Difference between revisions of "2024 AMC 10B Problems/Problem 9"
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If <math>\frac{a+b+c}{3} = 0</math>, that means <math>a+b+c=0</math>, and <math>(a+b+c)^2=0</math>. Expanding that gives <math>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc</math>. If <math>\frac{a^2+b^2+c^2}{3} = 10</math>, then <math>a^2+b^2+c^2=30</math>. Thus, we have <math>0 = 30 + 2ab + 2ac + 2bc</math>. Arithmetic will give you that <math>ac + bc + ac = -15</math>. To find the arithmetic mean, divide that by 3, so <math>\frac{ac + bc + ac}{3} = \boxed{\textbf{(A) }-5}</math> | If <math>\frac{a+b+c}{3} = 0</math>, that means <math>a+b+c=0</math>, and <math>(a+b+c)^2=0</math>. Expanding that gives <math>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc</math>. If <math>\frac{a^2+b^2+c^2}{3} = 10</math>, then <math>a^2+b^2+c^2=30</math>. Thus, we have <math>0 = 30 + 2ab + 2ac + 2bc</math>. Arithmetic will give you that <math>ac + bc + ac = -15</math>. To find the arithmetic mean, divide that by 3, so <math>\frac{ac + bc + ac}{3} = \boxed{\textbf{(A) }-5}</math> | ||
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+ | ~ARay10 [Feel free to clean this up!] | ||
==Solution 2== | ==Solution 2== |
Revision as of 20:23, 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
~ARay10 [Feel free to clean this up!]
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
Solution 3
Assume that and . Plugging into the second equation, , so . Observe that taking the positive or negative root won't matter, as c will be the opposite. If we let and , is -15, and dividing by 3 gives us
-aleyang
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
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.