1 = 1345x, then x is a digit greater than 5 and 2N1 = 2691y, a ontradi tion. Thus, N1 = 134x5 ≤ N . Finally, x 6= 7, hen e x = 8. Therefore, N1 = N and N is the smallest positive integer with the given property. [Ed.: Rolland Gaudet o ers the solution N = 6792 if initial zeroes are allowed, for then 2(6792) = 013584.℄ 7. The pizza toppings. At the Julio pizzeria, all the pizzas have heese and tomato sau e on them. The hoi e of toppings is limited to bla k olives, an hovies, and sausage. Of the 200 lients Julio had yesterday, 40 took an hovies, 80 took bla k olives, 120 took sausage, 60 took at the same time bla k olives and sausage, but none took at the same time an hovies and bla k olives or an hovies and sausage. How many lients took none of the three toppings?
Solution by the editor. Let t be the number of ustomers who took at least one topping. Any
ustomer who took an hovies took no other topping, so t = 40 + x where x is the number of ustomers who took bla k olives or sausage (or both). There were 60 ustomers who took both bla k olives and sausage, so 20 = 80 − 60 took just bla k olives and nothing else. Similarly, 60 = 120 − 60 ustomers took sausage and nothing else. Thus, t = 40+x = 40+(20+60+60) = 180, and the number of ustomers who took no toppings is 200 − t = 20.
11
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis, as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).
Mayhem Problems
Please send your solutions to the problems in this edition by 1 May 2009. Solutions re eived after this date will only be onsidered if there is time before publi ation of the solutions. Ea h problem is given in English and Fren h, the oÆ ial languages of Canada. In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8, Fren h will pre ede English. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. M376. Proposed by the Mayhem Sta .
Determine the value of x if
2 2 102009 + 25 − 102009 − 25 = 10x .
M377. Proposed by the Mayhem Sta . An arithmeti sequen e onsists of 9 positive integers. The sum of the terms in the sequen e is greater than 200 and less than 220. If the se ond term in the sequen e is 12, determine the sequen e. M378. Proposed by the Mayhem Sta . Points C and D are hosen on the semi- ir le with diameter AB so that C is loser to A. Segments CB and DA interse t at P ; segments AC and BD extended interse t at Q. Prove that QP extended is perpendi ular to AB . M379. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. The integers 27 + C , 555 + C , and 1371 + C are all perfe t squares, the square roots of whi h form an arithmeti sequen e. Determine all possible values of C .
12 M380. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Triangle ABC is right-angled at C and has BC = a and CA = b, with a ≥ b. Squares ABDE , BCF G, and CAHI are drawn externally to triangle ABC . The lines through F I and EH interse t at P , the lines through F I and DG interse t at Q, and the lines through DG and EH interse t at R. If triangle P QR is right-angled, determine the value of ab . M381. Proposed by Mihaly Ben ze, Brasov, Romania. Determine all solutions to the equation 1 1 1 1 + + + = x2 − 4x − 4 . x−1 x−2 x−6 x−7
................................................................. M376. Propose par l'Equipe de Mayhem. Determiner la valeur de x si
2 2 102009 + 25 − 102009 − 25 = 10x .
M377. Propose par l'Equipe de Mayhem. Determiner la suite arithmetique formee de 9 entiers positifs dont la somme se situe entre 200 et 220 et dont le se ond terme vaut 12. M378. Propose par l'Equipe de Mayhem. A partir du point A, on hoisit deux points C et D sur un demi- er le de diametre AB . Soit P l'interse tion des droites CB et DA, et Q elle des droites AC et BD. Montrer que la droite P Q est perpendi ulaire a AB . M379. Propose par John Grant M Loughlin, Universite du NouveauBrunswi k, Frederi ton, NB. Les entiers 27 + C , 555 + C et 1371 + C sont tous des arres parfaits dont les ra ines arrees forment une suite arithmetique. Trouver toutes les valeurs possibles de C . M380. Propose par Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL. Dans un triangle ABC d'angle droit en C , soit BC = a et CA = b, ave a ≥ b. Exterieurement au triangle ABC , on onstruit les arres ABDE , BCF G et CAHI . Soit respe tivement P , Q et R les interse tions des droites F I et EH , F I et DG, DG et EH . Determiner la valeur de ab pour que P QR soit un triangle re tangle.
13 M381. Propose par Mihaly Ben ze, Brasov, Roumanie. Determiner toutes les solutions de l'equation 1 x−1
+
1 x−2
+
1 x−6
+
1 x−7
= x2 − 4x − 4 .
Mayhem Solutions
M338. Proposed by the Mayhem Sta . Two students mis opy the quadrati equation x2 + bx + c = 0 that their tea her writes on the board. Jim opies b orre tly but mis opies c; his equation has roots 5 and 4. Vazz opies c orre tly, but mis opies b; his equation has roots 2 and 4. What are the roots of the original equation?
Solution by Taylor Thetford, student, Lakeview High S hool, San Angelo, TX, USA. The roots of the quadrati equation that Jim writes down are 5 and 4. His quadrati equation is thus (x − 5)(x − 4) = x2 − 9x + 20 = 0. Sin e Jim
opied b orre tly, we an on lude that in the original quadrati equation, b = −9. Similarly, sin e Vazz's roots are 2 and 4, his quadrati equation has the form (x − 2)(x − 4) = x2 − 6x + 8 = 0. Sin e Vazz opied c orre tly, then c = 8. Thus, the original equation was x2 − 9x + 8 = 0. Fa toring, we obtain (x − 1)(x − 8) = 0. Therefore, the roots of the original equation are 1 and 8. Also solved by CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; PETER CHIEN, student, Central Elgin Collegiate, St. Thomas, ON; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, IES Jakarta, Indonesia; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, \Abastos", Valen ia, Spain; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia International College, Hamilton, ON; and TITU ZVONARU, Comane sti, Romania.
M339. Proposed by the Mayhem Sta .
(a) Determine the number of integers between whi h ontain exa tly two equal digits.
100
and
199,
in lusive,
(b) An integer between 1 and 999 is hosen at random, with ea h integer being equally likely to be hosen. What is the probability that the integer has exa tly two equal digits?
14 Solutions by Taylor Thetford, student, Lakeview High S hool, San Angelo, TX, USA. 1xy .
(a) Ea h of the integers in the given range an be written in the form There are three ases to onsider.
Case 1. The rst and last digits are the same. Here, we are looking for integers 1x1 where the middle digit an take any value ex ept 1. This yields 9 possibilities. Case 2. The rst and se ond digits are the same. Here, we are looking for
integers 11y where the last digit an take any value ex ept 1. This again yields 9 possibilities.
Case 3. The se ond and last digits are the same. Here, we are looking for
integers 1xx where x is not 1. Again, there are 9 possibilities.
Adding the results from our three ases, we nd that there are 27 numbers between 100 and 199, in lusive, that ontain exa tly two equal digits. (b) We ount the number of integers in the range 1 to 999, in lusive, that have exa tly two equal digits. First, between 1 and 99, there are 9 of these, namely, 11, 22, . . . , 99. Next, between 100 and 199, we have ounted 27 in part (a). Using the same argument as in (a), we an show that there are 27 numbers between 200 and 299, in lusive, and for every other interval of one hundred numbers up to the range of 900 to 999. There are therefore 9 + 9 · 27 = 252 numbers between 1 and 999 whi h
ontain exa tly two equal digits. The probability that a randomly sele ted integer between 28 has exa tly two equal digits is thus 252 = . 999 111
1
and
999
Also solved by PETER CHIEN, student, Central Elgin Collegiate, St. Thomas, ON; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; and LUYAN ZHONG-QIAO, Columbia International College, Hamilton, ON. There were 3 in orre t solutions and 1 partial solution submitted.
M340. Proposed by the Mayhem Sta .
Let ABC be an isos eles triangle with AB = AC , and let M be the mid-point of BC . Let P be any point on BM . A perpendi ular is drawn to BC at P , meeting BA at K and CA extended at T . Prove that P K + P T is independent of the position of P (that is, the value of P K + P T is always the same, no matter where P is pla ed).
15 Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. T Sin e △ABC is isos eles with sides AB and AC of equal length, we have M A ⊥ BC . Also, sin e P T ⊥ BC , then M A||P T . A Sin e M A||P K , then △M BA is similar to △P BK sin e ea h is right-angled and they K share the angle at B . From this, we obtain P B · MA PK MA , hen e P K = M B . = PB MB Similarly, sin e M A||P T , then △CP T T MA is similar to △CM A, when e PP C = MC ...... ... .... .... ..... ... .... . ... ... ..... ... ... ... ... .... ... .......... .... ... ... ... ... ..... ...... . .... . ... ..... .... ...... ... .... .... ..... ... ... .... ... .... ... ..... .... ... ...... ... . . . . ... . .. ... . . . . ... .. ... .... ... . . . . ... .. .. .... . . ... . . ... .. ... .. . . . . ... .. ... ..... . ... . . ... . .. .. . . . . . ... .. .... .. ... . . . . ... ... .... .... . ... . . . .. ... ... . . . . .. . . .. . . . . . . ...............................................................................................................................................
MA and so P T = P CM· C . Sin e M B = M C =
PK + PT
B
1 BC 2
P M
C
, we an on lude that
=
(P B + P C) · M A P B · MA P C · MA + = MB MC MB
=
BC · M A = 2M A . MB
Thus, P K + P T is independent of the position of P , sin e it depends only on the length of M A. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The IES \Abastos", Philippines; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Comane sti, Romania. There was 1 in orre t solution submitted.
M341. Proposed by the Mayhem Sta . Let ABC be a right triangle with right angle at B . Sides BA and BC are in the ratio 3 : 2. Altitude BD divides CA into two parts that di er in length by 10. What is the length of CA?
Solution by Taylor Thetford, student, Lakeview High S hool, San Angelo, TX, USA. Let 2x and 3x be the lengths of B CB and AB , respe tively. Let y and y + 10 be the lengths of CD and DA, 3x respe tively. Let z be the length of 2x z BD . We wish to nd 2y + 10, whi h is the length of CA. By applying the Pythagorean A y Theorem in △ABC , we nd that y + 10 C D (2x)2 + (3x)2 = (2y + 10)2 and so 13x2 = 4y 2 + 40y + 100. .... .............. .... ... ............. ....... .... ... ....... .... ..... . . ....... . ....... .... .... ....... ... . . . . ....... .. .. . . ....... . . . .. .. ....... . . . . . ....... .. .. . . . ....... . . .. .. ....... . . . . . ....... .. . . . . . ....... . .. .. . ....... . . . . ....... .. .. . . . . . ....... .. ........ . . ....... . . . .. .. . ... . . . . . ......................................................................................................................................................................................................
16 Applying the Pythagorean Theorem to △BDC and △BDA, we nd that y2 + z 2 = 4x2 and z 2 + (y + 10)2 = 9x2 . Eliminating z in the last two equations gives 4x2 −y2 = 9x2 −(y+10)2 . Therefore, 5x2 = (y + 10)2 − y2 = 20y + 100 or x2 = 4y + 20, and so 13x2 = 52y + 260. Combining this result with 13x2 = 4y2 + 40y + 100, we nd that 52y + 260 4y − 12y − 160 y 2 − 3y − 40 (y + 5)(y − 8) 2
= = = =
4y 2 + 40y + 100 ; 0; 0; 0.
Sin e y > 0, then y = 8, and so CA = 2y + 10 = 26. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARD IES \Abastos", Valen ia, Spain; I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia International College, Hamilton, ON; and TITU ZVONARU, Comane sti, Romania. There was 1 in orre t solution submitted.
M342. Proposed by the Mayhem Sta . Quin y and Celine have to move 10 small boxes and 10 large boxes. The
hart below indi ates the time that ea h person takes to move ea h type of box. Celine Quin y small box 1 min. 3 min. large box 6 min. 5 min.
They start moving the boxes at 9:00 am. What is the earliest time at whi h they an be nished moving all of the boxes? Solution by Mayhem Sta . Let x represent the number of small boxes and y represent the number of large boxes that Celine moves. Sin e there are 10 small boxes and 10 large boxes, then Quin y moves 10 − x small boxes and 10 − y large boxes. Given the lengths of time that ea h takes, it takes Celine x+6y minutes and it takes Quin y 3(10 − x) + 5(10 − y) = 80 − 3x − 5y minutes. If x = 9 and y = 4, then Celine takes 33 minutes and Quin y takes 33 minutes. We show that it annot be done faster than this. If Quin y and Celine nish in fewer than 33 minutes, then ea h takes at most 32 minutes, so the total working time is at most 64 minutes, so x + 6y + (80 − 3x − 5y) = 80 − 2x + y ≤ 64 or 2x − y ≥ 16. Sin e x and y are nonnegative integers and ea h is less than 10, then the possible pairs (x, y) that satisfy this inequality are (8, 0), (9, 0), (9, 1), (9, 2), (10, 0), (10, 1), (10, 2), (10, 3), and (10, 4).
17 Sin e we want ea h of Celine's time and Quin y's time to be at most minutes, then we need x + 6y ≤ 32 and 80 − 3x − 5y ≤ 32. The rst inequality eliminates the pair (10, 4) from the list of possible pairs. The se ond inequality simpli es to 3x + 5y ≥ 48; none of the remaining pairs satisfy this inequality. Thus, none of these possibilities take any less time than 33 minutes. Therefore, the earliest possible nishing time is 9:33 a.m. 32
There were 4 in orre t and 3 in omplete solutions submitted. An expanded treatment of a similar problem appeared in the Problem of the Month
olumn in CRUX with MAYHEM, volume 34, number 2.
M343. Proposed by the Mayhem Sta . The Fibona
i numbers are de ned by f1 = f2 = 1 and, for n ≥ 2, by fn+1 = fn + fn−1 . The rst few Fibona
i numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . . . Find the sum of the rst 100 even Fibona
i numbers.
Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. Sin e f1 = f2 = 1, f3 = 2, and fn = fn−1 + fn−2 , then fm is even if and only if m is a multiple of 3. (This is be ause the parities of the terms will form the pattern Odd, Odd, Even, Odd, Odd, Even, and so on.) n P f3k , then If Sn = k=1
Sn =
n 1 X
2
(f3k +f3k ) =
k=1
n 1 X
2
(f3k−2 +f3k−1 )+f3k
k=1
=
3n 1 X
2
fk .
(1)
k=1
Next, we have f1 = f3 − f2 , and f2 = f4 − f3 , and also f3 = f5 − f4 , and so on until fr−1 = fr+1 − fr and fr = fr+2 − fr+1 . Sin e the right side of the sum of the n equations above \teles opes", it follows that r X fk = fr+2 − f2 = fr+2 − 1 . (2) k=1
From (1) and (2), we nd that Sn = 12 (f3n+2 − 1). In our parti ular
ase, S100 = 12 (f302 − 1). Maple omputes the value of S100 to be exa tly 290905784918002003245752779317049533129517076702883498623284700. For the re ord, by Binet's formula for Fibona
i numbers we have that √ √ 1 1 1 fm = √ (αm − β m ), where α = (1 + 5) and β = (1 + 5). Hen e 2 2 5
the required sum is also given by S100 =
1 1 √ α302 − β 302 − . 2 2 5
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; RICHARD IES \Abastos", Valen ia, Spain; I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, DIVYANSHU RANJAN, Delhi, India; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; and TITU ZVONARU, Comane sti, Romania.
