Difference between revisions of "2020 CIME I Problems/Problem 4"
(Created page with "==Problem 4== There exists a unique positive real number <math>x</math> satisfying <cmath>x=\sqrt{x^2+\frac{1}{x^2}} - \sqrt{x^2-\frac{1}{x^2}}.</cmath> Given that <math>x</ma...") |
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==Solution== | ==Solution== | ||
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| − | {{CIME box|year=2020|n=I|num-b= | + | We simply use the best technique of easy bash. |
| + | |||
| + | <math>x^2 = 2x^2 - 2\sqrt{x^4-\frac{1}{x^4}}</math> | ||
| + | |||
| + | <math>x^4 = 4x^4-\frac{4}{x^4}</math> | ||
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| + | <math>x^8 = \frac{4}{3}</math> | ||
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| + | <math>x = 2^{\frac{1}{4}}3^\frac{-1}{8}</math> | ||
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| + | The answer is then <math>14</math>. | ||
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| + | ==See also== | ||
| + | {{CIME box|year=2020|n=I|num-b=3|num-a=5}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAC Notice}} | {{MAC Notice}} | ||
Latest revision as of 19:43, 27 December 2021
Problem 4
There exists a unique positive real number 
satisfying ![\[x=\sqrt{x^2+\frac{1}{x^2}} - \sqrt{x^2-\frac{1}{x^2}}.\]](http://latex.artofproblemsolving.com/c/e/5/ce549c814c450bcb941089630021b4b0dd539a02.png)
Given that can be written in the form 
for integers 
with 
, find 
.
Solution
We simply use the best technique of easy bash.
The answer is then 
.
See also
| 2020 CIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
