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Regional Mathematical Olympiad 2015 (First Round)

Questions with Solutions1. Given 100 numbers a1, a2, ………. a100 such that 15

a1 – 3a2 + 2a3 0a2 – 3a3 + 2a4 0a3 – 3a4 + 2a5 0__________________a99 – 3a100 + 2a1 0a100 – 3a1 + 2a2 0

Prove that a1 = a2 = ……………… = a100.Sol. a1 – a2 > 2(a2 – a3)

.

.

.a100 – a1 > 2(a1 – a2) on multiplying ineqns, we get 1 > 2100 absurd. so, every where equality has to hold. a1 – a2 = 2(a2 – a3) & so on

If each of ai – aj is non-zero, on multiplication we get 1 = 2100 absurd so at least one of them has to be zero those two are equal, which in turn maves every one equal. Q.E.D

2. Prove that x12 – x9 + x4 – x + 1 > 0 for all real values of x 10

Sol. f(x) x12 – x9 + x4 – x + 1, in f(– x) no sign change is there No negative root.

If x > 1, x12 > x9, x4 – x > 0, f(x) > 0

If 0 < x < 1, 1 – x > 0, x4 – x9 > 0, x12 > 0 f(x) > 0 x.

(None Decarte’s rule)

(None f(x) < 0 clearly x < 0 as well.)__________________________________________________________________________________________________________________________________

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3. A student of 11th class found 11 distinct solutions of the equation . His teacher told him that being a student of 11th class he could have found one more solution. Justify with proof if his teacher was right.

15

Sol. Let for some , f() = 0 = 19x 19x2 – x – 96 = 0 D > 0 for .

Even no. of roots. one more root.4. In a class of 20 student no two of them have the same ordered pair (written and oral

examination) of scores in mathematics. We say that student A is better than B if his two scores are greater than or equal to the corresponding scores of B. The scores are integers between 1 and 10. Show that there exist 3 students A, B, and C such that A is better than B and B is better than C. 20

Sol. Let ordered no is (ai , bi) ai {1, 2, ……. , 10}

bi {1, 2, …… , 10}

Clearly for any particular ai, no more than two value of bi are possible or we are done. Similarly for any particular bj, no more than two value of ai s are possible, or we are done. So, exactly two students will have one particular. ai & exactly two students will have one particular bj. So clearly most differentiated case is

(10, 1), (10, 2), (9, 2), (9, 3), (8, 3), (8, 4), (7, 4) (7, 5), (6, 5), (6, 6) (5, 6), (5, 7), (4, 7), (4, 8), (3, 8), (3, 9) (2, 9), (2, 10)

Still two students with exactly 1 mark in written is left to be placed.

some 3 students will be in same group. For those three A, B, C , A B C Q.E.D

(Note:-) one more seq. with (1, 10), (1, 9), (2, 9), (2, 8) ….. etc. can be made)

5. Saurabh is walking up a stair that has 10 steps. With each stride (move) he goes up either one step or two steps. In how many different ways can he go up the stairs?

15Sol. 2x + y = 10, x, y 0 for x = 0, one way

When x = 1 9C8 case

x = 2 8C6 case

x = 3 7C4 case

x = 4 6C2 case

x = 5 5C0 case__________________________________________________________________________________________________________________________________




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total = 88 case

Ans: 88 ways + 1 way = 89 ways6. Determine all primes p for which the system of equations

10

p + 1 = 2x2

p2 + 1 = 2y2

Has a solution in integers x, y.

Sol. p + 1 = 2x2, p2 + 1 = 2y2 Let x, y > 0 I

p2 – p = 2 (y2 –x2) = p(p – 1) p & p – 1 are rel. prime factor.

p(p –1) = 2 (y – x) (y + x) should be split to rel. prime factor with diff = 1

case (a): 2y + 2x – (y – x) = 1 3x + 4 = 1

absurd

Case (b): (y + x) – 2(y – x) = 1 3x – 1 = y

p2 + 1 = 2y2 = 2(3x – 1)2 = 18x2 – 12x + 2

= 9 (p + 1) –12x + 2

12x = – (p2 – 10 – 9p) <

12x = 12 or 24 x = 1 or 2

x = 1 gives absurd answers x = 2, y = 5, p = 7

only p = 7 satisfy the given condition

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7. In an acute angled triangle ABC let a, b, c be its sides, ma, mb, mc the lengths of altitudes and from A, B, C respectively and da, db, dc the distances from the vertices A, B, C respectively to the orthocentre. Prove that 15

Sol. da = 2R cosA, ma =

( abc = 4R) = Q.E.D

Note

AE = c cos A

AH cos(900 – c) = AE = c cos A

AH = da =

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