IMO 1969 Problem 2

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Verified: no
Verdicts: PASS + FAIL
Solve time: 5m39s

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 $m$.

Exploration

The expression is a finite linear combination of shifted cosine functions with positive weights decreasing geometrically. The condition $f(x_1)=f(x_2)=0$ suggests comparing two trigonometric polynomials evaluated at different points.

A natural idea is to rewrite each cosine as the real part of a complex exponential:

$$\cos(a_k + x) = \Re\big(e^{i(a_k+x)}\big),$$

so that

$$f(x) = \Re\left(e^{ix}\sum_{k=1}^n \frac{e^{ia_k}}{2^{k-1}}\right).$$

This reduces the problem to the real part of a single complex number multiplied by $e^{ix}$.

Then zeros of $f(x)$ correspond to arguments where the real part of a fixed complex number rotated by $x$ vanishes, which happens when the rotated vector is purely imaginary. This suggests that all zeros occur at angles differing by $\pi$.

A potential difficulty is whether the inner complex sum could be zero, which would make $f(x)$ identically zero. That case must be ruled out or handled separately.

The core expectation is that $f(x)$ is of the form $R\cos(x+\varphi)$, forcing all zeros to differ by integer multiples of $\pi$.

Problem Understanding

This is a Type B problem: we must prove a structural statement about the zeros of a trigonometric expression.

We are given a function $f(x)$ that is a weighted sum of shifted cosines, and we must show that if it has two zeros at $x_1$ and $x_2$, then their difference is an integer multiple of $\pi$.

The core difficulty is to show that despite the different phase shifts $a_k$, the entire sum collapses to a single cosine wave with respect to $x$. A naive attempt to equate terms or analyze each cosine separately fails because the shifts $a_k$ are arbitrary and unrelated.

The correct intuition is that all dependence on $x$ factors through a single harmonic component $e^{ix}$, forcing a one-dimensional oscillation in $x$.

Proof Architecture

Lemma 1 states that $f(x)$ can be written as the real part of $e^{ix}z$, where $z=\sum_{k=1}^n 2^{1-k}e^{ia_k}$, and this follows from Euler’s formula and linearity of summation.

Lemma 2 states that if $z=0$, then $f(x)$ is identically zero, so any conclusion about differences of zeros is immediate.

Lemma 3 states that if $z\neq 0$, then $f(x)$ can be written as $R\cos(x+\varphi)$ for suitable real constants $R>0$ and $\varphi$, obtained from polar representation of $z$.

Lemma 4 states that if $R\cos(x+\varphi)$ vanishes at two points $x_1$ and $x_2$, then $x_2-x_1$ is an integer multiple of $\pi$, since cosine has period $2\pi$ and sign symmetry.

The key difficulty lies in Lemma 3, where the reduction to a single cosine must be fully justified.

Solution

Lemma 1

There exists a complex number

$$z=\sum_{k=1}^n \frac{e^{ia_k}}{2^{k-1}}$$

such that

$$f(x)=\Re\left(e^{ix}z\right).$$

Each term satisfies $\cos(a_k+x)=\Re(e^{i(a_k+x)})=\Re(e^{ix}e^{ia_k})$, and linearity of the real part over finite sums yields the identity.

This establishes that the entire function is governed by a single complex coefficient multiplying $e^{ix}$, and failure of this reduction would incorrectly treat independent phases as independent variables.

Lemma 2

If $z=0$, then $f(x)=0$ for all $x$.

The representation in Lemma 1 gives $f(x)=\Re(e^{ix}\cdot 0)=0$ for all $x$. Hence every pair $x_1,x_2$ satisfies $x_2-x_1=m\pi$ for some integer $m$, since the condition holds vacuously.

This removes the degenerate case where the amplitude of the resulting wave vanishes entirely.

Lemma 3

If $z\neq 0$, there exist real numbers $R>0$ and $\varphi$ such that

$$f(x)=R\cos(x+\varphi).$$

Write $z=Re^{i\varphi}$ with $R=|z|>0$ and $\varphi=\arg z$. Then

$$f(x)=\Re(e^{ix}Re^{i\varphi})=\Re(Re^{i(x+\varphi)})=R\cos(x+\varphi).$$

This shows the entire function is a single shifted cosine, so no hidden multi-frequency interaction survives the summation.

Lemma 4

If $R\cos(x+\varphi)=0$ at $x=x_1$ and $x=x_2$, then $x_2-x_1=m\pi$ for some integer $m$.

From $R>0$, both conditions imply $\cos(x_1+\varphi)=\cos(x_2+\varphi)=0$. The zeros of cosine are exactly

$$x+\varphi=\frac{\pi}{2}+k\pi,\quad k\in\mathbb{Z}.$$

Thus there exist integers $k_1,k_2$ such that

$$x_1+\varphi=\frac{\pi}{2}+k_1\pi,\quad x_2+\varphi=\frac{\pi}{2}+k_2\pi.$$

Subtracting gives

$$x_2-x_1=(k_2-k_1)\pi,$$

which is an integer multiple of $\pi$.

This completes the structural classification of all zeros once the single-frequency form is established.

Completion of the argument

If $z=0$, Lemma 2 applies and the conclusion holds for all pairs $x_1,x_2$. If $z\neq 0$, Lemma 3 reduces $f$ to a shifted cosine, and Lemma 4 yields the required relation between any two zeros.

This completes the proof.

∎

Verification of Key Steps

The representation $f(x)=\Re(e^{ix}z)$ is re-derived directly from $\cos t=\Re(e^{it})$, and each term contributes a factor $e^{ia_k}$ independent of $x$, so no cross-term or coupling between different $a_k$ can appear.

The reduction $z=Re^{i\varphi}$ is checked by confirming $R=|z|>0$ when $z\neq 0$, ensuring that no division by zero or undefined phase occurs.

The characterization of cosine zeros is verified by solving $\cos t=0$ directly via the unit circle, yielding exactly the arithmetic progression $\frac{\pi}{2}+k\pi$, so no additional solutions exist.

Alternative Approaches

A real-variable approach avoids complex numbers by proving inductively that any function of the form $A\cos x+B\sin x$ has all zeros differing by integer multiples of $\pi$, and then showing that the given weighted sum collapses into such a form by expanding each cosine using addition formulas.

A second approach uses vector geometry in $\mathbb{R}^2$, interpreting each cosine as the projection of a rotating unit vector, and combining weighted vectors into a single resultant vector whose projection is $f(x)$, again reducing the problem to a single oscillation.