1989 AMC 12 Problem 28

Attempt Problem 28 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

28.

Find the sum of the roots of tan⁡2x−9tan⁡x+1=0\tan^2x-9\tan x+1=0 that are between x=0x=0 and x=2πx=2\pi radians.
π2\frac\pi2
π\pi
3π2\frac{3\pi}2
3π3\pi
4π4\pi
Answer: D
Concepts:quadratictrigonometryalgebraic manipulation
Small hint:
Let the two positive roots in tan⁡x\tan x be rr and ss
Big hint:
Use rs=1rs=1 to relate arctan⁡r+arctan⁡s,\arctan r+\arctan s, then include the period of tangent
Solution:
The two roots r,sr,s of t2−9t+1=0t^2-9t+1=0 are positive and satisfy rs=1.rs=1. Therefore their acute arctangents add to π2.\frac{\pi}{2}. Each tangent value occurs twice between 00 and 2π,2\pi, with the second occurrence shifted by π.\pi. The sum of all four roots is 2(π2)+2π=3π. 2\left(\frac\pi2\right)+2\pi=3\pi.
Thus the correct answer is D.
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