Solutions, 2004 NCS/MAA TEAM COMPETITION

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1 Solutions, 004 NCS/MAA TEAM COMPETITION Each problem number is followed by an 11-tuple (a 10, a 9, a 8, a 7, a 6, a 5, a 4, a 3, a, a 1, a 0 ), where a k is the number of teams that scored k points on the problem 1 Quadrilateral area (51,4,3,1,,1,1,1,1,0,1) The area is 585 Let h be the altitude ED From the Pythagorean Theorem, BC = 16 From similar triangles, = 1; ie, h = 3, so h = 15 The area of h BD triangle ABC is 1 (1)(16) = 96, and the area of triangle DBE is 1(h)(10) = 75 Then the area of the quadrilateral ADEC is = 117 = 585 A C D E B Sequence sum (39,,5,1,4,5,4,0,1,0,5) The sum is 0 From the recursion we find the first eight terms to be a 1, a, a a 1, a 1, a, a 1 a, a 1, a, so a 7 = a 1 and a 8 = a It follows that a n+6 = a n for all n Let S n be the sum of the first n terms We see that S 6 = 0, and from the periodicity it follows that S 6n = 0 for all n In particular, S 004 = 0 3 Sum of cubes of roots (34,3,1,1,4,10,,0,5,0,6) We show that We have r 3 + s 3 = a3 3a r + s = a and rs = a 1 But (r + s) 3 = r 3 + 3rs(r + s) + s 3, so r 3 + s 3 = 3rs(r + s) + (r + s) 3 = 3(a 1) ( a) + ( a) 3 = a3 3a 1

2 Solutions, 004 NCS Team Competition 4 Integer linear combination (61,0,0,0,0,0,0,0,0,0,5) There do, as the example (130)(9) + (559)( ) = = 5 shows Divide the given equation by 13 to get the equivalent equation 10m 4 = 43( n) This tells us that we simply need 43n to end in the digit 6 (strictly speaking, to be 6 mod 10) This is true, eg, with n =, making m = 9 Another approach is to use Euclid s division algorithm: 559 = = Thus 13 = = 130 3( ) = = (130)(13) + (559)( 3) Multiplying by 4 we obtain the solution (130)(5) + (559)( 1) = 5 5 A polynomial in x 3 (10,0,0,0,0,0,3,0,,1,50) Yes; one such is Q(x) = x 6 +x 5 3x 3 x +x+4, making P (x)q(x) = x 9 4x 6 +7x 3 8 One finds quickly that there is no such Q(x) of degree three by trying Q(x) = x 3 +ax +bx+c To make the coefficients of x, x, x 4 and x 5 in the product all equal 0 places four conditions on the three quantities a, b, c, and there is no solution But with Q(x) = x 6 + a 5 x 5 + a 4 x 4 + a 3 x 3 + a x + a 1 x + a 0 we have six quantities a 0, a 1, a, a 3, a 4, a 5 at our disposal, and six conditions to make the coefficients of x, x, x 4, x 5, x 7 and x 8 vanish Denoting P (x)q(x) by c 0 + c 1 x + c x + + c 8 x 8 + x 9, we have c 1 = a 0 a 1 = 0 c = a 0 + a 1 a = 0 c 4 = a 1 a + a 3 a 4 = 0 c 5 = a a 3 + a 4 a 5 = 0 c 7 = a 4 a = 0 c 8 = a 5 1 = 0 The last two equations immediately give a 5 = 1 and a 4 = 0 Substituting these values in the equations for c 4 and c 5 and adding these two gives a 1 = Then the equation for c 1 gives a 0 = 4, after which we easily find a = 1 and a 3 = 3

