2019 CIME I Problems/Problem 10
Revision as of 15:21, 6 October 2020 by Icematrix2 (talk | contribs) (Created page with "Let <math>x</math>, <math>y</math>, and <math>z</math> be real numbers such that <math>x^2=3-2yz</math>, <math>y^2=4-2xz</math>, and <math>z^2=5-2xy</math>. The value of <math...")
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 $(y�-z)^2 = x^2 + y^2 + z^2 - �(x2 + 2yz)= n-�3$ (Error compiling LaTeX. Unknown error_msg). Similarly, $(x-�z)^2 = n-4$ (Error compiling LaTeX. Unknown error_msg) and . Now notice that since $y-�z = (x-z)+(y-x)$ (Error compiling LaTeX. Unknown error_msg), we have n^2-12n+36=4n^2-36n+803n^2-24n+44=0n=4 \pm \frac{2}{\sqrt 3}\boxed 9$.
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.