HMMT 十一月 2024 · GEN 赛 · 第 8 题

HMMT November 2024 — GEN Round — Problem 8

Let ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ f ( x ) = · · · || x | − 1 | − 2 − 3 − · · · − 10 . ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ Compute f (1) + f (2) + · · · + f (54) + f (55) .

解析

Let f ( x ) = · · · || x | − 1 | − 2 − 3 − · · · − 10 . Compute f (1) + f (2) + · · · + f (54) + f (55). Proposed by: Benjamin Shimabukuro Answer: 285 Solution: Let T denote the k -th triangular number 1 + 2 + · · · + k . k For any integer i , the function g ( x ) = | x − i | is a piecewise linear function with slopes ± 1. As i f ( x ) = g ( · · · ( g ( g ( x ))) · · · ), it is also piecewise linear with slopes ± 1. As g ( x ) has a cusp only 10 1 0 i where it evaluates to 0, the cusps of f occur precisely where g ( · · · ( g ( g ( x ))) · · · ) = 0 for some k 1 0 integer 0 ≤ k ≤ 10. Then, g ( · · · ( g ( x )) · · · ) = ± k , and since it is positive, it equals k . Similarly, k − 1 0 g ( · · · ( g ( x )) · · · ) = ( k − 1) ± k , so it must be k + ( k − 1). Continuing this argument, we see that k − 2 0 | x | = k + ( k − 1) + · · · + 1 = T , so the cusps occur precisely when | x | is a triangular between 0 and k 55, inclusive. As f (0) = 5 and f (1) = 6, the graph of f ( x ) on 0 ≤ x ≤ 55 looks as follows: 5 0 1 3 6 10 15 21 28 36 45 55 k Now observe that for 1 ≤ k ≤ 9, the ( k + 1) values g ( T ), g ( T + 1), . . . , g ( T − 1) are 5 − through k k k +1 2 k k k 5 + if k is even, and 5 . 5 + through 5 . 5 − if k is odd. Thus they average to 5 if k is even and 5 . 5 2 2 2 if k is odd. As f (55) = 0, the desired sum is 5 · (3 + 5 + 7 + 9) + 5 . 5 · (2 + 4 + 6 + 8 + 10) = 285 .