Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Northeastern WOOTers Mock AIME I Problems/Problem 5

Revision as of 21:45, 7 August 2021 by Skyguy88 (talk | contribs) (Created page with "== Problem 5 == Let <math>x</math>, <math>y</math>, and <math>z</math> be real numbers. Given that <math>x^2+2y^2+3z^2=1</math>, the maximum value of <math>\left( 3x+2y+z \ri...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 5

Let $x$, $y$, and $z$ be real numbers. Given that $x^2+2y^2+3z^2=1$, the maximum value of $\left( 3x+2y+z \right)^2$ can be represented $\frac{m}{n}$, where $m$ and $n$ are positive integers, where $m$ and $n$ are relatively prime. Find $m+n$.



Solution

By the Cauchy-Schwarz Inequality, \[(x^2 + 2y^2 + 3z^2)(9 + 2 + \frac{1}{3}) \geq \left( 3x + 2y + z \right)^2.\] It follows that $\left( 3x + 2y + z \right)^2 \leq \frac{34}{3}$. Since this is attainable when $\frac{x^2}{9} = \frac{2y^2}{2} = \frac{3z^2}{1/3} \Rightarrow x = \frac{9}{\sqrt{102}}, y = \frac{3}{\sqrt{102}}, z = \frac{1}{\sqrt{102}}$, the answer is $34 + 3 = \boxed{037}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Northeastern_WOOTers_Mock_AIME_I_Problems/Problem_5&oldid=159737"