Difference between revisions of "2023 AMC 12A Problems/Problem 23"
(→Solution) |
(→Solution 1: GM and AM inequality) |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}</math> | <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: GM | + | ==Solution 1: AM-GM Inequality== |
| + | |||
| + | Using AM-GM on the two terms in each factor on the left, we get <math>(1+2a)(2+2b)(2a+b) \ge 8\sqrt(2a \cdot 4b \cdot 2ab) = 32ab</math>, meaning the equality condition must be satisfied. This means <math>1 = 2a = b</math>, so we only have <math>\boxed{1}</math> solution. | ||
==See also== | ==See also== | ||
{{AMC12 box|ab=A|year=2023|num-b=22|num-a=24}} | {{AMC12 box|ab=A|year=2023|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:18, 10 November 2023
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 
, meaning the equality condition must be satisfied. This means 
, so we only have 
solution.
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. 
