IMAR 017 Problema 1 Fie P u puct situat î iteriorul uui triughi ABC Dreapta AP itersectează latura BC î puctul D ; dreapta BP itersectează latura CA î puctul E ; iar dreapta CP itersectează latura AB î puctul F Cercul de diametru BC itersectează cercul de diametru AD î puctele a şi a ; cercul de diametru CA itersectează cercul de diametru BE î puctele b şi b ; iar cercul de diametru AB itersectează cercul de diametru CF î puctele c şi c Arătaţi că puctele a, a, b, b, c, c sut cociclice Problema Fie u umăr atural eul Petru fiecare umăr atural eul k, otăm cu r k restul împărţirii lui la k Arătaţi că r k > Problema 3 O mulţime specială este o mulţime de umere aturale impare, astfel îcât iciu elemet al său u divide u alt elemet şi orice submulţime de cardial 3 coţie u elemet care divide suma celorlalte două O mulţime specială este maximală, dacă ea u este coţiută î icio altă mulţime specială Determiaţi umărul de elemete pe care îl poate avea o mulţime specială maximală Problema 4 Fie u umăr îtreg mai mare sau egal cu 3 şi fie P mulţimea tuturor -goaelor plaare simple), care u au icio pereche de laturi disticte paralele sau de-a lugul uei aceleiaşi drepte Petru fiecare poligo P di P, fie f P ) umărul miim de triughiuri formate de drepte-suport ale laturilor lui P, a căror reuiue acoperă poligoul P Determiaţi valoarea maximă pe care o poate lua f P ), câd P parcurge mulţimea P
IMAR 017 Solutios Problem 1 Let P be a poit i the iterior of the triagle ABC, ad let the lies AP, BP, CP meet the sides BC, CA, AB respectively at the poits D, E, F Let the circles o diameters BC ad AD meet at poits a ad a ; the circles o diameters CA ad BE meet at poits b ad b ; ad the circles o diameters AB ad CF meet at poits c ad c Show that the poits a, a, b, b, c, c lie o a circle Solutio Let A, B, C ad P have positio vectors a, b, c ad p = α α + + γ a + α + + γ b + γ α + + γ c, where α,, γ are the barycetric coordiates of P The D has positio vector + γ b + γ + γ c, so the circles o diameters BC ad AD have equatios r b) r c) = 0 ad r a) r respectively; alteratively, but equivaletly, the latter reads + γ b r a) r b) + γr a) r c) = 0 γ ) + γ c = 0, Ay liear combiatio of the two is the equatio of a circle or straight lie through a ad a I particular, the liear combiatio formed by multiplyig the first equatio by γ ad the secod by α, ad addig, is αr a) r b) + γr b) r c) + γαr c) r a) = 0 The symmetry of this equatio shows that this circle or, possibly, straight lie) also passes through b, b ad c, c Fially, otice that α + γ + γα is positive whe P lies iside the triagle ABC, so the locus is ideed a circle cetred at the poit whose barycetric coordiates are α + γ), γ + α), γα + ), respectively Remark The coditio for the locus to be a straight lie is α + γ + γα = 0, which implies that P lies o a ellipse through A, B ad C that has tagets at A, B ad C parallel to BC, CA ad AB, respectively Problem Let be a positive iteger For each positive iteger k, let r k deote the remaider leaves upo divisio by k Prove that r k > Solutio Let m =, ad split the set of itegers 1 through ito m + 1 pairwise disjoit subsets J 0, J 1,, J m, where J k cosists of all umbers of the form k l + 1), l = 0,, / k+1 1/
If j is a member of J k, the r j k, uless j = k i which case r j = 0 If k > 3 = m, the J k cosists of k aloe, so it cotributes othig to sum i questio Cosequetly, r k = r j k J k 1) = j J k = m + 1) 3 m +1 1) > k k+1 1 > 3 + 3 > k k+1 3 ) Remarks The lower boud i the statemet is rough for the sum S = r k uder cosideratio It has bee cojectured that S c for some suitable positive costat c, but it is ot eve kow whether this is the case for ifiitely may values of It has, however, bee show that S > /13 ) for ifiitely may values of Clearly, S 1++ + 1) = 1)/ It has also bee cojectured that S /4 for all but fiitely may values of, but this