REGIONAL MATHEMATICAL OLYMPIAD, 2017 (Maharashtra

RMO2017NATIONAL BOARD FOR HIGHER MATHEMATICSANDHOMI BHABHA CENTRE FOR SCIENCE EDUCATIONTATA INSTITUTE OF FUNDAMENTAL RESEARCHREGIONAL MATHEMATICAL OLYMPIAD, 2017(Maharashtra and Goa Region)TEST PAPER WITH SOLUTION& ANSWER KEYDate: 08th October, 2017 Duration: 3 HoursCorporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-1

MUMBAI & GOA REGION RMO – 2017 08-10-2017Time : 3 hoursOctober 08, 2017Total marks: 100Instruction1.Calculators (in any form) and protractors are not allowed.2.Rulers and compasses are allowed.3.Answer all the questions. Draw neat Geometry diagrams.4.Maximum marks are mentioned next to each question.5.Answerer to each question should start on a new page, clearly indicate the question number.6.Mathematical reasoning will be taken into consideration while assessing the answers.1.Consider a chessboard of size 8 units 8 units (i.e. each small square on the board has a sidelength of 1 unit). Let S be the set of all the 81 vertices of all the squares on the board. What isnumber of line segments whose vertices are in S1 and whose length is a positive integer ? (Thesegments need not be parallel to the sides of the board.)Sol.Number of line segment parallel to co-ordinate axis of 1 unit length equals to 8 9 2.Number of line segment parallel to co-ordinate axis of 2 unit length equals to 7 9 2.Number of line segment parallel to co-ordinate axis of 3 unit length equals to 6 9 2. and soon .Number of line segment parallel to co-ordinate axis of 8 unit length equals to 1 9 2.Number of line segment not parallel to co-ordinate axis of 5 unit length equals to 5 8 2 2.Number of line segment not parallel to co-ordinate axis of 10 unit length equals to 7 3 2 2.Total line segment equals to 780.2.For any positive integer n, let d(n) denotes the number of positive divisors of n; and let (n) denotethe number of elements from the set {1, 2, . n} that are coprime to n.(For example, d(12) 6 and (12) 4.)Find the smallest positive integer n such that d( (n)) 2017.Sol.d( (n)) 2017As 2017 is a prime number 1 1 n) n 1 1 pp1 2 1 . 1 pn where p1, P2 , Pn are the prime factors of nLet (n) tCorporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-2

MUMBAI & GOA REGION RMO – 2017 08-10-2017d(t) 2017 for minimum we have t is also minimum2016t 22016 (n) 2 1 1 1 2016n 1 1 . 1 2ppp1 2 m n 22017m 1 , p1 23.Let P(x) and Q(x) be polynomials of degree 6 and degree 3 respectively, such that:P(x) Q (x)2 Q(x) x2 – 6, for all x R.If all the roots of P(x) are real numbers, them prove that there exists two roots of P(x), any , such that 1.Sol.P(x) (Q(x))2 Q(x) x2 – 621 25 P(x) (Q(x) x2 –2 4 If all the roots of P(x) 0 are real then P(x) 0 between the roots. If P(x) 0 then21 25 2 (Q(x) 2 x – 4 0 2Now1 (Q(x) 2 is always positive then x 2 25 5 5 0 x , 4 2 2 5 5 Now let the roots of P(x) be 1, 2, 3, 4, 5, 6 and they all belongs to x , 2 2 So six roots be between the length of interval 5 so by pigeonhole principal there will be atleast onepair of roots such that 1Hence provedCorporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-3

MUMBAI & GOA REGION RMO – 2017 08-10-20174.Let 1, 2, 3, 40 be fourty parallel lines. As shown in the diagram, let m be another line thatintersects the lines 1 to 40 in the points A1, A2, A3, . A40 respectively. Similarly, let n beanother line that intersects the lines 1 to 40 in the points B1, B2, B3, . B40 respectively.Given that A1B1 1, A40B40 14, and the areas of the 39 trapeziums A1B1 B2A2 , A2B2B3A3, .,A39B39B40 A40 are all equal ; then count the number of segments A, B whose length is apositive integer; where i {1, 2, .40}B1A1A2A3I1I2I3B2B3B4I39 A4I40 A5Sol.B39B40Area A1B1C is1 2 sin sin x12sin Area A2B2C is1 2 sin sin x22sin . & so onCA1A2Now B1x2x2B21 2 sin sin 21 sin sin 2x1x 2 – x12 x12x 3 – x 222sin 2sin x12 , x 22 , x32 , . are in A.P.Now, x12 1. & x 290 196 common difference 5.2 x12 , x 24 16 , x82 36, x17 81, x 225 121, x 240 196 x1 1, x4 4, x8 6, x17 9, x25 11, x40 145.If a, b, c, d R such that a b c d 0 and a d b c; then prove that:(a b) (c d)2 a 2 b2 – c 2 d2Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-4

MUMBAI & GOA REGION RMO – 2017 08-10-2017Sol.Let a n k, b n, c m k, d m, d m , k 0 , m 0, n 0then bc – ad mn nk – mn – mk k (n – m) 0 (bc – ad) 0. b c a d 2acbd 2 22 2222 2 a c b c a d b d a c b d 2acbd 2222 (a b ) (c d ) (ac bd) 2(a 2 b2 )(c 2 d2 ) (ac bd)–2ac – 2bd – 2 (a 2 b2 ) c2 d2 a2 b2 c2 d2 – 2ac – 2bd (a2 b2) (c2 d2) –2 (a 2 b2 ) (a – c)2 (b – d)2 2(a – c)2 (a 2 b2 ) c 2 d2(a 2 b2 ) c 2 d2 c2 d2 222(a c) (a 2 b2 ) c 2 d2(a b) (c d)2 (a 2 b2 ) c 2 d2Hence proved 6.Let ABC be acute - angled; and let be its circumcircle, let D be a point on minor arc BC of .Let E and F be points on lines AD and AC respectively, such that BE AD and DF AC. Provethat EF BC if and only if D is the midpoint arc BC.Sol.Given: Let ABC be acute - angled; and let be its circum circle, let D be a point on minor arc BCof . Let E and F be points on lines AD and AC respectively, such that BE AD and DF AC.To Prove:(i) EF BC assuming D is mid point of arc BC(ii) D is mid point of BC assuming EF BCproof:(i) BAE EAF AE c cos AP A2(since arc BD arc DC)A22bcAcosb c2Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-5

MUMBAI & GOA REGION RMO – 2017 08-10-2017 B C B C 2R 2sin cos AB b c 2 2 AP2b2bNow BCD BAD A2A a 22DC cossince projection DC along BC isNow angle DFC C FDC .(i)B2A2B C 2 2AcosaA B C B C 2 OF DC cos sec cos A222222 sin2AFa AC 2b B C cos 2 2 Asin2from (i) and (ii) A B C 2R 2cos cos 2 2 2 2b .(ii)AE AF EF PCAP AC EF BCHence proveCorporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-6

MUMBAI & GOA REGION RMO – 2017 08-10-2017Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-7

MUMBAI & GOA REGION RMO – 2017 08-10-2017Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005Website : www.resonance.ac.in E-mail : contact@resonance.ac.inToll Free : 1800 258 5555 CIN: U80302RJ2007PLC024029RMO081017-8

REGIONAL MATHEMATICAL OLYMPIAD, 2017 (Maharashtra and Goa Region) TEST PAPER WITH SOLUTION & ANSWER KEY . Mathematical reasoning will be taken into consideration while assessin