Difference between revisions of "2023 AMC 12A Problems/Problem 23"

Revision as of 09:26, 10 November 2023

Problem

How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation $(1+2a)(2+2b)(2a+b) = 32ab?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$

Solution 1: AM-GM Inequality

Using AM-GM on the two terms in each factor on the left, we get $(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,$ meaning the equality condition must be satisfied. This means $1 = 2a = b$ , so we only have $\boxed{1}$ solution.

Solution 2: Sum Of Squares

Equation $(1+2a)(2+2b)(2a+b)=32ab$ is equivalent to $b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,$ where $a$ , $b>0$ . Therefore $2a-1=b-1=2a-b=0$ , so $(a,b)=\left(\tfrac12,1\right)$ . Hence the answer is $\boxed{\textbf{(B) }1}$ .

Video Solution 1 by OmegaLearn

https://youtu.be/LP4HSoaOCSU

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.

@@ Line 7: / Line 7: @@
 ==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.
+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==

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