No R.E. Woodrow. All communications about this column should be sent to Professor R.E.

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1 322 THE OLYMPIAD CORNER No. 192 R.E. Woodow All communications about this column should be sent to Pofesso R.E. Woodow, Depatment of Mathematics and Statistics, Univesity of Calgay, Calgay, Albeta, Canada. T2N 1N4. As a st Olympiad fo this issue we give the poblems of the 4 th Mathematical Olympiad of the Republic of China (Taiwan) witten Apil 13, 15, My thanks go to Bill Sands of the Univesity of Calgay, who collected these poblems when he was assisting with the 1995 Intenational Mathematical Olympiad held in Canada. 4 th MATHEMATICAL OLYMPIAD OF THE REPUBLIC OF CHINA (TAIWAN) Fist Day Taipei Apil 13, Let P (x) =a 0 +a 1 x++a n,1 x n,1 +a n x n be a polynomial with complexcoecients. Suppose the ootsofp(x)ae 1 ; 2 ;::: ; n with j 1 j > 1, j 2 j > 1;::: ;j j j > 1, and j j+1 j1;::: ;j n j1.pove: jy j i j p ja 0 j 2 +ja 1 j 2 ++ja n j 2 : i=1 2. Given a seuence of integes: x 1 ;x 2 ;x 3 ;x 4 ;x 5 ;x 6 ;x 7 ;x 8, one constucts a second seuence: jx 2, x 1 j; jx 3, x 2 j; jx 4, x 3 j; jx 5, x 4 j; jx 6, x 5 j; jx 7, x 6 j; jx 8, x 7 j; jx 1, x 8 j. Such a pocess is called a single opeation. Find all the 8-tem integal seuences having the following popety: afte nitely many applications of the single opeation the seuence becomes an integal seuence with all tems eual. 3. Suppose n pesons meet in a meeting, and evey one among them is familia with exactly 8 othe paticipants of that meeting. Futhemoe suppose that each pai of two paticipants who ae familia with each othe have 4 acuaintances in common at that meeting, and each pai of two paticipants who ae not familia with each othe have only 2 acuaintances in common. What ae the possible values of n?

2 323 Second Day Taipei Apil 15, Given n distinct integes m 1 ;m 2 ;::: ;m n,pove that thee exists a polynomial f (x) of degee n and with integal coecients which satises the following conditions: (1) f (m i )=,1, fo all i, 1 i n. (2) f (x) cannot be factoized into a poduct of two nonconstant polynomials with integal coecients. 5. Let P be a point on the cicumscibed cicle of 4A 1 A 2 A 3. Let H be the othocente of4a 1 A 2 A 3. Let B 1 (B 2 ;B 3 espectively) be the point of intesection of the pependicula fom P to A 2 A 3 (A 3 A 1, A 1 A 2 espectively). It is known that the thee points B 1, B 2, B 3 ae collinea. Pove that the line B 1 B 2 B 3 passes though the midpoint of the line segment PH. 6. Let a, b, c, d be integes such that ad, bc = k>0,(a; b) =1, and (c; d) =1.Pove that thee ae exactly k odeed pais of eal numbes (x 1 ;x 2 )satisfying 0 x 1, x 2 < 1 and fo which both ax 1 +bx 2 and cx 1 +dx 2 ae integes. As a second poblem set this issue fo you puzzling pleasue, we give the XI Italian Mathematical Olympiad witten May 5, 1995 at Cesenatico. Thanks go to Bill Sands of the Univesity of Calgay who collected them while at the 1995 IMO in Canada. XI ITALIAN MATHEMATICAL OLYMPIAD Cesenatico, May 5, 1995 Time: 4.5 hous 1. Detemine fo which values of the intege n it is possible to cove up, without ovelapping, a suae of side n with tiles of the type shown in the pictue whee each small suae of the tile has side 1.

