Difference between revisions of "2024 AMC 10B Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
− | + | Since <math>\frac{a+b+c}{3},</math> we have <math>a+b+c=0,</math> and | |
+ | <cmath>(a+b+c)^2= a^2 + b^2+c^2+2(ab+ac+bc)=0</cmath> | ||
− | <math>\ | + | From the second given, <math>\frac{a^2+b^2+c^2}{3} = 10</math>, so <math>a^2+b^2+c^3=30.</math> Substituting this into the above equation, |
+ | <cmath>2(ab+ac+bc) = (a+b+c)^2 -(a^2+b^2+c^2)=0-30 = -30. </cmath> | ||
+ | Thus, <math>ab+ac+bc=-15,</math> and their arithmetic mean is <math>\frac{-15}{3} = \boxed{\textbf{(A)}\ -5}.</math> | ||
− | + | ~laythe_enjoyer211, countmath1 | |
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− | ~laythe_enjoyer211 | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 12:22, 14 November 2024
Contents
[hide]Problem
Real numbers and have arithmetic mean 0. The arithmetic mean of and is 10. What is the arithmetic mean of and ?
Solution 1
If , that means , and . Expanding that gives . If , then . Thus, we have . Arithmetic will give you that . To find the arithmetic mean, divide that by 3, so
Solution 2
Since we have and
From the second given, , so Substituting this into the above equation, Thus, and their arithmetic mean is
~laythe_enjoyer211, countmath1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.