18 Problem of the Month
Ian VanderBurgh
Approximation is one of the most important on epts in mathemati s. Problem (2006 Canadian Open Mathemati s Challenge) Determine, with jus-
ti ation, the fra tion pq , where p and q are positive integers and q
< 100,
that is losest to, but not equal to, 37 .
While it is tempting to get out your al ulator, it an initially only help so mu h. If you al ulate 37 , you'll obtain 0.428571 . . .. This doesn't help in any obvious way to answer the question. A rst approa h after the al ulator might be to go for the fra tion with the largest possible denominator. This makes a lot of sense in some ways, as the fra tions with the largest denominators will be losest together and so would seem to have the best han e of being losest to 37 . In our ase, the largest possible denominator is q = 99. The given fra tion, 37 , is between 42 43 = 0.424242 . . . and = 0.434343 . . .. After a qui k look, we an tell 99 99 3 42 di er by that 7 is loser to 99 . From the de imal approximations, 37 and 42 99 about 0.004. Is this the losest of all possible fra tions? Another idea is to try to onvert 37 into the equivalent fra tion with the largest possible denominator and then adjust from there. Multiplying numerator and denominator by 14, we obtain 42 . We ould then add 1 or 98 41 43 1 −1 to the numerator to obtain or 98 , whi h di er from 37 by 98 . But this 98 means that the di eren e is bigger than 0.01, whi h is worse than before, so this approa h doesn't give a loser fra tion. Can we do better than 42 ? It is possible that, even though fra tions 99 with smaller denominators are further apart, they an be between some of the other fra tions that we've looked at, for example between 42 and 37 or 99 between 43 and 37 . 99 to approximate 37 . Let's al ulate their di eren e, whi h is what we want to minimize: Solution We want to use the fra tion
p q
p − 3 = 7p − 3q = |7p − 3q| q 7 7q 7q
.
19 What an we do to make this as small as possible? Two approa hes would be to make the numerator of the di eren e as small as possible or to make the denominator of the di eren e as large as possible. Let's fo us initially on the numerator. The numerator annot equal 0 be ause the fra tions pq and 37 are not equal. Thus, the smallest possible value for the numerator is 1, be ause p and q are integers. So let's try to nd values of p and q for whi h the numerator equals 1. In this ase, the 1 di eren e equals 7q whi h is minimized when q is largest. For the numerator to equal 1, we need 7p − 3q = ±1. Sin e we also want to maximize q, we rewrite this as 7p = 3q ±1 and work from the largest possible integer values of q to see when we also get an integer value for p. If q = 99, the equation be omes 7p = 3(99) ± 1 = 297 ± 1. Neither possibility is a multiple of 7. If q = 98, the equation be omes 7p = 3(98) ± 1 = 294 ± 1. Neither possibility is a multiple of 7. If q = 97, the equation be omes 7p = 3(97) ± 1 = 291 ± 1. Neither possibility is a multiple of 7. If q = 96, the equation be omes 7p = 3(96) ± 1 = 288 ± 1. Sin e 287 is a multiple of 7, then taking q = 96 and p = 41 gives a di eren e with numerator 1. 3 1 1 So we have 41 − = = and this is the smallest possible 96 7 7 · 96 672 di eren e with the numerator equal to 1. If the numerator equalled 2 or something larger, then the smallest possible di eren e o
urs when the numerator is as small as possible and the 2 denominator is as large as possible, so is 7 ·299 = 693 . This is the smallest possible di eren e with numerator at least 2. 1 Combining the ases, the smallest possible di eren e is indeed 672 , 3 and so the losest fra tion to 7 of all of the fra tions under onsideration is p 41 = . q 96 The approximation of fun tions with polynomials is often seen in rstyear university al ulus ourses. As part of these investigations, we learn how to estimate the amount of error when approximating, for example, sin x 1 5 with x − 16 x3 + 120 x . The te hniques used to estimate this type of error are not dissimilar to what we have seen above, and are very useful in many types of al ulations.
20
THE OLYMPIAD CORNER No. 275 R.E. Woodrow
The year has own by, and it has brought many hanges to Crux and to the Corner. Of ourse it has been overshadowed by the sudden and untimely loss of a great friend and a devoted olleague, Jim Totten, mid-way through the transition to a new Editor-in-Chief. I think Vazz Linek has done a wonderful job of stepping in and keeping the journal on tra k with only an understandable slowing of the produ tion pa e in the interim. Readers will have noti ed the announ ement at the end of the De ember Corner that Joanne Canape, who has transformed my s ribbles into a high quality tex le for many years, has de ided that two de ades is enough. As I ustomarily begin the year by thanking all those who ontributed to the Corner in the last year, I would be very remiss not to lead o with sin ere thanks to Joanne. It is also appropriate to thank those who submitted problem sets for our use as well as a spe ial thanks to the dedi ated readers who furnish their ni e solutions whi h we use. Hoping, as always, that I've not missed someone, here is the list for the 2008 members of the Corner. Robert Morewood Arkady Alt Andrea Munaro Miguel Amengual Covas Vedula N. Murty Jean-Claude Andrieux Felix Re io Houda Anoun Xavier Ros Ri ardo Barroso Campos D.J. Smeenk Mi hel Bataille Babis Stergiou Jose Luis Daz-Barrero Daniel Tsai J. Chris Fisher Panos E. Tsaoussoglou Kipp Johnson Geo rey A. Kandall George Tsapakidis Jan Verster Ioannis Katsikis Edward T.H. Wang R. Laumen Salem Maliki Luyan Zhong-Qiao Li Zhou Pavlos Maragoudakis Titu Zvonaru Our apologies to Svetoslav Sav hev for the misspelling of his name in the De ember 2008 Olympiad. For your problem solving pleasure in the new year we start o with the problems of the German Mathemati al Olympiad, Final Round, 2006. My thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia 2006, for olle ting them for our use.
21 German Mathemati al Olympiad Final Round, Grades 12{13 Muni h, April 29 { May 2, 2006
First Day
1. Determine all positive integers n for whi h the number zn = 101 · · 101} | ·{z 2n+1 digits
is a prime.
2. Five points are on the surfa e of a sphere of radius 1. Let amin denote the smallest distan e (measured along a straight line in spa e) between any two of these points. What is the maximum value for amin , taken over all arrangements of the ve points? 3. Find all positive integers n for whi h the numbers 1, 2, 3, . . . , 2n an be
oloured with n olours in su h a way that every olour appears twi e and every number 1, 2, 3, . . . , n appears exa tly on e as the di eren e of two numbers with the same olor. Se ond Day
4. Let D be a point inside the triangle ABC su h that AC − AD ≥ 1 and BC − BD ≥ 1. Prove that EC − ED ≥ 1 for any point E on the side AB . 5. Let x be a nonzero real number satisfying the equation ax2 + bx + c = 0. Furthermore, let a, b, and c be integers satisfying |a| + |b| + |c| > 1. Prove that 1 |x| ≥ . |a| + |b| + |c| − 1
6. Let a ir le through B and C of a triangle ABC interse t AB and AC in Y and Z , respe tively. Let P be the interse tion of BZ and CY , and let X be the interse tion of AP and BC . Let M be the point that is distin t from X and on the interse tion of the ir um ir le of the triangle XY Z with BC . Prove that M is the midpoint of BC
Our se ond problem set for this number is a set of sele ted problems from the Thai Mathemati al Olympiad Examinations 2005. Again, thanks go to Robert Morewood, team leader to the 47th IMO in Slovenia 2006, for
olle ting them for the Corner.
22 Thai Mathemati al Olympiad Examinations 2005 Sele ted Problems
1. Let P (x), Q(x), and R(x) be polynomials satisfying 2xP x3
+ Q −x − x2
=
Show that x − 1 is a fa tor of P (x) − Q(x).
1 + x + x2 R(x) .
2. Find all fun tions f : R → R su h that f x + y + f (xy) = f f (x + y) + xy
for all x, y ∈ R. 3. Let a, b, and c be positive real numbers. Prove that 1 +
3 6 ≥ ab + bc + ca a+b+c
.
4. Let n be a positive integer. Prove that n(n + 1)(n + 2) is not a perfe t square. 5. Find the least positive integer n su h that 2549| n2545 − 2541 . 6. Do there exist positive integers x, y, and z su h that
2548x + (−2005)y = (−543)z ?
7. Show that there exist positive integers m and n su h that gcd(m, n) = 1 and 2549| ((25 · 49)m + 25n − 2 · 49n ). 8. The median AM of a triangle ABC interse ts its in ir le ω at K and L. The lines through K and L parallel to BC interse t ω again at X and Y , respe tively. The lines AX and AY interse t BC at P and Q. Prove that BC = CQ. (Shortlist 2005) 9. Let ABC be an a ute-angled triangle with AB 6= AC , let H be its ortho entre and M the midpoint of BC . Points D on AB and E on AC are su h that AE = AD and D, H , and E are ollinear. Prove that HM is orthogonal to the ommon hord of the ir um ir les of triangles ABC and ADE . (Shortlist 2005) 10. Assume ABC is an isos eles triangle with AB = AC . Suppose that P is a point on the extension of side BC . X and Y are points on lines AB and AC su h that P XkAC and P Y kAB . Let T be the midpoint of ar BC . Prove that P T ⊥ XY . (Iran 2004)
23 As a third set of problems we give the 46th Ukrainian Mathemati al Olympiad Final Round 2006 - 11th form problems. Again, thanks go to Robert Morewood, team leader to the 47th IMO in Slovenia 2006, for olle ting them for our use. 46th Ukrainian Mathemati al Olympiad 2006 Final Round
th
11
Form
1. (V.V. Plakhotnyk) Prove that for any rational numbers a and b the graph of the fun tion f (x) = x3 − 6abx − 2a3 − 4b3 , x ∈ R has exa tly one point in ommon with the x-axis. 2. (O.A. Sarana) A ir le is divided into 2006 equal ar s by 2006 points. Baron Mun hausen laims that he an onstru t a losed polygonal urve with the set of verti es onsisting of these 2006 points su h that amongst its 2006 edges there are no two whi h are parallel to ea h other. Is his laim true or false? 3. (T.M. Mitelman) (a) Prove that for any rational number α ∈ (0, 1) exists an in nite there set of real numbers that satisfy the equation x x{x} = α and any two of them have the same fra tional part. (The fra tional part of a real number a is given by {a} = a − ⌊a⌋, where ⌊a⌋ is its integer part, that is, the greatest integer that does not ex eed a.) (b) Prove that for any rational number α ∈ (0, 1) exists an in nite there set of real numbers that satisfy the equation x x{x} = α and any two of them have di erent fra tional parts. 4. (V.A. Yasinskiy) Two ir les ω1 and ω2 interse t ea h other at two distin t points A and B . The tangent line of the ir le ω1 at the point A and the tangent line of the ir le ω2 at the point B meet at point C . The rst of these two lines interse ts the ir le ω2 for the se ond time at point T 6= A. The point X (distin t from A and B ) is on the ir le ω1 , and the line XA interse ts the ir le ω2 for the se ond time at point Y (distin t from A). The lines Y B and XC meet at point Z . Prove that T Z is parallel to XY . 5. (O.O. Kur henko) Prove that for any real numbers x and y | cos x| + | cos y| + | cos(x + y)| ≥ 1 . 6. (T.M. Mitelman) Find all fun tions f : R → R su h that f x3 + y 3
for all real numbers x and y.
= x2 f (x) + yf (y 2 )
24 7. (V.A. Yasinskiy) A point M lies inside a ube ABCDA1 B1 C1 D1 . Points A′ , B ′ , C ′ , D ′ , A′1 , B1′ , C1′ , and D1′ belong to the rays M A, M B , M C , M D , M A1 , M B1 , BC1 , and M D1 respe tively. Prove that if the polyhedron A′ B ′ C ′ D′ A′1 B1′ C1′ D1′ is a parallelepiped (that is, all of its fa es are parallelograms), then it is a ube. 8. (V.A. Yasinskiy) There are n ≥ 3 soldiers in aptain Petrenko's squad, no two of the same height. The aptain orders them to stand single- le (not ne essarily sorted by height). A \wave" is any subsequen e of (not ne essarily next to ea h other) soldiers in this line su h that the rst (leftmost) soldier in the wave is taller than the se ond soldier in it, but the se ond soldier in it is shorter than the third one, who is in turn taller than the fourth one, and so on. (For example, if n = 9, the soldiers are enumerated by height, and the aptain lines them up as 9, 3, 5, 7, 1, 2, 6, 4, 8 then a longest wave for this line-up is 9, 3, 7, 1, 6, 4, 8. However, if the aptain lines them up as 1, 2, 3, 4, 5, 6, 7, 8, 9, then a longest wave onsists of (any) one soldier.) For every n, onsider the number of possible lines with the longest waves of even lengths and the number of possible lines with the longest waves of odd lengths. Whi h of these numbers is bigger?
Continuing with problems for readers to solve we give the Cze h-PolishSlovak Mathemati s Competition written on June 26-28, 2006 at Z ilina, Slovakia. Thanks again go to Robert Morewood, Canadian team leader to the 47th IMO in Slovenia 2006, for olle ting them for our use. Cze h-Polish-Slovak Mathemati s Competition 2006 1. Five distin t points A, B , C , D, and E lie in this order on a ir le of radius r and satisfy AC = BD = CE = r. Prove that the ortho entres of the triangles ACD, BCD, and BCE are the verti es of a right-angled triangle. 2. There are n hildren sitting at a round table. Erika is the oldest among them and she has n andies. No other hild has any andy. Erika distributes the andies as follows. In every round, all the hildren with at least two
andies show their hands. Erika hooses one of them and he/she gives one
andy to ea h of the hildren sitting next to him/her. (So in the rst round Erika must hoose herself to begin the distribution.) For whi h n ≥ 3 is it possible to redistribute the andies so that ea h hild has exa tly one andy? 3. The sum of four real numbers is 9 and the sum of their squares is 21. Prove that these four numbers an be labelled as a, b, c, and d so that the inequality ab − cd ≥ 2 holds.
25 4. Prove that for every positive integer k there is a positive integer n su h that the de imal representation of 2n has a blo k of exa tly k onse utive zeros, that is, 2n = · · · a00 · · · 0b · · · , where a and b are nonzero digits with k zeros between them. 5. Find the number of integer sequen es (an )∞ n=1 su h that an 6= −1 and an+2 =
an + 2006 an+1 + 1
for every positive integer n. 6. Is there a onvex pentagon A1 A2 . . . A5 su h that for ea h i the lines and Ai+1 Ai+2 interse t in Bi and the points B1 , B2 , . . . , B5 are
ollinear? (By onvention A6 = A1 , A7 = A2 , and A8 = A3 .)
