Difference between revisions of "2023 AMC 12A Problems/Problem 10"
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| − | = Problem = | + | == 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>? | 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> | <math>\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math> | ||
| − | = Solution = | + | == 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>. | 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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| + | ==See also== | ||
| + | {{AMC12 box|year=2023|ab=A|before=num-a=9|num-a=11}} | ||
| + | |||
| + | [[Category:Rate Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:13, 9 November 2023
Problem
Positive real numbers 
and 
satisfy 
and 
. What is 
?
Solution
Because , set 
, 
(
). Put them in we get 
which implies 
. Solve the equation to get 
or 
. Since and are positive, and 
.
~plasta
See also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by num-a=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. 
