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Difference between revisions of "2024 AMC 10B Problems/Problem 9"
(More elegant solution :))
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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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 +
==Solution 2==
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 +
Given: <math>\frac{a+b+c}{3}=0</math>
 +
 +
<math>\Rightarrow a+b+c=0</math>
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Square both sides to get:<math>(a+b+c)^2=0</math>
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<math>a^2+b^2+c^2+2(ab+bc+ca)=0 \longrightarrow \raisebox{.5pt}{\textcircled{\raisebox{-.9pt}{1}}}</math>
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Also given:  <math>\frac{a^2+b^2+c^2}{3} = 10</math>
 +
 +
<math>\Rightarrow a^2+b^2+c^2 = 30</math>
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Substituting into equation <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math>,
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<math>30+2(ab+bc+ca)=0</math>
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<math>2(ab+bc+ca)=-30</math>
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<math>ab+bc+ca=-15</math>
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There are 3 terms, so the mean is: <math>\frac{ab+bc+ca}{3}</math>
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<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

Problem

Real numbers $a, b,$ and $c$ have arithmetic mean 0. The arithmetic mean of $a^2, b^2,$ and $c^2$ is 10. What is the arithmetic mean of $ab, ac,$ and $bc$?

$\textbf{(A) } -5 \qquad\textbf{(B) } -\dfrac{10}{3} \qquad\textbf{(C) } -\dfrac{10}{9} \qquad\textbf{(D) } 0 \qquad\textbf{(E) } \dfrac{10}{9}$

Solution 1

If $\frac{a+b+c}{3} = 0$, that means $a+b+c=0$, and $(a+b+c)^2=0$. Expanding that gives $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$. If $\frac{a^2+b^2+c^2}{3} = 10$, then $a^2+b^2+c^2=30$. Thus, we have $0 = 30 + 2ab + 2ac + 2bc$. Arithmetic will give you that $ac + bc + ac = -15$. To find the arithmetic mean, divide that by 3, so $\frac{ac + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

Solution 2

Given: $\frac{a+b+c}{3}=0$

$\Rightarrow a+b+c=0$

Square both sides to get:$(a+b+c)^2=0$

$a^2+b^2+c^2+2(ab+bc+ca)=0 \longrightarrow \raisebox{.5pt}{\textcircled{\raisebox{-.9pt}{1}}}$

Also given: $\frac{a^2+b^2+c^2}{3} = 10$

$\Rightarrow a^2+b^2+c^2 = 30$

Substituting into equation $\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}$,

$30+2(ab+bc+ca)=0$

$2(ab+bc+ca)=-30$

$ab+bc+ca=-15$

There are 3 terms, so the mean is: $\frac{ab+bc+ca}{3}$

$= \frac{-15}{3} = \boxed{\textbf{(A) }-5}$

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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