3 Solutions, 004 NCS Team Competition 6 Shuffling cards (41,0,0,,0,0,0,1,0,0,) The order after one shuffle was 7, 6, 4, 10, 8, K, 3, Q, 5,, A, J, 9 Let P represent the permutation executed by the shuffler As P in cycle notation is (A, 3, 10, 6, 9, 8, J, 7, 4,, K, 5, Q), a single cycle, it follows that P is a single cycle, and therefore that P 13 = I Then P = P 14 = (P ) 7, which one computes from P to be the cycle (A, 7, 3, 4, 10,, 6, K, 9, 5, 8, Q, J) This cycle applied to the cards in their original order puts them in the order 7, 6, 4, 10, 8, K, 3, Q, 5,, A, J, 9 (One may also find P by filling in the middle line of the table below by trial and error: n A J Q K P (n) P (n) 3 K 10 Q 9 4 J A 5 (1) After trying a value for P (A), the remainder of the table is determined) 7 Slanted asymptote (3,0,0,,4,0,0,0,9,3,45) The line is y = x, as one guesses from the fact that for x negative and x large, x + 4x + 5 = (x + ) + 1 is approximately x +, which is x We need to show that f(x) (x ) 0 as x Consider f(x) (x ) = x + (x + ) + 1 x + = x + + (x + ) + 1 = 1 x + + x (x + ) 3

4 Solutions, 004 NCS Team Competition For x <, x + = (x + ), and f(x) (x ) = (x + ) (x + ) 1 ( 1 + (x + ) = x = x + ( ) (x+) = (x+) < 1 x + 0 as x x + ( (x+) ) ) 1 (x + ) 8 Find the n-th term (11,1,0,1,0,1,0,0,0,0,5) The n-th term is a n = 3 n +668 n We see that a 0 = = and a 1 = 671 = A useful observation is that 671 = and 004 = Thus a = = ( ) = , suggesting the asserted formula To complete the proof by induction, assume that Then a k+ = 671a k+1 004a k a k = 3 k k and a k+1 = 3 k k+1 = ( )(3 k k+1 ) 3 668(3 k k ) = 3 k k k k+ 3 k k+1 = 3 k k+, so by induction the claim is established ALTERNATE SOLUTION We have a linear difference equation with constant coefficients The auxiliary equation is m 671m = (m 3)(m 668)0, with roots 3 and 668 Then the general solution has the form A 3 n + B 668 n, and the given initial values determine that A = B = 1 4

5 Solutions, 004 NCS Team Competition 9 Same fractonal parts (1,0,0,0,0,,0,0,1,3,59) We have x x = a and x n x = b, where a and b are integers We show first that if x is rational then it is an integer From x x a = 0 we have x = 1 ± 1 + 4a If this is rational then so is 1 + 4a, and if the square root of an integer is rational it is an integer Thus 1 + 4a is an odd integer, and x is an integer Thus it suffices to show that x is rational Now, x = x + a, so x 3 = x + ax = (x + a) + ax = (1 + a)x + a Suppose that x k = r k x + s k where r k and s k are integers Then x k+1 = r k x + s k x = r k (x + a) + s k x = r k+1 x + s k+1, where r k+1 = r k +s k and s k+1 = r k a It follows by induction the for every positive integer m, x m = r m x + s m for some integers r m and s m In particular, x n = r n x + s n This together with x n = x + b implies that (r n 1)x + (s n b) = 0, and thus x is rational, provided that r n 1 Note that a 0 because x is real If a = 0, then x = 0 or 1, so is an integer In the remaining case a 1 Now r = 1 and s = a 1; r 3 = a + 1 > 1 and s 3 = a 1 Suppose r k > 1 and s k 1 Then r k+1 = r k + s k > r k > 1 and s k+1 = r k a r k > 1 It follows by induction that r k > 1 for all k > Thus x is rational, and therefore an integer 5

6 Solutions, 004 NCS Team Competition 10 Limit of product of cosines (,0,0,0,,1,0,0,0,0,61) The limit is (sin x)/x Let Then from which it follows that Consequently, Since lim t 0 t sin t v n (x) = u n (x) sin x n = cos x cos x cos x n sin x n ( v n (x) = cos x cos x cos ( = cos x cos x cos x n 1 x n 1 = 1 u n 1(x) sin x n 1 = 1 v n 1(x), )(cos x n sin x n ) )( 1 sin x ) n 1 v n (x) = 1 v 1(x) = 1 n 1 cos x n 1 sin x = 1 sin x n = 1, we see that u n (x) = v n(x) sin x n = = x sin x n x sin x 1 sin x n sin x n x n = sin x x sin x n lim u n(x) = sin x n x n 6

Mathathon Round 1 (2 points each)

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