Difference between revisions of "2024 AIME II Problems/Problem 4"
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| − | + | ==Problem== | |
| + | |||
| + | 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> | ||
| + | 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>. | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | Denote <math>\log_2(x) = a</math>, <math>\log_2(y) = b</math>, and <math>\log_2(z) = c</math>. | ||
| + | |||
| + | Then, we have: | ||
| + | <math>a-b-c = \frac{1}{2}</math> | ||
| + | <math>-a+b-c = \frac{1}{3}</math> | ||
| + | <math>-a-b+c = \frac{1}{4}</math> | ||
| + | |||
| + | Now, we can solve to get <math>a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}</math>. Plugging these values in, we obtain <math>\mid 4a + 3b + 2c \mid = \frac{25}{8} \implies \boxed{033}</math>. ~akliu | ||
Revision as of 19:03, 8 February 2024
Problem
Let 
and 
be positive real numbers that satisfy the following system of equations:
![\[\log_2\left({x \over yz}\right) = {1 \over 2}\]](http://latex.artofproblemsolving.com/e/6/5/e65012d6b5c3081c7d2f9ec748f09b2b5cada208.png)
![\[\log_2\left({y \over xz}\right) = {1 \over 3}\]](http://latex.artofproblemsolving.com/a/8/e/a8e021d3f2a2db4d7a423d5ed27f93c58b19ea1b.png)
![\[\log_2\left({z \over xy}\right) = {1 \over 4}\]](http://latex.artofproblemsolving.com/3/a/8/3a806f801e846b4f7311aee3f5b30971dad5310b.png)
Then the value of 
is 
where 
and 
are relatively prime positive integers. Find 
.
Solution 1
Denote 
, 
, and 
.
Then, we have:
Now, we can solve to get 
. Plugging these values in, we obtain 
. ~akliu
