Bulgaria National Olympiad Problem 3

Hi,

Firstly,

${ \cos { \pi \left( a-x \right) } }^{ 2 }-2\cos { \pi \left( a-x \right) } +\cos { \frac { 3\pi x }{ 2a } } \cos { \left( \frac { \pi x }{ 2a } +\frac { \pi }{ 3 } \right) } +2=0\\ \Leftrightarrow { \left[ \cos { \pi \left( a-x \right) } -1 \right] }^{ 2 }+\frac { 1 }{ 2 } \left[ \cos { \left( \frac { 2\pi x }{ a } +\frac { \pi }{ 3 } \right) } +\cos { \left( \frac { \pi x }{ a } -\frac { \pi }{ 3 } \right) } \right] +1=0\\ \Leftrightarrow { \left[ \cos { \pi \left( a-x \right) } -1 \right] }^{ 2 }{ -\left[ \sin { \left( \frac { \pi x }{ a } +\frac { \pi }{ 6 } \right) } -\frac { 1 }{ 4 } \right] }^{ 2 }=-{ \left( \frac { 5 }{ 4 } \right) }^{ 2 }$

We can write ${ \left[ X-1 \right] }^{ 2 }-{ \left[ Y-\frac { 1 }{ 4 } \right] }^{ 2 }=-{ \left( \frac { 5 }{ 4 } \right) }^{ 2 }$ where $X\equiv \cos { \pi \left( a-x \right) }$ and $Y\equiv \sin { \left( \frac { \pi x }{ a } +\frac { \pi }{ 6 } \right) }$

Now, there is at least 2 ways to deal with the equation. The first one is to note that

${ \left[ X-1 \right] }^{ 2 }=\left( Y-\frac { 3 }{ 2 } \right) \left( Y+1 \right)$

$LHS\ge 0$ , meanwhile $RHS\le 0$ . Therefore, $LHS=RHS\Leftrightarrow LHS=RHS=0$

$\Leftrightarrow \begin{cases} X=\cos { \pi \left( a-x \right) } =1 \\ Y=\sin { \left( \frac { \pi x }{ a } +\frac { \pi }{ 6 } \right) } =-1 \end{cases}\\ \Leftrightarrow \begin{cases} \pi \left( a-x \right) =k\left( 2\pi \right) \quad (k\in Z) \\ \pi \left( \frac { x }{ a } +\frac { 1 }{ 6 } \right) =-\frac { \pi }{ 2 } +l\left( 2\pi \right) \quad (l\in Z) \end{cases}\\ \Rightarrow \left( 5+6l \right) a=6k\\ \Rightarrow 6\mid a$

The second method, divide the both hand sides by ${ \left( \frac { 5 }{ 4 } \right) }^{ 2 }$ , it now represents a hyperbola. At the same time, don't forger that $-1\le X,Y\le 1$ or $\left( X,Y \right)$ lies on the square whose center is at the origin and side length is $2$ . After some investigation, it turns out that the hyperbola intersects the square at just one point, $\left( 1,-1 \right) \equiv a\quad vertex\quad of\quad the\quad square\equiv a\quad vertex\quad of\quad the\quad hyperbola$ .

On ti-83 PLUS calculator plot
$\cos^2\pi(a-x)-2\cos\pi(a-x)+\cos\frac{3\pi x}{2a}\cos(\frac{\pi x}{2a}+\frac{\pi}{3})+2=0 \\Set~X_{min}=-500, ~~X_{max}=500,~~Y_{min},-1E-4~~Y_{max}=0.01.\\set~~ a=0 ~ENTER. ~~~1+a ~STR ~a:\\\text{Press GRAPH CLEAR ENTER. Comtinue pressing these keys in }\\ \text {that order, and get graph ploted for}~a=1, ~2,~...~,~6~when~Y\approx 0.01.\\So ~a=~~\huge \color{#D61F06}{6}$