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Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 13"

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==Solution==
 
==Solution==
 
<math>\displaystyle (x - x_1)(x - x_2) = x^2 + ax + 1 = 0</math>, so <math>a = -(x_1 + x_2) \displaystyle</math> and <math>x_1 \cdot x_2 = 1</math> (the same goes for <math>b,\ x_3,\ x_4</math>).
 
<math>\displaystyle (x - x_1)(x - x_2) = x^2 + ax + 1 = 0</math>, so <math>a = -(x_1 + x_2) \displaystyle</math> and <math>x_1 \cdot x_2 = 1</math> (the same goes for <math>b,\ x_3,\ x_4</math>).
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<math>\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4} = \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1 \cdot x_2}
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= x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = a^2 - 2</math>
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Similarly (for <math>b,\ x_3,\ x_4</math>) <math>\frac{x_3}{x_1x_2x_4} + \frac{x_4}{x_1x_2x_3} = b^2 - 2</math>
  
 
So
 
So
  
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<math>\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3} = \frac{x_1^2 + x_2^2 + x_3^2 + x_4^2}{x_1 \cdot x_2 \cdot x_3 \cdot x_4}</math><math> = (x_1 + x_2)^2 - 2x_1x_2 + (x_3 + x_4)^2 - 2x_3x_4 = a^2 + b^2 - 4 \Longrightarrow \mathrm{E}</math>
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<math>\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3} = a^2 + b^2 - 4 \Longrightarrow \mathrm{E}</math>
  
 
==See also==
 
==See also==
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{{CYMO box|year=2007|l=Lyceum|num-b=3|num-a=5}}
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{{CYMO box|year=2007|l=Lyceum|num-b=12|num-a=14}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 11:02, 8 May 2007

Problem

If $x_1,\ x_2$ are the roots of the equation $\displaystyle x^2+ax+1=0$ and $x_3,\ x_4$ are the roots of the equation $\displaystyle x^2+bx+1=0$, then the expression $\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3}$ equals to

$\mathrm{(A) \ } a^2+b^2-2\qquad \mathrm{(B) \ } a^2+b^2\qquad \mathrm{(C) \ } \frac{a^2+b^2}{2}\qquad \mathrm{(D) \ } a^2+b^2+1\qquad \mathrm{(E) \ } a^2+b^2-4$

Solution

$\displaystyle (x - x_1)(x - x_2) = x^2 + ax + 1 = 0$, so $a = -(x_1 + x_2) \displaystyle$ and $x_1 \cdot x_2 = 1$ (the same goes for $b,\ x_3,\ x_4$).


$\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4} = \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1 \cdot x_2} = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = a^2 - 2$

Similarly (for $b,\ x_3,\ x_4$) $\frac{x_3}{x_1x_2x_4} + \frac{x_4}{x_1x_2x_3} = b^2 - 2$

So

$\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3} = a^2 + b^2 - 4 \Longrightarrow \mathrm{E}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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