2024 AMC 10B Problems/Problem 9: Difference between revisions

Latest revision as of 20:29, 11 November 2025

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 $ab + bc + ac = -15$ . To find the arithmetic mean, divide that by 3, so $\frac{ab + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

~ARay10 [Feel free to clean this up!]

~Mr.Lightning [Cleaned it up a bit]

~small edits by RoyalPawn38

Solution 2 (Simple Algebra)

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 $b = -c$ . Since we know that the arithmetic mean of the three numbers is $10$ , this means $a^2+b^2+c^2=30$ . Using this 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 $c = -\sqrt{15}$ , $ab = 0\times\sqrt{15} = 0$ , $ac = 0\times-\sqrt{15} = 0$ , and $bc = \sqrt{15}\times-\sqrt{15} = -15$ , so $ab+ac+bc=-15$ . Doing $\frac{-15}{3}$ to get the arithmetic mean will give us $\boxed{\textbf{(A)}\ -5}$ .

-aleyang

~unpogged (cleaned it up, fixed some errors)

Solution 4 (interesting property)

From the definition of arithmetic means and some simple manipulation, we have $a+b+c=0$ and $a^2+b^2+c^2=30$ . Let $ab+ac+bc=k$ , we have:

$k=ab+ac+bc=(a+b)c+ab$

Substituting $c=-(a+b)$ into the expression, we have

$-c^2+ab=k$

Adding twice the equation above to $a^2+b^2+c^2=30$ , we have:

$a^2+b^2+c^2-2c^2+2ab=30+2k$

$a^2+b^2+2ab-c^2=30+2k$

$(a+b)^2-c^2=30+2k$

$(a+b+c)(a+b-c)=30+2k$

But we know that $a+b+c=0$ . So $30+2k=0$ and $k=-15$ . Therefore the arithmetic mean of $ab,ac,bc$ is $\frac{k}{3}=-5$ . So the answer is $\boxed{\textbf{(A)}\ -5}$

~IDKHowtoaddsolution

Video Solution by Number Craft (🔥✅ Includes important concepts)

https://youtu.be/IF3AINUVkVE

🎥✨ 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

Video Solution 3 by TheBeautyofMath

https://youtu.be/ZaHv4UkXcbs?t=880

~IceMatrix

@@ Line 1: / Line 1: @@
+==Problem==
+Real numbers <imath>a, b, </imath> and <imath>c</imath> have arithmetic mean <imath>0</imath>. The arithmetic mean of <imath>a^2, b^2, </imath> and <imath>c^2</imath> is <imath>10</imath>. What is the arithmetic mean of <imath>ab, ac, </imath> and <imath>bc</imath>?
+<imath>\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}</imath>
 ==Solution 1==

2024 AMC 10B (Problems • Answer Key • Resources)
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