Difference between revisions of "1973 IMO Problems/Problem 3"

Revision as of 23:50, 27 November 2017

Let $a$ and $b$ be real numbers for which the equation $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution. For all such pairs $(a, b)$ , find the minimum value of $a^2 + b^2$ .

Solution

Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$ . This equation has solutions $z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}$ . We also know that $|z|=|x+1/x| \geq 2$ . So,

$\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2$

$\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2$

$|a|+\sqrt{a^2+8-4b} \geq 4$

Rearranging and squaring both sides,

$a^2+8-4b \geq a^2-16|a|+16$

$2|a|-b \geq 2$

So, $a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}$ .

Therefore, the smallest possible value of $a^2+b^2$ is $\frac{4}{5}$ , when $a=\pm \frac{4}{5}$ and $b=\frac{-2}{5}$ .

Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html

@@ Line 6: / Line 6: @@
 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,
-<math>|\frac{-a \pm \sqrt{a^2+8-4b}}{2}| \geq 2</math>
+<math>\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2</math>
 <math>\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2</math>
 <math>|a|+\sqrt{a^2+8-4b} \geq 4</math>
@@ Line 13: / Line 15: @@
 <math>a^2+8-4b \geq a^2-16|a|+16</math>
 <math>2|a|-b \geq 2</math>

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Difference between revisions of "1973 IMO Problems/Problem 3"

Revision as of 23:50, 27 November 2017

Solution