questio is still ope as well Problem 3 A special set is a set of positive odd itegers o elemet of which divides aother, ad each 3-elemet subset of which has a member dividig the sum of the other two A special set is maximal if it is cotaied i o other special set Determie the umber of elemets a maximal special set may have Solutio Leavig aside the trivial case {1}, a maximal special set may have oly 3, 4 or 5 elemets Begi by oticig that if a < b are positive odd itegers, ad a does ot divide b, the a, b ad b a form a special set, so a maximal special set has at least three elemets At the other extreme, a special set i particular, oe that is maximal has at most five elemets The proof relies o the three facts below: 1) If a > b > c form a special set, the b + c is ot divisible by a This is because a is odd, ad b + c is a positive eve iteger less tha a ) If a > b are members of a special set S, the at most oe of the members of S less tha b does ot divide the sum a + b Suppose, if possible, c ad d are distict members of S less the b, either of which divides the sum a + b By 1), a divides either b + c or b + d, so a + c ad a + d are both divisible by b The c d = a + c) a + d) is a positive iteger less tha b ad divisible by b a cotradictio 3) If a, b, c, d form a special set, ad a + b ad a + c are both divisible by d, the b + c is ot divisible by d Otherwise, d would divide a + b) + a + c) b + c) = a, which is impossible, sice d is odd ad does ot divide a We are ow i a positio to prove that a special set has at most five elemets Suppose, if possible, a 1, a, a 3, b 1, b, b 3 are pairwise distict members of a special set We may ad will assume a 1 > a > a 3 > max b 1, b, b 3 ) Fix a pair of distict idices i ad j, ad write {i, j, k} = {1,, 3} By ), some b divides both a i + a k ad a j + a k, by 3), that b does ot divide a i + a j, so, with referece agai to ), it is the uique b ot dividig a i + a j Cosequetly, the three b s may be labelled so that b i ad b j both divide a i + a j, while b k does ot, {i, j, k} = {1,, 3}
By 1), a 1 does ot divide a 3 +b, ad sice b does ot divide a 1 +a 3, it follows that a 1 +b is divisible by a 3 Similarly, a + b 1 is divisible by a 3, ad hece so is a 1 + b ) + a + b 1 ) = a 1 +a )+b 1 +b ) Fially, sice a 1 +a is divisible by a 3, by ), so is b 1 +b, i cotradictio with 1) Cosequetly, a special set i particular, oe that is maximal has at most five elemets Next, we show that 5-elemet special sets actually exist Clearly, the umbers 3, 5, 7 form a special set To elarge this set to a 4-elemet special set by adjoiig a positive odd iteger k, otice that k divides o -term sum from {3, 5, 7}, to ifer that k satisfies oe of the two systems of liear cogrueces below: k + 1 0 mod ) k + 5 0 mod 3) k + 3 0 mod 7) k + 7 0 mod 5) or k + 1 0 mod ) k + 3 0 mod 5) k + 5 0 mod 7) k + 7 0 mod 3) By the Chiese remaider theorem, each of these systems has ifiitely may solutios; i each case, two solutios differ by a multiple of 3 5 7 = 10 The least positive solutio of the former is 193, ad the least positive solutio of the latter is 107 To elarge the set {3, 5, 7, 193} to a 5-elemet special set by adjoiig a positive odd iteger k, otice agai that k divides o -term sum from {3, 5, 7, 193}, to ifer that k satisfies the system of liear cogrueces k + 1 0 mod ) k + 3 0 mod 5) k + 5 0 mod 7) ; k + 7 0 mod 3) k + 7 0 mod 193) clearly, 3 ad 5 both divide k + 193 As before, the Chiese remaider theorem settles the case; icidetally, the least positive solutio is 3467, ad all five umbers are prime Similarly, the set {3, 5, 7, 107} exteds to a 5-elemet special set by adjoiig ay positive odd iteger k satisfyig the system of liear cogrueces k + 1 0 mod ) k + 5 0 mod 3) k + 3 0 mod 7) ; k + 7 0 mod 5) k + 7 0 mod 107) clearly, 3 ad 5 both divide k +107 I this case, the least positive solutio is 10693 = 17 37 Remark The 5-elemet special sets below are obtaied i the same way: {3, 5, 13, 17, 1767 = 31 557}, {3, 5, 17, 97, 14353 = 31 463}, {3, 7, 11, 35 = 5 47, 6309}, {3, 7, 11, 437 = 19 3, 6095 = 5 31 389} We ow show that the special set cosistig of 3, 5, 13, 17 is maximal Suppose, if possible, that k is a positive odd iteger such that 3, 5, 13, 17, k form a special set It is easily see that k divides o -term sum from {3, 5, 13, 17} 3
