HMMT 二月 2005 · TEAM1 赛 · 第 8 题
HMMT February 2005 — TEAM1 Round — Problem 8
题目详情
英文原题
- [25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of
2 3
1 x x x
2 x + + + + .
2 3 4
1 + x 1 + x 1 + x 1 + x
解析
英文解析
- [25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of
2 3
1 x x x
2 x + + + + .
2 3 4
1 + x 1 + x 1 + x 1 + x
Solution: Note that 3
2 6 4 4 6
x x x x x + x 1 + = + = = x = , and 4
2 3 2 2 5 2
1 + x 1 + x 1 + x x + x 1 + x x
3 5 4 4 5
1 x x x x + x 1 + = + = = x = .4
4 5
1 + x 1 + x 1 + x x + x 1 + x x
2 1
Therefore, the sum is just 2 x + . If x = 1, this is 4. Otherwise, let y = x + . Thenx xx satisfies
( ) ( )
1 1
2 3 4 2 2
0 = 1 + x + x + x + x = x + + 2 + x + − 1 = y + y − 1 ,
x x 2
√
√
− 1 ± 5
so solving this quadratic yields y = , or 2 y = − 1 ± 5. Since each value of y cancorrespond to only 2 possible values of x , and there are 4 possible values of x besides 2
√
1, both of these values for y are possible, which yields the answers, 4 and − 1 ± 5.