2023 AMC 12A Problems/Problem 10

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 1

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

Solution 2

Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$ . Solving the case where $y-x=+2y$ , we'd find that $x=-y$ . This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$ . This means that $x=3y$ . Inputting this value into the first equation, we find: $y^3 = (3y)^2$ $y^3 = 9y^2$ $y=9$ This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$

~lprado

Solution 2: Quadractic formula

first expand

$(y-x)^2 = 4y^2$

$y^2-2xy+x^2 = 4y^2$

$y^2-2yx+x^2 = 4y^2$

$x^2-2xy-3y^2 = 0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

consider a=1 b=-2y c=-3y^2

$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$

$x=\frac{2y\pm\sqrt{16y^2}}{2a}$

$\frac{2y+4y^{2}}{2}$ or $\frac{2y+4y^{2}}{2}$

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

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2023 AMC 12A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Solution 2: Quadractic formula

See also