Difference between revisions of "2023 AMC 12A Problems/Problem 23"
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| + | ==Problem== | ||
| + | How many ordered pairs of positive real numbers <math>(a,b)</math> satisfy the equation | ||
| + | <cmath>(1+2a)(2+2b)(2a+b) = 32ab?</cmath> | ||
| + | <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}</math> | ||
| + | |||
| + | ==Solution 1: AM-GM Inequality== | ||
| + | |||
| + | Using AM-GM on the two terms in each factor on the left, we get | ||
| + | <cmath>(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,</cmath> | ||
| + | meaning the equality condition must be satisfied. This means <math>1 = 2a = b</math>, so we only have <math>\boxed{1}</math> solution. | ||
| + | |||
| + | ==Solution 2: Sum Of Squares== | ||
| + | Equation <math>(1+2a)(2+2b)(2a+b)=32ab</math> is equivalent to | ||
| + | <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>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/bRQ7xBm1hFc | ||
| + | ~MathKatana | ||
| + | ==Video Solution 1 by OmegaLearn== | ||
| + | https://youtu.be/LP4HSoaOCSU | ||
| + | |||
| + | ==Video Solution by MOP 2024== | ||
| + | https://youtu.be/kkx7sm6-ZE8 | ||
| + | |||
| + | ~r00tsOfUnity | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/ZKdnv8MsEDI | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|ab=A|year=2023|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:20, 7 July 2024
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
https://youtu.be/bRQ7xBm1hFc ~MathKatana
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. 