3 In a class of 20 students no two of them have the same odeed pai (witten and oal examinations) of scoes in mathematics. We say that student A is bette than B if his two scoes ae geate than o eual to the coesponding scoes of B. The scoes ae integes between 1 and 10. (a) Show that thee exist thee students A, B and C such that A is bette than B and B is bette than C. (b) Would the same be tue fo a class of 19 students? 3. Inatown thee ae4pubs, A, B, C and D, connected as shown in the pictue. C A D B A dunkad wandes about the pubs stating with A and, afte having a dink, goes to any of the pubs diectly connected, with eual pobability. (a) What is the pobability that the dunkad is at pub C at his fth dink? (b) Whee is the dunkad moe likely to beafte n dinks? (n >5) 4. An acute-angled tiangle ABC is inscibed in a cicle with cente O. Let D be the intesection of the bisecto of A with BC, and suppose that the pependicula to AO though D meets the line AC in a point P inteio to the segments AC. Show that AB = AP. 5. Two non-coplana cicles in Euclidean space ae tangent at a point and have the same tangents at this point. Show that both cicles lie in some spheical suface. 6. Find all pais of positive integes x, y such that x =2 y : As a thid set of poblems fo you attention we give the Thid and Fouth Gade and IMO Team selection ounds of the Yugoslav Fedeal Competition fo Thanks again go to Bill Sands, the Univesity of Calgay, fo collecting them fo me.

4 325 YUGOSLAV FEDERAL COMPETITION 1995 Thid and Fouth Gade 1. Let p be a pime numbe. Pove that the numbe , is divisible by p, whee dots indicate that the coesponding digit appeas p times consecutively. 2. A polynomial P (x) with intege coecients is said to be divisible by a positive intege m if and only if the numbe P (k) is divisible by m fo all k 2 Z. If the polynomial P (x) = a 0 x n + a 1 x n,1 ++a n,1 x+a n is divisible by m, pove that a n n! is divisible by m. 3. A chod AB and a diamete CD? AB of a cicle k intesect at a point M. Let P lie on the ac ACB and let P 62 fa; B; Cg. Line PM intesects the cicle k at P and Q 6= P, and line PD intesects chod AB at R. Pove that RD > MQ. 4.Atetahedon ABCD is given. Let P and Q be midpoints of edges AB and CD, and let O and S be the incente and the cicumcente of the tetahedon, espectively. If points P, Q and S belong to the same line, pove that the point O also belongs to that line. Selection of the IMO Team 1. Find all the tiples (x; y; z) of positive ational numbes such that x y z and 1 x + y + z; x + 1 y + 1 ; xyz 2 Z: z 2. Let n be a positive intege having exactly 's in its binay epesentation. Pove that 2 n,1995 divides n!. 3. Let SABCD be a pyamid such that all of its edges ae of the same length. Let points M 2 BC and N 2 AS be such that the line MN is pependicula to line AD as well as to the line BC. Find the atios BM=MC and SN=NA.[Edito's note: ABCD is the base of the pyamid.] We now tun to eades' solutions to poblems of the Swedish Mathematics Contest, 1993 [1997: 196].

5 326 SWEDISH MATHEMATICS CONTEST 1993 Final Novembe The intege x is such that the sum of the digitsof3xis the same as the sum of the digits ofx.pove that 9 is a facto of x. Solutions by Jamie Batuwantudawe, student, Si Winston Chuchill High School, Calgay; by Michael Selby, Univesity of Windso, Windso, Ontaio; and by Enico Valeiano Cuba, National Univesity of Engineeing, Lima, Peu. We give the solution of Valeiano. Also Let S(n) be the sum of the digits ofn. Woking modulo 9 Since S(3x) =S(x), 3x S(3x) mod9: x S(x) mod9: 2x 0mod9; and since (2; 9) = 1, we have x 0mod9. 2.Aailway line is divided into 10 sections by the stations A, B, C, D, E, F, G, H, I, J and K. The distance between A and K is 56 km. A tip along two successive sections neve exceeds 12 km. A tip along thee successive sections is at least 17 km. What is the distance between B and G? z } { A B C D E F G H I J K Solutions by Jamie Batuwantudawe, student, Si Winston Chuchill High School, Calgay; by Michael Selby, Univesity of Windso, Windso, Ontaio; and by Enico Valeiano Cuba, National Univesity of Engineeing, Lima, Peu. We give Batuwantudawe's solution. A B C D E F G H I J K Now AK =56and AK = AD + DG + GJ + JK. We know that AD; DG; GJ 17. Thus JK 5to satisfy AK =56. We know HK 17, and since JK 5, HJ 12. But, we also know HJ 12. Thus HJ =12. Since HK 17 and HJ = 12, JK 5. The only possibility is that JK =5. Symmetically we nd that AB =5and BD =12. Now, DH = AK, AB, BD, HJ, JK = 56, 5, 12, 5, 12=22:

