2011 AIME I Problem 9

Question

Suppose

x

is in the interval

\left[0, \frac{\pi}{2}\right]

and

\log_{24 \sin x}(24 \cos x) = \frac{3}{2}.

Find

24 \cot^2 x.

Answer

Rewrite the equation in exponential form and square both sides to get

\cos^2 x = 24 \sin^3 x

Replace

\cos^2 x

by

1 - \sin^2 x

and look for a rational root of the resulting cubic in

\sin x

In exponential form the equation says

(24 \sin x)^{\frac{3}{2}} = 24 \cos x.

Squaring gives

24^3 \sin^3 x = 24^2 \cos^2 x,

so

\cos^2 x = 24 \sin^3 x.

Writing

s = \sin x

and using

\cos^2 x = 1 - s^2,

we get

24s^3 + s^2 - 1 = 0,

which factors as

(3s - 1)(8s^2 + 3s + 1) = 0.

The quadratic factor has negative discriminant, so

\sin x = \frac{1}{3}.

Then

\begin{aligned} 24 \cot^2 x &= 24 \cdot \frac{1 - \sin^2 x}{\sin^2 x} \\ &= 24 \cdot \frac{\frac{8}{9}}{\frac{1}{9}} \\ &= 24 \cdot 8 = 192. \end{aligned}

Problem 9 in other years