Difference between revisions of "2010 AIME I Problems/Problem 9"
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===Solution 3=== | ===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>. | 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>. | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/SpSuqWY01SE?t=1293 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== 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== | ||
− | https://youtu.be/LXct4j_rYfw | + | https://youtu.be/LXct4j_rYfw (Video unavailable as of 20240829) |
~Shreyas S | ~Shreyas S |
Latest revision as of 07:57, 29 August 2024
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
https://youtu.be/LXct4j_rYfw (Video unavailable as of 20240829)
~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.