Ai Ai+3
Our nal problem set for this issue is the XXI Olimpiadi Italiano della Matemati a, Cesenati o, written 5 May 2006. Thanks again go to Robert Morewood, Canadian team leader to the 47th IMO in Slovenia, for olle ting them for our use. XXI Olimpiadi Italiano della Matemati a Cesenati o May 5, 2006
1. Rose and Savino play a game with a de k of traditional Neapolitan playing
ards whi h onsists of 40 ards of four di erent suits, numbered 1 to 10. At the start ea h player has 20 ards. Taking turns, one shows a ard on the table. Whenever some ards on the table add to exa tly 15, these are then removed from the game (if the sum 15 an be obtained in more than one way, the player who last moved de ides whi h ards adding to 15 to remove). At the end of the game only one ard, a 9, is left on the table. Savino holds two
ards numbered 3 and 5, and Rose holds one ard. What is the number of Rose's ard? 2. Find all values of m, n, and p su h that pn + 144 = m2 ,
where m and n are positive integers and p is a prime number. 3. Let A and B be two points on a ir le Γ su h that AB is not a diameter. Let P be a point on Γ di erent from A and B , and let H be the ortho entre of the triangle ABP . Find the lo us of H as P varies over all points of Γ di erent from A and B .
26 4. On an in nite hessboard all the positive integers are written in as ending order along a spiral, starting from 1 and pro eeding anti lo kwise; a portion of the hessboard is shown in the gure. A "right half-line" of the hessboard is the set of squares given by a square C and by all squares in the same row as C and to the right of C .
....................................................................................................................................................................................... ... 17 ...... 16 ...... 15 ...... 14 ...... 13 ... ...................................................................................................................................................................... ...... ..... ..... ..... .... .... ..... 18 ..... 5 .... 4 .... 3 ..... 12 .... ................................................................................................................................ ...... ...... ..... ...... ..... ..... ..........19 ... 6 ... 1 .. 2 ... 11 . ... .............................................................................................................................................. ...... 20 ....... 7 ....... 8 ....... 9 ....... 10 ...... ................................................................................................................................................................. ...... 21 ........ 22 ........ 23 ........ 24 ........ 25 ........ ...................................................................................................................................................................................
(a) Prove that there exists a right half-line none of whose squares ontains a multiple of 3. (b) Determine if there exist in nitely many pairwise disjoint right half-lines none of whose squares ontains a multiple of 3. 5. Consider the inequality (x1 + · · · + xn )2 ≥ 4(x1 x2 + x2 x3 + · · · + xn x1 ) .
(a) Determine for whi h n ≥ 3 the inequality holds true for all possible
hoi es of positive real numbers x1 , x2 , . . . , xn . (b) Determine for whi h n ≥ 3 the inequality holds true for all possible
hoi es of any real numbers x1 , x2 , . . . , xn . 6. Albert and Barbara play a game. At the start there are some piles of oins on the table, not all ne essarily with the same number of oins. The players move in turn and Albert starts. At ea h turn a player may either take a oin from a pile or divide a pile into two piles with ea h pile ontaining at least one oin (a player may exer ise only one of these options). The one who takes the last oin wins the game. In terms of the number of piles and the number of oins in ea h pile at the start, determine whi h of the players has a winning strategy.
Now we turn to our le of solutions from the readers to problems from the Mar h 2008 number of the Corner and the Estonian IMO Sele tion Contest 2004-2005, given at [2008 : 79-80℄. 3. Find all pairs (x, y) of positive integers satisfying (x + y)x = xy .
Solution by Konstantine Zelator, University of Toledo, Toledo, OH, USA. We show that there are exa tly two su h pairs, (x, y) = (2, 6), (3, 6). We will make use of two basi fa ts from elementary number theory.
27 (a) If a, b ∈ Z and gcd(a, b) = 1 , then gcd(am , bn ) = 1 for any positive integers m and n. (b) If a|b, a is a positive integer, b ∈ Z, and gcd(a, b) = 1, then a = 1.
x and y be positive integers that satisfy (x + y)x = xy . Write d = gcd(x, y) and let x = dx1 , y = dy1 , where x1 and y1 are positive integers that are relatively prime. Substituting for x and y yields
Let
dx1 d · (x1 + y1 )x1 d = x1y1 d · dy1 d .
(1)
dd(x1 −y1 ) · (x1 + y1 )x1 d = xy11 d .
(2)
We make ases by omparing the sizes of x1 and y1 . Case 1. Suppose that x1 = y1 . Sin e gcd(x1 , y1 ) = 1, we have x1 = y1 = 1. Thus, equation (1) be omes dd · 2d = dd , whi h is impossible sin e d ≥ 1. Case 2. Suppose that y1 < x1 . Then x1 − y1 is a positive integer and from equation (1) we obtain Sin e gcd(x1 , y1 ) = 1 it follows that gcd(x1 + y1 , x1 ) = 1. By (a) above, we have gcd (x1 + y1 )x d , xy1 d = 1. However, by equation (2), the positive integer (x1 +y1 )x d is a divisor of xy1 d . Sin e these two integers are relatively prime, it follows by (b) that (x1 + y1 )x d = 1, whi h is impossible sin e x1 + y1 ≥ 2 and x1 · d ≥ 1. Case 3. Suppose that x1 < y1 . From (1) we obtain 1
1
1
1
1
(x1 + y1 )x1 d = x1y1 d · dd(y1 −x1 ) . (3) Sin e gcd(x1 , y1 ) = 1 we have gcd (x1 + y1 )x1 d , xy11 d = 1 and from equation (3) we see that xy11 d is a divisor of (x1 + y1 )x1 d , whi h implies that xy11 d = 1. Sin e y1 d is a positive integer this means that x1 = 1. Going ba k to equation (3) we see that (1 + y1 )d = dd(y1 −1) , hen e 1 + y1 = dy1 −1 .
(4)
Note that d 6= 1; otherwise equation (4) be omes 1 + y1 = 1, ontrary to the fa t that y1 is a positive integer. Thus, d ≥ 2. Sin e y1 = 1 does not satisfy equation (4), we also have y1 ≥ 2. Setting k = y1 − 1 equation (4) then be omes k + 2 = dk . By Indu tion (or the Binomial Theorem) we obtain 2k > k + 2 for all integers k ≥ 3. Sin e dk ≥ 2k , it follows from k + 2 = dk that k = 1 or k = 2. For k = 2 we have 4 = d2 , hen e d = 2. From 2 = k = y1 − 1 we then have y1 = 3. Re all that x1 = 1. Going ba k, we have x = x1 d = 1 · 2 = 2 and y = y1 d = 3 · 2 = 6. This is the solution (x, y) = (2, 6). Similarly, for k = 1 we have d = 3. Then y1 = k + 1 = 2 and sin e x1 = 1 we obtain x = dx1 = 3 · 1 = 3 and y = dy1 = 3 · 2 = 6. This is the other solution (x, y) = (3, 6).
28 4. Find all pairs (a, b) of real numbers su h that all roots of the polynomials 6x2 − 24x − 4a and x3 + ax2 + bx − 8 are non-negative real numbers.
Solution by Titu Zvonaru, Comane sti, Romania. Let β1 , β2 , and β3 be the roots of the polynomial x3 + ax2 + bx − 8, so that x3 + ax2 + bx − 8 = (x − β1 )(x − β2 )(x − β3 ). Comparing oeÆ ients yields β1 + β2 + β3 = −a and β1 β2 β3 = 8. Sin e β1 , β2 , and β3 are nonnegative real numbers, by the AM-GM Inequality we have β1 + β2 + β3 ≥ 3
p 3
β1 β2 β3 ,
hen e −a ≥ 6 or a ≤ −6. The equation 6x2 − 24x − 4a = 0 has real roots if and only if 242 − 4 · (−4a) · 6 ≥ 0, whi h implies 242 + 24 · 4a ≥ 0 and hen e a ≥ −6. Therefore, a = −6. √ Now we have β1 + β2 + β3 = 6 = 3 β1 β2 β3 , from whi h it follows that β1 = β2 = β3 = 2 and b = 12. Thus, the only pair satisfying the
ondition is (a, b) = (−6, 12). 3
Next we turn to a solution to a problem of the Trentieme Olympiad Mathematique Belge Maxi Finale, Mer redi 20 avril 2005 given at [2008 : 80℄. 3. Dans le triangle ABC , les droites AE et CD sont les bisse tri es interieures des angles ∠BAC et ∠ACB respe tivement ; E appartient a BC et D appartient a AB . Pour quelles amplitudes de l'angle ∠ABC a-t-on
ertainement (b) |AD| + |EC| > |AC|? (a) |AD| + |EC| = |AC|? ( ) |AD| + |EC| < |AC|?
Solution by Titu Zvonaru, Comane sti, Romania As usual let a = BC , b = CA, and c = AB . The rst equation below is the Angle Bise tor Theorem; the following equations are equivalent to it: BE EC BE EC BE + EC EC
EC
Similarly, AD =
= = = = bc . a+b
AB ; AC c ; b b+c b ab b+c
A
... ........ .. ... .... .. .. ... .. .... ....... . . .... .. .... .... .. ... .... .. .... ... .. .... .. .. . .... ... . .... ... ... .... . . ... . .... ... .. .... . ... .... . . .... ... ... .... .. . . .... ... . .... ........... ... .... . ......... . .. .......... .... . ... .......... ... .... . .......... ... .... . ........... . .... .......... ... .... . . . . . . . . . . .... ... ........... . . .... . . ... . ... .......... .... . . . . . . . . .......... ... .... . . . . . ... .......... .... .. . . . .... . . . . . .......... .. . .... .......... ... ... . .......... . ... . .......... ....... . .. .......... .... ... ... ........ ... ... ..............................................................................................................................................................................................................................
D
; .
B
E
C
29 We have |AD| + |EC| − |AC| = = b
bc a+b
+
ab b+c
− b
c(b + c) + a(a + b) − (a + b)(b + c) (a + b)(b + c)
.
By the Law of Cosines a2 + c2 − b2 = 2ac cos B , so the equation above an be rewritten as |AD| + |EC| − |AC| =
2abc cos(B) −
1 2
(a + b)(b + c)
.
Hen e, |AD| + |EC| = |AC| ⇐⇒ |AD| + |EC| > |AC| ⇐⇒ |AD| + |EC| < |AC| ⇐⇒
∠ABC = 60◦ , ∠ABC < 60◦ , ∠ABC > 60◦ .
Next we turn to solutions from our readers to problems of the 2005 Vietnam Mathemati al Olympiad given at [2008 : 81℄. 1. Find the smallest and largest values of√the expression √ P = x + y , where x and y are real numbers satisfying x − 3 x + 1 = 3 y + 2 − y .
Solved by Arkady Alt, San Jose, CA, USA; and Mi hel Bataille, Rouen, Fran e. We give the solution of Bataille. √ √ and Pmax = 9 + 3 15. We show that Pmin = 21 9 + 3 21 √ First, x = −1 and y = 12 11 + 3 21 satisfy the onstraint equation √ √ √ 2 √ √ x−3 x + 1 = 3 y + 2−y (easily he ked using 10+2 21 = 3+ 7 ) √ √ √ and P = 12 9+3 21√. Similarly, for x = 12 10+3 15 , y = 21 8+3 15 , we have P = 9 + 3 15 and the onstraint√is satis ed.√ Now, let x and y satisfy the onstraint equation. Then P = 3 x + 1 + 3 y + 2, so that p P 2 = 9 P + 3 + 2 (x + 1)(y + 2)
.
(1)
It follows that P ≥ 0 and P 2 − 9P − 27 ≥ 0. Thus, P is not less than 2 the positive solution √ of the quadrati x − 9x − 27 = 0 and we dedu e that 1 P ≥ 2 9 + 3 21 . From the AM-GM Inequality and (1), we obtain P 2 ≤ 9(P + 3 + x + 1 + y + 2) = 9(2P + 6) = 18P + 54 , √ P 2 − 18P − 54 ≤ 0, whi h implies that P ≤ 9 + 3 15. The proof
or
omplete.
is
30 4. Find all real-valued fun tions f de ned on R that satisfy the identity f f (x − y) = f (x)f (y) − f (x) + f (y) − xy .
Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; and Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. We give the write up of Bataille. It is readily he ked that the fun tion f (x) = −x for all x is a solution. We show that there are no other solutions. Let f satisfy f f (x − y) = f (x)f (y) − f (x) + f (y) − xy
for all x, y ∈ R and let a = f (0). Taking x = y and then taking y = x shows that
=0
(1)
in (1) gives f (a) = a2
f (x)2 = x2 + a2 2 2 4 2 for all real √ x. In parti ular f (a) √ = 2a , that is, a = 2a , hen e √ numbers a ∈ {0,√ 2, − 2}. Assume √ that a = 2. Then for any given x we have f (x) = x2 + 2 or f (x) = − x2 + 2 and taking y = 0 in (1), we obtain
√ 2 − 1 f (x) + 2. (2) √ √ √ Now, f 2 = f (a) = a2 = 2 so that f f 2 = f (2) = ± 6, but then √ √ 2 taking x = 2 in (2) yields a ontradi tion sin e 6 6= 3 2 − 2 . Similarly, √ the assumption a = − 2 leads to a ontradi tion. It follows that a = 0, hen e f (x) = x or f (x) = −x for any givenx. However, if f (x0 ) = x0 for some nonzero real number x0 , then f f (x0 ) = f (x0 ) = x0 , while from (1) we have f f (x0 ) = f f (x0 − 0) = −f (x0 ) = −x0 . This is impossible sin e x0 6= 0, hen e f (x) = −x for all real x. f f (x) =
√
5. Find all triples of non-negative integers (x, y, n) su h that (with the onvention 0! = 1).
x! + y! = 3n n!
Solved by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan; and Konstantine Zelator, University of Toledo, Toledo, OH, USA. We give the solution of Tsai. Let S be the set of all ordered triples (x, y, n) of nonnegative integers + y! su h that x! n! = 3n , or equivalently x! + y! = 3n n!. For n = 0, it is seen at on e that there is no orresponding (x, y, n) ∈ S . For n = 1, let (x, y, n) ∈ S ; then if x ≥ 3 or y ≥ 3 we have x! + y! ≥ 7 > 3 = 31 1!, thus x, y < 3 and simple he king yields that {(0, 2, 1), (1, 2, 1), (2, 0, 1), (2, 1, 1)} ⊂ S . Lemma Let (x, y, n) ∈ S with n ≥ 2. Then x, y ≥ n and x > n or y > n.
Proof: If x, y ≤ n, then x! + y! ≤ 2n! < 3n n!, so x > n or y > n. If x! + y! x! y! x! x > n and y < n, then = + = 3n , a ontradi tion sin e is n! n! n! n!
31 y! an integer but n! is not. Thus, if x > n, then y y > n we have x ≥ n by symmetry.
≥ n.
