Difference between revisions of "2023 AMC 12A Problems/Problem 23"
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<cmath>b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,</cmath> | <cmath>b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,</cmath> | ||
where <math>a</math>, <math>b>0</math>. Therefore <math>2a-1=b-1=2a-b=0</math>, so <math>(a,b)=\left(\tfrac12,1\right)</math>. Hence the answer is <math>\boxed{\textbf{(B) }1}</math>. | where <math>a</math>, <math>b>0</math>. Therefore <math>2a-1=b-1=2a-b=0</math>, so <math>(a,b)=\left(\tfrac12,1\right)</math>. Hence the answer is <math>\boxed{\textbf{(B) }1}</math>. | ||
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+ | ==Video Solution 1 by OmegaLearn== | ||
+ | https://youtu.be/LP4HSoaOCSU | ||
==Video Solution by MOP 2024== | ==Video Solution by MOP 2024== | ||
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~r00tsOfUnity | ~r00tsOfUnity | ||
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==Video Solution== | ==Video Solution== |
Revision as of 18:42, 13 November 2023
Contents
Problem
How many ordered pairs of positive real numbers satisfy the equation
Solution 1: AM-GM Inequality
Using AM-GM on the two terms in each factor on the left, we get
meaning the equality condition must be satisfied. This means
, so we only have
solution.
Solution 2: Sum Of Squares
Equation is equivalent to
where
,
. Therefore
, so
. Hence the answer is
.
Video Solution 1 by OmegaLearn
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.