Trigonometric Polynomial Sum
If $ \arccos x + \arccos 2x + \arccos 3x = \pi, $ then $ x $ satisfies a cubic polynomial of the form
\[ax^3 + bx^2 + cx + d = 0,\]where $ a, $ $ b, $ $ c, $ and $ d $ are integers, and $ a \neq 0 $. Find the smallest possible value of $ |a| + |b| + |c| + |d| $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
From the equation $ \arccos x + \arccos 2x + \arccos 3x = \pi, $ $ \arccos x + \arccos 2x = \pi - \arccos 3x, $ so
\[\cos (\arccos x + \arccos 2x) = \cos (\pi - \arccos 3x).\]From the angle addition formula, the left-hand side becomes
\begin{align*}
\cos (\arccos x + \arccos 2x) &= \cos (\arccos x) \cos (\arccos 2x) - \sin (\arccos x) \sin (\arccos 2x) \\
&= (x)(2x) - (\sqrt{1 - x^2})(\sqrt{1 - 4x^2}) \\
&= 2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)}.\end{align*}The right-hand side becomes
\[\cos (\pi - \arccos 3x) = -\cos (\arccos 3x) = -3x,\]so
\[2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)} = -3x.\]Then $ \sqrt{(1 - x^2)(1 - 4x^2)} = 2x^2 + 3x $. Squaring both sides, we get
\[(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2.\]This simplifies to $ 12x^3 + 14x^2 - 1 = 0 $. Thus, the smallest possible value of $ |a| + |b| + |c| + |d| $ is $ 12 + 14 + 0 + 1 = \boxed{27} $.
Freeflows