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Regional Mathematical Olympiad- 7 Solutions Time: 3 hours October 8, 7 Instructions: Calculators (in any form) and protractors are not allowed. Rulers and compasses are allowed.

1 Regional Mathematical Olympiad- 7 Solutions Time: 3 hours October 8, 7 Instructions: Calculators (in any form) and protractors are not allowed. Rulers and compasses are allowed. Answer all the questions. All questions carry equal marks. Maximum marks:. Answer to each question should start on a new page. Clearly indicate the question number.. Let AOB be a given angle less than 8 and let P be an interior point of the angular region determined by AOB. 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 = :. 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. CPX ~ DPO CP PX DP PO PC : PD = :. Show that the equation a 3 + (a + ) 3 + (a + ) 3 + (a + 3) 3 + (a + 4) 3 + (a + 5) 3 + (a + 6) 3 = b 4 + (b + a) 4 has no solutions in integers a, b. Since 7 consecutive numbers appear on left side, it s a good idea to try modulo ( ) + + ( ) + ( ) (mod 7) So LHS is always divisible by 7. Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page of 6

2 Or, a (a )...(a 6) r (mod 7) = Now RHS, b 4 + (b + ) 4 for any integral b, 7 r b r(mod 7), where r is,,, 3, 4, 5, 6 b 4 + (b + ) 4 r 4 + (r + ) 4 (mod 7) for any r =,,, 3, 4, 5, 6 Hence, no solution. 3. Let P(x) = x 7(7 ) (mod 7) = (8) (mod 7) = (mod 7) 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)) = x P x x b Q(x) = x + cx + d P(x). Q(x) = Q(P(x)) let us consider P(x) = = Q() = d (from (i)) d = Now P(x). Q(x) = [P(x)] + cp(x) Q(x) = [P(x)] + c x cx x b c x c and b c b (i) x Px x and Q x x x Now since P(Q(x)) =, Q(x) is a root of P(x) = x i.e. x x x x, Q(x) has to be either or Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page of 6

3 Case : Q(x) = x x 5i x 4 Case : Qx x x x, So total 4 roots, real and imaginary. Real roots are & 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 whenever i j n and k 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? Here i i n; k 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 Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page 3 of 6

4 condition. There will be at least n colors. Now let us prove n colours are sufficient. 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. 5. Let be a circle with a chord AB which is not a diameter. Let T be a circle on one side of AB such that it is tangent to AB at C internally tangent to at D. Likewise, let T be a circle on the other side of AB such that it is tangent to AB at E and internally tangent to at F. Suppose the line DC intersects at X D and the line FE intersects at Y F. Prove that XY is a diameter of. Let O be centre of, O of and Oof. Join OX and extend it to meet AB at M. Join OO it will pass through D (as both circles are tangent) Now let DCB θ or OCD 9 θ ODC 9 θ ODX 9 θ OXD ODX 9 θ MXC 9 θ Also MCX DCB θ(voa) In XCM CMX = 8 MXC MCX = 8 (9 θ) θ = 9 Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page 4 of 6

5 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. 6. Let x, y, z be real numbers, each greater than. Prove that x y z x y z y z x y z x x y z x y z y z x y z x x x y y z z 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 y y z z x x i.e., x y z y z x y z x y z x Now, without loss of generality, let z y x z y x x y z Now, applying rearrangement inequality, x y z x y z x y z y z x x y z x y z x y z y z x Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page 5 of 6

6 Alternate : x y y z z x y z x let without loss of generality, x y z x y z x y z z y x a, b, c z y x let a b c It is suffices to show that x y y z x z y z x i.e. b (x y) + a(y z) c(x z) which is true by adding the following b(x y) c(x y) a(y z) c(y z) Pioneer Education SCO 3, Sector 4 D, Chandigarh , Page 6 of 6

ALLEN. Pre Nurture & Career Foundation Division For Class 6th to 10th, NTSE & Olympiads. å 7 3 REGIONAL MATHEMATICAL OLYMPIAD-2017 SOLUTION

ALLEN. Pre Nurture & Career Foundation Division For Class 6th to 10th, NTSE & Olympiads. å 7 3 REGIONAL MATHEMATICAL OLYMPIAD-2017 SOLUTION 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,