Difference between revisions of "2020 CIME I Problems/Problem 4"
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<math>x = 2^{\frac{1}{4}}3^\frac{-1}{8}</math> | <math>x = 2^{\frac{1}{4}}3^\frac{-1}{8}</math> | ||
| − | The answer is then 14. | + | 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}} | {{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. 
