HMMT 二月 2005 · GEN2 赛 · 第 2 题
HMMT February 2005 — GEN2 Round — Problem 2
题目详情
- Find three real numbers a < b < c satisfying: a + b + c = 21 / 4 1 /a + 1 /b + 1 /c = 21 / 4 abc = 1 .
解析
- Find three real numbers a < b < c satisfying: a + b + c = 21 / 4 1 /a + 1 /b + 1 /c = 21 / 4 abc = 1 . Solution: 1 / 4 , 1 , 4 By inspection, one notices that if b is a number such that b +1 /b = 17 / 4, then a = 1 , c = 2 1 /b will work. Again by inspection (or by solving the quadratic b − 17 b/ 4 + 1 = 0), one finds b = 1 / 4 or 4, so the numbers are 1 / 4, 1, and 4. Alternative Solution: Note that ab + bc + ca = abc (1 /a + 1 /b + 1 /c ) = 21 / 4, so 3 2 a , b , and c are the roots of the polynomial x − 21 x / 4 + 21 x/ 4 − 1, which factors as 1 ( x − 1)( x − 4)(4 x − 1), giving the same answer. 4