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Difference between revisions of "1959 IMO Problems/Problem 3"

 
(IMO box)
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== Problem ==
 
== Problem ==
  
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Let <math>\displaystyle a,b,c</math> be real numbers. Consider the quadratic equation in <math>\displaystyle \cos{x}</math> :
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Let <math>a,b,c</math> be real numbers. Consider the quadratic equation in <math>\cos{x}</math> :
  
 
<center>
 
<center>
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<math>\displaystyle a\cos ^{2}x + b\cos{x} + c = 0.</math>
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<math>a\cos ^{2}x + b\cos{x} + c = 0.</math>
 
</center>
 
</center>
  
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Using the numbers <math>\displaystyle a,b,c</math>, form a quadratic equation in <math>\displaystyle \cos{2x}</math>, whose roots are the same as those of the original equation. Compare the equations in <math>\displaystyle \cos{x}</math> and <math>\displaystyle \cos{2x}</math> for <math>\displaystyle a=4, b=2, c=-1</math>.
+
Using the numbers <math>a,b,c</math>, form a quadratic equation in <math>\cos{2x}</math>, whose roots are the same as those of the original equation. Compare the equations in <math>\cos{x}</math> and <math>\cos{2x}</math> for <math>a=4, b=2, c=-1</math>.
  
 
== Solution ==
 
== Solution ==
  
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Let the original equation be satisfied only for <math>\displaystyle \cos{x}=m, \cos{x}=n </math>.  Then we wish to construct a quadratic with roots <math>\displaystyle 2m^2 -1, 2n^2 -1</math>.
+
Let the original equation be satisfied only for <math>\cos{x}=m, \cos{x}=n </math>.  Then we wish to construct a quadratic with roots <math>2m^2 -1, 2n^2 -1</math>.
  
 
Clearly, the sum of the roots of this quadratic must be
 
Clearly, the sum of the roots of this quadratic must be
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<center>
 
<center>
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<math> \displaystyle a^2 \cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\cos{2x} + (a+2c)^2 - 2b^2 = 0 </math>
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<math>a^2 \cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\cos{2x} + (a+2c)^2 - 2b^2 = 0 </math>
 
</center>
 
</center>
  
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Now, when we let <math>\displaystyle a=4, b=2, c= -1</math>, our equations are  
+
Now, when we let <math>a=4, b=2, c= -1</math>, our equations are  
  
 
<center>
 
<center>
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<math>\displaystyle 4 \cos^2 {x} + 2 \cos {x} - 1 = 0 </math>
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<math>4 \cos^2 {x} + 2 \cos {x} - 1 = 0 </math>
 
</center>
 
</center>
  
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<center>
 
<center>
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<math>\displaystyle 16 \cos^2 {2x} + 8 \cos {2x} - 4 = 0, </math>
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<math>16 \cos^2 {2x} + 8 \cos {2x} - 4 = 0, </math>
 
</center>
 
</center>
  
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i.e., they are multiples of each other.  The reason behind this is that the roots of the first equation are <math> \cos {x} = \frac{-1 \pm \sqrt{5}}{4}</math>, which implies that <math>\displaystyle x</math> is one of two certain multiples of <math>\frac{\pi}{5}</math>, and when <math> \displaystyle 5 \nmid n</math>, <math> \left|\cos\left(\frac{n\pi}{5}\right)\right|</math> can only assume two distinct values.  Q.E.D.
+
i.e., they are multiples of each other.  The reason behind this is that the roots of the first equation are <math> \cos {x} = \frac{-1 \pm \sqrt{5}}{4}</math>, which implies that <math>x</math> is one of two certain multiples of <math>\frac{\pi}{5}</math>, and when <math>5 \nmid n</math>, <math> \left|\cos\left(\frac{n\pi}{5}\right)\right|</math> can only assume two distinct values.  Q.E.D.
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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{{IMO box|year=1959|num-b=2|num-a=4}}
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* [[1959 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341512#p341512 Discussion on AoPS/MathLinks]
 
−
 
 
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 19:22, 25 October 2007

Problem

Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ :

$a\cos ^{2}x + b\cos{x} + c = 0.$

Using the numbers $a,b,c$, form a quadratic equation in $\cos{2x}$, whose roots are the same as those of the original equation. Compare the equations in $\cos{x}$ and $\cos{2x}$ for $a=4, b=2, c=-1$.

Solution

Let the original equation be satisfied only for $\cos{x}=m, \cos{x}=n$. Then we wish to construct a quadratic with roots $2m^2 -1, 2n^2 -1$.

Clearly, the sum of the roots of this quadratic must be

$2 (m^2 + n^2) - 2 = 2 \left(\frac{b^2-2ac}{a^2}\right) - 2 = \frac{2b^2 - 4ac - 2a^2}{a^2},$

and the product of its roots must be

$4m^2 n^2 - 2(m^2 + n^2) + 1 = \frac{4c^2}{a^2}+\frac{4ac - 2b^2}{a^2} + \frac{a^2}{a^2} = \frac{(a+2)^2 - 2b^2}{a^2}$

Thus the following quadratic fulfils the conditions:

$a^2 \cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\cos{2x} + (a+2c)^2 - 2b^2 = 0$

Now, when we let $a=4, b=2, c= -1$, our equations are

$4 \cos^2 {x} + 2 \cos {x} - 1 = 0$

and

$16 \cos^2 {2x} + 8 \cos {2x} - 4 = 0,$

i.e., they are multiples of each other. The reason behind this is that the roots of the first equation are $\cos {x} = \frac{-1 \pm \sqrt{5}}{4}$, which implies that $x$ is one of two certain multiples of $\frac{\pi}{5}$, and when $5 \nmid n$, $\left|\cos\left(\frac{n\pi}{5}\right)\right|$ can only assume two distinct values. Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1959 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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