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>. | ||
+ | |||
+ | 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== | ||
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Add three times the first equation to the second equation and factor to get <math>(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375</math>. Taking the cube root yields <math>xy(x+y)=15</math>. Noting that the first equation is <math>(xy)^3\cdot(xy(x+y))=810</math>, we find that <math>(xy)^3=\frac{810}{15}=54</math>. Plugging this into the second equation and dividing yields <math>x^3+y^3 = \frac{945}{54} = \frac{35}{2}</math>. Thus the sum required, as noted in Solution 1, is <math>54+\frac{35}{2}\cdot2 = \boxed{089}</math>. | Add three times the first equation to the second equation and factor to get <math>(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375</math>. Taking the cube root yields <math>xy(x+y)=15</math>. Noting that the first equation is <math>(xy)^3\cdot(xy(x+y))=810</math>, we find that <math>(xy)^3=\frac{810}{15}=54</math>. Plugging this into the second equation and dividing yields <math>x^3+y^3 = \frac{945}{54} = \frac{35}{2}</math>. Thus the sum required, as noted in Solution 1, is <math>54+\frac{35}{2}\cdot2 = \boxed{089}</math>. | ||
+ | |||
+ | ==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>; 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 | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Factor the first and second equations as <math>(xy)^4(x+y)=810</math> and <math>(xy)^3(x+y)(x^2-xy+y^2)=945</math>, respectively. Dividing them (allowed, since neither are <math>0</math>), we have | ||
+ | <cmath>\frac{xy}{x^2-xy+y^2}=\frac67</cmath> | ||
+ | or | ||
+ | <cmath>x^2-\frac{13}{6}xy+y^2=0.</cmath> | ||
+ | Plugging into the quadratic formula and solving for <math>x</math> in terms of <math>y,</math> we have | ||
+ | <cmath>x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .</cmath> | ||
+ | WLOG, let <math>x=\frac{3y}2 .</math> | ||
+ | Plugging into our first equation, we have | ||
+ | <cmath>\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .</cmath> | ||
+ | Plugging this result (and the one for <math>x</math> in terms of <math>y</math>) into our desired expression, we have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2x^3+(xy)^3+2y^3 &= \frac{27}{4}y^3 + \left(\frac{3}{2} y^2\right)^3 +2y^3 \\ | ||
+ | &= \frac{35}{4}y^3 +\frac{27}{8}(y^3)^2 \\ | ||
+ | &= 35+54 \\ | ||
+ | &= \boxed{089} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | ~ASAB | ||
==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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 16:37, 17 March 2022
Contents
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
Solution 6
Factor the first and second equations as and , respectively. Dividing them (allowed, since neither are ), we have or Plugging into the quadratic formula and solving for in terms of we have WLOG, let Plugging into our first equation, we have Plugging this result (and the one for in terms of ) into our desired expression, we have
~ASAB
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.