Revision as of 19:10, 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

@@ Line 1: / Line 1: @@
-Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
+= Problem =
+Positive real numbers <math>x</math> and <math>y</math> satisfy <math>y^3=x^2</math> and <math>(y-x)^2=4y^2</math>. What is <math>x+y</math>?
+<math>\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math>
--Jonathan Yu
+= Solution =
+Because <math>y^3=x^2</math>, set <math>x=a^3</math>, <math>y=a^2</math> (<math>a\neq 0</math>). Put them in <math>(y-x)^2=4y^2</math> we get <math>(a^2(a-1))^2=4a^4</math> which implies <math>a^2-2a+1=4</math>. Solve the equation to get <math>a=3</math> or <math>-1</math>. Since <math>x</math> and <math>y</math> are positive, <math>a=3</math> and <math>x+y=3^3+3^2=\boxed{\textbf{(D)} 36}</math>.
+~Plasta

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

Revision as of 19:10, 9 November 2023

Problem

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