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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>
 +
 +
~ARay10 [Feel free to clean this up!]
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:23, 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 $\frac{ac + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

~ARay10 [Feel free to clean this up!]

Solution 2

Since $\frac{a+b+c}{3},$ we have $a+b+c=0,$ and \[(a+b+c)^2= a^2 + b^2+c^2+2(ab+ac+bc)=0\]

From the second given, $\frac{a^2+b^2+c^2}{3} = 10$, so $a^2+b^2+c^3=30.$ Substituting this into the above equation, \[2(ab+ac+bc) = (a+b+c)^2 -(a^2+b^2+c^2)=0-30 = -30.\] Thus, $ab+ac+bc=-15,$ and their arithmetic mean is $\frac{-15}{3} = \boxed{\textbf{(A)}\ -5}.$

~laythe_enjoyer211, countmath1

Solution 3

Assume that $a = 0$ and $b = -c$. Plugging into the second equation, $b^{2} + b^{2} = 30$, so $b^2 = 15$. Observe that taking the positive or negative root won't matter, as c will be the opposite. If we let $b = \sqrt{15}$ and $c = -\sqrt{15}$, $0\times\sqrt{15} + 0\times-\sqrt{15} + \sqrt{15}-\sqrt{15}$ is -15, and dividing by 3 gives us $\boxed{\textbf{(A)}\ -5}.$

-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

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