2023 AMC 12A Problems/Problem 10: Difference between revisions

Revision as of 19:16, 9 November 2023

Problem

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ ? $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution

Because $y^3=x^2$ , set $x=a^3$ , $y=a^2$ ( $a\neq 0$ ). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$ . Solve the equation to get $a=3$ or $-1$ . Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$ .

~plasta

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ==See also==
-{{AMC12 box|year=2023|ab=A|num-a=10}}
+{{AMC12 box|year=2023|ab=A|num-b=9|num-a=11}}
 [[Category:Rate Problems]]
 {{MAA Notice}}

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2023 AMC 12A Problems/Problem 10: Difference between revisions

Revision as of 19:16, 9 November 2023

Problem

Solution

See also