Difference between revisions of "2015 AIME II Problems/Problem 14"
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==Solution 4== | ==Solution 4== | ||
− | As with the other solutions, factor. But this time, let <math>a=xy</math> and <math>b=x+y</math>. Then <math>a^4b=810</math>. Notice that <math>x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)</math>. Now, if we divide the second equation by the first one, we get <math>7/6 = \frac{b^2-3a}{a}</math>; then <math>\frac{b^2}{a}=\frac{25}{6}</math>. Therefore, <math>a = \frac{6}{25}b^2</math>. Substituting <math>a</math> into <math>b</math> in equation 2 gives us <math>b^3 = \frac{5^3}{2}</math>; the value we wanted to find is <math>2b(b^2-3a)+a^3</math>. With a conversion, we get <math>35+54=\boxed{089}</math>. | + | As with the other solutions, factor. But this time, let <math>a=xy</math> and <math>b=x+y</math>. Then <math>a^4b=810</math>. Notice that <math>x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)</math>. Now, if we divide the second equation by the first one, we get <math>7/6 = \frac{b^2-3a}{a}</math>; then <math>\frac{b^2}{a}=\frac{25}{6}</math>. Therefore, <math>a = \frac{6}{25}b^2</math>. Substituting <math>a</math> into <math>b</math> in equation 2 gives us <math>b^3 = \frac{5^3}{2}</math>; the value we wanted to find is <math>2b(b^2-3a)+a^3</math>. With a conversion, we get <math>35+54=\boxed{089}</math>. The key was realizing the commonness of <math>x+y</math> and <math>xy</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 21:42, 28 August 2017
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
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
; the value we wanted to find is
. With a conversion, we get
. The key was realizing the commonness of
and
.
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.