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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Given: <math>\frac{a+b+c}{3}=0</math> | ||
+ | |||
+ | <math>\Rightarrow a+b+c=0</math> | ||
+ | |||
+ | Square both sides to get:<math>(a+b+c)^2=0</math> | ||
+ | |||
+ | <math>a^2+b^2+c^2+2(ab+bc+ca)=0 \longrightarrow \raisebox{.5pt}{\textcircled{\raisebox{-.9pt}{1}}}</math> | ||
+ | |||
+ | Also given: <math>\frac{a^2+b^2+c^2}{3} = 10</math> | ||
+ | |||
+ | <math>\Rightarrow a^2+b^2+c^2 = 30</math> | ||
+ | |||
+ | Substituting into equation <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math>, | ||
+ | |||
+ | <math>30+2(ab+bc+ca)=0</math> | ||
+ | |||
+ | <math>2(ab+bc+ca)=-30</math> | ||
+ | |||
+ | <math>ab+bc+ca=-15</math> | ||
+ | |||
+ | There are 3 terms, so the mean is: <math>\frac{ab+bc+ca}{3}</math> | ||
+ | |||
+ | <math>= \frac{-15}{3} = \boxed{\textbf{(A) }-5}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:44, 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
Given:
Square both sides to get:
Also given:
Substituting into equation ,
There are 3 terms, so the mean is:
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 |
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