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

Difference between revisions of "2024 AIME II Problems/Problem 4"
(→‎Problem)
Line 3: Line 3:
 
Let <math>x,y</math> and <math>z</math> be positive real numbers that satisfy the following system of equations:
 
Let <math>x,y</math> and <math>z</math> be positive real numbers that satisfy the following system of equations:
 
<cmath>\log_2\left({x \over yz}\right) = {1 \over 2}</cmath><cmath>\log_2\left({y \over xz}\right) = {1 \over 3}</cmath><cmath>\log_2\left({z \over xy}\right) = {1 \over 4}</cmath>
 
<cmath>\log_2\left({x \over yz}\right) = {1 \over 2}</cmath><cmath>\log_2\left({y \over xz}\right) = {1 \over 3}</cmath><cmath>\log_2\left({z \over xy}\right) = {1 \over 4}</cmath>
−
Then the value of <math>\left|\log_2(x^4y^3z^2)\right|</math> is <math>{m \over n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
Then the value of <math>\left|\log_2(x^4y^3z^2)\right|</math> is <math>\tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution 1==
 
==Solution 1==

Revision as of 13:54, 10 February 2024

Problem

Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: \[\log_2\left({x \over yz}\right) = {1 \over 2}\]\[\log_2\left({y \over xz}\right) = {1 \over 3}\]\[\log_2\left({z \over xy}\right) = {1 \over 4}\] Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Denote $\log_2(x) = a$, $\log_2(y) = b$, and $\log_2(z) = c$.

Then, we have: $a-b-c = \frac{1}{2}$ $-a+b-c = \frac{1}{3}$ $-a-b+c = \frac{1}{4}$

Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$. Plugging these values in, we obtain $|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}$. ~akliu

Solution 2

$\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}$

$\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}$

$\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}$

$\log_2(x) = -\frac{7}{24}$

$\log_2(y) = -\frac{3}{8}$

$\log_2(z) = -\frac{5}{12}$

$4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8$

$25 + 8 = \boxed{033}$

~Callisto531

Video Solution

https://youtu.be/SUie2Jlo-pg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME II (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 AIME Problems and Solutions

[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_II_Problems/Problem_4&oldid=215187"