42nd International Mathematical Olympiad

Transcription

1 nd Interntionl Mthemticl Olympid Wshington, DC, United Sttes of Americ July 8 9, 001 Problems Ech problem is worth seven points. Problem 1 Let ABC be n cute-ngled tringle with circumcentre O. Let P on BC be the foot of the ltitude from A. Suppose tht BCA ABC 0. Prove tht CAB COP 90. Problem Prove tht 8 b c b b 8 c c c 8 b 1 for ll positive rel numbers, b nd c. Problem Twenty-one girls nd twenty-one boys took prt in mthemticl contest. Ech contestnt solved t most six problems. For ech girl nd ech boy, t lest one problem ws solved by both of them. Prove tht there ws problem tht ws solved by t lest three girls nd t lest three boys. Problem Let n be n odd integer greter thn 1, nd let k 1, k,,k n be given integers. For ech of the n permuttions 1,,, n of 1,,,n, let n S k i i. i 1 Prove tht there re two permuttions b nd c, b c, such tht n is divisor of S b S c.

2 IMO 001 Competition Problems Problem 5 In tringle ABC, let AP bisect BAC, with P on BC, nd let BQ bisect ABC, with Q on CA. It is known tht BAC 60 nd tht AB BP AQ QB. Wht re the possible ngles of tringle ABC? Problem 6 Let, b, c, d be integers with b c d 0. Suppose tht c b d b d c b d c. Prove tht b c d is not prime.

3 IMO 001 Competition Problems Problems with s Problem 1 Let ABC be n cute-ngled tringle with circumcentre O. Let P on BC be the foot of the ltitude from A. Suppose tht BCA ABC 0. Prove tht CAB COP Let Α CAB, Β ABC, Γ BCA, nd COP. Let K nd Q be the reflections of A nd P, respectively, cross the perpendiculr bisector of BC. Let R denote the circumrdius of ABC. Then OA OB OC OK R. Furthermore, we hve QP KA becuse KQPA is rectngle. Now note tht AOK AOB KOB AOB AOC Γ Β 60. It follows from this nd from OA OK R tht KA R nd QP R. Therefore, using the Tringle Inequlity, we hve OP R OQ OC QC QP PC R PC. It follows tht OP PC, nd hence in COP, PCO. Now since Α 1 BOC PCO 90 PCO, it indeed follows tht Α 90. As in the previous solution, it is enough to show tht OP PC. To this end, recll tht by the (Extended) Lw of Sines, AB RsinΓ nd AC RsinΒ. Therefore, we hve BP PC ABcosΒ A C cos Γ R sin Γ cos Β sin Β cos Γ R sin Γ Β. It follows from this nd from 0 Γ Β Γ 90 tht BP PC R. Therefore, we obtin tht R OP BO OP BP R PC, from which OP OC, s desired.

4 IMO 001 Competition Problems We first show tht R CP CB. To this end, since CB RsinΑ nd CP ACcosΓ RsinΒcosΓ, it suffices to show tht 1 sinαsinβcosγ. We note tht 1 sinα sin Γ Β sinγcosβ sinβcosγ nd 1 sin Γ Β sinγcosβ sin ΒcosΓ since 0 Γ Β 90. It follows tht 1 sinβcosγ nd tht 1 sinαsinβcosγ. Now we choose point J on BC so tht CJ CP R. It follows from this nd from R CP CB tht CJ CB, so tht OBC OJC. Since OC CJ PC CO nd JCO OCP, we hve JCO OCP nd OJC POC. It follows tht OBC 90 Α or Α 90. On the one hnd, s in the third solution, we hve R CP CB. On the other hnd, the power of P with respect to the circumcircle of ABC is BP PC R OP. From these two equtions we find tht OP R BP PC PC CB BP PC PC, from which OP PC. Therefore, s in the first solution, we conclude tht Α 90. Problem Prove tht 8 b c b b 8 c c c 8 b 1 for ll positive rel numbers, b nd c. First we shll prove tht 8 b c, b c or equivlently, tht b c 8 b c. The AM-GM inequlity yields b c b c b c b c 8 b c. b 1 c 1 Thus

