Difference between revisions of "2019 CIME I Problems/Problem 10"
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=Solution 1= | =Solution 1= | ||
| − | Let <math>n</math> be our answer. Notice <math>( | + | Let <math>n</math> be our answer. Notice <math>(y-z)^2 = x^2 + y^2 + z^2 - (x^2 + 2yz)= n-3</math>. Similarly, <math>(x-z)^2 = n-4</math> and <math>(y-x)^2 = n-5</math>. Now notice that since <math>y-z = (x-z)+(y-x)</math>, we have <math>\sqrt{n-3}=\sqrt{n-4}+\sqrt{n-5} \implies n^2-12n+36=4n^2-36n+80</math> so <math>3n^2-24n+44=0</math> and <math>n=4 \pm \frac{2}{\sqrt 3}</math>. The answer is <math>\boxed 9</math>. |
==See also== | ==See also== | ||
Latest revision as of 20:52, 19 February 2021
Let 
, 
, and 
be real numbers such that 
, 
, and 
. The value of 
can be written as 
for positive integers 
, where 
is not divisible by the square of any prime. Find 
.
Solution 1
Let 
be our answer. Notice 
. Similarly, 
and 
. Now notice that since 
, we have 
so 
and 
. The answer is 
.
See also
| 2019 CIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
