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

Revision as of 12:22, 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 $\frac{ac + bc + ac}{3} = \boxed{\textbf{(A) }-5}$

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

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 ==Solution 2==
-Given: <math>\frac{a+b+c}{3}=0</math>
+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>
-<math>\Rightarrow a+b+c=0</math>
+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>
-Square both sides to get:<math>(a+b+c)^2=0</math>
+~laythe_enjoyer211, countmath1
-<math>a^2+b^2+c^2+2(ab+bc+ca)=0 \longrightarrow \raisebox{.5pt}{\textcircled{\raisebox{-.9pt}{1}}}</math>
-Also given:  <math>\frac{a^2+b^2+c^2}{3} = 10</math>
-<math>\Rightarrow a^2+b^2+c^2 = 30</math>
-Substituting into equation <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math>,
-<math>30+2(ab+bc+ca)=0</math>
-<math>2(ab+bc+ca)=-30</math>
-<math>ab+bc+ca=-15</math>
-There are 3 terms, so the mean is: <math>\frac{ab+bc+ca}{3}</math>
-<math>= \frac{-15}{3} = \boxed{\textbf{(A) }-5}</math>
-~laythe_enjoyer211
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

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