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>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Factor the given equations as: | ||
+ | <cmath>x^4y^4(x+y)=810</cmath> | ||
+ | <cmath>x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945</cmath> | ||
+ | We note that these expressions (as well as the desired expression) can be written exclusively in terms of <math>x+y</math> and <math>xy</math>. We make the substitution <math>s=x+y</math> and <math>p=xy</math> (for sum and product, respectively). | ||
+ | |||
+ | <cmath>x^4y^4(x+y)=p^4s=810</cmath> | ||
+ | <cmath>x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945</cmath> | ||
+ | |||
+ | We see that <math>p^4s</math> shows up in both equations, so we can eliminate it and find <math>sp</math>, after which we can get <math>p^3</math> from the first equation. If you rewrite the desired expression using <math>s</math> and <math>p</math>, it becomes clear that you don't need to actually find the values of <math>s</math> and <math>p</math>, but I will do so for the sake of completion. | ||
+ | |||
+ | <cmath>s^3p^3=945+3p^4s</cmath> | ||
+ | <cmath>s^3p^3=945+3(810)=3375</cmath> | ||
+ | <cmath>sp=15</cmath> | ||
+ | |||
+ | <cmath>p^3=\frac{810}{sp}=54</cmath> | ||
+ | <cmath>p=3\cdot2^{1/3}</cmath> | ||
+ | <cmath>s=\frac{15}{p}=5\cdot2^{-1/3}</cmath> | ||
+ | |||
+ | The desired expression can be written as: | ||
+ | <cmath>2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3</cmath> | ||
+ | <cmath>2(s)(s^2-3p)+p^3=2s^3-6sp+p^3</cmath> | ||
+ | |||
+ | Plugging in <math>s</math> and <math>p</math>, we get: | ||
+ | <cmath>2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}</cmath> | ||
+ | |||
+ | - gting | ||
==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 21:27, 17 February 2021
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 .
Solution 5
Factor the given equations as: We note that these expressions (as well as the desired expression) can be written exclusively in terms of and . We make the substitution and (for sum and product, respectively).
We see that shows up in both equations, so we can eliminate it and find , after which we can get from the first equation. If you rewrite the desired expression using and , it becomes clear that you don't need to actually find the values of and , but I will do so for the sake of completion.
The desired expression can be written as:
Plugging in and , we get:
- gting
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.