Difference between revisions of "2023 AMC 12A Problems/Problem 23"
R00tsofunity (talk | contribs) |
|||
| Line 15: | Line 15: | ||
<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>. | ||
| + | |||
| + | ==Video Solution by MOP 2024== | ||
| + | https://youtu.be/kkx7sm6-ZE8 | ||
| + | |||
| + | ~r00tsOfUnity | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
Revision as of 12:54, 11 November 2023
Contents
Problem
How many ordered pairs of positive real numbers 
satisfy the equation
![\[(1+2a)(2+2b)(2a+b) = 32ab?\]](http://latex.artofproblemsolving.com/8/4/a/84a5218ec979bd23c38a0c1d454f8fefab75d9b2.png)
Solution 1: AM-GM Inequality
Using AM-GM on the two terms in each factor on the left, we get
![\[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\]](http://latex.artofproblemsolving.com/f/2/e/f2e389fb9c2a257c18f4728e972399ebdea00faa.png)
meaning the equality condition must be satisfied. This means 
, so we only have 
solution.
Solution 2: Sum Of Squares
Equation 
is equivalent to
![\[b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,\]](http://latex.artofproblemsolving.com/d/e/4/de4f6744ca2710a7c03968ac3d6c157a825d2ffa.png)
where 
, 
. Therefore 
, so 
. Hence the answer is 
.
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution 1 by OmegaLearn
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. 
