Difference between revisions of "1973 IMO Problems/Problem 3"
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| + | ==Problem== | ||
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Let <math>a</math> and <math>b</math> be real numbers for which the equation | Let <math>a</math> and <math>b</math> be real numbers for which the equation | ||
<math>x^4 + ax^3 + bx^2 + ax + 1 = 0</math> | <math>x^4 + ax^3 + bx^2 + ax + 1 = 0</math> | ||
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Substitute <math>z=x+1/x</math> to change the original equation into <math>z^2+az+b-2=0</math>. This equation has solutions <math>z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}</math>. We also know that <math>|z|=|x+1/x| \geq 2</math>. So, | Substitute <math>z=x+1/x</math> to change the original equation into <math>z^2+az+b-2=0</math>. This equation has solutions <math>z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}</math>. We also know that <math>|z|=|x+1/x| \geq 2</math>. So, | ||
| − | < | + | <cmath>\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2</cmath> |
| − | < | + | |
| − | < | + | <cmath>\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2</cmath> |
| + | |||
| + | <cmath>|a|+\sqrt{a^2+8-4b} \geq 4</cmath> | ||
Rearranging and squaring both sides, | Rearranging and squaring both sides, | ||
| − | < | + | <cmath>a^2+8-4b \geq a^2-16|a|+16</cmath> |
| − | < | + | |
| + | <cmath>2|a|-b \geq 2</cmath> | ||
So, <math>a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}</math>. | So, <math>a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}</math>. | ||
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Therefore, the smallest possible value of <math>a^2+b^2</math> is <math>\frac{4}{5}</math>, when <math>a=\pm \frac{4}{5}</math> and <math>b=\frac{-2}{5}</math>. | Therefore, the smallest possible value of <math>a^2+b^2</math> is <math>\frac{4}{5}</math>, when <math>a=\pm \frac{4}{5}</math> and <math>b=\frac{-2}{5}</math>. | ||
| − | Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html | + | Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html] |
| + | |||
| + | == See Also == {{IMO box|year=1973|num-b=2|num-a=4}} | ||
Latest revision as of 04:01, 21 July 2022
Problem
Let 
and 
be real numbers for which the equation
has at least one real solution. For all such pairs 
, find the minimum value of 
.
Solution
Substitute 
to change the original equation into 
. This equation has solutions 
. We also know that 
. So,
![\[\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2\]](http://latex.artofproblemsolving.com/6/8/6/6866914d2cecb3b0a96fc9791fdc66632a07baae.png)
![\[\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2\]](http://latex.artofproblemsolving.com/8/0/a/80a4f03aed3d782c0c9feac07728cb2a9df6085f.png)
![\[|a|+\sqrt{a^2+8-4b} \geq 4\]](http://latex.artofproblemsolving.com/9/c/c/9ccbbfd2a314b3dbdf426e566a21f3a21ba4ddd9.png)
Rearranging and squaring both sides,
![\[a^2+8-4b \geq a^2-16|a|+16\]](http://latex.artofproblemsolving.com/a/3/5/a358e6e5e390dc6b5ecf53ae770a3cfbca0a5a9c.png)
![\[2|a|-b \geq 2\]](http://latex.artofproblemsolving.com/8/2/4/824d4573973eab394b4ec6814faf7ed2b8f9f264.png)
So, 
.
Therefore, the smallest possible value of 
is 
, when 
and 
.
Borrowed from [1]
See Also
| 1973 IMO (Problems) • Resources | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
| All IMO Problems and Solutions | ||
