Difference between revisions of "1963 IMO Problems/Problem 1"
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Find all real roots of the equation | Find all real roots of the equation | ||
<center><math>\sqrt{x^2-p}+2\sqrt{x^2-1}=x</math>,</center>where <math>p</math> is a real parameter. | <center><math>\sqrt{x^2-p}+2\sqrt{x^2-1}=x</math>,</center>where <math>p</math> is a real parameter. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/N499P8lsb7U?si=BH8letxqFGEVpq34 [Video Solution by little-fermat] | ||
+ | |||
==Solution== | ==Solution== | ||
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<cmath>4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2</cmath>. | <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 | 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)} | + | <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 | We must have <math>4 - 2p > 0 \iff p < 2</math>, so we have | ||
<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> |
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− | + | 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> | |
==See Also== | ==See Also== | ||
{{IMO box|year=1963|before=First Question|num-a=2}} | {{IMO box|year=1963|before=First Question|num-a=2}} |
Latest revision as of 00:49, 15 September 2023
Contents
Problem
Find all real roots of the equation
![$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$](http://latex.artofproblemsolving.com/4/c/5/4c50288029a0bd6789e2dcec7a9350ad9a6c605f.png)
where is a real parameter.
Video Solution
https://youtu.be/N499P8lsb7U?si=BH8letxqFGEVpq34 [Video Solution by little-fermat]
Solution
Assuming , square the equation, obtaining
.
If we have
, we can square again, obtaining
We must have , so we have
However, this is only a solution when
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 |