Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
<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>, or <cmath>x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}</cmath>
+
<cmath>x^2 = \frac {(p - 4)^2}{4(4 - 2p)} or 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
Line 14: Line 14:
 
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>.
+
so we have <math>p\leq 0</math> or <math>p \geq \frac {4}{3}</math>
  
−
But if <math>p < 0</math>, then <math>\sqrt {x^2 - p} > x</math>, contradiction.
+
But if <math>p < 0</math>, then <math>\sqrt {x^2 - p} > x</math> contradiction.
  
 
So we have <math>x = \frac {4 - p}{2\sqrt {4 - 2p}}</math> for <math>p = 0, \frac {4}{3}\leq p < 2</math>.
 
So we have <math>x = \frac {4 - p}{2\sqrt {4 - 2p}}</math> for <math>p = 0, \frac {4}{3}\leq p < 2</math>.

Revision as of 14:42, 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 \[x^2 = \frac {(p - 4)^2}{4(4 - 2p)} or x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}\]

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 $p = 0, \frac {4}{3}\leq p < 2$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1963_IMO_Problems/Problem_1&oldid=112924"