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^3+y^3)=945</math>, respectively. Dividing the latter by the former equation yields <math>\frac{x^2-xy+y^2}{xy} = \frac{945}{810}</math>. Adding 3 to both sides and simplifying yields <math>\frac{(x+y)^2}{xy} = \frac{25}{6}</math>. Solving for <math>x+y</math> and substituting this expression into the first equation yields <math>\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810</math>. Solving for <math>xy</math>, we find that <math>xy = 3\sqrt[3]{2}</math>, so <math>x^3y^3 = 54</math>. Substituting this into the second equation and solving for <math>x^3+y^3</math> yields <math>x^3+y^3=\frac{35}{2}</math>. So, the expression to evaluate is equal to <math>2 \times \frac{35}{2} + 54 = \boxed{089}</math>. | Factor the given equations as <math>x^4y^4(x+y) = 810</math> and <math>x^3y^3(x^3+y^3)=945</math>, respectively. Dividing the latter by the former equation yields <math>\frac{x^2-xy+y^2}{xy} = \frac{945}{810}</math>. Adding 3 to both sides and simplifying yields <math>\frac{(x+y)^2}{xy} = \frac{25}{6}</math>. Solving for <math>x+y</math> and substituting this expression into the first equation yields <math>\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810</math>. Solving for <math>xy</math>, we find that <math>xy = 3\sqrt[3]{2}</math>, so <math>x^3y^3 = 54</math>. Substituting this into the second equation and solving for <math>x^3+y^3</math> yields <math>x^3+y^3=\frac{35}{2}</math>. So, the expression to evaluate is equal to <math>2 \times \frac{35}{2} + 54 = \boxed{089}</math>. | ||
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+ | Note that since the value we want to find is <math>2(x^3+y^3)+x^3y^3</math>, we can convert <math>2(x^3+y^3)</math> into an expression in terms of <math>x^3y^3</math>, since from the second equation which is <math>x^3y^3(x^3+y^3)=945</math>, we see that <math>2(x^3+y^3)=1890+x^6y^y,</math> and thus the value is <math>\frac{1890+x^6y^6}{x^3y^3}.</math> Since we've already found <math>x^3y^3,</math> we substitute and find the answer to be 89. | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:36, 20 October 2020
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
.
Note that since the value we want to find is , we can convert
into an expression in terms of
, since from the second equation which is
, we see that
and thus the value is
Since we've already found
we substitute and find the answer to be 89.
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
Solution 3
Add three times the first equation to the second equation and factor to get . Taking the cube root yields
. Noting that the first equation is
, we find that
. Plugging this into the second equation and dividing yields
. Thus the sum required, as noted in Solution 1, is
.
Solution 4
As with the other solutions, factor. But this time, let and
. Then
. Notice that
. Now, if we divide the second equation by the first one, we get
; then
. Therefore,
. Substituting
into
in equation 2 gives us
; we are looking for
. Finding
, we get
. Substituting into the first equation, we get
. Our final answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.