Similarly, in the ase
We shall prove that for n ≥ 2 there is no orresponding (x, y, n) by onsidering ases on n modulo 3.
∈S
Case 1. n ≡ 0 (mod 3). Let (x, y, n) ∈ S and assume without loss of generality that x ≤ y. By the Lemma, n ≤ x ≤ y and one of these two + y! inequalities is stri t. If x > n, then from x! n! = 3n it follows that (n+1)|3n . However, n+1 has a prime divisor other than 3, a ontradi tion. Therefore, n = x < y, and onsequently x! + y! n!
=
x! n!
+
y! n!
= 1 + (n + 1)(n + 2) · · · y = 3n .
Thus, 3 divides 1+(n+1)(n+2) · · · y and (n+1)(n+2) · · · y ≡ 2 (mod 3), whi h implies that y = n + 2. However, 1 + (n + 1)(n + 2) < 3n for n ≥ 3 (by indu tion) and 1 + (2 + 1)(2 + 2) 6= 32 , ontradi ting the fa t that (x, y, n) ∈ S . Case 2. n ≡ 1 (mod 3). Let (x, y, n) ∈ S and assume without loss of generality that x ≤ y. By reasoning similar to that of Case 1 it follows that x = n and y = n + 1. However, 1 + (n + 1) < 3n for ea h integer n ≥ 2,
ontradi ting the fa t that (x, y, n) ∈ S . Case 3. n ≡ 2 (mod 3). Let (x, y, n) ∈ S and assume without loss of generality that x ≤ y. By the Lemma, n ≤ x ≤ y and one of these two + y! inequalities is stri t. If x ≥ n + 2, then from x! n! = 3n it follows that (n + 2)|3n . However, n + 2 has a prime divisor other than 3, a ontradi tion, hen e n ≤ x < n + 2. If x = n, then n = x < y and x! + y! n!
=
x! n!
+
ontradi ting n + 1 ≡ 0 If x = n + 1, then
y! n!
= 1 + (n + 1)(n + 2) · · · y = 3n ,
(mod 3).
x! + y! x! y! = + = (n + 1) + (n + 1)(n + 2) · · · y = 3n . n! n! n!
If furthermore y = n + 1, then 2(n + 1) = (n + 1) + (n + 1)(n + 2) · · · y = 3n
is even, a ontradi tion. Thus, y > n+1 and (n+1) 1+(n+2) . . . y = 3n . It follows that 1 + (n + 2) · · · y ≡ 0 (mod 3) and (n + 2) · · · y ≡ 2 (mod 3), whi h implies that y = n + 3. However, (n + 1) 1 + (n + 2)(n + 3) 6= 3n
32 for 1 ≤ n ≤ 5 and by indu tion (n + 1) 1 + (n + 2)(n + 3)
ontradi ting the fa t that (x, y, n) ∈ S .
< 3n
for n ≥ 6,
6. Let the sequen e x1 , x2 , x3 , . . . , be de ned by x1 = a, where a is a real number, and the re ursion xn+1 = 3x3n − 7x2n + 5xn for n ≥ 1. Find all values of a for whi h the sequen e has a nite limit as n tends to in nity, and nd this limit.
Solved by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e; and Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. We give Bataille's write-up. Let f (x) = 3x3 − 7x2 + 5x and g(x) = f (x) − x = x(x − 1)(3x − 4). The sequen e {xn }, whi h satis es xn+1 = f (xn ) for all positive integers n,
an only onverge to a root of g(x) = 0. Thus, the only possible nite limits of {xn } are 0, 1, and 43 . We show that the sequen e is onvergent if and only if 0 ≤ a ≤ 43 , in whi h ase the limit is 1 ex ept if a = 0 and n→∞ lim xn = 0 or 4 4 if a = 3 and n→∞ lim xn = 3 . Suppose rst a < 0. Sin e g(x) < 0 when x < 0, it follows that xn < x1 = a < 0 for all positive integers n. If {xn } had a nite limit, ℓ, we would have ℓ ≤ a, ontradi ting the fa t that ℓ ∈ {0, 1, 43 }. Thus, {xn } is divergent when a < 0. Using the fa t that g(x) > 0 for x > 43 , similar reasoning shows that {xn } is divergent when a > 43 . If a ∈ {0, 1, 43 }, then the sequen e {xn } is onstant. If a ∈ (1, 34 ), then using f (x) − 1 = (x − 1)2 (3x − 1) an easy indu tion shows that 1 < xn+1 < xn for all positive integers n. Thus, {xn } is de reasing and bounded, hen e onvergent. Its limit ℓ satis es ℓ ≥ 1 and ℓ ∈ {0, 1, 43 }, that is, ℓ = 1. If a ∈ [ 13 , 1) then x2 = f (a) ≥ 1 and x2 < 43 , as the maximum of f on < 43 . From the previous ase, we see that lim xn = 1. [0, 1] is f 59 = 275 243 n→∞ 1 1 It remains to study the ase a ∈ 0, 13 . Then, 3m+1 ≤ a < m for 3 some unique positive integer m. If any of the numbers x2 , x3 , . . . , xm is not less than 13 , let xk be the one with the smallest index. Then 13 ≤ xk < 43 and by the previous ases {xn }n≥k onverges to 1 and n→∞ lim xn = 1. Otherwise, noting that f (x) − 3x = x(x − 2)(3x − 1) is positive for x ∈ 0, 13 , we have
x2
=
f (x1 ) > 3x1 = 3a ≥
x3
=
f (x2 ) > 3x2 ≥
··· xm
=
1 3m−1
f (xm−1 ) > 3xm−1 ≥
1
3m
,
, 1 32
,
and nally xm+1 > 13 . So {xn }n≥m+1 onverges to 1 and again n→∞ lim xn = 1.
33 To nish this number of the Corner we give solutions from the readers to problems of the 2005 German Mathemati al Olympiad, given at [2008 : 82℄. 1. Determine all pairs (x, y) of reals, whi h satisfy the system of equations x3 + 1 − xy 2 − y 2 y 3 + 1 − x2 y − x2
= =
0, 0.
Solved by George Apostolopoulos, Messolonghi, Gree e; Konstantine Zelator, University of Toledo, Toledo, OH, USA; and Titu Zvonaru, Comane sti, Romania. We give the write-up of Apostolopoulos. We subtra t the two equations of the system to obtain x3 − y 3
+ xy(x − y) + x2 − y 2 = 0 ,
whi h upon fa toring be omes (x − y)(x + y)(x + y + 1) = 0 .
Thus, y = x or y = −x or y = −x − 1. If y = x, then x = ±1, hen e (x, y) = (1, 1) or (x, y) = (−1, −1). If y = −x, then again x = ±1, hen e (x, y) = (1, −1) or (x, y) = (−1, 1). If y = −x − 1 we substitute into the rst equation to obtain x3 + 1 − x(x + 1)2 − (x + 1)2 = −3x(x + 1) = 0 ,
hen e x = 0 or x = −1 and (x, y) = (0, −1) or (x, y) = (−1, 0). 2. Let A, B , and C be three distin t points on the ir le k. Let the lines h and g ea h be perpendi ular to BC with h passing through B and g passing through C . The perpendi ular bise tor of AB meets h in F and the perpendi ular bise tor of AC meets g in G. Prove that the produ t |BF | · |CG| is independent of the hoi e of A, whenever B and C are xed.
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi hel Bataille, Rouen, Fran e; Konstantine Zelator, University of Toledo, Toledo, OH, USA; and Titu Zvonaru, Comane sti, Romania. We give the solution of Amengual Covas. Let A′ be the point diametri ally opposite to A. Let M and N be the midpoints of the segments AB and AC , respe tively. Let B ′ and C ′ be the se ond points of interse tion of the lines g and h with the ir le k, respe tively.
34 . . .... .... ... .... .... .... .... .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . ... . . . .......... ................................................... . . . . . . . . . . . . . . .... . . . . . . ... ....... ...... ... .... .... ............. ............... . . . . . . . . . . . . .... . . ... . . ........ ........ . ... ... ...... ...... . . . . . . . . . . . . . . . . . . . .... .... . . . . . ....... ........ . .... ... ...... ..... . . . . . . . . . . . ... . . . . . . . . .... . . . . ........ ...... . ... ... ..... .... . . . . . . . . . . . . . . .... . . . . . . .. ........ .... ... . ... .. ........ ..... .... ... .... ........ .................. . . . ... ........ ... .. . . . . . . . . . . . . . . . . . . . . ........ ... ... ′ ... .... ........... ′ ... .. . .... . . . . ............... . . . . ........... ... ......... ... ... ...... ... .............. .. .. ................ .. .. ........... ..... ..... .. . ... ..... .... ............ . . . . . . . . . . . . . .. .. ... .. .. .. .. ...... ....... .. ... .. ... ... ... .. ................ ............ ...... .... .... .... ... ... ....... ... . ..... ....... ... .. .. .. .. .... ... .. ..... ................. .. ............. .. ....................... ... . .. .. ... ......... ............ ... . . .. . . . . . . . . . . . . .. .. ........... ............ ....... .. ......... .. .... .... ... ... ........ ....... .. ....... .... ...................... . . ... . . . . . . . . .... . . . ... . . . ......... ...... ... .......... .. .... . .. ... . . . . . . . . . . . . ... ... . . . . ... ... .. ....... . ... ... . . . . . . . . . . . . . ... .... . . . . .. . . ....... . . ... ... ... .... . . . . ... .. . . . . . . . .... . . .... . . . ....... . ...... .... ... .. ... ... . . ....... .. ....... ... . . . .. .... . ... ... .............. . . .. ..... . .... . . ...... ............. . .... . . . . . . . ... . .... . ... . . ... . .. ..... .... ........... . ... .. . ...... . .. . . . . . . . . . ....... . ... . ... . .. . . . .. . . . . . . . . . . . . . . . . . . . . . ....... .. . ... ... . .. .. ....... .. ... .... ....... .... .... ... ... ... .. ....... .... ... ............ ... ... .. ... ... ... .. ........ ... ......... .. .. ... .. ... .... .... .. ............. .. .. .. . . .. . . . . . . . . . . . . . . . . . . ....... ... . .. ... . .. .. .. ....... .. ... ... ..... ....... .... ... .... ... .. ....... ... .. . ....... ... .. ... ... . .. ....... ... .......... .... ............ .. ... .. ....... ............... .... ... .... ...... ......... .. ... . . . . . . . . . . . . . . . . . . . . . . . . ....... ... ... .. ... ... .. ....... .. ....... .... ... ... ... .. .. ....... ... ... ........ .. ............... ... . ..................................................................................................................................................................................................................................................... ............ . ... .. .......... . . . . . . . .... ......... . . . . ... . ...... ...... . .. . . .... ......... . . . . . . . . . .. ... .... ........ .. .. ......... ........ ..... . ... .. .. ......... . ....... ................ ......... ... ..... .... ...... . ....... ......... ....... ........ ................ ... ... .. ......... ....... . . . . . . ........ ......... . . . . . . . . . . . . . . . . ..... .......... ..... . . . . . . . . . ............ ................ ............. ................. .......................................... ............ ..................................................
A
h
B
......... ...............
............ .........
C
g
F
G
............ .........
.......... ............
N
M
B
k
... .. ...... .....
C
A′
Sin e C ′ BC and BCB ′ are right angles, the segments BB ′ and CC ′ are diameters of the ir le k. Thus, the two quadrilaterals AC ′ A′ C and ABA′ B ′ are re tangles, and hen e parallelograms, so we have |C ′ A′ | = |CA|
and
|A′ B ′ | = |AB| .
Sin e the right triangles BM F and A′ C ′ A are similar, as are the right triangles CN G and A′ B ′ A, we have BF BM = = ′ AA C ′ A′
and from whi h we obtain
CG CN = = ′ AA A′ B ′ BF · CG =
1 AB 2
CA 1 CA 2
AB
,
1 |AA′ |2 4
as the square of the radius of k, whi h is independent of the hoi e of A. 3. A lamp is pla ed at ea h latti e point (x, y) in the plane (that is, x and y are both integers). At time t = 0 exa tly one lamp is swit hed on. At any integer time t ≥ 1, exa tly those lamps are swit hed on whi h are at a distan e of 2005 from some lamp whi h is already swit hed on. Prove that every lamp will be swit hed on at some time.
Solution by Titu Zvonaru, Comane sti, Romania. Assume that at time t = 0 the lamp at O(0, 0) is swit hed on. Sin e √ 2005 = 13572 + 14762 then at some time the lamps at the following latti e
35 points will be swit hed on: A1 (1357, 1476) , A2 (3 · 1357, 1476) ,
.. .
Ak (2k − 1) · 1357, 1476 ,
O1 (2 · 1357, 0) , O2 (4 · 1357, 0) ,
.. .
Ok (2k · 1357, 0) ;
and then the lamps at these latti e points will be swit hed on: B1 (2k · 1357 − 2005, 0) , B2 (2k · 1357 − 2 · 2005, 0) , B3 (2k · 1357 − 3 · 2005, 0) , .. . Bt (2k · 1357 − t · 2005, 0) . The equation 2k · 1357 − 2005t = 1 is the same as 2714k − 2005t = 1, whi h has a solution in positive integers k, t be ause gcd(2714, 2005) = 1, for example, 2714 · 1134 − 2005 · 1535 = 1. Thus the lamp at (1, 0) will be swit hed on at some time. It follows (by symmetry) that every lamp will be swit hed on at some time. 4. LetQ(n) denote thesum of the digits of the positive integer n. Prove that Q Q Q(20052005 ) = 7.
Solution by Titu Zvonaru, Comane sti, Romania. It is well known that Q(n) ≡ n (mod then k ≡ 20052005 (mod 9) and we have 20052005
≡
≡
9).
Q Q(20052005 ) ,
Let k = Q
(−2)2005 = −2 · 22004 = −2 · 23
−2(−1)668 = −2 ≡ 7 (mod 9)
,
668
so that k ≡ 7 (mod 9). The number 20052005 has at most 4 · 2005 = 8020 digits. Hen e, 2005 Q 2005 is at most 9 · 8020 = 72180. This implies that
Q Q 20052005 ≤ 5 · 9 = 45 and hen e k = Q Q Q(20052005 ) is at most Q(39) = 12. Altogether, k is an integer satisfying 0 ≤ k ≤ 12 and k ≡ 7 (mod 9), hen e k = 7, as desired.
That ompletes the Corner for this issue. Send me your ni e solutions, generalizations, and Olympiad problem sets.
