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

Latest revision as of 16:29, 15 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 $30 + 2ab + 2ac + 2bc = 0$ 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 $c = -\sqrt{15}$ , $0\times\sqrt{15} + 0\times-\sqrt{15} + \sqrt{15}-\sqrt{15}$ 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

@@ Line 1: / Line 1: @@
+==Problem==
+Real numbers <math>a, b, </math> and <math>c</math> have arithmetic mean 0. The arithmetic mean of <math>a^2, b^2, </math> and <math>c^2</math> is 10. What is the arithmetic mean of <math>ab, ac, </math> and <math>bc</math>?
+<math>\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}</math>
+==Solution 1==
+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 <cmath>(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc</cmath> 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 <cmath>30 + 2ab + 2ac + 2bc = 0</cmath> 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!]
+~Mr.Lightning [Cleaned it up a bit]
+==Solution 2==
+Since <math>\frac{a+b+c}{3},</math> we have <math>a+b+c=0,</math> and
+<cmath>(a+b+c)^2= a^2 + b^2+c^2+2(ab+ac+bc)=0</cmath>
+From the second given, <math>\frac{a^2+b^2+c^2}{3} = 10</math>, so <math>a^2+b^2+c^3=30.</math> Substituting this into the above equation,
+<cmath>2(ab+ac+bc) = (a+b+c)^2 -(a^2+b^2+c^2)=0-30 = -30. </cmath>
+Thus, <math>ab+ac+bc=-15,</math> and their arithmetic mean is <math>\frac{-15}{3} = \boxed{\textbf{(A)}\ -5}.</math>
+~laythe_enjoyer211, countmath1
+==Solution 3==
+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>0\times\sqrt{15} + 0\times-\sqrt{15} + \sqrt{15}-\sqrt{15}</math> is -15, and dividing by 3 gives us <math>\boxed{\textbf{(A)}\ -5}.</math>
+-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==
+{{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}}
+{{MAA Notice}}

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