PUMaC 2019 · 代数（A 组） · 第 8 题
PUMaC 2019 — Algebra (Division A) — Problem 8
题目详情
- For real numbers a and b , define the sequence { x ( n ) } as follows: x (1) = a , x (2) = b , a,b a,b a,b 2 2 and for n > 1, x ( n + 1) = ( x ( n − 1)) + ( x ( n )) . For real numbers c and d , define a,b a,b a,b the sequence { y ( n ) } as follows: y (1) = c , y (2) = d , and for n > 1, y ( n + 1) = c,d c,d c,d c,d 2 ( y ( n − 1)+ y ( n )) . Call ( a, b, c ) a good triple if there exists d such that for all n sufficiently c,d c,d 2 large, y ( n ) = ( x ( n )) . For some ( a, b ) there are exactly three values of c that make ( a, b, c ) c,d a,b a good triple. Among these pairs ( a, b ), compute the maximum value of b 100( a + b ) c . 1
解析
- For real numbers a and b , define the sequence { x ( n ) } as follows: x (1) = a , x (2) = b , a,b a,b a,b 2 2 and for n > 1, x ( n + 1) = ( x ( n − 1)) + ( x ( n )) . For real numbers c and d , define a,b a,b a,b the sequence { y ( n ) } as follows: y (1) = c , y (2) = d , and for n > 1, y ( n + 1) = c,d c,d c,d c,d 2 ( y ( n − 1)+ y ( n )) . Call ( a, b, c ) a good triple if there exists d such that for all n sufficiently c,d c,d 2 large, y ( n ) = ( x ( n )) . For some ( a, b ) there are exactly three values of c that make ( a, b, c ) c,d a,b a good triple. Among these pairs ( a, b ), compute the maximum value of b 100( a + b ) c . Proposed by: Eric Neyman Answer: 120 2 Define ( a, b, c, d ) to be good if for n large enough, y ( n ) = ( x ( n )) . Fix a good quadruple c,d a,b ( a, b, c, d ). For brevity of notation, we will denote x ( n ) as x and y ( n ) as y . a,b n c,d n 3 We claim that y = x 2 for all n ≥ 3. Suppose for contradiction that this is not the case, and n n 2 2 let k ≥ 3 be such that y = x for all n > k , but y 6 = x . We have n k n k 2 y = x k +2 k +2 2 2 2 2 ( y + y ) = ( x + x ) k k +1 k k +1 2 2 y + y = ± ( x + x ) k k +1 k k +1 2 y + y = ± ( x + y ) . k k +1 k +1 k 2 We can’t choose the plus sign because then y = x , which we assumed to not be the case. k k 2 Thus, y + y = − x − y , so k k +1 k +1 k 2 2 2 y = − x − 2 y = − x − 2( y + y ) ≤ 0 . k k +1 k − 1 k k k 2 2 But y = ( y + y ) ≥ 0, so y = 0. This means that x = 0, so x = 0, contradicting our k k − 2 k − 1 k k k 2 2 assumption that y 6 = x . Therefore, y = x for all n ≥ 3. k n k n Suppose that ( a, b, c, d ) is good. We have 2 2 2 2 2 2 x = a, x = b, x = a + b , x = b + ( a + b ) 1 2 3 4 and 2 2 2 y = c, y = d, y = ( c + d ) , y = ( d + ( c + d ) ) . 1 2 3 4 2 2 Since y = x and y = x , we have the equations 3 4 3 4 2 2 c + d = ± ( a + b ) (1) and 2 2 2 2 2 d + ( c + d ) = ± ( b + ( a + b ) ) . (2) 2 2 2 2 Plugging in ( a + b ) for ( c + d ) in (2), we have 2 2 2 2 2 2 2 d + ( a + b ) = ± ( b + ( a + b ) ) . 2 2 2 2 2 This gives two possibilities: d = b or d = − b − 2( a + b ) . 2 2 2 2 2 2 2 Suppose that d = b . Then (1) gives c + b = ± ( a + b ), so c is either a or − a − 2 b . 2 2 2 2 Suppose that d = − b − 2( a + b ) . Then (2) gives 2 2 2 2 2 2 c − b − 2( a + b ) = ± ( a + b ) , 2 2 2 2 2 2 2 2 2 so c is either 2( a + b ) − a or a + 2 b + 2( a + b ) . Note that all four of the values of c that are listed work, because all our steps can be reversed 2 2 2 and if x = y and x = y , then x = y for all n ≥ k . k k +1 n k k +1 n We want exactly two of the four listed values of c to be equal. Note that if a = 0 then 2 4 2 4 the four values of c are 0, − 2 b , 2 b , and 2 b + 2 b , which are all different unless b = 0, in which case they are all the same. Thus, we may assume that a 6 = 0. This means that 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2( a + b ) − a < a + 2 b + 2( a + b ) , a < a + 2 b + 2( a + b ) , − a − 2 b < a , and 2 2 2 2 2 2 − a − 2 b < 2( a + b ) − a . Thus, for two of the values of c to be the same, we must have ( ) 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2( a + b ) − a = a , i.e. ( a + b ) = a . Thus, a + b = ± a , so a ± + b = . 2 4 ( ) 1 1 This means that ( a, b ) is a point on either the circle with radius centered at , 0 or the 2 2 ( ) 1 − 1 circle with radius centered at , 0 . a + b is maximized at the point where the rightmost 2 2 circle is tangent to a line with slope − 1 that is ”furthest right.” This happens at the point ( ) √ √ √ √ 1 2 2 1 2 1+ 2
- , , where a + b = + = . 2 4 4 2 2 2 √ Thus, our answer is b 50 + 50 2 c = 50 + 70 = 120 . 4