HOMCS2014 sol CS

Hanoi Open Mathematical Competition 2014 Senior Section Answers and solutions Q1 Let a and b satisfy the conditions a3 − 6a2 + 15a = b3 − 3b2 + 6b = −1 The value of (a − b)2014 is (A): 1; (B): 2; (C): 3; (D): 4; (E) None of the above Answer (A) Write the system in the form (a − 2)3 = − 3a (b − 1)3 = −2 − 3b (1) (2) Subtracting (2) from (1) we find (a − 2)3 − (b − 1)3 = − 3(a − b) and then (a − b − 1)[(a − 2)2 + (a − 2)(b − 1) + (b − 1)2 + 3] = It follows a − b = and A = Q2 How many integers are there in {0, 1, 2, , 2014} such that x 999 C2014 ≥ C2014 (A): 15; (B): 16; (C): 17; (D): 18; (E) None of the above Answer (C) Note that x 2014−x C2014 = C2014 x+1 x and C2014 < C2014 for x = 0, 1, , 1006 x 999 These imply C2014 ≥ C2014 that is equivalent to 999 ≤ x ≤ 1015 Q3 How many 0’s are there in the sequence x1 , x2 , , x2014 , where n n+1 − √ , n = 1, 2, , 2014 xn = √ 2015 2015 (A): 1128; (B): 1129; (C): 1130; (D): 1131; (E) None of the above Answer (E) It is easy to check that ≤ xn ≤ Hence xn ∈ {0, 1} and the number of all in the √ 2015 given sequence is x1 + x2 + · · · + x2014 = √ − √ = 2015 It easy 2015 2015 √ to check that 144 < 2015 < 145 Hence, the number of all in the given sequence is 144 Thus, the number of all in the given sequence 2014 − 144 > 1130 Q4 Find the smallest positive integer n such that the number 2n + 28 + 211 is a perfect square (A): 8; (B): 9; (C): 11; (D): 12; (E) None of the above Answer (D) 2 For n > we find 28 + 211 + 2n = (24 ) (1 + + 2n−8 ) = (24 ) (9 + 2n−8 ) Hence, we find n such that 2n−8 + is a perfect square Putting 2n−8 + = k , we get (k − 3)(k + 3) = 2n−8 It follows k − and k + are the powers of and their k+3=8 then k = and n = 12 difference is So k−3=2 Indded, 28 + 211 + 212 = 802 is a perfect square For n ≤ 8, n ∈ {1; 2; 3; 4; 5; 6; 7; 8} The corresponding values 28 + 211 + 2n are not perfect squares Q5 The first two terms of a sequence are and Each next term thereafter is the sum of the nearest previous two terms if their sum is not greater than 10 and is otherwise The 2014th term is (A): 0; (B): 8; (C): 6; (D): 4; (E) None of the above Answer (B) The corresponding sequence is 2, 3, 5, 8, 0, 8, 8, 0, 8, 8, if n − = 1(mod 3) otherwise Hence 2014th term is for 2014 − = 2010 = 0(mod 3) So an = Q6 Let S be area of the given parallelogram ABCD and the points E, F belong to BC and AD, respectively, such that BC = 3BE, 3AD = 4AF Let O be the intersection of AE and BF Each straightline of AE and BF meets that of CD at points M and N, respectively Determine area of triangle M ON Solution Let a = AB Note that EAB ∼ EM C It follows CM CE = = AB BE This implies CM = 2a By the same argument as above we obtain DN = a/3 Hence, M N = a/3 + a + 2a = 10a/3 (1) Let h denote the height of the paralelogram, and let h1 and h2 the heights of triangles OM N and OAB, respectively We have h1 MN 10 = = h2 AB We deduce h1 h1 10 = = h2 h1 + h2 3 We then have h1 = 10 h (2) Combining (1) and (2) we receive 10a 10h 50 50 = ah = S SM ON = M N.h1 = 2 13 39 39 The answer is 50 S 39 Q7 Let two circles C1 , C2 with different radius be externally tangent at a point T Let A be on C1 and B be on C2 , with A, B = T such that ∠AT B = 900 (a) Prove that all such lines AB are concurrent (b) Find the locus of the midpoints of all such segments AB Solution Let O1 , O2 denote the centres of C1 , C2 whose radii are r1 , r2 , respectively Without loss of generality we can assume that r1 < r2 We have ∠AT B = 900 It implies ∠O1 T A + ∠O2 T B = 900 We deduce ∠O1 AB + O2 BA = ∠O1 AT + ∠T AB + ∠ABT + ∠O2 BT = 1800 Hence, O1 A O2 B Let X denote the intersection of AB and O1 O2 We then have XO1 r1 = XO2 r2 which proves the conclusion b) Let M, N denote the midpoints of AB, O1 O2 , respectively We have r1 + r2 O1 A + O2 B = 2 MN = We deduce that M is on the circle whose centre is N and the radius equals to except for O1 , O2 r1 + r2 , Q8 Determine the integral part of A, where A= 1 + + ··· + 672 673 2014 Solution Consider the sum 3n+1 S= k k=n+1 Note that there are 2n + terms in the sum and the middle term is can write the sum in the form n 1 + + S= 2n + k=1 2n + + k 2n + − k n = + 2n + 2n + k=1 k 1− 2n + On the other hand, using the inequalities 1+a< 1 < + 2a for < a < , 1−a we get 1− and k 2n + k 1− 2n + k 2n + >1+ k 2n +