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

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(→‎Solution 3:)
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Let <math>\alpha=x+y, \beta=xy</math>, it becomes <math>(1+\alpha+\beta)\alpha=8\beta</math>,
 
Let <math>\alpha=x+y, \beta=xy</math>, it becomes <math>(1+\alpha+\beta)\alpha=8\beta</math>,
  
−
notice we have <math>\alpha^2-4\beta\ge \beta</math>, now <math>\beta= \alpha(1+\alpha)/(8-\alpha)</math>
+
notice we have <math>\alpha^2-4\beta=(x-y)^2\ge 0</math>, now <math>\beta= \frac{\alpha(1+\alpha)}{(8-\alpha)}</math>
  
−
<math>\alpha^2\ge \beta=\alpha(1+\alpha)/(8-\alpha)</math>, <math>(\alpha-2)^2\le 0</math>, <math>\alpha=2</math>.
+
<math>\alpha^2\ge \beta=\alpha(1+\alpha)/(8-\alpha)</math>, <math>(\alpha-2)^2\le 0</math>, <math>\alpha=2</math>,
 +
and <math>\beta=1</math>, so <math>x=y=1</math> and <math>a=\frac12, b=1</math> is the only solution.
 +
 
 +
answer is <math>\box (C) </math>
 +
 
 +
~szhangmath
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:01, 12 September 2024

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 $\boxed{\textbf{(B) }1}$.


Solution 3:

$(1+2a)(1+b)(2a+b)=16ab$,

let $x=2a, y=b$, then it becomes $(1+x)(1+y)(x+y)=8xy$, or $(1+x+y+xy)(x+y)=8xy$.

Let $\alpha=x+y, \beta=xy$, it becomes $(1+\alpha+\beta)\alpha=8\beta$,

notice we have $\alpha^2-4\beta=(x-y)^2\ge 0$, now $\beta= \frac{\alpha(1+\alpha)}{(8-\alpha)}$

$\alpha^2\ge \beta=\alpha(1+\alpha)/(8-\alpha)$, $(\alpha-2)^2\le 0$, $\alpha=2$, and $\beta=1$, so $x=y=1$ and $a=\frac12, b=1$ is the only solution.

answer is $\box (C)$ (Error compiling LaTeX. Unknown error_msg)

~szhangmath

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

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

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