36
BOOK REVIEWS Amar Sodhi
Putnam and Beyond By Razvan Gel a and Titu Andrees u, Springer S ien e + Business Media LLC, New York, 2007 ISBN-13: 978-0-387-25765-5, soft over, 798+xvi pages, US$69.95 e-ISBN-13: 978-0-387-68445-1 Reviewed by Jeff Hooper, A adia University, Wolfville, NS One of my favourite problem books is the one by Andrees u and Gel a, Mathemati al Olympiad Challenges (Birkhauser, 2000), in whi h the authors
olle t numerous problems entred mainly around past Olympiads, and group them together by similar topi . In Putnam and Beyond, the authors again
ombine to deliver a similar problem book, based this time around on the Putnam Competition. Stru turally, the book onsists of six major hapters: Methods of Proof, Algebra, Real Analysis, Geometry and Trigonometry, Number Theory, and Combinatori s and Probability. Ea h hapter is divided into numerous se tions and subse tions, ea h of whi h fo uses on an important problem idea. For instan e, the initial se tion of the Algebra hapter is Identities and Inequalities, and here we nd the topi divided into a number of important ideas: Algebrai Identities, the equation x2 ≥ 0, the Cau hy-S hwartz Inequality, the Triangle Inequality, the AM{GM Inequality, and so on. Ea h se tion ontains explanations of the key ideas and several worked problems, along with a number of problems to solve. Full solutions are provided at the ba k of the book, along with sour es and/or helpful referen es where ne essary. The initial hapter is a parti ularly ni e introdu tion for students. There is some overlap, of ourse, with the topi s overed in the authors' previous book, and I was expe ting there to be mu h in ommon. To their
redit, the authors have avoided that sort of mild heating: for the topi s in
ommon with their rst book, the examples and problems they o er are new. This new book is also a mu h larger and far more extensive e ort. In addition to all of the examples provided, the book ontains more than 900 problems. And these are Putnam-level problems, so they are mainly a level up from the earlier book. In fa t the topi s in the book go very deep, and over most of the major ideas found in the undergraduate mathemati s urri ulum. So there's lots here. The book is not error-free, however, and a tea her or oa h who uses this book should be a little areful. I ertainly haven't worked through every problem in the book, but did dip into them in a number of se tions. Just to give an example, Problem 33 asks the solver, "Given 50 distin t positive integers stri tly less than 100, prove that some two of them sum to 99." A little thought shows that one an take the numbers 50, 51, . . . , 99 and the statement fails.
37 But these problems are of the minor variety. This wonderful book is an ex ellent problems resour e and should be ome a part of any serious library for problem solving. By olle ting together problems by topi , the authors provide readers the han e to study ea h of these important problem-solving te hniques and ideas in isolation, and help them begin to see the inherent patterns. This should be one of the rst books onsidered as a resour e by anyone oa hing groups of problem solvers. Mathemati al Conne tions: A Companion for Tea hers and Others By Al Cuo o, Mathemati al Asso iation of Ameri a, 2005 ISBN 978-0-88385-739-7, hard over, 239+xix pages, US$54.95 Reviewed by Peter S. Brouwer, State University of New York, Potsdam, NY, USA Al Cuo o is the Dire tor of the Center for Mathemati s Edu ation at the Edu ation Development Center in Newton, MA, where he works in the areas of urri ulum development and professional development of tea hers. This book joins a number of re ent others in addressing se ondary s hool mathemati s ontent topi s from an advan ed (or deeper) perspe tive. The primary audien e is high s hool mathemati s tea hers, and the book is based on the assumption that providing a more advan ed treatment of some of the mathemati al topi s taught at that level is a valuable form of professional development. The author's emphasis is on making onne tions between topi s and developing mathemati al habits of mind. The hoi e of topi s is somewhat idiosyn rati , and re e ts the interrelated topi s that Cuo o is interested in exploring. The hapter titles are: 1. Di eren e Tables and Polynomial Fits, 2. Form and Fun tion: The Algebra of Polynomials, 3. Complex Numbers, Complex Maps, and Trigonometry, 4. Combinations and Lo ks, and 5. Sums of Powers. The strength of this book is that it is essentially a problems book (on the above topi s). There are many problems, in luding 90 in the rst hapter alone, and the reader is asked to work them sequentially while reading through the text. These are grouped by themes, whi h aids oheren e. In addition, there are many problems given as exer ises. The author in ludes helpful notes on sele ted problems at the end of ea h hapter. Many of the problems in this book are quite hallenging, but its in remental and themati approa h helps. As polynomial algebra (and patterns in polynomial oeÆ ients) appear in every hapter, the reader must be omfortable manipulating rather ompli ated algebrai expressions. I would re ommend this book for serious, mathemati s-based professional development programs as well as for experien ed independent readers who would enjoy pursuing a fruitful intelle tual journey through sele ted advan ed se ondary mathemati s topi s.
38 Velo ity Analysis: an Approa h to Solving Geometry Problems
Peng YuChen
1
Introduction
We introdu e velo ity analysis for solving otherwise omplex geometry problems, then we give two examples of the use of this method: (1) to give a very brief proof of the opti al property of the ellipse, and (2) to nd the length of a logarithmi spiral. At the end of this note we leave some problems for the reader; the solutions will be very brief if velo ity analysis is used.
2
Velocity Analysis
In analyti geometry, a urve is a set of points whose oordinates (x, y) satisfy a ertain formula. For example, the set of points {(x, y) : x2 + y2 = R2 } is a ir le or radius R. As we all know, the tra e of a moving point is a urve. In the analyti notion of a urve above, we mainly pay attention to the position of a moving point, rather than its velo ity. In fa t, either the position or the velo ity of a moving point an des ribe the ourse of its motion, and in some ases it is more onvenient to study the velo ity rather than the position. If instead of studying the oordinates of a moving point we study its velo ity, then it is very simple for us to dedu e ertain geometri properties of the tra ed
urve without engaging in omplex mathemati s, espe ially when seeking the tangent or ar length. (As we know, the velo ity ve tor of a moving point gives a tangent to the urve tra ed by the point.) Now we introdu e the basi idea of velo ity analysis in solving geometry problems. → − A point P is moving under a ertain restri tion on its velo ity V . For → − −→ −→ example, if V · OP ≡ 0 (here OP is the position ve tor of P ), then obviously the point P tra es out a ir le whose entre is O. That is to say, we nd an equivalent way of de ning the ir le by studying the velo ity rather than the
oordinates of a moving point. When studying velo ity, usually it is more onvenient to study omponents. In our example of the ir le, we break the velo ity of P into two −→ −→ − → dire tions: along OP and orthogonal to OP . Name the two omponents V1 Copyright
c 2009
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 1
39 − →
− →
−→
− →
and V2 respe tively; if the omponent V1 along OP satis es V1 ≡ 0, then the point P tra es out a ir le. This is the basi idea of velo ity analysis: if we de ompose the velo ity of a moving point into two appropriate dire tions, giving ertain restri tions to the omponents of the velo ity, the tra e of the moving point will be ome a ertain urve. In this setting it is very easy to analyse the tangent to the
urve, for we already know the restri tion on the velo ity of the point. Example 1 The opti al property of the ellipse. To show the power of the ve tor analysis approa h, we use it to solve this lassi al geometry problem. In an ient times people found that oni se tions have very spe ial and beautiful opti al properties. One example is this: if a ray of light leaves one fo us of an ellipse and strikes the ellipse, it will be re e ted to the other fo us of the ellipse (see Figure 1). L1
...
. ..........
.. ................................................................................ . . . . . . . . . . .. ... .... .... ..... ............... ...... .. ... .. ...... ...... .... ......... ... ............. . . . ... . . . ...... . . ... ..... . . . . . . . . . . . .. ..... .. . . . . . . . .. . . . . ... . ... ........ ..... ... ........ . . . . . . . . . . . .................... ................................r..........................................................................................................................................................r................................................. . ... F1 F2 .. ..... ... .... . .... . . ...... .... . ..... . . . ........ . ... ............... ......... ..................... .............................
. .........
.... .
Figure 1
............ ............ ............ ............ ............ ............ .... ........................ .. ........ .. ................ ... ... ............. ........ . . . ............ . . . ........ . . ............ ........ ... ............ ........ . . . ..... . ... . . . ..... . . . . . . . . . .. ..... . . . . . . . . .. ..... . . . . . . . . ... ... ........ ... ........ . ........ ...........................................................................................................................................................................................................................................
.....................α...................... ............. ........ . . . . . . . .. .... . . β ....... . . ... .... .. ... r r .... ... ... F1 F2 ... .... . ...... .... . . . . ........ . ............... ....... ........................................
Figure 2
Using al ulus to prove this property would be ompli ated, so we use the approa h of analysing the velo ity of a moving point to give a mu h shorter proof. As in Figure 2, to prove the opti al property of the ellipse, we need to prove that α = β , whi h is enough to satisfy the Law of Re e tion. Here L1 is tangent to the ellipse. We usually de ne an ellipse like this: two points F1 and F2 in the plane − → V2 are xed. A point P moves so that → − −−→ −−→ ......... V .. |P F1 |+|P F2 | is always a onstant. We γ
all the tra e of P an ellipse. In a
or................................r..P . . . . . . . . . . . . . . . . ......... α dan e with the idea of velo ity analy........ .... ...... . . sis, we des ribe an ellipse as follows: β ...... . .... ... . → ......... − ... V1 De nition. Two points F1 , F2 in the ... ... r r . ... plane are xed (see Figure 3). A mov. . ... F1 → − F2 .. ing point, P , has velo ity ve tor V . .... . − → − → → − ...... .... . . . . Let V1 and V2 be the omponents of V ........ . ....... ................ towards F1 and away from F2 , respe ...................................... − → − → tively. If |V1 | ≡ |V2 |, then the tra e of Figure 3 P is an ellipse. .. ..........
.. ... . ...... .... .. ... ...... .. ... ...... . ... . . . . . .. ............ ...... ... ............ .. .. ...................... ... . ............ .. ............ .. ............ ... .. ............ ............ ........... .............. ... .. ... .................. . ........... ... .............. . .. . . .. .... ................. ..... . ....... ............ ........ ............ ... ........ ............ ... ........ . ... . . ... . . . . ..... . . . . . . . . . .. .. ......... ... ........ . . . . . . . . . . .. .. ........ ... ........ . ........ ............................................................................................................................................................................................................................................ ......
40 It is obvious that the two de nitions are equivalent. For when P moves towards F1 a short distan e, it must then move away from F2 by that same −−→ −−→ distan e to ensure that |P F1 | + |P F2 | is always a onstant. → − − → − → Now V is a tangent ve tor of the ellipse. Sin e |V1 | ≡ |V2 |, then α = γ . Sin e β = γ , we also have α = β . That is all. Example 2 The length of the logarithmi spiral. As another example, we nd
the length of the logarithmi spiral ρ = ρ0 eaθ between ρ = ρ1 and ρ = ρ2 . Usually we annot gure out this problem without using a lot of al ulus, so we introdu e a physi al model. As in Figure 4, three √ points A, B , and C are ea h at a vertex of an equilateral triangle of side 3ρ0 . Point A moves towards B , B moves towards C , and C moves towards A. The speed of ea h point is s. Where will the three points meet and how far will they travel before meeting? − → V2........ A
A
....
Figure 4
....
.................
V
O r
....
B
.. ............ ...
....
.. . . . ... − →
O r
... ... ... .
........................................... .. .. .. .. .. ... .. . .. ..... .... ..... . . .. .. ... .... ..... ..... ... .. .. ... ... .... ..... ... ..................... ... .. . ... . . . 1 ....... ... . ... .. . ... .. ... ... ... ... . ... . .. ... . . ... .. . . ... . . . ... ... ... . ... .. . ... .. . ... . . . ... ... ... . ... .. . ... . . . ... .. . . ... ... ... . . ... . . ... .. . . . ...........................................................................................................................................................................................................
C
B
... ... ... .
....
...... ... .... ... ... ... ... . ... ... ... .. ... . . ... . . ... ... . ... .. ... . . . ... . ... ... . ... .. . ... . . . ... .. . ... . . ... . . ... .. . . ... .. . ... . . . ... . ... ... . ... .. . ... .. ... . . . ... . ... ... . ... .. . ... . . ... . ... ... . ... .. . ... . . . ... .. . .. . ...........................................................................................................................................................................................................
....
.. . . . ...
.................
Figure 5
C
The solution of this problem is easy: obviously the three points will meet at the point O, whi h is the entre of the triangle. We de ompose the → − velo ity V of A at ea h instant into two omponents: one towards the point −→ − → O and the other perpendi ular to AO , and we all these two √ omponents V1 − → − → → − and V2 , respe tively. We have that |V1 | = | V | cos 30◦ = 23 s (see Figure 5), ρ0 ρ0 so the time until they meet is t = − and the distan e traveled → = √ is
2ρ d = st = √ 0 . 3
|V 1 |
( 3/2)s
Now we onsider the urve tra ed by A. We use a polar oordinate − → system with the origin at O and with the ve tor V2 pointing in the dire tion − → → − − → of in reasing θ. At ea h instant, |V2 | = | V | sin 30◦ = |V1 | tan 30◦ . Sin e − → − → ρdθ |V 2 | = and |V1 | = − dρ , we have dρ = aρdθ, the di erential equation dt dt √ of the spiral ρ = ρ0 eaθ with a = cot 150◦ = − 3. Thus, the length of − ρ2 | 2 the urve from ρ = ρ1 to ρ = ρ2 is |ρ1 − · s = √ |ρ1 − ρ2 |. We leave → |V 1 |
3
41
........
the problem of nding a suitable physi al model for other values of a to the reader. We leave three more problems for the reader to ... Figure 6 solve. They ould be solved by using analyti ge... ometry and al ulus, but are more onveniently . solved using the approa h of velo ity analysis. ..... ..... A 1. The pursuant traje tory problem. As in Figure .. ..r 1 . . 6, suppose that an obje t A starts from the point .. . (0, 1), and moves with a onstant speed s in the . B . dire tion of the positive y{axis. At the same time .. . . another obje t B starts from the point (−1, 0), .r..... and moves with speed 2s and always in the dire 0 −1 tion of the obje t A. When will the obje ts meet? 2. The opti al property of a hyperbola. If a light ray leaves one fo us of a hyperbola and strikes the hyperbola, then the (reverse) extended line of its re e tion will pass through the other fo us of the hyperbola (see Figure 7). 3. The opti al property of a parabola. If a light ray leaving the fo us is re e ted in the parabola, then its re e tion is parallel to the axis of symmetry (see Figure 8).
.......
. .... .... ... .... ... .... .... .... ... .... ... .... .... .... ... .... .... .... ... .... ... .... .... .... ... .... .... .............................................................................................................................
. .. ... ..... . ... . . . .. ...
F2
Figure 7
.............
F1...
.. ............... . . . . . . . . . . . . . ......... ............ . . . . . . . . . ...... ........ . . . . . .. ..... ... ... r ... F ..... ....... ........ ......... ........... ............ .............. Figure 8 ................ ......
... .... .. ..... ... .... .... ........ .... .... .... .... .... .... .... .... ..................................................................... .... .... .... .... ... . . .... . .. .... ... ... .... .... . . . ... .. ... .................................................................................................................................................................................................................................. ... .. ..... ... .... .... ... .... .... ... .... .... .... .... ... ... .
......... .
. ...........
...... ... .......... .. .......... ... . ....... .. ........ ...... ........ .... ..... .... ... .. .... .... ... .. .... .. ... .. . ... ... ... .. ... ... .. . ..........................................................................r................................................. ................................................................r .