6 327 Now GJ 17 but HJ =12. Hence GH 5. Since DG 17 and DH = DG + GH =22, we obtain DG =17and GH =5. Now BG = BD + DG = 12+17=29: 3. Assume that a and b ae integes. Pove that the euation a 2 + b 2 + x 2 = y 2 has an intege solution x, y if and only if the poduct ab is even. Solutions by Bob Pielipp, Univesity of Wisconsin{Oshkosh, Wisconsin, USA; Michael Selby, Univesity of Windso, Windso, Ontaio; by See Sanyal, student, Westen Canada High School, Calgay, Albeta; by Eniue Valeiano Cuba, National Univesity of Engineeing, Lima, Peu; and by Michael Lebedinsky, student, Heny Wise Wood High School, Calgay, Albeta. We give Selby's solution. Fist, we pove that this condition is necessay. Suppose ab is odd. Then a, b ae odd and a 2 b 2 1mod4. Now x 2 0 o 1mod4, and y 2 0 o 1mod4. Theefoe a 2 + b 2 + x 2 = y 2 is not possible, since if we conside this modulo 4, 2+x 2 y 2 mod 4, which is impossible since 2+x 2 2o 3mod4. Theefoe ab must be even. If ab is even, then, without loss of geneality, a =2k. Conside 4k 2 + b 2 + x 2 = y 2. If 4k 2 +b 2 =2t+1, t an intege, then set y,x =1and y +x =2t+1, 2y =(t+ 1)2, y = t +1and x = t. Then 2t +1+t 2 =(t+1) 2.Weae done. If 4k 2 + b 2 is even, then b =2sand 4k 2 + b 2 =4(k 2 +s 2 )=4m. Again, y 2, x 2 =4m. Set y,x =2and y+x =2m. Then y = m+1and x = y,2 =m,1. Now 4m +(m,1) 2 = (m +1) 2, and again we ae done. Hence a 2 + b 2 + x 2 = y 2 always has a solution when ab is even. 4. To each pai of eal numbes a and b, whee a 6= 0and b 6= 0, thee is a eal numbe a b such that Solve the euation x 36 = 216. a (b c) =(ab)c, aa=1: Solutions by See Sanyal and Aliya Walji, students, Westen Canada High School, Calgay, Albeta; and by Eniue Valeiano Cuba, National Univesity of Engineeing, Lima, Peu.

7 328 Now a (a a) =(aa)a,so a1=1a=a: Also a (b b) =(ab)band a = a 1=(ab)bso a b = a b : Finally x 36 = 216 =) x = 7776 : 5. A tiangle with peimete 2p has sides a, b and c. If possible, a new tiangle with the sides p, a, p, b and p, c is fomed. The pocess is then epeated with the new tiangle. Fo which oiginal tiangles can the pocess be epeated indenitely? Solutions by Michael Selby, Univesity of Windso, Windso, Ontaio; by Eniue Valeiano, National Univesity of Engineeing, Lima, Peu; and by Sonny Chan, student, Westen Canada High School, Calgay, Albeta. We give Valeiano's solution. Let a b c and be the dieence between the longest and the shotest side. Oiginal Tiangle Peimete =2p =c,a New Tiangle Peimete (1) =3p,(a+b+c)=p (1) =(p,a),(p,c)=c,a We can see that the peimete of the new tiangle is half the pevious peimete, but is the same. Then, if > 0, epeating this pocess we can obtain Peimete (k) = 2p 2 k < (k) = c, a: If c (k) is the longest side we obtain the absud elation c (k) < Peimete (k) < (k) < c (k). Finally, only with an euilateal tiangle as the oiginal tiangle ( = 0) can we epeat the pocess indenitely. 6. Let a and b be eal numbes and let f(x) =(ax+b),1. Fo which a and b ae thee thee distinct eal numbes x 1, x 2, x 3 such that f(x 1 )=x 2, f(x 2 )=x 3 and f (x 3 )=x 1? Solutions by Filip Cnogoac and Sonny Chan, students, Westen Canada High School, Calgay, Albeta; and by Michael Selby, Univesity of Windso, Windso, Ontaio. We give Selby's wite-up. Conside the functions of the fom g(x) = x + x + :