We first show that k + 5 is divisible by 3 if ad oly if k + 3 is divisible by 5; sice oe holds, so does the other If k + 5 is divisible by 3, the k + 13 is ot, so k + 3 is divisible by 13 It follows that k + 5 is ot divisible by 13, so k + 13 is divisible by 5, showig that k + 3 is ideed divisible by 5 Coversely, if k + 3 is divisible by 5, the k + 17 is ot, so k + 5 is divisible by 17 It follows that k + 3 is ot divisible by 17, so k + 17 is divisible by 3, showig that k + 5 is ideed divisible by 3 By the precedig, k + 3 is divisible by 13, ad k + 5 is divisible by 17 The former implies that k + 17 is ot divisible by 13, ad the latter implies that k + 13 is ot divisible by 17 To derive a cotradictio, recall that k divides o -term sum from {3, 5, 13, 17}; i particular, it does ot divide the sum 13 + 17 Fially, we show that the special set cosistig of 11, 15, 51 is maximal Suppose, if possible, that k is a positive odd iteger such that 11, 15, 51, k form a special set, ad otice that k divides o -term sum from {11, 15, 51} Cosideratio of 15, 51 ad k shows that k is divisible by 3 It the follows that of the two umbers 15 ad 51, oe divides k + 11, so k + 15 ad k + 51 are both divisible by 11, which is clearly impossible Problem 4 Let be a iteger greater tha or equal to 3, ad let P be the collectio of all plaar simple) -gos o two distict sides of which are parallel or lie alog some lie For each member P of P, let f P ) be the least cardial a cover of P by triagles formed by lies of support of sides of P may have Determie the largest value f P ) may achieve, as P rus through P Solutio The required maximum is This follows from the fact that the image of f cosists of the first positive itegers Iduct o to show that f P ) for all P i P The base case = 3 is clear If P is covex, the f P ) = 1, sice P has three sides whose lies of support form a triagle that covers P simply exted two o-adjacet sides till they meet to obtai a polygo with fewer sides ad proceed iductively If P is ot covex, cosider a vertex v iterior to the covex hull of P Exted oe of the sides through v beyod v till it first meets agai the boudary, to split P ito a 1 -go P 1 ad a -go P, where 1 + + Hece f P ) f 1 P 1 )+f P ) 1 + Give a positive iteger m, we ow describe a polygo P = a 0 a 1 a 1 i P such that f P ) = m Cosider a parallelogram aa 0 a 1 a m+1, let a,, a m be iterior poits of the triagle a 0 a 1 a m+1 formig, alog with a 1 ad a m+1, a m + 1)-poit cofiguratio i strictly covex positio, ad let a m+,, a 1 be iterior poits of the triagle aa 0 a m+1 formig, alog with a 0 ad a m+1, a m)-poit cofiguratio i strictly covex positio; clearly, it is always possible to choose the latter so that P has o parallel sides For coveiece, a triagle formed by the lies of support of three sides of P will be called a peritriagle To show that f P ) m, otice that, for each idex i i the rage 1 through m, coverig the midpoit of the side a i a i+1 of P requires a peritriagle i lyig i the same half-plae as a 0 relative to the lie a i a i+1, oe side of which cotais the lie-segmet a i a i+1 Sice, for distict idices i ad j i the rage 1 through m, the lie a i a i+1 separates a 0 ad the midpoit of the lie-segmet a j a j+1, the k are pairwise distict, so f P ) m To coclude that f P ) = m, otice that P is covered by the m peritriagles formed by the lies a 0 a 1, a i a i+1 ad a i+1 a i+, where i rus from 1 through m 4