Referen es
[1℄ Cai Suilin, Ordinary Di erential Equations, ZheJiang University Press, 1988, pp. 46-48. [2℄ R.T. Co man and C.S. Ogilvy, The \Re e tion Property" of the Coni s, Mathemati s Magazine, Vol. 36, No. 1 (Jan., 1963), pp. 11-12. Peng YuChen Lantian-1 4083 Zi Jingang Campus, ZheJiang University 388 Yu Hangtang Road, Hangzhou, China
[email protected]
42
PROBLEMS Solutions to problems in this issue should arrive no later than 1 August 2009. An asterisk (⋆) after a number indi ates that a problem was proposed without a solution. Ea h problem is given in English and Fren h, the oÆ ial languages of Canada. In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8, Fren h will pre ede English. In the solutions' se tion, the problem will be stated in the language of the primary featured solution. The editor thanks Rolland Gaudet of the University College of Saint Bonifa e and Jean-Mar Terrier of the University of Montreal for translations of the problems. 3401. Proposed by Tigran Sloyan, Basi Gymnasium of SEUA, Yerevan, Armenia. Let ABCDE be a onvex pentagon su h that ∠BAC = ∠EAD and ∠BCA = ∠EDA, and let the lines CB and DE interse t in the point F . Prove that the midpoints of CD, BE , and AF are ollinear. 3402. Proposed by Mihaly Ben ze, Brasov, Romania. Let D and E be the midpoints of the sides AB and AC in triangle ABC , respe tively. Prove that CD is perpendi ular to BE if and only if 5BC 2 = AC 2 + AB 2 .
3403. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. The ir les Γ1 and Γ2 interse t at P and Q. A line ℓ through P interse ts Γ1 and Γ2 for the se ond time at A and B , respe tively. The tangents to Γ1 and Γ2 at A and B interse t at C . If O is the ir um entre of △ABC determine the lo us of O when ℓ rotates about P . 3404. Proposed by Mi hel Bataille, Rouen, Fran e. Let Q be a y li quadrilateral. The perpendi ulars to ea h diagonal through its endpoints form a parallelogram, P . Chara terize the entre of P and show that opposite sides of Q interse t on a diagonal of P . 3405. Proposed by Mi hel Bataille, Rouen, Fran e. Find the minimum value of | cos α| + | cos β| + | cos γ| + | cos(α − β)| + | cos(β − γ)| + | cos(γ − α)| ,
where α, β , and γ are real numbers.
43 3406. Proposed by Jose Luis Daz-Barrero and Miquel Grau-San hez, Universitat Polite ni a de Catalunya, Bar elona, Spain. Find " # n 1 Y
lim ln
2n
n→∞
2+
k=1
k
.
n2
3407. Proposed by Roy Barbara, Lebanese University, Fanar, Lebanon. Let S be a set of positive integers ontaining the integer 2007 and su h that
(a) If x, y ∈ S and x 6= y, then |x − y| ∈ S , and
(b) If x ∈ S , then
x3 − 1007x + 3007 ∈ S .
Prove that S is the set of all positive integers.
3408. Proposed by Slavko Simi , Mathemati al Institute SANU, Belgrade, Serbia. Let {ci }∞ i=1 be a sequen e of distin t positive integers, and let |q| < 1. Prove that the inequality ∞ X
ci q ci
i=1
1+
∞ X
q ci
≤
q 1−q
i=1
holds for all su h sequen es {ci }∞ i=1 if and only if q ∈
1 0, 2 .
3409. Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let a, b, c, and d be positive real numbers. Prove that ab + bc + ca ab + bd + da ac + cd + da bc + cd + db + 3 + 3 + 3 a 3 + b3 + c3 a + b3 + d3 a + c3 + d3 b + c3 + d3
≤ min
a 2 + b2 c2 + d2 a 2 + c2 b2 + d2 a 2 + d2 b2 + c2 + , + , + 3/2 3/2 3/2 3/2 3/2 (ab) (cd) (ac) (bd) (ad) (bc)3/2
.
3410. Proposed by Joe Howard, Portales, NM, USA. Let a, b, and c be the sides of triangle ABC , let R be its ir umradius, and let F be its area. Prove that X bc sin2 A/2
y li
b+c
≥
F 2R
.
44 3411. Proposed by Mihaly Ben ze, Brasov, Romania. Let a, b, and c be positive real numbers su h that a6 + b6 + c6 <
32 33
a3 + b3 + c3
Prove that at least one of the quadrati s cx2 + ax + b has no real roots.
2
.
ax2 + bx + c, bx2 + cx + a,
or
3412. Proposed by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. Let a, b, and c be positive real numbers su h that abc = 1. Prove that X
y li
1 ≤ 1. √ 3 a + 2b3 + 6
3413. Proposed by Vo Quo Ba Can, Can Tho University of Medi ine and Pharma y, Can Tho, Vietnam. Let a, b, c, and d be real numbers in the interval [1, 2]. Prove that a+b c+d
+
c+d a+b
−
a+c b+d
≤
3 2
.
................................................................. 3401. Propose par Tigran Sloyan, Ly ee de la SEUA, Erevan, Armenie. Soit ABCDE un pentagone onvexe tel que ∠BAC = ∠EAD et ∠BCA = ∠EDA. Soit F le point d'interse tion des droites CB et DE . Montrer que les milieux des CD, BE et AF sont olineaires. 3402. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit le triangle ABC , ou D et E sont les milieux des ot ^ es AB et AC , respe tivement. Montrer que CD est perpendi ulaire a BE si et seulement si 5BC 2 = AC 2 + AB 2 . 3403. Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. Les er les Γ1 et Γ2 interse tent a P et Q. Une ligne ℓ passant par P interse te une se onde fois Γ1 et Γ2 a A et B , respe tivement. Les tangentes de Γ1 et Γ2 a A et B interse tent a C . Si O est le entre du er le ir ons rit du △ABC determiner le lieu de O lorsque ℓ tourne autour de P .
45 3404. Propose par Mi hel Bataille, Rouen, Fran e. Soit Q un quadrilatere y lique. Les perpendi ulaires a haque diagonale issues de ses sommets forment un parallelogramme P . Cara teriser le ^ es opposes de Q se oupent sur une dia entre de P et montrer que les ot gonale de P . 3405. Propose par Mi hel Bataille, Rouen, Fran e. Determiner la valeur minimum de | cos α| + | cos β| + | cos γ| + | cos(α − β)| + | cos(β − γ)| + | cos(γ − α)| ,
ou α, β et γ sont des nombres reels.
3406. Propose par Jose Luis Daz-Barrero et Miquel Grau-San hez, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Cal uler " # n 1 Y
lim ln
2n
n→∞
2+
k=1
k
n2
.
3407. Propose par Roy Barbara, Universite Libanaise, Fanar, Liban. Soit S un ensemble d'entiers ontenant l'entier 2007 et tel que (a) Si x, y ∈ S et x 6= y , alors |x − y| ∈ S , et (b) Si x ∈ S , alors x3 − 1007x + 3007 ∈ S . Montrer que S est l'ensemble de tous les entiers positifs. 3408. Propose par Slavko Simi , Institut de Mathematiques SANU, Belgrade, Serbie. Soit {ci }∞ i=1 une suite d'entiers positifs distin ts, et soit |q| < 1. Montrer que l'inegalit e ∞ X
ci q ci
i=1
1+
∞ X
q ci
≤
q 1−q
i=1
vaut pour toutes les suites de e type si et seulement si q ∈
1 0, 2 .
3409. Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit a, b, c et d des nombres reels positifs. Montrer que ab + bc + ca ab + bd + da ac + cd + da bc + cd + db + 3 + 3 + 3 a 3 + b3 + c3 a + b3 + d3 a + c3 + d3 b + c3 + d3
≤ min
a 2 + b2 c2 + d2 a 2 + c2 b2 + d2 a 2 + d2 b2 + c2 + , + , + (ab)3/2 (cd)3/2 (ac)3/2 (bd)3/2 (ad)3/2 (bc)3/2
.
46 3410. Propose par Joe Howard, Portales, NM, E-U. Soit a, b et c les ot ^ es du triangle ABC , soit R le rayon de son er le
ir ons rit et F son aire. Montrer que X bc sin2 A/2 F ≥ b+c 2R
y lique
.
3411. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit a, b et c des nombres reels positifs tels que a6 + b6 + c6 <
2 32 3 a + b3 + c3 33
.
Montrer que au moins une des quadratiques ax2 + bx + c, bx2 + cx + a et cx2 + ax + b n'a au une ra ine reelle. 3412. Propose par Cao Minh Quang, College Nguyen Binh Khiem, Vinh Long, Vietnam. positifs tels que abc = 1. Montrer Soit a, b et c trois nombres reels que X 1 ≤ 1. 3 3
y lique
a + 2b + 6
3413. Propose par Vo Quo Ba Can, Universite de Mede ine et Pharma ie de Can Tho, Can Tho, Vietnam. Soit a, b, c et d quatre nombres reels dans l'intervalle [1, 2]. Montrer que a+b c+d a+c 3 + − ≤ . c+d
a+b
b+d
2
47
SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems. Last year we re eived a bat h of orre t solutions from Steven Karp, student, University of Waterloo, Waterloo, ON, to problems 3289, 3292, 3294, 3296, 3297, 3298, and 3300, whi h did not make it into the De ember issue due to being mis led. Our apologies for this oversight. 3301. [2008 : 44, 46℄ Proposed by Ovidiu Furdui, University of Toledo, Toledo, OH, USA. Prove that 1 1 1 1 1 ∞ ∞ ln 2 − + + ··· + 1 + + ··· + X X n+1 n+2 2n 2 n = n (2n + 1)(2n + 2) n=1 n=1
.
What is this ommon value ?
Solution by Manuel Benito, Os ar Ciaurri, and Emilio Fernandez, Logrono, ~ Spain, expanded by the editor. Let A and B denote the summations on the left side and the right side of the proposed equality, respe tively. Also, let Hn = 1 + 21 + · · · + n1 . Then 1 n+1 =
1 1 + ··· + = H2n − Hn n+2 2n 2n X 1 1 1 (−1)j−1 H2n − 2 + + ··· + = 2 4 2n j j=1 +
Sin e it is well known that
∞ P (−1)j −1
of the double summation we have A
=
∞ X 1
n=1
=
∞ X
∞ X
k=1
j−1
(−1)
j
H⌊ j−1 ⌋ = 2
Hk (2k + 1)(2k + 2)
=
j
j=2n+1
∞ X (−1)j−1
j=3
=
n
= ln 2,
j
j=1
∞ X
Hk
k=1
=
j
1 2k + 1
∞ X 1+
n=1
by hanging the order
∞ X (−1)j−1
j=3
1 2
.
−
⌊ j−1 2 ⌋
X 1
n=1
1
2k + 2
+ ··· +
1 n
(2n + 1)(2n + 2)
n
= B.
48 To nd the ommon value of the two absolutely onvergent series, let ∞ X (−1)j−1
f (x) =
j
j=3
H⌊ j−1 ⌋ xj , 2
where the power series for f onverges for all x ∈ (−1, 1). Then f ′ (x)
= =
∞ X
(−1)j−1 H⌊ j−1 ⌋ xj−1 =
j=3 ∞ X
2
∞ X
(−1)j H⌊ j ⌋ xj 2
j=2
(Hn x2n − Hn x2n+1 ) = (1 − x)
n=1
∞ X
Hn x2n .
(1)
n=1
Now, it is well known that 1 1−x
[Ed : Multiply 1 −1 x
ln
1 1−x
= 1+x+x2 +· · ·
=
∞ X
Hn xn .
(2)
n=1
with − ln(1−x) = x+ 12 x2 + 13 x3 +· · ·
and observe that the oeÆ ient of xn is 1 + 12 + · · · + n1 .℄ From (2) we have (1 − x) Thus, (1 − x2 )
∞ P
n=1
(1 − x)
∞ P
n=1
Hn xn = − ln(1 − x).
Hn x2n = − ln(1 − x2 ),
∞ X
n=1
Hn x2n = −
1 1+x
and it follows that
ln(1 − x2 ) .
(3)
From (1) and (3) we obtain f ′ (x) = −
1 ln(1 − x2 ) . 1+x
Sin e f (0) = 0 and the last improper integral below is onvergent, by applying Abel's Continuity Theorem for power series we have A =
lim f (x) =
x→1−
Z
1
′
f (x) dx = −
0
Z
1 0
ln 1 − x2 dx . 1+x
(4)
It remains to evaluate the last integral in (4). With the hange of variable x = 2u − 1, we have Z
1 0
ln 1 − x2 1+x
dx =
Z
1
1/2
ln 4u(1 − u) u
du = I1 + I2 ,
(5)
49 where I1
1 ln(4u) 1 2 1 du = ln (4u) = (ln 4)2 − (ln 2)2 u 2 2 1/2 1/2 1 3 1 (ln 4 + ln 2)(ln 4 − ln 2) = (ln 8)(ln 2) = (ln 2)2 . 2 2 2
Z
= =
1
(6)
On the other hand, using integration by parts and then making the
hange of variable u = 1 − t, we have I2
1 Z 1 ln(1 − u) ln u = du = (ln u) ln(1 − u) + du u 1/2 1/2 1 − u 1/2 Z 1/2 Z 1 ln(1 − t) ln(1 − t) = −(ln 2)2 + dt = −(ln 2)2 + dt − I2 , t t 0 0 Z
1
from whi h we obtain
2I2 = −(ln 2)2 +
Z
1 0
ln(1 − t) dt . t
(7)
[Ed : Sin e the integrals in the omputations above are improper, are must be taken ; e.g., the evaluation of (ln u) ln(1 − u) at u = 1 must be done by
omputing lim (ln u) ln(1 − u) using L'Hospital's ^ Rule.℄ u→1 Finally, −
Z 1 1 1 ln(1 − t) dt = − ln dt t 1−t 0 t 0 ! Z 1 ∞ ∞ Z 1 n−1 X 1 X tn t = − dt = − dt t n n 0 n=1 n=1 0 Z
= −
1
∞ X
n=1
! ∞ X tn 1 1 π2 = − = − n2 0 n2 6 n=1
From (4) { (8), we on lude that A = B =
.
(8)
π2 − (ln 2)2 . 12
Also solved by PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy ; and the proposer. There were also three in omplete solutions, all of whi h only demonstrated that the two given summations are equal.
3302. [2008 : 44, 47℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let s, r, and R denote the semiperimeter, the inradius, and the
ir umradius of a triangle ABC , respe tively. Show that (s2 + r 2 + 4Rr)(s2 + r 2 + 2Rr) ≥ 4Rr(5s2 + r 2 + 4Rr) ,
and determine when equality holds.
50 Solution by Mi hel Bataille, Rouen, Fran e. First, using the two known formulas s2 + r2 + 4Rr = ab + bc + ca and abc = 4Rrs, where a, b, and c are the sides of the triangle, we dedu e that (a + b)(b + c)(c + a) = = =
(a + b + c)(ab + bc + ca) − abc 2s s2 + r 2 + 4Rr − 4Rrs 2s s2 + r 2 + 2Rr .
