{ks=kh; xf.kr vksfyafi;km & 2014

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Oscar Hugo Rogers
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1 REGIONAL MATHEMATICAL OLYMPIAD 0 {ks=kh; xf.kr vksfyafi;km & 0 Time : 3 Hours December 07, 0 Instructions Caculators (in any form) and protractors are not allowed fd lhhkhrjg d sx.kd rfkkpkankd siz;ksx d hvuqefr ughagsa Rulers and compasses are allowed isekukrfkkijd kj d siz;ksx d hvuqefr gsa Answer all the questions lhkhiz'uksad smùkj nhft, A All questions are carry equal marks. Maximum marks : 0 lhkhiz'uksad svad leku gsa]vf/kd re vad %0 Answer to each question should start on a new page. Clearly indicate the question number. izr;sd iz'u d kmùkj u, ist lsizkjahkd hft, A iz'u Ø ekad Li"V : i lsbafxr d hft, A. Let ABC be an acute angled triangle and suppose ABC is the largest angle of the triangle. Let R be its circumcentre. Suppose the circumcircle of triangle ARB cuts AC again in X. Prove that RX is perpendicular to BC. Ekku yhft, fd ABC,d U;wu d ks.kf=khkqt gsrfkk ABC f=khkqt d klclscm+kdks.kgsa eku yhft, fd R bld k ifjd suæ gsa ;g Hkhekfu;sfd f=khkqt ARB d kifjo`ùk AC d ksiqu%x ij d kvrkgsa fl) d hft, fd RX, BC d s yecor gsa Lets prove that a From the figure we can see that r from R to BC cuts the circle through R, A & B on the line AC.

2 In the figure AD = DB & P is centre of circumcircle of ARB APR = 80 & RXA = 90 Because APR = RXA X must lie on circumference Hence proved. Find all real numbers x and y such that. og lhkhoklrfod la[;k, x rfkky Kkr d hft, ft ud sfy, x + y + x (y + ) x + y xy + x 0 x (y + ). x + (y + ) 0 If Roots of the equation, x (y + ). x + (y + ) = 0 are &, x (, ) On solving the given equation, (y ) (y ) 8(y ) (y ) 6y 6y (y ) (y ) x to be real, y = & (x, y) =, 3. Prove that there does not exist any positive integer n < 30 such that n (30 n) is a multiple of 30. fl) d hft, fd,slsd ksbz/kukred iw.kkzad n < 30 d k vkflrro ughagsft ld sfy, 30 d k,d xq.kd n (30 n) gsa If n (30 n) is multiple of 30. Then n (30 n) = 30 k n 30 n + 30 k = 0 30 (30) (30)(k) n = 30 30(30 k) = 30 contains all prime factors for being 30 (30 k) a perfect square (30 k) 30 k 0 n 30 Therefore there does not exist any positive integer n < 30 such that n (30 n) is a multiple of 30. Find all positive real numbers x, y, z such that. og lkjh/kukred la[;k, x, y, z Kkr d hft, ft ld sfy, x y + z = 0, y z + 0, z x + y = 0 Because of Sym y = z y = z = 0

3 5. Let ABC be a triangle. Let X be on the segment BC such that AB = AX. Let AX meet the circumcircle of triangle ABC again at D. Show that the circumcentre of BDX lies on. Eku yhft, d habc,d f=khkqt gs]eku yhft, fd BC d k,d vuqhkkx X bl izd kj gsfd AB = AX A eku yhft, fd AX f=khkqt ABC d sifjo`ùk ij iqu% D ij feyrh gsa fn[kkb;sfd BDX d k ifjd suæ d h ifjf/k ij fufgr gsa Lets assume that circumcentre of BXD lies on circumcircle of ABC Our target is to prove that FEX = BDX Here XEF = 90 = FEB & BEA = BDA = 90 BEA = FEX = 90 = BDX 6. For any natural number n, let S(n) denote the sum of the digits of n. Find the number of all 3 digit numbers n such that S (S(n)) =. fd lhizkd `r la[;k n d sfy,]eku yhft, fd S (n), n d svad ksad s;ksx d ksizd V d jrkgsa mu lhkhrhu vad ksad h la[;kn d hd qy la[;kkkr d hft, ft ud sfy, S (S(n)) = A Let n = abc (a three digit number), 9 a Where 9 b 0 9 c 0 S (n) = a + b + c Now a + b + c can be a two digit number or a single digit number Lets assume a + b + c = de (either a two digit number or single digit number) Now S (S(n)) = d + e S (S(n)) = d + e = (d, e) can be (0, ), (, ) or (, 0) a + b + c = Or, a + b + c = Or, a + b + c = 0 No of solution of the above equation will be, (coefficient of x + coefficient of x + coefficient of x 0 ) in the expansion of (x + x + + x 9 ). (x 0 + x + x + + x 9 ). Therefore number of three digit numbers = number of the solution of the above equation = 00 3

