2024 AMC 10B Problems/Problem 9

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 $30 + 2ab + 2ac + 2bc = 0$ 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!]

~Mr.Lightning [Cleaned it up a bit]

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 $c = -\sqrt{15}$ , $0\times\sqrt{15} + 0\times-\sqrt{15} + \sqrt{15}\times-\sqrt{15}$ will result in -15, and dividing that by $3$ will give us $\boxed{\textbf{(A)}\ -5}$ .

-aleyang ~unpogged (minor error fix)

🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)

https://youtu.be/T_QESWAKUUk?si=yFqIs-XfQ878b9bg&t=520

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

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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