AIME 2006 I · 第 12 题

Solution

Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$, we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$. But $a+b = 2\cos 4x\cos x$, so we require $\cos x = 0$, $\cos 3x = 0$, $\cos 4x = 0$, or $\cos 5x = 0$.

Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$.

Solution 2 (complex nuke)

Let $z=\cos(x)+i \sin(x)$. Then, we have that $|z|=1$, so $\overline{z}=\frac{1}{z}$. Thus, $\frac{1}{z}=\cos(x)-i \sin(x)$, and we have $\cos(x)=\frac{z+\frac{1}{z}}{2}$. Now, this means we also have

$$z^{3}=\cos(3x)+i\sin(3x) \rightarrow z^{3}+\frac{1}{z^{3}} = 2\cos(3x) \rightarrow \frac{z^{3}+\frac{1}{z^{3}}}{2}=\cos(3x)$$ $$\frac{z^{5}+\frac{1}{z^{5}}}{2}=\cos(5x)$$ Then, we seek the solutions to

$$\frac{1}{8} \cdot \left(z^{3}+\frac{1}{z^{3}}\right)^{3} + \frac{1}{8} \cdot \left(z^{5}+\frac{1}{z^{5}}\right)^{3} = 8 \cdot \frac{1}{64} \cdot \left(z+\frac{1}{z}\right)^{3}\left(z^{4}+\frac{1}{z^{4}}\right)^{3}$$ $$\rightarrow \left(z^{3}+\frac{1}{z^{3}}\right)^{3}+\left(z^{5}+\frac{1}{z^{5}}\right)^{3}=\left(z^{3}+\frac{1}{z^{3}}+z^{5}+\frac{1}{z^{5}}\right)^{3}$$ Now, if we let $a=z^{3}+\frac{1}{z^{3}}$ and we let $b=z^{5}+\frac{1}{z^{5}}$, we actually obtain the equation

$$a^{3}+b^{3}=(a+b)^{3} \rightarrow ab(a+b)=0$$ which means we either have

$$\textbf{Case 1:} \hspace{3pt} z^{3}+\frac{1}{z^{3}}=0$$ $$\textbf{Case 2:} \hspace{3pt} z^{5}+\frac{1}{z^{5}}=0$$ $$\textbf{Case 3:} \hspace{5pt} z^{3}+\frac{1}{z^{3}}=z^{5}=\frac{1}{z^{5}}$$ In Case 1, we have $z^{12}=1$ but $z^{6} \neq 1$. Thus, we have that the possibilities for $\arg[z]$ are $30^\circ$, $90^\circ$, $\underline{150}^\circ$, $210^\circ$, $270^\circ$, and $330^\circ$.

In Case 2, we have $z^{20}=1$ but $z^{10} \neq 1$. Thus, we have that the set of possible $z$ are 10th roots of unity shifted by $18^\circ$ counterclockwise, so we have

$$\arg[z] \in \{18^\circ, 54^\circ, 90^\circ, \underline{126}^\circ, \underline{162}^\circ, \underline{198}^\circ, \dots, 342^\circ \}$$ so we obtain $3$ values. In case 3, we have

$$z^{3}+\frac{1}{z^{3}}=z^{5}+\frac{1}{z^{5}} \rightarrow z^{10}+z^{8}+z^{2}+1=0 \rightarrow (z^{8}+1)(z^{2}+1)=0$$ so either $z \in \{i, -i\}$ or $z$ is a primitive 8th root of unity shifted by $22.5^\circ$, which gives

$$\arg[z] \in \{22.5^\circ, 67.5^\circ, \underline{112.5}^\circ, \underline{157.5}^\circ, 202.5^\circ, 247.5^\circ, 292.5^\circ, 337.5^\circ \}$$ The sum of all underlined values is $150+126+162+198+112.5+157.5=\fbox{906}$