4 REGIONAL MATHEMATICAL OLYMPIAD BIHAR-0 {ks=kh; xf.kr vksfyafi;km] fcgkj&0 Time : 3 Hours December 07, 0 Instructions Caculators (in any form) and protractors are not allowed fd lhhkhrjg d sx.kd rfkkpkankd siz;ksx d hvuqefr ughagsa Rulers and compasses are allowed isekukrfkkijd kj d siz;ksx d hvuqefr gsa Answer all the questions lhkhiz'uksad smùkj nhft, A All questions are carry equal marks. Maximum marks : 0 lhkhiz'uksad svad leku gsa]vf/kd re vad %0 Answer to each question should start on a new page. Clearly indicate the question number. izr;sd iz'u d kmùkj u, ist lsizkjahkd hft, A iz'u Ø ekad Li"V : i lsbafxr d hft, A. Let ABC be an acute angled triangle and suppose ABC is the largest angle of the triangle. Let R be its circumcentre. Suppose the circumcircle of triangle ARB cuts AC again in X. Prove that RX is perpendicular to BC. Ekku yhft, fd ABC,d U;wu d ks.kf=khkqt gsrfkk ABC f=khkqt d klclscm+kdks.kgsa eku yhft, fd R bld k ifjd suæ gsa ;g Hkhekfu;sfd f=khkqt ARB d kifjo`ùk AC d ksiqu%x ij d kvrkgsa fl) d hft, fd RX, BC d s yecor gsa Lets prove that a From the figure we can see that r from R to BC cuts the circle through R, A & B on the line AC.

5 In the figure AD = DB & P is centre of circumcircle of ARB APR = 80 & RXA = 90 Because APR = RXA X must lie on circumference Hence proved. Find all real numbers x and y such that. og lhkhoklrfod la[;k, x rfkky Kkr d hft, ft ud sfy, x + y + x (y + ) x + y xy + x 0 x (y + ). x + (y + ) 0 If Roots of the equation, x (y + ). x + (y + ) = 0 are &, x (, ) On solving the given equation, (y ) (y ) 8(y ) (y ) 6y 6y (y ) (y ) x to be real, y = & (x, y) =, 3. Prove that there does not exist any positive integer n < 30 such that n (30 n) is a multiple of 30. fl) d hft, fd,slsd ksbz/kukred iw.kkzad n < 30 d k vkflrro ughagsft ld sfy, 30 d k,d xq.kd n (30 n) gsa If n (30 n) is multiple of 30. Then n (30 n) = 30 k n 30 n + 30 k = 0 30 (30) (30)(k) n = 30 30(30 k) = 30 contains all prime factors for being 30 (30 k) a perfect square (30 k) 30 k 0 n 30 Therefore there does not exist any positive integer n < 30 such that n (30 n) is a multiple of 30. Find all positive real numbers x, y, z such that. og lkjh/kukred la[;k, x, y, z Kkr d hft, ft ld sfy, x y + z = 0, y z + 0, z x + y = 0 Because of Sym y = z y = z = 0

6 5. Let ABC be a triangle. Let X be on the segment BC such that AB = AX. Let AX meet the circumcircle of triangle ABC again at D. Show that the circumcentre of BDX lies on. Eku yhft, d habc,d f=khkqt gs]eku yhft, fd BC d k,d vuqhkkx X bl izd kj gsfd AB = AX A eku yhft, fd AX f=khkqt ABC d sifjo`ùk ij iqu% D ij feyrh gsa fn[kkb;sfd BDX d k ifjd suæ d h ifjf/k ij fufgr gsa Lets assume that circumcentre of BXD lies on circumcircle of ABC Our target is to prove that FEX = BDX Here XEF = 90 = FEB & BEA = BDA = 90 BEA = FEX = 90 = BDX 6. For any natural number n, let S(n) denote the sum of the digits of n. Find the number of all 3 digit numbers n such that S (S(n)) =. fd lhizkd `r la[;k n d sfy,]eku yhft, fd S (n), n d svad ksad s;ksx d ksizd V d jrkgsa mu lhkhrhu vad ksad h la[;kn d hd qy la[;kkkr d hft, ft ud sfy, S (S(n)) = A Let n = abc (a three digit number), 9 a Where 9 b 0 9 c 0 S (n) = a + b + c Now a + b + c can be a two digit number or a single digit number Lets assume a + b + c = de (either a two digit number or single digit number) Now S (S(n)) = d + e S (S(n)) = d + e = (d, e) can be (0, ), (, ) or (, 0) a + b + c = Or, a + b + c = Or, a + b + c = 0 No of solution of the above equation will be, (coefficient of x + coefficient of x + coefficient of x 0 ) in the expansion of (x + x + + x 9 ). (x 0 + x + x + + x 9 ). Therefore number of three digit numbers = number of the solution of the above equation = 00 3

Regional Mathematical Olympiad -2011

Regional Mathematical Olympiad -2011 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