Difference between revisions of "2024 AMC 10B Problems/Problem 9"
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==Solution 1== | ==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 < | + | 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== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:29, 15 November 2024
Contents
[hide]Problem
Real numbers and have arithmetic mean 0. The arithmetic mean of and is 10. What is the arithmetic mean of and ?
Solution 1
If , that means , and . Expanding that gives If , then . Thus, we have Arithmetic will give you that . To find the arithmetic mean, divide that by 3, so
~ARay10 [Feel free to clean this up!]
~Mr.Lightning [Cleaned it up a bit]
Solution 2
Since we have and
From the second given, , so Substituting this into the above equation, Thus, and their arithmetic mean is
~laythe_enjoyer211, countmath1
Solution 3
Assume that and . Plugging into the second equation, , so . Observe that taking the positive or negative root won't matter, as c will be the opposite. If we let and , is -15, and dividing by 3 gives us
-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
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 |
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