5 IMO 001 Competition Problems 5 b c 8 b c 8 b c, so 8 b c b c. Similrly, we hve b b 8 c b b c nd c c 8 b c b c. Adding these three inequlities yields 8 b c b b 8 c c c 8 b 1. Comment. It cn be shown tht for ny, b, c 0 nd Λ 8, the following inequlity holds: Λ b c b b Λ c c c Λ b 1 Λ. Problem Twenty-one girls nd twenty-one boys took prt in mthemticl contest. Ech contestnt solved t most six problems. For ech girl nd ech boy, t lest one problem ws solved by both of them. Prove tht there ws problem tht ws solved by t lest three girls nd t lest three boys. 1 We introduce the following symbols: G is the set of girls t the competition nd B is the set of boys, P is the set of problems, P g is the set of problems solved by g G, nd P b is the set of problems solved by b B. Finlly, G p is the set of girls tht solve p P nd B p is the set of boys tht solve p. In terms of this nottion, we hve tht for ll g G nd b B, P g 6, P b 6, b P g P b. We wish to prove tht some p P stisfies G p nd B p. To do this, we shll ssume the contrry nd rech contrdiction by counting (two wys) ll ordered triples p, q, r such tht p P g P b. With T p, g, b : p P g P b, condition (b) yields

6 6 IMO 001 Competition Problems T g G b B P g P b G B 1. (1) Assume tht no p P stisfies G p nd B p. We begin by noting tht G p P g 6 G nd B p 6 B. p P g G p P () (Note. The equlity in () is obtined by stndrd double-counting technique: Let Χ g, p 1 if g solves p nd Χ g, p 0 otherwise, nd interchnge the orders of summtion in p P g G Χ g, p.) Let P p P : G p, P p P : G p. Clim. p P G p G ; thus p P G p 5 G. Also p P B p B ; thus p P B p 5 B. Proof. Let g G be rbitrry. By the Pigeonhole Principle, conditions () nd (b) imply tht g solves some problem p tht is solved by t lest 1 6 boys. By ssumption, B p implies tht p P, so every girl solves t lest one problem in P. Thus G p G. p P () In view of () nd () we hve p P G p p P G p p P G p 5 G. Also, ech boy solves problem tht is solved by t lest four girls, so ech boy solves problem p P. Thus p P B p B, nd the clcultion proceeds s before using (). Using the clim just estblished, we find T p P G p B p p P G p B p p P G p B p p P G p p P B p 10 G 10 B 0 1. This contrdicts (1), so the proof is complete.

7 IMO 001 Competition Problems 7 Let us use some of the nottion given in the first solution. Suppose tht for every p P either G p or B p. For ech p P, color p red if G p nd otherwise color it blck. In this wy, if p is red then G p nd if p is blck then B p. Consider chessbord with 1 rows, ech representing one of the girls, nd 1 columns, ech representing one of the boys. For ech g G nd b B, color the squre corresponding to g, b s follows: pick p P g P b nd ssign p's color to tht squre. (By condition (b), there is lwys n vilble choice.) By the Pigeonhole Principle, one of the two colors is ssigned to t lest 1 1 squres, nd thus some row hs t lest blck squres or some column hs t lest 11 red squres. Suppose the row corresponding to g G hs t lest 11 blck squres. Then for ech of 11 squres, the blck problem tht ws chosen in ssigning the color ws solved by t most boys. Thus we ccount for t lest 11 6 distinct problems solved by g. In view of condition (), g solves only these problems. But then t most 1 boys solve problem lso solved by g, in violtion of condition (b). In exctly the sme wy, contrdiction is reched if we suppose tht some column hs t lest 11 red squres. Hence some p P stisfies G p nd B p. Problem Let n be n odd integer greter thn 1, nd let k 1, k,,k n be given integers. For ech of the n permuttions 1,,, n of 1,,,n, let n S k i i. i 1 Prove tht there re two permuttions b nd c, b c, such tht n is divisor of S b S c. Let S be the sum of S over ll n permuttions 1,,, n. We compute S mod n two wys, one of which depends on the desired conclusion being flse, nd rech contrdiction when n is odd. First wy. In S, k 1 is multiplied by ech i 1,,n totl of n 1 times, once for ech permuttion of 1,,n in which 1 i. Thus the coefficient of k 1 in S is n 11 n n 1. The sme is true for ll k i, so n 1 S k i. n i 1 Second wy. If n is not divisor of S b S c for ny b c, then ech S must hve different reminder mod n. Since there re n permuttions, these reminders must be precisely the numbers 0, 1,,, n 1. Thus (1) S n 1 n mod n. Combining (1) nd (), we get n n 1 k i n 1 n mod n. i 1 () ()

