Difference between revisions of "2015 AIME II Problems/Problem 14"
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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>; we are looking for <math>2b(b^2-3a)+a^3</math>. Finding <math>a</math>, we get <math>35</math>. Substituting into the first equation, we get <math>b=54</math>. Our final answer is <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>; we are looking for <math>2b(b^2-3a)+a^3</math>. Finding <math>a</math>, we get <math>35</math>. Substituting into the first equation, we get <math>b=54</math>. Our final answer is <math>35+54=\boxed{089}</math>. | ||
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==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}} |
Latest revision as of 19:50, 1 December 2019
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 ; 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.