Difference between revisions of "2010 AIME I Problems/Problem 9"
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== Remark == | == 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>. | 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== | ==Video Solution== |
Revision as of 09:40, 29 August 2021
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
.
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.