For Class 6th to 0th, NTSE & Olympiads REGIONAL MATHEMATICAL OLYMPIAD-07. Let AOB be a given angle less than 80 and let P be an interior point of the angular region determined by ÐAOB. Show, with proof, how to construct, using only ruler and compasses, a line segment CD passing through P such that C lies on the ray OA and D lies on the ray OB, and CP : PD = :. A C X Sol. O P D B Just extend OP to X, such that OP : PX = :. Draw a line through X parallel to OB which meets OA at C. Extend CP to meet OB at D. CD is the required line. Q DCPX ~DDPO CP PX = = DP PO \ PC : PD = :. Show that the equation 3 3 3 3 3 3 3 4 4 a +(a+) +(a+) +(a+3) +(a+4) +(a+5) +(a+6) =b +(b +) has no solutions in integers a, b. Sol. Since 7 consecutive numbers appear on left side, it s a good idea to try modulo 7. 3 + 3 + 3 3 + 4 3 + 5 3 + 6 3 + 7 3 º + + ( ) + + ( ) + ( ) º 0 (mod 7) So LHS is always divisible by 7. å 7 3 r= Or,a 3 + (a + ) 3 +... (a + 6) 3 º r ( mod 7 ) Now RHS, b 4 + (b + ) 4 for any integral b, b º r (mod 7), where r is 0,,, 3, 4, 5, 6 ( + ) é7 7 ù =ê ú ë û ( mod 7) = (8) (mod 7) = 0 (mod 7) b 4 + (b + ) 4 º r 4 + (r + ) 4 º/ 0 (mod 7) for any r = 0,,, 3, 4, 5, 6 Hence, no solution. 3. Let P(x) = x + x+b and Q(x) = x + cx + d be two polynomials with real coefficients such that P(x)Q(x) = Q(P(x)) for all real x. Find all the real roots of P(Q(x)) =0
For Class 6th to 0th, NTSE & Olympiads REGIONAL MATHEMATICAL OLYMPIAD-07 P x = x + + b Sol. ( ) x Q(x) = x + cx + d P(x) Q(x) = Q(P(x)) let us consider P(x) = 0 0 = Q(0) = d (from (i)) d = 0 Now P(x) Q(x) = [P(x)] + cp(x) Q(x) = [P(x)] + c x + cx = x x + + b+ c c = and b + c = 0 b =- x P( x) x...(i) = + - and Q(x) = x + x Now since P(Q(x)) = 0, Q(x) is a root of P(x) = 0 i.e. x x 0 + - = ( ) \ Q(x) has to be either or Case : Q(x) = x + x+ = 0 - ± 5i x = 4 Case : Q( x) = x + x- = 0 æ ö x+ ç x- = 0 è ø x =,- So total 4 roots, real and imaginary. x =-, Real roots are &-
For Class 6th to 0th, NTSE & Olympiads REGIONAL MATHEMATICAL OLYMPIAD-07 4. Consider n unit squares in the xy-plane centred at point (i, j) with integer coordinates, i n, j n. It is required to colour each unit square in such a way that when ever i < j n and k < l n, the three squares with centres at (i,k), (j,k), (j,l) have distinct colours. what is the least possible number of colours needed? Sol. Here i < i n; k < l n n l (j,) l k (,) ik (j,) k i j n as per given condition, (i, k), (j,k), (j, l) are of distinct colours. No two square of k th row will be of same color as (i, k) and (j, k) are of distinct colors and no to squares of i th column of same color as (j, l) and (j, k) are of distinct color all square of type (x, ) for x=,...n, are of distinct color total n distinct colours Similarly all square of type (n, y) for y =,...n, are of distinct colors, also no square of (x, ) and (n, y) with same color otherwise take square (n, n) with (x, ) and (n, y), we will get contradiction to given condition. There will be at least n colors. Now let us prove n colours are sufficient. y y x (x,y) x+ (x+,y ) x+y = c = (i + k) We can see that (x, y) and (x +, y ) can have same colours. So paint all squares with same colour for which x + y is same. Here minimum x + y will be (for x =, y = ) and maximum x + y = n As from to n there n values, so n colours will be sufficient. 3
For Class 6th to 0th, NTSE & Olympiads REGIONAL MATHEMATICAL OLYMPIAD-07 5. Let W be a circle with a chord AB which is not a diameter. Let T be a corcle on one side of AB such that it is tangent to AB at C internally tangent to W at at D. Likewise, let T be a circle on the other side of AB such that it is tangen to AB at E and internally tangent to W at F. Suppose the line DC intersects W at X ¹ D and the line FE intersects W at Y ¹ F. Prove that XY is a diameter of W. Sol. Let O be centre of G, O of W and O of G. Join OX and extend it to meet AB at M. Join OO it will pass through D (as both circles are tangent) Now let Ð D C B = q ÐO CD = 90 q ÐO DC = 90 q A E Y M O q C D q 90 q B ÐODX = 90 q ÐOXD = ÐODX = 90 q F or ÐMXC = 90 q Also ÐMCX = ÐDCB = q (VOA) In DXCM ÐCMX = 80 ÐMXC ÐMCX = 80 (90 q) q = 90 O O X 90 q X is mid point of arc AB» (not containing D) similarly Y is mid point of arc AB» (containing D) XY is a diameter of W. 6. Let x,y,z be real numbers, each greater than. Prove that Sol. x + + + - - - + y + z x + y + z y+ z+ x+ y- z- x- x - - - + + + + y + z ³ x + y + z y- z- x- y+ z+ x+ x- x+ y- y+ z- z+ - + - + - ³ 0 y- y+ z- z+ x- x+ (x -y) (y -z) (z -x) y - z - x - x- y y-z z-x y - z - x - x y y z z x i.e., - + - + - ³ 0 y - y - z - z - x - x - x + y + z ³ y + z + x y - z - x - y - z - x - 4
For Class 6th to 0th, NTSE & Olympiads REGIONAL MATHEMATICAL OLYMPIAD-07 Now, without loss of generality, let z ³ y ³ x > z ³ y ³ x > 0 ³ ³ > 0 x - y - z - Now, applying rearrangement inequility, é x y z ù é x y z ù ê ú ê ú ê ú ê ú êëx - y - z -úû êëy - z - x -úû x y z x y z + + + + x - y - z - y - z - x - Alternate : x-y y-z z-x y - z - x - let without loss of generality, x ³ y ³ z x ³ y ³ z x ³ y ³ z ³ ³ z - y - x - let = a, = b, = c z - y - x - \ a ³ b ³ c It is suffices to show that x - y - - + y z ³ x z y - z - x - i.e. b (x y) + a(y z) ³ c(x z) which is true by adding the following b(x y) ³ c(x y) a(y - z) ³ c(y z) 5