2011 AIME I Problem 9

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9.

Suppose xx is in the interval [0,π2]\left[0, \frac{\pi}{2}\right] and log24sinx(24cosx)=32.\log_{24 \sin x}(24 \cos x) = \frac{3}{2}. Find 24cot2x.24 \cot^2 x.

Answer: 192
Concepts:logarithmtrigonometric identityfactoring
Difficulty rating: 2650
Solution:

In exponential form the equation says (24sinx)3/2=24cosx.(24 \sin x)^{3/2} = 24 \cos x. Squaring gives 243sin3x=242cos2x,24^3 \sin^3 x = 24^2 \cos^2 x, so cos2x=24sin3x.\cos^2 x = 24 \sin^3 x.

Writing s=sinxs = \sin x and using cos2x=1s2,\cos^2 x = 1 - s^2, we get 24s3+s21=0,24s^3 + s^2 - 1 = 0, which factors as (3s1)(8s2+3s+1)=0.(3s - 1)(8s^2 + 3s + 1) = 0. The quadratic factor has negative discriminant, so sinx=13.\sin x = \frac{1}{3}.

Then 24cot2x=241sin2xsin2x=248/91/9=248=192. \begin{aligned} 24 \cot^2 x &= 24 \cdot \frac{1 - \sin^2 x}{\sin^2 x} \\ &= 24 \cdot \frac{8/9}{1/9} \\ &= 24 \cdot 8 = 192. \end{aligned}

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