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Difference between revisions of "2024 AMC 10B Problems/Problem 9"

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==Solution 3==
 
==Solution 3==
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Assume that <math>a = 0</math> and <math>b = -c</math>. Plugging into the second equation, <math>b^{2} + b^{2} = 30</math>, so <math>b^2 = 15</math>. Observe that taking the positive or negative root won't matter as c will be the opposite. If we let <math>b = \sqrt{15}</math> and <math>c = -\sqrt{15}</math>, <math>ab = 0\times\sqrt{15} = 0</math>, <math>ac = 0\times-\sqrt{15} = 0</math>, and <math>bc = \sqrt{15}\times-\sqrt{15} = -15</math>, so <math>ab+ac+bc=-15</math>. Doing <math>\frac{-15}{3}</math> to get the arithmetic mean will give us <math>\boxed{\textbf{(A)}\ -5}</math>.
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Assume that <math>a = 0</math> and <math>b = -c</math>. Since we know that the arithmetic mean of the three numbers is <math>10</math>, this means <math>a^2+b^2+c^2=30</math>. Using this equation, <math>b^{2} + b^{2} = 30</math>, so <math>b^2 = 15</math>. Observe that taking the positive or negative root won't matter as <math>c</math> will be the opposite. If we let <math>b = \sqrt{15}</math> and <math>c = -\sqrt{15}</math>, <math>ab = 0\times\sqrt{15} = 0</math>, <math>ac = 0\times-\sqrt{15} = 0</math>, and <math>bc = \sqrt{15}\times-\sqrt{15} = -15</math>, so <math>ab+ac+bc=-15</math>. Doing <math>\frac{-15}{3}</math> to get the arithmetic mean will give us <math>\boxed{\textbf{(A)}\ -5}</math>.
  
 
-aleyang
 
-aleyang
  
 
~unpogged (cleaned it up, fixed some errors)
 
~unpogged (cleaned it up, fixed some errors)
 +
 +
== Video Solution by Number Craft (🔥✅ Includes important concepts)==
 +
 +
https://youtu.be/IF3AINUVkVE
  
 
==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)==
 
==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)==

Latest revision as of 23:08, 19 January 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 $ac + bc + ac = -15$. 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 $b = -c$. 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)

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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