Difference between revisions of "2015 AIME II Problems/Problem 14"
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− | Factor the given equations as <math>x^4y^4(x+y) = 810</math> and <math>x^3y^3(x+y)(x^ | + | 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> |
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=13|num-a=15}} | {{AIME box|year=2015|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:57, 27 March 2015
Contents
[hide]Problem
Let and be real numbers satisfying and . Evaluate .
Solution
The expression we want to find is .
Factor the given equations as and , respectively. Dividing the latter by the former equation yields . Adding 3 to both sides and simplifying yields . Solving for and substituting this expression into the first equation yields . Solving for , we find that , so . Substituting this into the second equation and solving for yields . So, the expression to evaluate is equal to .
Solution 2
Factor the given equations as and , respectively. By the first equation, . Plugging this in to the second equation and simplifying yields . Now substitute . Solving the quadratic in , we get or As both of the original equations were symmetric in and , WLOG, let , so . Now plugging this in to either one of the equations, we get the solutions , . Now plugging into what we want, we get
See also
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 |
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