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Difference between revisions of "1969 IMO Problems/Problem 2"

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==Solution==
 
==Solution==
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{{solution}}
+
Because the period of <math>\cos(x)</math> is <math>2\pi</math>, the period of <math>f(x)</math> is also <math>2\pi</math>.
 +
<cmath>f(x_1)=f(x_2)=f(x_1+x_2-x_1)</cmath>
 +
We can get <math>x_2-x_1 = 2k\pi</math> for <math>k\in N^*</math>. Thus, <math>x_2-x_1=m\pi</math> for some integer <math>m.</math>
 +
==Solution 2 (longer)==
 +
By the cosine addition formula,
 +
<cmath>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}</cmath>
 +
This implies that if <math>f(x_1)=0</math>,
 +
<cmath>\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}}</cmath>
 +
Since the period of <math>\tan{x}</math> is <math>\pi</math>, this means that <math>\tan{x_1}=\tan{(x_1+\pi)}=\tan{(x_1+m\pi)}</math> for any natural number <math>m</math>. That implies that every value <math>x_1+m\pi</math> is a zero of <math>f(x)</math>.
 +
 
 +
==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 <math>f(x)</math> has period <math>2\pi</math> and <math>f(x_1) = f(x_2)</math> then <math>x_2 - x_1 = m\pi</math> for
 +
some integer <math>m</math>.  This is plainly wrong (think of <math>\sin{\pi/3} = \sin{2\pi/3}</math>).
 +
There is an obvious "red flag" as far as solutions go, namely the solution did
 +
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 (incorrectly) prove
 +
the converse of the given problem, namely that if <math>f(x_1) = 0</math> then
 +
<math>f(x_1 + m\pi) = 0</math> for any <math>m</math>.
 +
 
 +
Below, I will give a solution to the problem.
 +
 
 +
==Solution==
 +
 
 +
For simplicity of writing, denote <math>b_k = 1/2^{k - 1}</math>.
 +
<math>f(x) = b_1\cos{(a_1 + x)} + b_1\cos{(a_1 + x)} + \cdots + b_n\cos{(a_n + x)}</math>.
 +
First, we want to prove that it is not the case that <math>f(x) = 0</math> for all <math>x</math>.
 +
To prove this, we will prove that the maximum value of <math>f</math> is at least
 +
<math>b_n = 1/2^n</math>.  This will ensure that <math>f \ne 0</math>.
 +
 
 +
We do this by induction.  The statement is clear for <math>n = 1</math>: <math>f</math> has a maximum
 +
value of <math>b_k = 1</math>.  Assume that we have <math>n - 1</math> terms in <math>f</math>, and the maximum
 +
value of <math>f</math> is at lease <math>b_{n - 1} = 1/2^{n - 1}</math>.  Now add the term
 +
<math>b_n\cos{(a_n + x)}</math> (and remember that <math>b_n = 1/2^n</math>).  This additional term
 +
has values in <math>[-1/2^n, 1/2^n]</math>, so it can decrease the maximum of <math>f</math> by
 +
subtracting at most <math>1/2^n</math> from the previous maximum, which was at least
 +
<math>1/2^{n - 1}</math>.  So, the new maximum is at least <math>1/2^n</math>.
 +
 
 +
Now we will the formula
 +
<math>\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}</math>.
 +
 
 +
We get
 +
<math>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}</math>.
 +
 
 +
If both <math>b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} = 0</math>
 +
and <math>b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n} = 0</math> then
 +
<math>f(x) = 0</math> for all <math>x</math>.  So at least one of these sums is <math>\ne 0</math>.
 +
I will give the proof for the case
 +
<math>b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n} \ne 0</math>.
 +
The other case is proven similarly.
 +
 +
Using <math>f(x_1) = 0</math>, we get
 +
<math>(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}</math>,
 +
and similarly for <math>x_2</math>.
 +
 
 +
If <math>b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} = 0</math>, then
 +
<math>\cos{x_1} = 0</math> and <math>\cos{x_2} = 0</math>.  It follows that both <math>x_1</math> and
 +
<math>x_2</math> are odd multiples of <math>\pi/2</math>, so they differ by <math>m\pi</math> for some
 +
integer <math>m</math>.
 +
 
 +
If <math>b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} \ne 0</math>, then
 +
we can divide by this quantity, and we get
 +
 
 +
<math>\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}}</math>.
 +
 
 +
Thinking of the graph of <math>y = \tan{x}</math> would be enough for many people
 +
to conclude that <math>x_1 - x_2 = m\pi</math> for some integer <math>m</math>.  If we want
 +
to be more formal, we proceed by writing <math>\tan{x_1} - \tan{x_2} = 0</math>.
 +
Some easy computations yield
 +
<math>\tan{x_1} - \tan{x_2} = \frac{\sin{(x_1 - x_2)}} {\cos{x_1}\cos{x_2}} = 0</math>.
 +
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>.
 +
 
 +
[Solution by pf02, August 2024]
 +
 
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1969|num-b=1|num-a=3}}
 
{{IMO box|year=1969|num-b=1|num-a=3}}
−
[[Category:Olympiad Geometry Problems]]
 

Latest revision as of 18:32, 10 November 2024

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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