Difference between revisions of "2023 AMC 12A Problems/Problem 10"

Latest revision as of 19:35, 10 June 2024

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 3: Quadratic 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 $\frac{2y-4y^{2}}{2}$

$x=y+2y$ and $x=y-2y$

$x=3y$ and $x=-y$

we can see both x and y will be postive in $x=3y$

now do same as solution 2 $y^3 = (3y)^2$ $y^3 = 9y^2$ $y=9$ This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$

Solution 4: Substitution

Since $a^2 = |a|^2$ , we can rewrite the second equation as $(x-y)^2=4y^2$

Let $u=x+y$ . The second equation becomes

$(u-2y)^2 = 4y^2$ $u^2 - 4uy = 0$ $u = 4y$ $x+y = 4y$ $x = 3y.$

Substituting this into the first equation, we have

$y^3 = (3y)^2,$ so $x = 9$ .

Hence $x = 27$ and $x + y = \boxed{\textbf{(D)} 36}.$

-Benedict T (countmath1)

Solution 5: Difference of Squares

We will use the difference of squares in the second equation.

$(y-x)^2=4y^2$ $(y-x)^2-(2y)^2=0$ $(y-x-2y)(y-x+2y)=0$ $-(x+y)(3y-x)=0$

Since x and y are positive, x+y is non-zero. Thus, $3y=x$ .

Substituting into the first equation:

$y^3=x^2$ $y^3=9y^2$ $y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}$

Video Solution (easy to digest) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279

Video Solution (⚡ Under 3 minutes ⚡)

https://youtu.be/G63hRr9bkzI

~Education, the Study of Everything

Video Solution

https://youtu.be/OqlerYo-uPU

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 12: / Line 12: @@
 <cmath>y^3 = (3y)^2</cmath>
 <cmath>y^3 = 9y^2</cmath>
-<cmath>y=9</cmath>.
+<cmath>y=9</cmath>
 This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math>
 ~lprado
+== Solution 3: Quadratic formula==
+first expand
+<math>(y-x)^2 = 4y^2</math>
+<math>y^2-2xy+x^2 = 4y^2</math>
+<math>y^2-2yx+x^2 = 4y^2</math>
+<math>x^2-2xy-3y^2 = 0</math>
+<math>x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>
+consider
+a=1
+b=-2y
+c=-3y^2
+<math>x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}</math>
+<math>x=\frac{2y\pm\sqrt{16y^2}}{2a}</math>
+<math>\frac{2y+4y^{2}}{2}</math> or <math>\frac{2y-4y^{2}}{2}</math>
+<math>x=y+2y</math> and <math>x=y-2y</math>
+<math>x=3y</math> and <math>x=-y</math>
+we can see both x and y will be postive in <math>x=3y</math>
+now do same as solution 2
+<cmath>y^3 = (3y)^2</cmath>
+<cmath>y^3 = 9y^2</cmath>
+<cmath>y=9</cmath>
+This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math>
+== Solution 4: Substitution==
+Since <math>a^2 = |a|^2</math>, we can rewrite the second equation as <math>(x-y)^2=4y^2</math>
+Let <math>u=x+y</math>. The second equation becomes
+<cmath>(u-2y)^2 = 4y^2</cmath>
+<cmath>u^2 - 4uy = 0</cmath>
+<cmath>u = 4y</cmath>
+<cmath>x+y = 4y</cmath>
+<cmath>x = 3y.</cmath>
+Substituting this into the first equation, we have
+<cmath>y^3 = (3y)^2,</cmath>
+so <math>x = 9</math>.
+Hence <math>x = 27</math> and <math>x + y = \boxed{\textbf{(D)} 36}.</math>
+-Benedict T (countmath1)
+== Solution 5: Difference of Squares==
+We will use the difference of squares in the second equation.
+<cmath>(y-x)^2=4y^2</cmath>
+<cmath>(y-x)^2-(2y)^2=0</cmath>
+<cmath>(y-x-2y)(y-x+2y)=0</cmath>
+<cmath>-(x+y)(3y-x)=0</cmath>
+Since x and y are positive, x+y is non-zero. Thus, <cmath>3y=x</cmath>.
+Substituting into the first equation:
+<cmath>y^3=x^2</cmath>
+<cmath>y^3=9y^2</cmath>
+<cmath>y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}</cmath>
+==Video Solution (easy to digest) by Power Solve==
+https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279
+==Video Solution (⚡ Under 3 minutes ⚡)==
+https://youtu.be/G63hRr9bkzI
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/OqlerYo-uPU
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See also==

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

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