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1969 IMO Problems/Problem 2: Difference between revisions

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not use that <math>f(x_1) = 0</math> and <math>f(x_2) = 0</math>.
not use that <math>f(x_1) = 0</math> and <math>f(x_2) = 0</math>.


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


Below, I will give a solution to the problem.
Below, I will give a solution to the problem.
Line 83: Line 83:
It follows that <math>\sin{(x_1 - x_2)} = 0</math>, which implies that
It follows that <math>\sin{(x_1 - x_2)} = 0</math>, which implies that
<math>x_1 - x_2 = m\pi</math> for some integer <math>m</math>.
<math>x_1 - x_2 = m\pi</math> for some integer <math>m</math>.
[Solution by pf02, August 2024]
The above solution is great, but why don't we consider another elegant answer relying on geometric sense- \\
Consider the following polyline, such that it starts with <math>O=A_{0}</math> and satisfy <math>A_{j-1}A_{j}=2^{j-1}</math> and the angle <math>A_{j-1}A_{j}</math> makes with positive direction of x-axis is simply <math>a_{j}</math>, to reach a destination of <math>A_{n}=A</math>.\\
First, it is trivial to see that <math>OA>=|OA_{1}|-|OA_{2}|-|OA_{3}|-...=1-1/2-1/4-...=1/2^{n}</math>, so the point <math>A_{n}</math> does not coincide with the origin.\\
Second, the horizontal coordinate of <math>A</math> is just given by the sum of horizontal projection of each segment <math>A_{j-1}A_{j}</math>, such that <math>x(A_{n}=x(A_{n})-x(O)=\sum_{j=1}^{n}(x(A_{j})-x(A_{j-1}))=\sum_{j=1}^{n}cos(a_{j})</math>.\\
More importantly, when we rotate this polyline with center O and angle x, we can see that each term in the summation above is replaced with <math>cos(a_{j}+x)</math>. So f(x) is just the horizontal coordinate of <math>OA</math> rotated by angle x.\\
Now the simple observation is that, OA needs to be vertical if <math>f(x)=0</math>, and in order to rotate from a vertical position into another one, you must rotate by an integer multiple times of <math>\pi</math>.
[Soultion by KROSA1910, Sep 2025]


==See Also==
==See Also==


{{IMO box|year=1969|num-b=1|num-a=3}}
{{IMO box|year=1969|num-b=1|num-a=3}}

Latest revision as of 20:14, 13 September 2025

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]

The above solution is great, but why don't we consider another elegant answer relying on geometric sense- \\

Consider the following polyline, such that it starts with $O=A_{0}$ and satisfy $A_{j-1}A_{j}=2^{j-1}$ and the angle $A_{j-1}A_{j}$ makes with positive direction of x-axis is simply $a_{j}$, to reach a destination of $A_{n}=A$.\\ First, it is trivial to see that $OA>=|OA_{1}|-|OA_{2}|-|OA_{3}|-...=1-1/2-1/4-...=1/2^{n}$, so the point $A_{n}$ does not coincide with the origin.\\ Second, the horizontal coordinate of $A$ is just given by the sum of horizontal projection of each segment $A_{j-1}A_{j}$, such that $x(A_{n}=x(A_{n})-x(O)=\sum_{j=1}^{n}(x(A_{j})-x(A_{j-1}))=\sum_{j=1}^{n}cos(a_{j})$.\\ More importantly, when we rotate this polyline with center O and angle x, we can see that each term in the summation above is replaced with $cos(a_{j}+x)$. So f(x) is just the horizontal coordinate of $OA$ rotated by angle x.\\ Now the simple observation is that, OA needs to be vertical if $f(x)=0$, and in order to rotate from a vertical position into another one, you must rotate by an integer multiple times of $\pi$.

[Soultion by KROSA1910, Sep 2025]

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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