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

(The "solution 2" was a complete balooney. The two polynimials are not equivalent by any means)
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<math> 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} </math>
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<math> 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+2c)^2 - 2b^2}{a^2} </math>
 
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<math>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>
 
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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>. The roots of the second equation are <math>\cos(2x) = \frac{1 \pm \sqrt{5}}{4}</math>. It is straightforward to verify that they result in the same values of <math>x</math>.  
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This simplifies to previous equation.
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The first root of the first equation <math> \cos {x} = \frac{-1 + \sqrt{5}}{4}</math> corresponds to <math> \cos {2x} = \frac{-1 - \sqrt{5}}{4}</math> and the second root of the first equation <math> \cos {x} = \frac{-1 - \sqrt{5}}{4}</math> corresponds to <math> \cos {2x} = \frac{-1 + \sqrt{5}}{4}</math>. These are relatively well-known trigonometric values, yielding roots <math>x = \boxed{72^\circ, 144^\circ}</math>.
  
 
{{alternate solutions}}
 
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== See Also ==
 
{{IMO box|year=1959|num-b=2|num-a=4}}
 
{{IMO box|year=1959|num-b=2|num-a=4}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Latest revision as of 22:47, 1 June 2024

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

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

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

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

This simplifies to previous equation. The first root of the first equation $\cos {x} = \frac{-1 + \sqrt{5}}{4}$ corresponds to $\cos {2x} = \frac{-1 - \sqrt{5}}{4}$ and the second root of the first equation $\cos {x} = \frac{-1 - \sqrt{5}}{4}$ corresponds to $\cos {2x} = \frac{-1 + \sqrt{5}}{4}$. These are relatively well-known trigonometric values, yielding roots $x = \boxed{72^\circ, 144^\circ}$.

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

See Also

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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