REGIONAL MATHEMATICAL OLYMPIAD-2010

Transcription

1 REGIONAL MATHEMATICAL OLYMPIAD-00 Time: 3 hrs December 05, 00 Instructions: Calculator (in any form) and protractors are not allowed. Rulers and compasses are allowed. Answer all the questions. Maximum marks: 00. Answer to each question should start on a new page. Clearly indicate the question number.. Let ABCDEF be a convex hexagon in which the diagonals AD, BE, CF are concurrent at O. Suppose the area of triangle OAF is the geometric mean of those of OAB and OEF; and the area of the triangle OBC is the geometric mean of those of OAB and OCD. Prove that the area of triangle OED is the geometric mean of those of OCD and OEF. [6] Alternate-: If A, A are on other sides of BC and AA meets BC at K, then [ABC] AK = [A'BC] KA' which can be proved easily by projecting A, A on BC and using similarity for two right triangles. Now, in the hexagon, let AC meet OB at M, GE meet OD at N and EA meet OF at P. By Ceva s theorem for triangle AGE, we have And from the fact, we have a c e,, b d f = where AM CN EP.. MC NE PA = a = [OAB]; b = [OBC] ; c = [OCD] ; d = [ODE] ; e = [OEF]; f = [OFA] So, a c e b d f = We also have ae = f ; ac = b

2 ce = d Alternate-: For any XYZ,[XYZ] = sinx.yz Now [OAF] = [OAB][OEF] sin γ.a f = sin β.sin α.afbe 4 4 sin γ.af = sin α.sin β.be () Similarly we get using [OBC] = [OAB] [OCD] sin β.bc = sin α.sin γ.ad () Multiplying these equations, we get, sin β.sin γ.abcf = sin αsinβsin γ.abde sin β.sin γ.cf = sin α.de Multiplying by de 4, we get sin β.ef sin.cd sin.de γ = α [OCD][OEF] = [OED]. Let P(x) = ax bx c, P(x) = bx cx a, P3(x) = cx ax b be three quadratic polynomials where a, b, c are non-zero real numbers. Suppose there exists a real

3 numbers such that P( ) = P( )=P3( ). Prove that a = b = c. [7] Alternate-: Adding the three equalities, we get (a + b + c) (k k ) = 0 ) a + b + c = 0 The system is then : ak bk + a + b = 0 bk + (a + b)k b = 0 eliminating k between these two lines we get (a + ab + b )(k ) = 0 k = would imply a = b = 0, impossible (a, b, c are non zero ) So a + ab + b = 0, impossible too if a, b are non zero. ) k k = 0 The system is then : k = k + (a b) k + (a c) = 0 (b c) k + (b a) = 0 And so Q.E.D. (a c) (b c) = (b a) (a b) (a b c) + 3(b c) = 0andsoa = b = c Alternate-: () : ax bx c = d () : bx cx a = d (3) : cx ax b = d ()-() : ()-(3) : (a b)x (b c)x (c a) = 0 (b c)x (c a)x (a b) = 0 Setting a b = u and b c = v (4) : (5) : ux vx + (u+ v) = 0 vx + (u+ v)x u = 0 (4)u+(5)(u+v) : (u + uv + v )x(x + ) = 0

4 If x = 0, (5) implies u = 0 and (4) implies u + v = 0and so u = v = 0 and so a = b = c If x =. (5) implies u = 0 and (4) implies u + v = 0and so u = v = 0 and so a = b = c if u + uv + v = 0, then (u + v) + 3v = 0and so u = v = 0and so u = v = 0and so a = b = c Q.E.D. Alternate-3: If we let P( α ) = P( α ) = P( 3 α), we get px qx r = 0; qx r x p = 0; r x px q = 0 To have a common real root α where p = a b; q = b c; r = c a. If a = b, then some manipulating gives b = c and we are done. If we let p, q, r 0, we have q r x x = 0 p p q r If this equations s roots are α, α, we get α+ α αα = + = and so, p p (α ) (α ) =. If the roots of so, r p x x = 0 are α, α,wehave( α )( α ) = which forces α = a q q px qx r qx rx p rx px q p:q:r= q: r: p = r: q p = q = r Let a b = b c a = k a = b + k,b = c + k,c = a + k (a+ b + c) = (a+ b + c) + 3k k = 0 a = b = c Alternate-4: Q (x) = P(x) P(x) = (a b)x (b c)x (c a); Q (x) = P(x) P(x) 3 = (b c)x (c a)x (a b); Q 3(x) P(x) 3 P(x) (c a)x (a b)x (b c) = =. Then α is real root of the equation Qi(x); so that Q i(x) 0 i=,,3;where f(x)

5 denoted the Discriminant of a quadratic function f in x. Now, using this we have, (b c) + 4(a b)(c a) 0; (c a) + 4(a b)(b c) 0; (a b) + 4(b c)(c a) 0. Summing these up and using the identity (a b) + (b c) + (c a) + (a b)(c a) We obtain, 0 (a b)(c a) = (ca+ ab bc a ) cyc cyc cyc = [(a b) + (b c) + (c a) ]; Possible if and only if = [(a b) + (b c) + (c a)] = 0; (a b) + (b c) + (c a) = 0 a = b = c. 3. Find the number of 4-digit numbers (in base 0) having non-zero digits and which are divisible by 4 but not by 8. [6] Mathod-: Let A be the set of -digit multiples of 4 composed of non zero digits. If we let B to be the set of -digit numbers which are multiples of 4 but not 8. Let C = A B. We can find that B = C = 9. We can prefix any of the digits, 4, 6, 8 to all elements of B to get the set B and any of the digits, 3, 5, 7, 0 to any of the elements of C to get the set C. Let A be their union. So, B' + C' = = 8and if abcd is the required 4-digit number, bcd belongs to A and so, the number of 4 digit numbers satisfying the required property is 9 8 = 79 Method-: We first find no. of nos. divisible by 4. Skeleton is (abcd)0 a, b are nonzero digits. and (cd)0 is two digit no. s.td 0,4 (cd) 0

