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

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==Problem==
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How many ordered pairs of positive real numbers <math>(a,b)</math> satisfy the equation
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<cmath>(1+2a)(2+2b)(2a+b) = 32ab?</cmath>
  
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}</math>
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==Solution 1: AM-GM Inequality==
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Using the AM-GM inequality on the two terms in each factor on the left-hand side, we get
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<cmath>(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,</cmath>
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This means the equality condition must be satisfied. Therefore, we must have <math>1 = 2a = b</math>, so the only solution is <math>\boxed{\textbf{(B) }1}</math>.
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~ semistevehan
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==Solution 2: Sum Of Squares==
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Equation <math>(1+2a)(2+2b)(2a+b)=32ab</math> is equivalent to
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<cmath>b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,</cmath>
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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>.
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==Solution 3: ==
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<math>(1+2a)(1+b)(2a+b)=16ab</math>,
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let <math>x=2a, y=b</math>, then it becomes <math>(1+x)(1+y)(x+y)=8xy</math>, or <math>(1+x+y+xy)(x+y)=8xy</math>.
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Let <math>\alpha=x+y, \beta=xy</math>, it becomes <math>(1+\alpha+\beta)\alpha=8\beta</math>,
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notice we have <math>\alpha^2-4\beta=(x-y)^2\ge 0</math>, now <math>\beta= \frac{\alpha(1+\alpha)}{(8-\alpha)}</math>
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<math>\alpha^2\ge 4\beta=4\alpha(1+\alpha)/(8-\alpha)</math> (notice we must have <math>8-\alpha>0</math>), <math>(8-\alpha)\alpha\ge 4(1+\alpha)</math>, <math>(\alpha-2)^2\le 0</math>, <math>\alpha=2</math>,
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and <math>\beta=1</math>,
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so <math>x=y=1</math> and <math>a=\frac12, b=1</math> is the only solution.
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answer is <math>\boxed{\textbf {(B)}} </math>
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~szhangmath
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==Video Solution==
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https://youtu.be/bRQ7xBm1hFc
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~MathKatana
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==Video Solution 1 by OmegaLearn==
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https://youtu.be/LP4HSoaOCSU
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==Video Solution by MOP 2024==
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https://youtu.be/kkx7sm6-ZE8
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 +
~r00tsOfUnity
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==Video Solution==
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 +
https://youtu.be/ZKdnv8MsEDI
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See also==
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{{AMC12 box|ab=A|year=2023|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 13:49, 2 November 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 the AM-GM inequality on the two terms in each factor on the left-hand side, we get \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\] This means the equality condition must be satisfied. Therefore, we must have $1 = 2a = b$, so the only solution is $\boxed{\textbf{(B) }1}$.

~ semistevehan

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 $\beta= \frac{\alpha(1+\alpha)}{(8-\alpha)}$

$\alpha^2\ge 4\beta=4\alpha(1+\alpha)/(8-\alpha)$ (notice we must have $8-\alpha>0$), $(8-\alpha)\alpha\ge 4(1+\alpha)$, $(\alpha-2)^2\le 0$, $\alpha=2$, and $\beta=1$,

so $x=y=1$ and $a=\frac12, b=1$ is the only solution.

answer is $\boxed{\textbf {(B)}}$

~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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