8 329 Lemma. g(x) has at least 3 distinct xed points if and only if = =0,=6=0. Poof. If = =0,=6=0,g(x)=xand it clealy has at least 3 distinct points x 1 ;x 2 ;x 3 such that g(x i ) = x i, i = 1;2;3. Convesely conside the euation fo a xed point x, g(x) =x. This implies x 2 +x = x + o x 2 +(,)x, =0. Suppose this has thee distinct oots. Then the uadatic must be identically 0, o==0and =. Now, if f (x) = 1 ax+b, then ff(x) = ax + b abx + b 2 + a and fff(x) = abx + a + b2 a(a + b 2 )x + ab + b(b 2 + a) : The poblem implies fff has thee distinct eal xed points x 1 ;x 2 ;x 3. By the above lemma, this is tue if and only if a + b 2 = a(a + b 2 )=0 and ab + b(a + b 2 )=ab 6= 0: This is tue if and only if a =,b 2 and ab 6= 0. To complete this numbe of the Olympiad Cone we tun to eades' solutions to poblems of the Dutch Mathematical Olympiad, second ound, Septembe 1993 [1997: 197]. DUTCH MATHEMATICAL OLYMPIAD Second Round Septembe, Suppose that V = f1; 2; 3;::: ;24; 25g. Pove that any subset of V with 17 o moe elements contains at least two distinct numbes the poduct of which is the suae of an intege. Solution by Sonny Chan and Filip Cnogoac, students, Westen Canada High School, Calgay. The set of numbes A = f1; 2;::: ;24; 25g contains a total of ve pefect suaes f1; 4; 9; 16; 25g. The poduct of any two of these will also be a pefect suae. Thee is one tiplet, the poduct of any two of its elements will esult in a pefect suae: f2; 8; 18g. The only othe pais of numbes fom A whose poduct is a pefect suae aef3;12g, f5; 20g, f6; 24g. The othe eleven elements of the set A ae f7; 10; 11; 13; 14; 15; 17; 19; 21; 22; 23g and they cannot fom a pefect suae when multiplied with any othe element of set A. Goup the elements of set Aas follows: f1; 4; 9; 16; 25g; f2; 8; 18g; f3; 12g; f5; 20g; f6; 24g; f7g; f10g; f11g; f13g; f14g; f15g; f17g; f19g; f21g; f22g; f23g:

9 330 If moe than one numbe is chosen fom a given goup, a pefect suae will esult. Thee is a total of 16 goups, so 16 numbes can be chosen without ceating a pefect suae poduct. Howeve, if any 17 numbes ae chosen, then two must be contained within the same goup, and theefoe will fom a pefect suae poduct. 2. Given is a tiangle ABC, \A =90.Dis the midpoint of BC, F is the midpoint of AB, E the midpoint of AF and G the midpoint of FB. AD intesects CE, CF and CG espectively in P, Q and R. Detemine the atio PQ QR. C D P Q R A E F G B Solution by Filip Cnogoac, student, Westen Canada High School, Calgay. C J K P Q R D A E F G B We know that two medians in a tiangle divide each othe in 2:1atio, o in othe wods the point of intesection is 2 the way fom the vetex. 3 Since CF and AD ae both medians in 4ABC, then AQ QD = 2, whee 1 Q is the point of intesection. Also, since D is the midpoint of the hypotenuse in the ight tiangle ABC, then it is the cente of the cicumscibed cicle with adius DA = DC = DB. Dop a pependicula fom D onto sides AB and CA. The feet of the pependiculas will be F and J, espectively, whee J is the midpoint of AC, since DF and DJ ae altitudes in isosceles tiangles 4ADB and 4ADC, espectively. Now conside 4CFB. The segments CG and FDae medians and theefoe intesect at H say in the atio 2:1so, HD FD = 1.Fom hee 3 it can be seen that 4ARC and 4DRH ae simila, since thei angles ae