6 90 no. of two digit nos. divisible by 4 = = 4 but 0, 40, 60, 80 are counted and they have a 0. so no. of possible values for ( cd)0 = 4 = 8 Hence no. of possible values for (abcd)0 = = 458 Now nos. divisible by 8. Skeleton is (pqrs)0 P is nonzero digit and (qrs)0 is 3-digit no. s.t. r,s 0,8 (qrs) No. of 3-digit nos. divisible by 8 = = 8 But 0, 60,.., 960 are counted which have a 0. There are such nos. So no. of possible values for (qrs)0 = = 90 hence no. of possible values for (pqrs)0 = 9.90 = 80 Hence reqd no. = = Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer. [7] Alternate-: Let 0 < x < y < z such integers (x, y, z) = (, 3, 6) is a solution. Let us show that there is no solution with x + y + z 0 x >, else we should have + = n, possible y z so x + y + z = 9 and we have very few cases to check :

7 n x + y + z = 9 (x,y,z) = (,3,4)but then + = is wrong 3 4 n x + y + z = 0 (x,y,z) = (,3,5)but then + = is wrong 3 5 Hence the answer (x, y, z) = (, 3, 6) Alternate-: Have + = a, + = b, + = c. This leads to y z x z x y x y z a + b + c + = = = k, and then to x y z + + =. Since x, y, z need be a + b + c + distinct, so are a, b, c, hence the only solution in, 3, 6. For minimum sum, take k =, and x, y, z to be, 3, Let ABC be a triangle in which angle A = 60 o. Let BE and CF be the bisectors of the angles B and C with E on AC and F on AB. Let M be the reflection of A in the line EF. Prove that M lies on BC. [7] Alternate-: Let D be the point on BC such that BFD = C. So we have BFD BCA, and hence AC DF =.BF = AF. Also note that BC AB BD =.BF= BC c a + b. Similarly, let D be the point on BC such that CED = B, then we have D E = AE and CD = b a + c. By cosine s law we get a = b + c bc, therefore BD + CD = c b a(b + c ) + b + c + = a b a c a ab bc ca

8 a(a + bc) + (b+ c)(b + c bc) a(a + bc) + (b + c)a = = = a a + ab + bc + ca a + ab + bc + ca Hence D = D. Now we have AF = FD and AE = ED, so AEDF is a kite, which implies that M = D. Therefore M lies on BC. Alternate-: I BE CF is the incentre of ABC and AI cuts BC and EF at D, P respectively. Because of FIE= 0, it follows that A, F, I, E lie on a circle with center U BC' is the polar of PWRT (U) U P is perpendicular to BC through a point M. Since cross ratio (I, A, P, D) is harmonic and M P M D, we deduce that M PU and BC bisect IM' A internally and externally, then UA = UI implies that AUIM is cyclic. But, since P lies on the diagonal FE of the rhombus FUEI formed by equilateral UEIand UFI, it follows that PU = PI AUIM is an isosceles trapezoid with symmetry axis EF M is the reflection of A about EF, i.e. M M'. Alternate-3: AFIE is cyclic. Circle AFC cut BC at D. (BF. BA)=(BI.BE)=(BD. BC). IDCE is cyclic, angle EIC = angle EDC=60.

9 ABDE is cyclic. AF=FD, AE=ED (angle ACF = angle FCD, ABE = EBD). AFDE is kite. Thus, D is the mirror of A in EF. Alternate-4: Let AD be the foot of perpendicular from A onto EF and let it meet BC in M Join FM. Now, note that FAE= 60 BIC= 0.ie FAE+ FIE= 80 Hence AFIE is cyclic quadrilateral. Now note that FAI = FEI = EAI= IFE leads to IF = IE,ie IFE = IEM= 30 Now, on the other hand we have FAD = 90 AFD = 90 AIE = 90 ( ABI + IAB) = ACB = FCM; Therefore, AF MC is also cyclic. So we get, AFM = 80 ACB = 90 ACB = AFD, ie DFM = DFA Therefore considering DFA and DFM, they are congruent from the SAS rule of congruency, leading to AD = DM. Hence M is the reflection of A onto EF which lies on BC..

10 n 6. For each integer n, define a n =, where [x] denotes the largest integer not [ n] exceeding x, for any real numbers x. Find the number of all n in the set {,, 3, 00} for which an > an+. [7] Alternate-: Let pn p n + p n = n p + = pn = p for some n, then xn+ = If n n + n = x p p n So we need P n + > Pn and since And so pn n < pn + = p n + n + n <, we need pn+ = pn + n + and so n + is a perfect square. Let then n = m : m xn = = m + m x n+ m = = m < x m n And so xn+ < xn n + is a perfect square

11 And so the requested quantity is the number of perfect squares in { },3,...0 andsois 0 Hence the result : 43 Alternate-: Let n = a + b for natural a and whole b such that b < a + n b = a + n a For b [0,a ]U[a,a ]wehavean+ = an For n+ n b = a,a,wehavea = a + For n+ n b = a,wehavea = a. So, we need n = a + a = (a + ) The required n are, 3, 4,. 44 constituting 43 possibilities.