Difference between revisions of "2024 AIME II Problems/Problem 4"
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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>|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}</math>. ~akliu | 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>|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}</math>. ~akliu | ||
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+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=3|num-a=5|n=II}} | ||
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+ | [[Category:]] | ||
+ | {{MAA Notice}} |
Revision as of 20:52, 8 February 2024
Problem
Let and be positive real numbers that satisfy the following system of equations: 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
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.