HMMT 十一月 2014 · THM 赛 · 第 1 题

HMMT November 2014 — THM Round — Problem 1

Let f ( n ) denote the real part of the complex number z . Assume the parabola p ( n ) = an + bn + c intersects f ( n ) four times, at n = 0 , 1 , 2 , 3. Assuming the smallest possible value of k , find the largest possible value of a .

解析

Let f ( n ) denote the real part of the complex number z . Assume the parabola p ( n ) = an + bn + c intersects f ( n ) four times, at n = 0 , 1 , 2 , 3. Assuming the smallest possible value of k , find the largest possible value of a . 2 πj 1 ℜ z Answer: Let r = | z | , θ = arg z , and C = = cos θ = cos for some j with gcd( j, k ) = 1. The 3 | z | k condition of the four consecutive points lying on a parabola is equivalent to having the finite difference f (3) − 3 f (2) + 3 f (1) − f (0) = 0 . This implies f (3) − f (0) = 3 [ f (2) − f (1)] ( ) 3 2 ⇐⇒ r cos(3 θ ) − 1 = 3 r cos(2 θ ) − r cos( θ ) 3 3 2 2 ⇐⇒ r (4 C − 3 C ) − 1 = 3( r (2 C − 1) − rC ) . Now we simply test the first few possible values of k . 3 2 3 k = 1 implies C = 1, which gives r − 1 = 3( r − r ) = ⇒ ( r − 1) = 0 = ⇒ r = 1. This is not allowed since r = 1 implies a periodic function. 3 2 3 k = 2 implies C = − 1, which gives − r − 1 = 3 r + r = ⇒ ( r + 1) = 0, again not allowed since r > 0. 1 − 3 1 3 2 k = 3 implies C = − . This gives r − 1 = ( r − r ) = ⇒ ( r − 1)( r + )( r + 2) = 0. These roots are 2 2 2 either negative or 1, again not allowed. 1 1 2 √ √ k = 4 implies C = 0. This gives − 1 = − 3 r = ⇒ r = ± . r = is allowed, so this will generate 3 3 our answer. Again by finite differences (or by any other method of interpolating with a quadratic), we get 2 a = 2 1 f (0) + f (2) − 2 f (1) = , so a = . 3 3 Theme Round