Difference between revisions of "2010 AIME I Problems/Problem 9"
m (→Solution) |
Pi is 3.14 (talk | contribs) (→Solution 3) |
||
(11 intermediate revisions by 8 users not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | Add the three equations to get <math> | + | |
+ | ===Solution 1=== | ||
+ | Add the three equations to get <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Now, let <math>abc = p</math>. <math>a = \sqrt [3]{p + 2}</math>, <math>b = \sqrt [3]{p + 6}</math> and <math>c = \sqrt [3]{p + 20}</math>, so <math>p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})</math>. Now [[cube]] both sides; the <math>p^3</math> terms cancel out. Solve the remaining [[quadratic]] to get <math>p = - 4, - \frac {15}{7}</math>. To maximize <math>a^3 + b^3 + c^3</math> choose <math>p = - \frac {15}{7}</math> and so the sum is <math>28 - \frac {45}{7} = \frac {196 - 45}{7}</math> giving <math>151 + 7 = \fbox{158}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | This is almost the same as solution 1. Note <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Next, let <math>k = a^3</math>. Note that <math>b = \sqrt [3]{k + 4}</math> and <math>c = \sqrt [3]{k + 18}</math>, so we have <math>28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22</math>. Move 28 over, divide both sides by 3, then cube to get <math>k^3-6k^2+12k-8 = k^3+22k^2+18k</math>. The <math>k^3</math> terms cancel out, so solve the quadratic to get <math>k = -2, -\frac{1}{7}</math>. We maximize <math>abc</math> by choosing <math>k = -\frac{1}{7}</math>, which gives us <math>a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}</math>. Thus, our answer is <math>151+7=\boxed{158}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | We have that <math>x^3 = 2 + xyz</math>, <math>y^3 = 6 + xyz</math>, and <math>z^3 = 20 + xyz</math>. Multiplying the three equations, and letting <math>m = xyz</math>, we have that <math>m^3 = (2+m)(6+m)(20+m)</math>, and reducing, that <math>7m^2 + 43m + 60 = 0</math>, which has solutions <math>m = -\frac{15}{7}, -4</math>. Adding the three equations and testing both solutions, we find the answer of <math>\frac{151}{7}</math>, so the desired quantity is <math>151 + 7 = \fbox{158}</math>. | ||
+ | |||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/SpSuqWY01SE?t=1293 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Remark == | ||
+ | It is tempting to add the equations and then use the well-known factorization <math>x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-xz-yz)</math>. Unfortunately such a factorization is just a red herring: it doesn't give much information on <math>a^3+b^3+c^3</math>. | ||
+ | == Another Remark == | ||
+ | The real problem with adding the equations is that <math>x, y, z</math> are real numbers based on the problem, but the adding trick only works when <math>x, y, z</math> are integers. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/LXct4j_rYfw | ||
+ | |||
+ | ~Shreyas S | ||
== See Also == | == See Also == |
Latest revision as of 07:35, 4 November 2022
Contents
Problem
Let be the real solution of the system of equations
,
,
. The greatest possible value of
can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
Add the three equations to get . Now, let
.
,
and
, so
. Now cube both sides; the
terms cancel out. Solve the remaining quadratic to get
. To maximize
choose
and so the sum is
giving
.
Solution 2
This is almost the same as solution 1. Note . Next, let
. Note that
and
, so we have
. Move 28 over, divide both sides by 3, then cube to get
. The
terms cancel out, so solve the quadratic to get
. We maximize
by choosing
, which gives us
. Thus, our answer is
.
Solution 3
We have that ,
, and
. Multiplying the three equations, and letting
, we have that
, and reducing, that
, which has solutions
. Adding the three equations and testing both solutions, we find the answer of
, so the desired quantity is
.
Video Solution by OmegaLearn
https://youtu.be/SpSuqWY01SE?t=1293
~ pi_is_3.14
Remark
It is tempting to add the equations and then use the well-known factorization . Unfortunately such a factorization is just a red herring: it doesn't give much information on
.
Another Remark
The real problem with adding the equations is that are real numbers based on the problem, but the adding trick only works when
are integers.
Video Solution
~Shreyas S
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.