Difference between revisions of "2015 AIME II Problems/Problem 14"

Revision as of 10:59, 28 March 2015

Problem

Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ .

Solution

The expression we want to find is $2(x^3+y^3) + x^3y^3$ .

Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$ , respectively. Dividing the latter by the former equation yields $\frac{x^2-xy+y^2}{xy} = \frac{945}{810}$ . Adding 3 to both sides and simplifying yields $\frac{(x+y)^2}{xy} = \frac{25}{6}$ . Solving for $x+y$ and substituting this expression into the first equation yields $\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810$ . Solving for $xy$ , we find that $xy = 3\sqrt[3]{2}$ , so $x^3y^3 = 54$ . Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\frac{35}{2}$ . So, the expression to evaluate is equal to $2 \times \frac{35}{2} + 54 = \boxed{089}$ .

Solution 2

Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$ , respectively. By the first equation, $x+y=\frac{810}{x^4y^4}$ . Plugging this in to the second equation and simplifying yields $(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}$ . Now substitute $\frac{x}{y}=a$ . Solving the quadratic in $a$ , we get $a=\frac{x}{y}=\frac{2}{3}$ or $\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$ , WLOG, let $\frac{x}{y}=\frac{2}{3}$ , so $x=\frac{2}{3}y$ . Now plugging this in to either one of the equations, we get the solutions $y=\frac{3(2^{\frac{2}{3}})}{2}$ , $x=2^{\frac{2}{3}}$ . Now plugging into what we want, we get $8+54+27=\boxed{089}$

Solution 3

Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$ . Taking the cube root yields $xy(x+y)=15$ . Noting that the first equation is $(xy)^3 * (xy(x+y))=810$ , we find that $(xy)^3=\frac{810}{15}=54$ . Plugging this into the second equation and dividing yields $x^3+y^3 = \frac{945}{54} = \frac{35}{2}$ . Thus the sum required, as noted in Solution 1, is $54+\frac{35}{2} * 2 = \boxed{089}$ .

@@ Line 10: / Line 10: @@
 ==Solution 2==
 Factor the given equations as <math>x^4y^4(x+y) = 810</math> and <math>x^3y^3(x+y)(x^2-xy+y^2)=945</math>, respectively. By the first equation, <math>x+y=\frac{810}{x^4y^4}</math>. Plugging this in to the second equation and simplifying yields <math>(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}</math>. Now substitute <math>\frac{x}{y}=a</math>. Solving the quadratic in <math>a</math>, we get <math>a=\frac{x}{y}=\frac{2}{3}</math> or <math>\frac{3}{2}</math> As both of the original equations were symmetric in <math>x</math> and <math>y</math>, WLOG, let <math>\frac{x}{y}=\frac{2}{3}</math>, so <math>x=\frac{2}{3}y</math>. Now plugging this in to either one of the equations, we get the solutions <math>y=\frac{3(2^{\frac{2}{3}})}{2}</math>, <math>x=2^{\frac{2}{3}}</math>. Now plugging into what we want, we get <math>8+54+27=\boxed{089}</math>
+==Solution 3==
+Add three times the first equation to the second equation and factor to get <math>(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375</math>. Taking the cube root yields <math>xy(x+y)=15</math>. Noting that the first equation is <math>(xy)^3 * (xy(x+y))=810</math>, we find that <math>(xy)^3=\frac{810}{15}=54</math>. Plugging this into the second equation and dividing yields <math>x^3+y^3 = \frac{945}{54} = \frac{35}{2}</math>. Thus the sum required, as noted in Solution 1, is <math>54+\frac{35}{2} * 2 = \boxed{089}</math>.
 ==See also==
 {{AIME box|year=2015|n=II|num-b=13|num-a=15}}
 {{MAA Notice}}

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search