10 331 the same. Also, since we know that FD = JA, and 2JA = AC then HD = 1 CA and 4ARC is 6 times bigge than 4DRH. Now we can see 6 that AR RD = 6 RD and since AR + RD = AD, then 1 AD = 1. 7 Similaly 4AP E 4KPD, whee medians DJ and CE meet at K. We know that AE = 1 AB, so then 4 JK = 1 JD, since JD is paallel to 4 AB. It now follows that AE, and fom the similaity of the tiangles AP PD = 2 3 KD = 2 3 AP. Also, since AP + PD = AD, then AD esults wehaveap = 2 AD, 5 AQ = 2 AD, 3 QD = 1 3 AD and RD = 1 AD. 7 Thus PQ = AQ, AP = 2 3 AD, 2 5 AD = 4 15 AD = 2. Combining these 5 and QR = QD, RD = 1 3 AD, 1 7 AD = 4 21 AD: Fom these PQ QR = A seies of numbes is dened as follows: u 1 = a, u 2 = b, u n+1 = 1 2 (u n + u n,1 ) fo n 2. Pove that lim n!1 u n exists. Expess the value of the limit in tems of a and b. Solution by the Editos. Fo the ecuence u n+1 = 1 2 (u n + u n,1 ) we obtain the associated euation 2 2,, 1 = 0, which has oots =, 1 ; 1. Thus we seek a 2 solution of the fom u n = X,, 1 2 n + Y.Fom u1 = a and u 2 = b we get, 1 2 X + Y = a 1 4 X + Y = b so that X + 4 a+2b (b, a) and Y =. It is now easy to check by induction 3 3 that u n = 4 3 (b,, a), 2 1 n + 1+2b and as n!1,u 3 n! 1 3 a+2b In a plane V a cicle C is given with cente M. P is a point not on the cicle C. P B M V A

11 332 (a) Pove that fo a xed point P, AP 2 + BP 2 is a constant fo evey diamete AB of the cicle C. (b) Let AB be any diamete of C and P a point on a xed sphee S not intesecting V. Detemine the point(s) P on S such that AP 2 + BP 2 is minimal. Solution by Jamie Batuwantudawe, student, Si Winston Chuchill High School, Calgay. (a) With 4PAB,wecan join P and M to ceate two new tiangles, 4PMA and 4PMB. Let \PMA =. Then \PMB = 180,. Because M is the cente of cicle C and A and B both lie on cicle C, wehave MA = MB =, the adius of the cicle. and By the Law of Cosines, BP 2 = MP 2 + 2,2MPcos(180, ) = MP MPcos AP 2 so AP 2 + BP 2 =2MP = MP 2 + 2, 2MPcos The ight hand side is a constant depending only on the adius of the cicle and the distance of P fom the cente. (b) Fom (a), we know that AP 2 + BP 2 =2MP Fo any point P on sphee S, the adius of the cicle will emain constant. Theefoe the only vaiable aecting the sum AP 2 + BP 2 is MP, the distance fom the point P to the cente of the cicle. AP 2 +BP 2 will be a minimum when MP is minimum. Theefoe we ae looking fo the point on the sphee closest to M. Let T be the cente of the sphee S, D be the point on the segment MT that lies on the sphee, and D 0 be any othe point on S. We know that MD+DT < MD 0 +D 0 T because the shotest distance between M and T is a staight line. We know that DT = D 0 T. Thus MD < MD 0. Thus D is the point on the sphee which minimizes the sum. That completes the Cone fo this issue. Send me you nice solutions and genealizations as well as Olympiad contests.