8 8 IMO 001 Competition Problems Now, for n odd, the left side of () is congruent to 0 modulo n, while for n 1 the right side is not congruent to 0 (n 1 is odd). For n 1 nd odd, we hve contrdiction. Problem 5 In tringle ABC, let AP bisect BAC, with P on BC, nd let BQ bisect ABC, with Q on CA. It is known tht BAC 60 nd tht AB BP AQ QB. Wht re the possible ngles of tringle ABC? Denote the ngles of ABC by Α 60, Β, nd Γ. Extend AB to P so tht BP BP, nd construct P on AQ so tht AP AP. Then BP P is n isosceles tringle with bse ngle Β. Since AQ QP AB BP AB BP AQ QB, it follows tht QP QB. Since AP P is equilterl nd AP bisects the ngle t A, we hve PP PP. Clim. Points B, P, P re colliner, so P coincides with C. Proof. Suppose to the contrry tht BPP is nondegenerte tringle. We hve tht PBQ PP B PP Q Β. Thus the digrm ppers s below, or else with P is on the other side of BP. In either cse, the ssumption tht BPP is nondegenerte leds to BP PP PP, thus to the conclusion tht BPP is equilterl, nd finlly to the bsurdity Β 60 so Α Β

9 IMO 001 Competition Problems 9 Thus points B, P, P re colliner, nd P C s climed. Since tringle BCQ is isosceles, we hve 10 Β Γ Β, so Β 80 nd Γ 0. Thus ABC is degree tringle. Problem 6 Let, b, c, d be integers with b c d 0. Suppose tht c b d b d c b d c. Prove tht b c d is not prime. 1 Suppose to the contrry tht b c d is prime. Note tht b c d d c b c m gcd d, b c for some positive integer m. By ssumption, either m 1 or gcd d, b c 1. We consider these lterntives in turn. Cse (i): m 1. Then gcd d, b c b c d b c d b c d d c 1 b c 1 gcd d, b c, which is flse. Cse (ii): gcd d, b c 1. Substituting c b d d b b c for the left-hnd side of c b d b d c b d c, we obtin d c d b c b c d.

10 10 IMO 001 Competition Problems In view of this, there exists positive integer k such tht c d k b c, b c d k d. Adding these equtions, we obtin b k b c d nd thus k c d k 1 b. Recll tht b c d. If k 1 then c d, contrdiction. If k then k k 1 b c d contrdiction., Since contrdiction is reched in both (i) nd (ii), b c d is not prime. The equlity c b d b d c b d c is equivlent to c c b b d d. (1) Let ABCD be the qudrilterl with AB, BC d, CD b, AD c, BAD 60, nd BCD 10. Such qudrilterl exists in view of (1) nd the Lw of Cosines; the common vlue in (1) is BD. Let ABC Α, so tht CDA 180 Α. Applying the Lw of Cosines to tringles ABC nd ACD gives d d cos Α A C b c b c cos Α. Hence cos Α d b c d b c, nd A C d d d b c d b c Becuse ABCD is cyclic, Ptolemy's Theorem gives A C B D b c d It follows tht b c d c b d d b c. c b d c c b c d d b c. (Note. Strightforwrd lgebr cn lso be used obtin () from (1).) Next observe tht b c d c b d d b c. The first inequlity follows from d b c 0, nd the second from b c d 0. () () Now ssume tht b c d is prime. It then follows from () tht b c d nd c b d re reltively prime. Hence, from (), it must be true tht c b d divides d b c. However, this is impossible by (). Thus b c d must not be prime. Note. Exmples of -tuples, b, c, d tht stisfy the given conditions re 1, 18, 1, 1 nd 65, 50,, 11.