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Difference between revisions of "2020 CIME I Problems/Problem 4"
(→‎Solution)
 
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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>.
  
 
==See also==
 
==See also==

Latest revision as of 19:43, 27 December 2021

Problem 4

There exists a unique positive real number $x$ satisfying \[x=\sqrt{x^2+\frac{1}{x^2}} - \sqrt{x^2-\frac{1}{x^2}}.\] Given that $x$ can be written in the form $x=2^\frac{m}{n} \cdot 3^\frac{-p}{q}$ for integers $m, n, p, q$ with $\gcd(m, n) = \gcd(p, q) = 1$, find $m+n+p+q$.

Solution

We simply use the best technique of easy bash.

$x^2 = 2x^2 - 2\sqrt{x^4-\frac{1}{x^4}}$

$x^4 = 4x^4-\frac{4}{x^4}$

$x^8 = \frac{4}{3}$

$x = 2^{\frac{1}{4}}3^\frac{-1}{8}$

The answer is then $14$.

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

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