Difference between revisions of "1963 IMO Problems/Problem 1"
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<cmath>x = \frac {4 - p}{2\sqrt {4 - 2p}}</cmath> | <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)\ | + | However, this is only a solution when <cmath>p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\geq(p - 4)^2 \iff 0\geq p(3p - 4)</cmath> |
| − | so we have <math>p\ | + | so we have <math>p\geq 0</math> and <math>p \leq \frac {4}{3}</math> |
| − | + | and <math>x = \frac {4 - p}{2\sqrt {4 - 2p}}</math> | |
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==See Also== | ==See Also== | ||
{{IMO box|year=1963|before=First Question|num-a=2}} | {{IMO box|year=1963|before=First Question|num-a=2}} | ||
Revision as of 02:36, 26 April 2021
Problem
Find all real roots of the equation
,where 
is a real parameter.
Solution
Assuming 
, square the equation, obtaining
![\[4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2\]](http://latex.artofproblemsolving.com/f/6/a/f6a7f5002129ee9be9a3b492e9e870df24e951c4.png)
.
If we have 
, we can square again, obtaining
![\[x^2 = \frac {(p - 4)^2}{4(4 - 2p)} \implies x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}\]](http://latex.artofproblemsolving.com/b/a/1/ba1fb6f92b506b248195bfa0a86fb4f4c2e7632c.png)
We must have 
, so we have
![\[x = \frac {4 - p}{2\sqrt {4 - 2p}}\]](http://latex.artofproblemsolving.com/f/8/e/f8e838d14aaa9deec26dfc3d84ff2649dad8aed6.png)
However, this is only a solution when ![\[p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\geq(p - 4)^2 \iff 0\geq p(3p - 4)\]](http://latex.artofproblemsolving.com/b/7/9/b79eac0cfaec6c666aecf8dfe934cb6216259521.png)
so we have 
and 
and 
See Also
| 1963 IMO (Problems) • Resources | ||
| Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
