HMMT 二月 2012 · 冲刺赛 · 第 19 题
HMMT February 2012 — Guts Round — Problem 19
题目详情
英文原题
[11] Given that Pis a real polynomial of degree at most 2012 such that P(n) = 2nforn= 1,2, . . . , 2012, what choice(s) of P(0) produce the minimal possible value of P(0)2+P(2013)2?
解析
英文解析
- [ 11 ] Given that P is a real polynomial of degree at most 2012 such that P ( n ) = 2 for n = 1 , 2 , . . . , 2012,
2 2
what choice(s) of P (0) produce the minimal possible value of P (0) + P (2013) ?
2012 1 i i − 1 i − 1
Answer: 1 − 2 Define ∆ ( n ) = P ( n + 1) − P ( n ) and ∆ ( n ) = ∆ ( n + 1) − ∆ ( n ) for
2012
i > 1. Since P ( n ) has degree at most 2012, we know that ∆ ( n ) is constant. Computing, we
1 i i − 1
obtain ∆ (0) = 2 − P (0) and ∆ (0) = 2 for 1 < i ≤ 2012. We see that continuing on gives
2012 2012 i 2013 − i
∆ (0) = ∆ (1) = P (0) and ∆ (2012 − i ) = 2 for 1 ≤ i ≤ 2011. Then, P (2013) =
1 1 2012 2013
P (2012) + ∆ (2012) = . . . = P (2012) + ∆ (2011) + . . . + ∆ (0) = P (0) + 2 − 2. Now, we want
2 2 2 2013 2013 2
to minimize the value of P (0) + P (2013) = 2 P (0) + 2 P (0)(2 − 2) + (2 − 2) , but this occurs
2013 20121
simply when P (0) = − (2 − 2) = 1 − 2 . 2