Difference between revisions of "1973 USAMO Problems/Problem 4"
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+ | == Solution 4 == | ||
+ | |||
+ | We are going to use Intermediate Algebra Techniques to solve this equation. | ||
+ | |||
+ | Let's start with the first one: <math>x+y+z=3</math>. This will be referred as the FIRST equation. | ||
+ | |||
+ | We are going to use the first equation to relate to the SECOND one (<math>x^2+y^2+z^2=3</math>) and the THIRD one (<math>x^3+y^3+z^3=3)</math>. | ||
+ | |||
+ | Squaring this equation: <math>x^2+y^2+z^2+2xy+2yz+2xz=9</math> | ||
+ | |||
+ | Subtracting this equation from the 2nd equation in the problem, we have <math>2xy+2yz+2xz=6</math>, so <math>xy+xz+yz=3</math>. | ||
+ | |||
+ | Now we try the same idea with the cubed terms. Cube the first equation: | ||
+ | |||
+ | <math>x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27</math>. Plug in <math>x^3+y^3+z^3=3</math> and factor partially: | ||
+ | |||
+ | <math>3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27</math>. | ||
+ | |||
+ | Now here is the key step. Note that <math>z=3-x-y, y=3-x-z, x=3-y-z</math>. So we are going to substitute <math>y+z, x+z, x+y</math> for each of the expressions and we get: <math>-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz</math> (I rearranged it a bit). | ||
+ | |||
+ | Resubstituting in the second and third equation: <math>-3+3(3)=8-2xyz</math>. So <math>xyz=1</math>. | ||
+ | |||
+ | So now we have three equations for the elementary symmetric sums of <math>x,y,z</math>: | ||
+ | |||
+ | Equation 4: <math>x+y+z=3</math> (this is also equation 1) | ||
+ | |||
+ | Equation 5: <math>xy+yz+xz=3</math> | ||
+ | |||
+ | Equation 6: <math>xyz=1</math>. | ||
+ | |||
+ | If we call the solutions of <math>t^3-3t^2+3t-1=0</math> (Equation 7) <math>a,b,c</math>, then <math>x,y,z</math> are the three roots <math>a,b,c</math> but in some order. Notice that Equation 7 can be factored as <math>(t-1)^3=0</math>, which means that <math>t=1</math>. Therefore <math>(x,y,z)</math> are permutations of <math>(1,1,1)</math> in some order, which can be only <math>(1,1,1)</math>. (This step uses Vieta's formulas) | ||
+ | |||
+ | Therefore, the only solution is <math>\boxed{(x,y,z)=(1,1,1)}</math>. | ||
+ | |||
+ | ~hastapasta | ||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 13:39, 30 November 2022
Problem
Determine all the roots, real or complex, of the system of simultaneous equations
,
.Solution
Let , , and be the roots of the cubic polynomial . Let , , and . From this, , , and . Solving each of these, , , and . Thus , , and are the roots of the polynomial . Thus , and there are no other solutions.
Solution 2
Let have roots x, y, and z. Then using our system of equations, so . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let . Then we can use the system of equations to find that as well, and so is the only solution to the system of equations.
Solution 3
Let and Then We have Then one of and has to be 0, and easy to prove the other two are also 0. So is the only solution to the system of equations.
J.Z.
Solution 4
We are going to use Intermediate Algebra Techniques to solve this equation.
Let's start with the first one: . This will be referred as the FIRST equation.
We are going to use the first equation to relate to the SECOND one () and the THIRD one (.
Squaring this equation:
Subtracting this equation from the 2nd equation in the problem, we have , so .
Now we try the same idea with the cubed terms. Cube the first equation:
. Plug in and factor partially:
.
Now here is the key step. Note that . So we are going to substitute for each of the expressions and we get: (I rearranged it a bit).
Resubstituting in the second and third equation: . So .
So now we have three equations for the elementary symmetric sums of :
Equation 4: (this is also equation 1)
Equation 5:
Equation 6: .
If we call the solutions of (Equation 7) , then are the three roots but in some order. Notice that Equation 7 can be factored as , which means that . Therefore are permutations of in some order, which can be only . (This step uses Vieta's formulas)
Therefore, the only solution is .
~hastapasta
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1973 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.