1969 IMO Problems/Problem 2

Problem

Let $a_1, a_2,\cdots, a_n$ be real constants, $x$ a real variable, and $f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).$ Given that $f(x_1)=f(x_2)=0,$ prove that $x_2-x_1=m\pi$ for some integer $m.$

Solution

Because the period of $\cos(x)$ is $2\pi$ , the period of $f(x)$ is also $2\pi$ . $f(x_1)=f(x_2)=f(x_1+x_2-x_1)$ We can get $x_2-x_1 = 2k\pi$ for $k\in N^*$ . Thus, $x_2-x_1=m\pi$ for some integer $m.$

Solution 2 (longer)

By the cosine addition formula, $f(x)=(\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}\cos{a_n})\cos{x}-(\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}}\sin{a_n})\sin{x}$ This implies that if $f(x_1)=0$ , $\tan{x_1}=\frac{\cos{a_1}+\frac{1}{2}\cos{a_2}+\frac{1}{4}\cos{a_3}+\cdots+\frac{1}{2^{n-1}}\cos{a_n}}{\sin{a_1}+\frac{1}{2}\sin{a_2}+\frac{1}{4}\sin{a_3}+\cdots+\frac{1}{2^{n-1}}\sin{a_n}}$ Since the period of $\tan{x}$ is $\pi$ , this means that $\tan{x_1}=\tan{x_1+\pi}=\tan{x_1+m\pi}$ for any natural number $m$ . That implies that every value $x_1+m\pi$ is a zero of $f(x)$ .

1969 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1969 IMO Problems/Problem 2

Contents

Problem

Solution

Solution 2 (longer)

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