Regional Mathematical Olympiad -2011

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Virginia Stephens
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1 RMO-011 EXMINTION REER POINT Regional Mathematical Olmpiad -011 Time : 3 Hours December 04, 011 Instructions : alculators (In an form) and protractors are not allowed. fdlh Hkh rjg ds dsydqsvj ;k dks.kekid dk iz;ksx oftzr gs Rulers and compasses are allowed. :j vksj deikl dk iz;ksx fd;k tk ldrk gs nswer all the questions. Maximum marks 100. lhkh iz'uksa ds mùkj nsa dq fu/kkzfjr vad 100 gsa nswer to each question should start on a new page. learl indicate the question number. izr;sd u, iz'u dk mùkj u, i`" ls izkjahk djsa iz'u la[;k dk Li"V mys[k djsa 1. Let be a triangle. Let D, E, F be points respectivel on the segments,, such that D, E, F concur at the point K. Suppose D/D = F/F and D = F. Prove that E = D.,d f=khkqt gs ekukfd D, E, F Øe'k% [k.m,, ij bl rjg gs fd D, E, F fcunq K ij laxkeh gsa eku hft, fd D/D = F/F rfkk D = F. fl) dhft, fd E = D. [1] Q D = F 1 = = x (let) 3 = (180 x) (linear propert). FK + KD = 180 DKF is cclic 5 = KF but 5 = 6 5 = 6 = ( 7) D F lso = D F F K D E (i) (ii) {in KF, exterior angle} (iii) (same segment) (iv) b Thalse theorem FD REER POINT, P Tower, Road No.1, IPI, Kota (Raj.) Ph.: / 5

2 RMO-011 EXMINTION REER POINT 4 = = D (corresponding) (v) Now in D D = D = x (b (i)) = D = + 7 D = 7 b equation (iv) and (vi) 5 = D E = D Proved. (b eqn. (iv) & (v)) (vi). Let (a 1, a, a 3,, a 011 ) be a permutation (that is a rearrangement) of the numbers 1,, 3,., 011. Show that there exist two numbers j, k such that 1 j < k 011 and a j j = a k k. ekukfd (a 1, a, a 3,, a 011 ) 1,, 3,., 011 la[;kvksa dk,d Øep; gs (vfkkzr~ iqufozu;kl gs) fn[kkb;s fd nks la[;k,a j, k bl rjg gs fd 1 j < k 011 rfkk a j j = a k k. [19] Let j, k taken all x axis and rearrangement (a 1, a, a 3,, a 011 ) be on -axis. a k = a j + a j 's a k = a j x O j k 011 Let k j = Hence for an j & k at diff. of '' Two points on -axis can be taken on difference of ''. 3. natural number n is chosen strictl between two consecutive perfect squares. The smaller of these two squares is obtained b subtracting k from n and the larger one is obtained b adding l to n. Prove that n kl is a perfect square.,d izkñfrd la[;k n nks Øekxr iw.kzoxks± ds chp esa pquh xbz gs bu nks oxks± esa NksVh okh la[;k n esa ls k dks?kvkus ij izkir gksrh gs rfkk cm+h okh la[;k n esa l dks tksm+us ij izkir gksrh gs fl) dhft, fd n kl,d iw.kz oxz gs [1] REER POINT, P Tower, Road No.1, IPI, Kota (Raj.) Ph.: / 5

3 RMO-011 EXMINTION REER POINT Let λ and (λ + 1) are the consecutive perfect squares. So n k = λ (i) n + l = (λ +1) (ii) Now, n kl = λ + k k [(λ + 1) n] [Using (i) & (ii)] = λ + k k [(λ + 1) λ k] = λ + k k [1 + λ k] = λ λk + k = (λ k) which is a perfect square 4. onsider a 0-sided convex polgon K, with vertices 1,,..., 0 in that order. Find the number of was in which three sides of K can be chosen so that ever pair among them has at least two sides of K between them. (For example ( 1, 4 5, 11 1 ) is an admissible triple while ( 1, 4 5, 19 0 ) is not.) [19] ekuk fd K,d 0-Hkqtk okk voeq[k cgqhkqt gs ftlds 'kh"kz 1,,..., 0 mlh Øe esa gs Kkr dhft, fd,sls fdrus rjhds gsa ftlesa K dh rhu Hkqtk, bl rjg pquh tk ldrh gsa fd muesa ls izr;sd ;qxe ds chp esa K dh de ls de nks Hkqtk,a gksa (mnkgj.kkfkz ( 1, 4 5, 11 1 ) xzkg; f=kd gs tcfd ( 1, 4 5, 19 0 ) ugha gs) Total was to select 3 sides = 0 3 Now, No. of was of selection when exactl three sides are common = 0 i.e. [( 1, 3, 3 4 ), ( 3, 3 4, 4 5 )] etc Now, No. of was when exactl two sides are common. No. of selection = 0 16 [If we select ( 1, 3 ) then third cannot be selected from 0 1 & 3 4 ] Now no. of was of selection when the selected two sides are let 1 & 3 4 (i.e. a gap of 1 side) = REER POINT, P Tower, Road No.1, IPI, Kota (Raj.) Ph.: / 5

4 RMO-011 EXMINTION REER POINT Here two cases arise for third side (i) Third side selected at a gap of two from above selected two sides was of selection of third side = 13 Required was = 0 13 (ii) Third side is at a gap of one side from an of above selected side Total was = 0 Hence total was of selection = = Let be a triangle and let 1, 1 be respectivel the bisectors of, with 1 on and 1 on. Let E, F be the feet of perpendiculars drawn from onto 1, 1 respectivel. Suppose D is the point at which the incircle of touches. Prove that D = EF. [19] ekuk fd,d f=khkqt gs rfkk 1, 1 Øe'k%, ds n~fohkktd gs] ij fcunq 1 gs rfkk ij fcunq 1 gs ekuk fd E, F Ec ds pj.k gsa tks fd Øe'k% ls 1, 1 ij [khaps x, gsa eku hft, fd D og fcunq gs ftl ij dk var% o`ùk dks Li'kZ djrk gs fl) dhft, D = EF + 1 π 1 F E I F = b sin E = c sin + FE = cosine rule (EF) = (F) + (E) + (F) (E) cos = b sin + c sin bc sin sin sin = (s a) EF = (s a) D = r cot REER POINT, P Tower, Road No.1, IPI, Kota (Raj.) Ph.: / 5

5 RMO-011 EXMINTION REER POINT = 4R sin sin sin cot = (s a) D = (s a) Hence D = EF x + x+ 6. Find all pairs (x, ) of real numbers such that = 1 [19] x + x+ oklrfod la[;kvksa ds lhkh ;qxe (x, ) bl rjg Kkr dhft, tgk = 1 x + x = 1 Q.M G.M 16 x + x x + x x + x+ + 1 ( x + x+ + ) x + x x x True if 1 1 x = ; = Ordered pair is 1 1, REER POINT, P Tower, Road No.1, IPI, Kota (Raj.) Ph.: / 5

2009 (Odd) xzqi&a ds lhkh 20 ç'uksa ds mùkj nsaa ¼izR;sd ds 1 vad gsaa½

2009 (Odd) xzqi&a ds lhkh 20 ç'uksa ds mùkj nsaa ¼izR;sd ds 1 vad gsaa½ 009 (Odd) Time : Hrs. Full Marks : 80 Pass Marks : 6 SemI-G Engg. Math.-I GROUP-A.(A) Write down the most correct answer for the following question from four given alternatives : = Answer all 0 questions