For onvenien e, let e1 = a + b + c, e2 = ab + bc + ca, and e3 It follows that the required inequality is su
essively equivalent to (a + b)(b + c)(c + a)e2 (a + b)(b + c)(c + a)e2 ab + a b + bc2 + b2 c + ca2 + c2 a e2 2
2
a2 b3 + a3 b2 + b2 c3 + b3 c2 + c2 a3 + c3 a2
≥ ≥ ≥ ≥
= abc.
8Rrs(5s2 + r 2 + 4Rr) , 2e3 e21 + e2 , 2e3 e21 ,
2a2 b2 c + 2a2 bc2 + 2ab2 c2 ,
and nally to a2 (b − c)2 (b + c) + b2 (c − a)2 (c + a) + c2 (a − b)2 (a + b) ≥ 0 .
The last inequality is obviously true, whi h ompletes the proof. Equality holds if and only if a = b = c, that is, if and only if the triangle ABC is equilateral. Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Gree e ; SEFKET ARSLANAGIC, Herzegovina ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; OLIVER GEUPEL, Bruhl, NRW, Germany ; JOE HOWARD, Portales, NM, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ; KEE-WAI LAU, Hong Kong, China ; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e ; SALEM student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; ANDREA MUNARO, MALIKIC, student, University of Trento, Trento, Italy ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer.
3303. [2008 : 44, 47℄ Proposed by Mihaly Ben ze, Brasov, Romania.
Let a, b, and c be positive real numbers. Show that Y
y li
2(a + b)3
≥
Y
y li
(a + s1 )(bc + s2 ) ,
where s1 = a + b + c and s2 = ab + bc + ca.
51 Composite of similar solutions by Manuel Benito, Os ar Ciaurri, Emilio Fernandez, and Luz Ron al, Logrono, ~ Spain ; and Chip Curtis, Missouri Southern State University, Joplin, MO, USA. By straightforward omputations, we have 2(a + b)(b + c)(c + a) − (a + s1 )(bc + s2 ) = 2(a + b)(b + c)(c + a) − (2a + b + c)(ab + 2bc + ca) = 2(a2 b + ab2 + b2 c + bc2 + c2 a + ca2 + 2abc) − (2a2 b + ab2 + 2b2 c + 2bc2 + c2 a + 2ca2 + 6abc) = ab2 + ac2 − 2abc = a(b − c)2 ≥ 0 .
Hen e,
2(a + b)(b + c)(c + a) ≥ (a + s1 )(bc + s2 ) .
Similarly, we have
2(a + b)(b + c)(c + a) ≥ (b + s1 )(ca + s2 ) ; 2(a + b)(b + c)(c + a) ≥ (c + s1 )(ab + s2 ) .
The result now follows by multiplying a ross the last three inequalities. Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, University of Sarajevo, Sarajevo, Bosnia and HerMessolonghi, Gree e ; SEFKET ARSLANAGIC, zegovina ; DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; MICHEL BATAILLE, Rouen, Fran e ; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam ; OLIVER GEUPEL, Bruhl, NRW, Germany ; student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; ANDREA SALEM MALIKIC, MUNARO, student, University of Trento, Trento, Italy ; NGUYEN MANH DUNG, High s hool student, HUS, Hanoi, Vietnam ; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy ; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer.
3304. [2008 : 45, 47℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let a1 , a2 , . . . , an be positive real numbers and identify an+1 with a1 . Prove that n n X X a3k ≥ ak a2k+1 . k=1
k=1
Similar solutions by Mi hel Bataille, Rouen, Fran e ; Walther Janous, Ursulinengymnasium, Innsbru k, Austria ; Steven Karp, student, University of Waterloo, Waterloo, ON ; Xavier Ros, student, Universitat Polite ni a de Catalunya, Bar elona, Spain ; and Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. Reorder the numbers a1 , a2 , . . . , an from smallest to largest and rename them x1 , x2 , . . . , xn . Then 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn and if yi = x2i
52 for ea h i then we also have 0 ≤ y1 ≤ y2 ≤ · · · ≤ yn . The Rearrangen n P P ment Inequality states that xi yi ≤ xi yσ(i) for any permutation σ of {1, 2, . . . , n}.
n P
i=1
i=1
n P
n P
Sin e = xi yi and ak a2k+1 k=1 i=1 k=1 appropriate permutation σ, the result follows. a3k
=
n P
xi yσ(i)
i=1
for an
Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; ARKADY ALT, San Jose, UniCA, USA ; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; SEFKET ARSLANAGIC, versity of Sarajevo, Sarajevo, Bosnia and Herzegovina ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam ; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; OVIDIU FURDUI, University of Toledo, Toledo, OH, USA ; OLIVER GEUPEL, Bruhl, NRW, Germany ; student, Sarajevo College, Sarajevo, BosJOE HOWARD, Portales, NM, USA ; SALEM MALIKIC, nia and Herzegovina ; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam ; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy ; HENRY RICARDO, Medgar Evers College (CUNY ), Brooklyn, NY, USA ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer. Ri ardo omments that this problem appears as Problem 11.7 on p. 148 of Elementary Inequalities by D.S. Mitrinovi (P. Nordho , 1964), but that no solution is provided there.
3305. [2008 : 45, 47℄ Proposed by Stanley Rabinowitz, MathPro Press, Chelmsford, MA, USA. Prove that tan
6π 2π + 4 sin 13 13
4π + 4 sin 13 5π tan + 4 sin 13
= tan =
π 13 q √ 2π = 13 + 2 13 . 13
Solution by Manuel Benito, Os ar Ciaurri, Emilio Fernandez, and Luz Ron al, Logrono, ~ Spain, in memory of Jim Totten. We will prove that tan
2π 6π + 4 sin 13 13
5π + 4 sin 13 6π − 4 sin = tan 13 = tan
and tan
4π 13
+ 4 sin
π 13
= − tan = − tan
π 13 3π 13
+ 4 sin + 4 sin
2π 13 q √ 5π = 13 + 2 13 13
(1)
3π 13 4π 13
=
q √ 13 − 2 13 .
(2)
We will make use of two elegant results due to K.F. Gauss and in luded in the Se tio VII of the Disquisitiones Arithmeti (DA).
53 Lemma (DA, art. 362, II). Let n > 1 be an odd number and ω = k
is any of the numbers 1, 2, . . . , n − 1. Then,
2kπ , n
where
tan ω = 2 sin(2ω) − sin(4ω) + sin(6ω) + · · · ∓ sin (n − 1)ω .
Theorem (DA, art. 356). Let n > 1 be an odd prime number, R be the set of
the (positive and less than n) quadrati residues modulo n, and N be the set of the (positive and less than n) quadrati non-residues modulo n. Then,
and
√ X 2πm 2πr n if − cos = cos 0 if n n r∈R m∈N X
X
r∈R
sin
X 2πr 2πm √0 − sin = n n n m∈N
For n = 13 with ω =
2kπ n
if if
n ≡ 1 (mod 4) n ≡ 3 (mod 4)
, ,
n ≡ 1 (mod 4) n ≡ 3 (mod 4)
, .
and 1 ≤ k ≤ 12 the Lemma yields
tan ω = 2 sin(2ω) − sin(4ω) + sin(6ω) − sin(8ω) + sin(10ω) − sin(12ω) .
We ompute with di erent values of k in this identity as follows. If k = 1 and ω = 2π , then the Lemma yields 13
. (3)
6π π 2π 3π 4π 5π 6π tan = 2 sin + sin + sin + sin + sin + sin
. (4)
tan
2π 4π 5π π 3π 6π 2π = 2 sin − sin + sin + sin − sin + sin 13 13 13 13 13 13 13
If k = 3 and ω = 13
then the Lemma yields
13
If k = 4 and ω = tan
6π , 13
8π , 13
13
then tan 5π 13
13
= − tan
13
8π 13
13
13
and the Lemma yields
. (5)
. (6)
. (7)
5π 3π 6π 4π π 2π 5π = 2 sin + sin + sin + sin − sin − sin 13 13 13 13 13 13 13
By omparing the equations (3), (4), and (5) we see that the rst three expressions in equation (1) are equal. If k = 2 and ω = 4π , then the Lemma yields 13 tan
4π 5π 3π 2π 6π π 4π = 2 sin + sin − sin − sin − sin + sin 13 13 13 13 13 13 13
If k = 5 and ω = tan
10π , 13
then tan 3π 13
= − tan
10π 13
and the Lemma yields
6π π 5π 2π 4π 3π 3π = 2 sin − sin − sin + sin + sin − sin 13 13 13 13 13 13 13
54 If k = 6 and ω = tan
12π , 13
π then tan 13
= − tan
12π 13
and the Lemma yields
2π 4π 6π 5π 3π π π = 2 sin − sin + sin − sin + sin − sin 13 13 13 13 13 13 13
. (8)
By omparing the equations (6), (7), and (8) we see that the rst three expressions in equation (2) are equal. 6π π and B = tan 4π Now we take A = tan 2π + 4 sin + 4 sin . 13 13 13 13 Clearly A and B are positive numbers. From (3) and (6) it follows that π 3π 4π A + B = 4 sin + sin + sin 13
and
13
13
5π 6π 2π − sin + sin A − B = 4 sin 13
Then,
13
13
.
π 3π 4π 2π 5π 6π A2 − B 2 = 16 sin + sin + sin sin − sin + sin
.
π 2π 3π 4π 5π 6π A2 − B 2 = 8 cos + cos + cos − cos − cos + cos
.
13
13
13
13
13
13
Applying the identity 2 sin a sin b = cos(a − b) − cos(a + b), we have 13
13
13
13
13
13
However, for n = 13 ≡ 1 (mod 4), the sets R and N in the Theorem are R = {1,4,9,3,12,10} and N = {2,8,6,11,5,7} ; thus, by the Theorem √ 2π 6π 5π π 3π 4π 2 cos + cos − cos + cos + cos − cos = 13 , 13
13
and therefore
13
13
13
13
√ A2 − B 2 = 4 13 .
Similarly, using the identity 2 sin2 a = 1 − cos(2a) we dedu e that AB
=
= =
π 2π 3π 4π 5π 6π 4 sin + sin + sin + sin − sin + sin 13 13 13 13 13 13 π 2π 3π 4π 5π 6π × sin − sin + sin + sin + sin − sin 13 13 13 13 13 13 2π 3π 4π 5π 6π π + cos + cos − cos − cos + cos 6 cos 13 13 √ √ 6 13 = 3 13 . 2
13
13
13
13
For positive real numbers A and B with √ √ A > B , the solutions of the equations A2 − B 2 = 4 13 and AB = 3 13 are A =
q √ 13 + 2 13
and
B =
q √ 13 − 2 13 .
55 This ompletes the proof of the identities (1) and (2). The following similar identities an be dedu ed when n = 11 : tan
3π π + 4 sin 11 11
2π 5π + 4 sin 11 11 3π 2π 4π π tan + 4 sin = tan + 4 sin 11 11 11 11 5π 4π tan − 4 sin 11 11
= − tan = = =
√ 11 .
University of Sarajevo, Sarajevo, Bosnia and HerAlso solved by SEFKET ARSLANAGIC, zegovina ; MICHEL BATAILLE, Rouen, Fran e ; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e (2 solutions) ; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA ; and PETER Y. WOO, Biola University, La Mirada, CA, USA. q √ 4π π All solvers noted that tan 13 +4 sin 13 6= 13 + 2 13, as did George Apostolopoulos, Messolonghi, Gree e ; and Luyan Zhong-Qiao, Columbia International College, Hamilton, ON. Wagon used Mathemati a to he k that the rst and third identities are orre t and the se ond is in orre t. The proposer o ered a partially orre t solution. π π π Woo wondered if similar results hold for π5 , π7 , 11 or 17 . For the ase of 11 Benito et al. answered (above) in the aÆrmative. The interested reader may want to investigate the other
ases. Woo also hallenges the readers to nd geometri proofs for the equalities in (1) and (2).
3306. [2008 : 45, 47℄ Proposed by Stanley Rabinowitz, MathPro Press, Chelmsford, MA, USA. Find a real number t and polynomials f (x), g(x), and h(x) with integer
oeÆ ients, su h that f (t) =
√ 2,
g(t) =
√ 3,
and
h(t) =
√ 7.
Solution by Roy Barbara, Lebanese University, Fanar, Lebanon. √ √ √ √ √ √ Set θ = 2 + 3 + 7 and ψ = 2 3 7. Computing θ3 , θ5 , and θ7 , we obtain √ √ √ 2 + 3 + 7 √ √ √ 16 2 + 15 3 + 11 7 + 3ψ
This is a
=
θ,
1 3 θ , 2 √ √ √ 1 5 281 2 + 241 3 + 161 7 + 60ψ = θ , 4 √ √ √ 1 7 4796 2 + 3975 3 + 2611 7 + 1043ψ = θ . 8 √ √ √ √ √ linear system for 2, 3, 7, and ψ. Solving for 2, 3, =
and
√ 7
56 yields √ 2 √ 3 √ 7
59 1 1 5 θ − θ3 + θ , 20 2 80 313 297 3 67 5 3 7 = θ − θ + θ − θ , 80 160 320 640 469 377 3 71 5 3 7 = − θ + θ − θ + θ . 80 160 320 640 =
√
√
√
√
√
2+ 3+ 7 θ Finally, setting t = 80 = we obtain 2 = f (t), 3 = g(t), 80 √ and 7 = h(t), where the polynomials f (x), g(x), and h(x) have integer
oeÆ ients :
1
f (x)
= 236x −
g(x)
= 313x −
h(x)
= −469x +
2 1 2
(80)3 x3 + (80)4 x5 , (80)2 · 297x3 + 1 2
1 4
(80)2 · 377x3 −
(80)4 · 67x5 − 1 4
3 8
(80)6 x7 ,
(80)4 · 71x5 +
3 8
(80)6 x7 .
Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ; and the proposer.
3307. [2008 : 45, 47℄ Proposed by D.E. Prithwijit, University College Cork, Republi of Ireland. Eliminate θ from the system λ cos(2θ) λ sin(2θ)
= cos(θ + α) ,
= 2 sin(θ + α) .
Similar solutions by Manuel Benito, Os ar Ciaurri, Emilio Fernandez, and Luz Ron al, Logrono, ~ Spain ; Joe Howard, Portales, NM, USA ; and George Tsapakidis, Agrinio, Gree e. The given system an be rewritten as a linear system in cos α and sin α : cos θ cos α − sin θ sin α sin θ cos α + cos θ sin α
= λ(cos2 θ − sin2 θ) , = λ sin θ cos θ .
Its solution is then cos α = λ cos3 θ ,
and
sin α = λ sin3 θ ;
57 when e,
(cos α)2/3 + (sin α)2/3 = λ2/3 .
