Difference between revisions of "1963 IMO Problems/Problem 1"

Revision as of 14:43, 15 December 2019

Problem

Find all real roots of the equation

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ ,

where $p$ is a real parameter.

Solution

Assuming $x \geq 0$ , square the equation, obtaining $4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2$ . If we have $p + 4 \geq 4x^2$ , we can square again, obtaining $x^2 = \frac {(p - 4)^2}{4(4 - 2p)} \implies x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}$

We must have $4 - 2p > 0 \iff p < 2$ , so we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$

However, this is only a solution when $p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)$

so we have $p\leq 0$ or $p \geq \frac {4}{3}$

But if $p < 0$ , then $\sqrt {x^2 - p} > x$ contradiction.

So we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$ for $p = 0, \frac {4}{3}\leq p < 2$ .

@@ Line 7: / Line 7: @@
 <cmath>4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2</cmath>.
 If we have <math>p + 4 \geq 4x^2</math>, we can square again, obtaining
-<cmath>x^2 = \frac {(p - 4)^2}{4(4 - 2p)} or x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}</cmath>
+<cmath>x^2 = \frac {(p - 4)^2}{4(4 - 2p)} \implies x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}</cmath>
 We must have <math>4 - 2p > 0 \iff p < 2</math>, so we have
 <cmath>x = \frac {4 - p}{2\sqrt {4 - 2p}}</cmath>
-However, this is only a solution when <cmath>p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)</cmath>,
+However, this is only a solution when <cmath>p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)</cmath>
 so we have <math>p\leq 0</math> or <math>p \geq \frac {4}{3}</math>

1963 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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Difference between revisions of "1963 IMO Problems/Problem 1"

Revision as of 14:43, 15 December 2019

Problem

Solution

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