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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 $f(x)$.

Remarks (added by pf02, August 2024)

Both solutions given above are incorrect.

The first solution is hopelessly incorrect. It states that (and relies on it) if $f(x)$ has period $2\pi$ and $f(x_1) = f(x_2)$ then $x_2 - x_1 = m\pi$ for some integer $m$. This is plainly wrong (think of $\sin{\pi/3} = \sin{2\pi/3}$). There is an obvious "red flag" as far as solutions go, namely the solution did not use that $f(x_1) = 0$ and $f(x_2) = 0$.

The second solution starts promising, but then it goes on to (incorrectly) prove the converse of the given problem, namely that if $f(x_1) = 0$ then $f(x_1 + m\pi) = 0$ for any $m$.

Below, I will give a solution to the problem.

Solution

For simplicity of writing, denote $b_k = 1/2^{k - 1}$. $f(x) = b_1\cos{(a_1 + x)} + b_1\cos{(a_1 + x)} + \cdots + b_n\cos{(a_n + x)}$. First, we want to prove that it is not the case that $f(x) = 0$ for all $x$. To prove this, we will prove that the maximum value of $f$ is at least $b_n = 1/2^n$. This will ensure that $f \ne 0$.

We do this by induction. The statement is clear for $n = 1$: $f$ has a maximum value of $b_k = 1$. Assume that we have $n - 1$ terms in $f$, and the maximum value of $f$ is at lease $b_{n - 1} = 1/2^{n - 1}$. Now add the term $b_n\cos{(a_n + x)}$ (and remember that $b_n = 1/2^n$). This additional term has values in $[-1/2^n, 1/2^n]$, so it can decrease the maximum of $f$ by subtracting at most $1/2^n$ from the previous maximum, which was at least $1/2^{n - 1}$. So, the new maximum is at least $1/2^n$.

Now we will the formula $\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}$.

We get $f(x) = (b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n})\cos{x} - (b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n})\sin{x}$.

If both $b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} = 0$ and $b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n} = 0$ then $f(x) = 0$ for all $x$. So at least one of these sums is $\ne 0$. I will give the proof for the case $b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n} \ne 0$. The other case is proven similarly.

Using $f(x_1) = 0$, we get $(b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n})\cos{x_1} = (b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n})\sin{x_1}$, and similarly for $x_2$.

If $b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} = 0$, then $\cos{x_1} = 0$ and $\cos{x_2} = 0$. It follows that both $x_1$ and $x_2$ are odd multiples of $\pi/2$, so they differ by $m\pi$ for some integer $m$.

If $b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} \ne 0$, then we can divide by this quantity, and we get

$\tan{x_1} = \tan{x_2} = \frac{b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n}} {b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n}}$.

Thinking of the graph of $y = \tan{x}$ would be enough for many people to conclude that $x_1 - x_2 = m\pi$ for some integer $m$. If we want to be more formal, we proceed by writing $\tan{x_1} - \tan{x_2} = 0$. Some easy computations yield $\tan{x_1} - \tan{x_2} = \frac{\sin{(x_1 - x_2)}} {\cos{x_1}\cos{x_2}} = 0$. It follows that $\sin{(x_1 - x_2)} = 0$, which implies that $x_1 - x_2 = m\pi$ for some integer $m$.

[Solution by pf02, August 2024]


See Also

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