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Difference between revisions of "1969 IMO Problems/Problem 2"
Line 34: Line 34:
  
 
Do the proof for the slightly more general function
 
Do the proof for the slightly more general function
−
<math>f(x) = b_1\cos{(a_1 + x)} + b_1\cos{(a_1 + x)} + \cdots + b_n\cos{(a_n + x)}</math>
+
<math>f(x) = b_1\cos{(a_1 + x)} + b_1\cos{(a_1 + x)} + \cdots + b_n\cos{(a_n + x)}</math>.
 +
We assume that <math>b_k \ne 0</math>, and no two of them are equal.  (This is where we
 +
make use of the fact that in the problem <math>b_k = 1/2^{k - 1}</math>.)  It follows that
 +
the function <math>f(x)</math> is not identical to <math>0</math>.
  
−
Just like the previous attempt to solve the problem, start by using the formula
+
Start by using the formula
 
<math>\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}</math>.
 
<math>\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}</math>.
  
Line 43: Line 46:
 
(b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n})\sin{x}</math>.
 
(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
 
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} =
 
<math>(b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n})\cos{x_1} =
Line 63: Line 73:
 
to conclude that <math>x_1 - x_2 = m\pi</math> for some integer <math>m</math>.  If we want
 
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>.
 
to be more formal, we proceed by writing <math>\tan{x_1} - \tan{x_2} = 0</math>.
−
Using the formula <math>\tan{x_1} - \tan{x_2} = \frac{\sin{(x_1 - x_2)}}
+
Some easy computations yield
−
{\cos{x_1}\cos{x_2}} = 0</math>. It follows that <math>\sin{(x_1 - x_2)} = 0</math>,
+
<math>\tan{x_1} - \tan{x_2} = \frac{\sin{(x_1 - x_2)}} {\cos{x_1}\cos{x_2}} = 0</math>.
−
which implies that <math>x_1 - x_2 = m\pi</math> for some integer <math>m</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>.
−
Note that in the course of the proof, I used that if
 
−
<math>b_1\sin{a_1} + b_2\sin{a_2} + \cdots + b_n\sin{a_n} = 0</math>, then
 
−
<math>b_1\cos{a_1} + b_2\cos{a_2} + \cdots + b_n\cos{a_n} \ne 0</math>.
 
−
 
 
−
 
 
−
 
 
−
 
 
−
TO BE CONTINUED. I AM JUST SAVING NOW SO THAT I DON'T LOSE WORK DONE SO FAR.
 
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1969|num-b=1|num-a=3}}
 
{{IMO box|year=1969|num-b=1|num-a=3}}

Revision as of 18:05, 8 August 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 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. I feel a little uneasy about it, it has an "orange flag", namely I make no use of the fact that the coefficients if $\cos{(a_k + x)}$ are the given powers of $1/2$.

Solution

Do the proof for the slightly more general function $f(x) = b_1\cos{(a_1 + x)} + b_1\cos{(a_1 + x)} + \cdots + b_n\cos{(a_n + x)}$. We assume that $b_k \ne 0$, and no two of them are equal. (This is where we make use of the fact that in the problem $b_k = 1/2^{k - 1}$.) It follows that the function $f(x)$ is not identical to $0$.

Start by using 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 $m$.

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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