Difference between revisions of "1969 IMO Problems/Problem 2"
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<cmath>f(x_1)=f(x_2)=f(x_1+x_2-x_1)</cmath> | <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> | 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>. | ||
==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 00:55, 9 August 2024
Contents
Problem
Let be real constants,
a real variable, and
Given that
prove that
for some integer
Solution
Because the period of is
, the period of
is also
.
We can get
for
. Thus,
for some integer
Solution 2 (longer)
By the cosine addition formula,
This implies that if
,
Since the period of
is
, this means that
for any natural number
. That implies that every value
is a zero of
.
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 has period
and
then
for
some integer
. This is plainly wrong (think of
).
There is an obvious "red flag" as far as solutions go, namely the solution did
not use that
and
.
The second solution starts promising, but then it goes on to (incorrectly) prove
the converse of the given problem, namely that if then
for any
.
Below, I will give a solution to the problem.
Solution
For simplicity of writing, denote .
.
First, we want to prove that it is not the case that
for all
.
To prove this, we will prove that the maximum value of
is at least
. This will ensure that
.
We do this by induction. The statement is clear for :
has a maximum
value of
. Assume that we have
terms in
, and the maximum
value of
is at lease
. Now add the term
(and remember that
). This additional term
has values in
, so it can decrease the maximum of
by
subtracting at most
from the previous maximum, which was at least
. So, the new maximum is at least
.
Now we will the formula
.
We get
.
If both
and
then
for all
. So at least one of these sums is
.
I will give the proof for the case
.
The other case is proven similarly.
Using , we get
,
and similarly for
.
If , then
and
. It follows that both
and
are odd multiples of
, so they differ by
for some
integer
.
If , then
we can divide by this quantity, and we get
.
Thinking of the graph of would be enough for many people
to conclude that
for some integer
. If we want
to be more formal, we proceed by writing
.
Some easy computations yield
.
It follows that
, which implies that
for some integer
.
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 |