HMMT 十一月 2011 · GEN 赛 · 第 7 题

英文解析

7. [ 5 ] Determine the number of angles θ between 0 and 2 π , other than integer multiples of π/ 2, such that the quantities sin θ , cos θ , and tan θ form a geometric sequence in some order.
   Answer: 4 If sin θ, cos θ, and tan θ are in a geometric progression, then the product of two mustequal the square of the third. Using this criterion, we have 3 cases.
   2 2
   2 3 2
   • Case 1: sin θ · tan θ = cos θ . This implies that (sin θ ) = (cos θ ). Writing sin θ as 1 − cos θ
   3 2
   and letting cos θ = x , we have that x + x − 1 = 0. We wish to find the number of solutions of
   2 3 2
   this where | x | ≤ 1. Clearly − 1 is not a root. If − 1 < x ≤ 0, we have that x + x ≤ x < 1 so
   3 2 3 2
   x + x − 1 < 0 and there are no roots. If 0 < x ≤ 1, then x + x − 1 is a strictly increasingfunction. Since it has value − 1 at x = 0 and value 1 and x = 1, there is exactly one root between
   0 and 1, non-inclusive. There are 2 values of θ such that cos θ equals this root, and thus, twosolutions in this case.
   2 3
   • Case 2: sin θ · cos θ = tan θ . This implies that cos θ = sin θ . To find the number of solutions inthis case, we can analyze the graphs of the functions in different ranges. Note that from θ = 0
   3πto θ = , cos θ decreases strictly from 1 to 0 while sin θ increases strictly from 0 to 1. Hence,
   there is one solution in this range. By a similar argument, a solution exists between θ = π and 2
   3 π π 3 πθ = . In the intervals [ , π ] and [ , 2 π ], we have that one function is negative and the other
   2 2 2
   is positive, so there are no solutions. Thus, there are two solutions in this case.
   2 2
   • Case 3: cos θ · tan θ = sin θ . This implies that sin θ = sin θ, so sin θ = 0 , 1. Clearly the onlyπsolutions of these have θ as an integer multiple of . Thus, there are no pertinent solutions inthis case.2
   We can see that the solutions for the first two cases are mutually exclusive. Hence, there are 4 solutionsin total.