(1)
Comments from the Spanish team. Note that the original system has a solution if and only if α and λ satisfy (1). In parti ular, letting x = cos α and y = sin α, we see that for 1 ≤ |λ| ≤ 2 the solutions an be represented by the interse tion points of the unit ir le x2 + y2 = 1 with the astroid x2/3 + y 2/3 = λ2/3 . Thus for ea h λ with absolute value between 1 and 2, (1) will be satis ed for eight values of α ; for λ ∈ {±1, ±2}, it will be satis ed by four values of α. There an be no real solutions for other values of λ. Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Gree e ; SEFKET ARSLANAGIC, Herzegovina ; MICHEL BATAILLE, Rouen, Fran e ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e ; JOSE LUIS D I AZ-BARRERO, Universitat Polite ni a de Catalunya, Bar elona, Spain ; OLIVER GEUPEL, Bruhl, NRW, Germany ; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA ; ANDREA MUNARO, student, University of Trento, Trento, Italy ; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy ; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain ; BOB SERKEY, Leonia, NJ, USA ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; and the proposer. There was one in orre t submission. Our readers produ ed solutions in a variety of formats. Here are a few of the ni est. Instead of (1), Alt, Bataille, and the proposer independently obtained the equivalent equation sin2 (2α) =
3 4 λ2 − 1 27λ2
.
Geupel found that in terms of a real parameter t, the solutions of the given system satisfy θ = arctan t + mπ ,
α = arctan(t ) + nπ , 3
and
m+n
λ = (−1)
s
(1 + t2 )3 1 + t6
,
for integers m and n. In addition, there were several impli it solutions where √ the solver simply presented an equation for θ in terms of α or λ ; for example, θ = arctan 3 tan α + kπ ame from Arslanagi and from Ros.
3308. [2008 : 45, 48℄ Proposed by D.E. Prithwijit, University College Cork, Republi of Ireland. √ Given △ABC√, let AD be the altitude to BC . If AB : AC = 1 : 3, prove that AD ≤ 23 BC . When does equality hold ?
I. Solution by Joe Howard, Portales, NM, USA. √ It suÆ es to take AB = 1 and AC = 3 ; onsequently, AD = sin B . Writing a = BC , we therefore must show that √ 3 sin B ≤ a. 2
58 We start with the inequality 1 ≥ =
sin(60◦ + A) = sin 60◦ cos A + sin A cos 60◦ √ 3 1 cos A + sin A , 2 2
whi h is equivalent to √ 3 cos A ≥ sin A . (1) √ By the osine law, a2 = 4 − 2 3 cos A, so the inequality (1) is equivalent to 2−
a2 ≥ 2 sin A .
By the sine law,
sin A sin B = √ , a 3
(2)
or
a =
√ 3 sin A sin B
,
when e (2) is equivalent to a
√ 3 sin A sin B
≥ 2 sin A ,
whi h redu es immediately to what we were to show.√Equality o
urs when 60◦ + A = 90◦ , in whi h ase A = 30◦ and a2 = 4 − 2 3 cos 30◦ = 1. Thus, equality holds if and only if △ABC is isos eles with A = C = 30◦ . II. Solution by Mi hel Bataille, Rouen, Fran e. We are given that vertex A belongs to the lo us of those points P for B 1 whi h P = √ . We re ognize this lo us to be a ir le Γ alled the ir le of PC 3 Apollonius ; it interse ts symmetri ally the line joining B to C in points that √ divide the segment BC internally and externally in the ratio 1 : 3. As an −−→ − − → easy onsequen e, the entre K of Γ satis es BK = − 21 BC , and its radius is
N
√
3BC . 2
Let N N ′ be the diameter of Γ that is perpendi ular to BC . Then and N ′ are the points of Γ that are farthest from the line BC , hen e √ 3BC AD = d(A, BC) ≤ d(N, BC) = 2
.
Equality holds if and only if A is situated at N or N ', in whi h ase △ABC is isos eles with BA = BC and ∠ABC = 120◦ . Also solved by MOHAMMED AASSILA, Strasbourg, Fran e ; ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; SEFKET ARSLANAGIC, University of Sarajevo, Sarajevo, Bosnia and Herzegovina ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CHIP CURTIS, Missouri Southern State University, Joplin,
59 MO, USA ; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA ; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines ; FRANCISCO IES Alvarez JAVIER GARC I A CAPITAN, Cubero, Priego de Cordoba, Spain ; OLIVER GEUPEL, Bruhl, NRW, Germany ; STEVEN KARP, student, University of Waterloo, Waterloo, ON ; Y, Big Rapids, MI, USA ; KEE-WAI LAU, Hong Kong, China ; THANOS V ACLAV KONECN student, Sarajevo MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e ; SALEM MALIKIC, College, Sarajevo, Bosnia and Herzegovina ; ANDREA MUNARO, student, University of Trento, Trento, Italy ; JOEL SCHLOSBERG, Bayside, NY, USA ; PANOS E. TSAOUSSOGLOU, Athens, Gree e ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer. q More generally, Kone ny proved that AD ≤ 2 BC if AC : AB = q with q > 1. q −1 His argument was mu h like that of solution II above.
3309. [2008 : 45, 48℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let α, β , and γ be xed non-zero real numbers. Show that the system αx + βy + γz xy + yz + zx
= =
1, 1,
has a unique solution for (x, y, z) if and only if α2 + β 2 + γ 2 + 1 = 2(αβ + βγ + γα) ,
and, in this ase nd that unique solution. Solution by George Tsapakidis, Agrinio, Gree e, modi ed by the editor. Substituting αx = 1 − βy − γz into α(xy + yz + zx) = α we obtain (y + z)(1 − βy − γz) + αyz = α, whi h upon simplifying yields the following quadrati equation in y βy 2 − 1 + (α − β − γ)z y + γz 2 − z + α = 0 .
(1)
Equation (1) has a unique solution in y if and only if
[1 + (α − β − γ)z] − 4β(γz 2 − z + α) = 0 , 2
whi h an be written as the following quadrati equation in z (α − β − γ)2 − 4βγ z 2 + 2(α + β − γ)z + 1 − 4αβ = 0 .
(2)
Equation (2) has a unique solution in z if and only if
(α + β − γ)2 − (α − β − γ)2 − 4βγ (1 − 4αβ) = 0 ,
whi h, by straightforward omputations, redu es to
α2 + β 2 + γ 2 + 1 = 2(αβ + βγ + γα) .
(3)
60 Now, suppose (3) holds. Then we have (α − β − γ)2 − 4βγ = α2 + β 2 + γ 2 − 2αβ − 2βγ − 2γα = −1 .
Thus, (2) redu es to −z 2 + 2(α + β − γ)z + 1 − 4αβ = 0, the unique solution of whi h is given by z = α + β − γ . Using this and (3), we nd that y
= =
1 + (α − β − γ)(α + β − γ)
=
1 + (α − γ)2 − β 2
2β 2β −2β 2 + 2αβ + 2βγ 1 + α2 − β 2 + γ 2 − 2αγ = = γ +α−β. 2β 2β
Finally, using (3) we have αx = =
1 − βy − γz = 1 − β(γ + α − β) − γ(α + β − γ) −α2 + αβ + γα ,
from whi h it follows that x = β + γ − α. Therefore, the unique solution to the given system is (x, y, z) = (β + γ − α, γ + α − β, α + β − γ) .
Also solved by ARKADY ALT, San Jose, CA, USA ; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; ROY BARBARA, Lebanese University, Fanar, Lebanon ; MICHEL BATAILLE, Rouen, Fran e ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; OLIVER GEUPEL, Bruhl, NRW, Germany ; STEVEN KARP, student, University of Waterloo, Waterloo, ON ; PAOLO PERFETTI, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy ; DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer. There was also one partially in orre t solution submitted.
3310. [2008 : 46, 48℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let a, b, and c denote, as usual, the lengths of the sides BC , CA, and AB , respe tively, in △ABC . Let s be the semiperimeter of △ABC , r the inradius, ha the altitude to side BC , and ra , rb , and rc the exradii to A, B , and C , respe tively.
(a) Show that for x > 0, we have ha or x = rc .
=
(b) Show that for x > 0, we have ha x = r or x = ra .
=
2s(s − a)x x2 + s(s − a)
if and only if x = rb
2(s − b)(s − c)x |x2 − (s − b)(s − c)|
if and only if
61 Similar solutions by Mi hel Bataille, Rouen, Fran e ; Oliver Geupel, Bruhl, NRW, Germany ; Titu Zvonaru, Comane sti, Romania ; and the proposer. It is well known that the area F of △ABC p an be variously expressed as 12 aha , rs, ra (s − a), rb (s − b), rc (s − c), or s(s − a)(s − b)(s − c). We have rb + rc
=
F
s−b
+
F
s−c
=
F (2s − b − c)
(s − b)(s − c)
=
a · F · (s − a) as(s − a) = s(s − a)(s − b)(s − c) F
=
2as(s − a) aha
2s(s − a)
=
(1)
ha
and rb rc
=
F2 (s − b)(s − c)
=
s(s − a)(s − b)(s − c) (s − b)(s − c)
= s(s − a) .
(2)
It follows from (1) and (2) that x = rb and x = rc are the solutions of ha x2 − 2s(s − a)x + ha s(s − a) = 0 ,
hen e these are the solutions of the equation in (a). We also have ra − r
= =
F s−a
−
F s
=
aF s(s − a)
2(s − b)(s − c) a · F · (s − b)(s − c) = 2 F ha
and ra r =
F2 = (s − b)(s − c) . s(s − a)
(3) (4)
By (3) and (4) it follows that the equation
ha x2 − 2(s − b)(s − c)x − (s − b)(s − a)ha = 0
has the solutions x = ra and x = −r and that the equation
ha x2 + 2(s − b)(s − c)x − (s − b)(s − a)ha = 0
has the solutions x = −ra and x = r. Thus, the only positive solutions of the pre eding two equations are r and ra , and it follows that these are the only positive solutions to the equation in (b). Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; GEORGE TSAPAKIDIS, Agrinio, Gree e ; and PETER Y. WOO, Biola University, La Mirada, CA, USA.
62 3311. [2008 : 46, 48℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let n be an integer with n ≥ 2. Suppose that for k = 0, 1, . . . , n − 2 we have n−2 ≡ (−1)k (k + 1) (mod n) . k
Show that n is a prime.
Solution by Oliver Geupel, Bruhl, NRW, Germany. It is suÆ ient to prove that if n is a omposite integer with n ≥ 2, then there exists an integer k with 0 ≤ k ≤ n − 2 su h that n−2 k
6≡ (−1)k (k + 1) (mod n)
.
(1)
Toward that end, let p be the least prime fa tor of n. If (1) holds for some k < p, then we are done. Otherwise the given ongruen e holds in parti ular for k = p − 2, and we have n−2 n−2 n−p (p − 1) = · · (n − p − 1) p p−2 p n ≡ (−1)p−2 (p − 1) · − 1 · (−p − 1) p n ≡ (p − 1) · (−1)p p + 1 − (mod n) . p Be ause p − 1 is oprime to the modulus n, we an divide both sides by it and on lude that (1) holds with k = p : n−2 p
n ≡ (−1)p p + 1 − 6≡ (−1)p (p + 1) (mod n) p
.
Also solved by MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA ; STEVEN KARP, student, University of Waterloo, Waterloo, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina ; DAVID ON ; SALEM MALIKIC, E. MANES, SUNY at Oneonta, Oneonta, NY, USA ; JOEL SCHLOSBERG, Bayside, NY, USA ; and the proposer. The onverse is a known result : If p is a prime, then p−2 ≡ (−1)k (k + 1) (mod p) k for k = 0, 1, . . . , p − 2. Karp provided a simple proof (by indu tion on k) ; Bataille provided a referen e : E. Lu as, Theorie des nombres, A. Blan hard (1961), p. 420.
3312. [2008 : 46, 48℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let n be a positive integer ongruent to 1 modulo 6. Show that 3/n an be expressed as 1
a1
+
1
a2
+ ··· +
1
ak
for some distin t positive integers a1 , a2 , . . . , ak , and nd the minimal value of k.
63 Solution by Steven Karp, student, University of Waterloo, Waterloo, ON. We solve the problem for all positive integers n. If n ≡ 0 or 3 (mod 6), 1 then n3 = n/3 . If n ≡ −1 (mod 6), then 3 = n
If n ≡ ±2
1 n+1 3
1
+
n(n+1) 3
(mod 6), then 3 1 1 = n + n n 2
If n ≡ 1
.
(mod 6)
.
and n has a fa tor c su h that c ≡ −1 3 = n
1 n+c 3
1
+
n(n+c) 3c
(mod 6),
then
.
We see that k is minimal in all of these ases, sin e n3 = a1 for some positive 1 integer a1 only if n ≡ 0 (mod 3). Now, if n > 1, n ≡ 1 (mod 6) and all divisors of n are ongruent to 1 modulo 6, then 3 = n
1 n+1 2
+
1 + n
1 n(n+1) 2
,
and we laim that k = 3 is minimal in this ase. To prove this suppose for the sake of ontradi tion that 3 n
=
1 a
+
1 b
for distin t positive integers a and b. Then n =
b(3a − n) a
,
and we have a|b(3a − n). Therefore, a = pq where p and q are positive integers su h that p|b and q|(3a − n). Then pb n, when e pb ≡ 1 (mod 6). Sin e q|a and q|(3a − n), we also have q|n, so q ≡ 1 (mod 6). We now have 1 ≡ qn =
b p
(3a − n) ≡ 3a − n ≡ −1 or 2 (mod 6)
,
a ontradi tion. Finally, for n = 1 we have 3 = 1+
1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + + 2 3 4 5 6 7 8 9 10 15 230 57960
A omputer sear h shows that k = 13 is minimal in this ase.
.
64 Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon ; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logrono, ~ Spain ; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA ; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA ; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria ; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA ; JOEL SCHLOSBERG, Bayside, NY, USA ; PETER Y. WOO, Biola University, La Mirada, CA, USA ; TITU ZVONARU, Comane sti, Romania ; and the proposer. Benito et al. refer to the note by Thomas R. Hagedorn, A proof of a onje ture on Egyptian fra tions, Amer. Math. Monthly, 107 (2000) 62-63, where it is proved that for ea h odd integer n ≥ 3 not divisible by 3 there exist distin t odd, positive integers a, b, and c su h that 3 1 1 1 = + + n a b c
.
The result had been onje tured by R. Hardin and N. Sloane. When n = 6p+ 1 Hagedorn gives the de omposition 3 3 1 1 1 = = + + n 6p + 1 2p + 1 (2p + 1)(4p + 1) (4p + 1)(6p + 1)
.
Janous refers to a paper by Andrzej S hinzel, Sur quelques propriet es des nombres 3/n et 4/n, ou n est un nombre impair, Mathesis 65 (1956) 219-222, for a treatment of our problem. Sin e the improper fra tion 3 = 3/1 arises, he asks if any CRUX readers know the minimum number ℓ(n) of distin t Egyptian fra tions needed to represent the positive integer n.
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