Difference between revisions of "1973 USAMO Problems/Problem 4"
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<math>x^3+y^3+z^3=3</math>.</center> | <math>x^3+y^3+z^3=3</math>.</center> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>x</math>, <math>y</math>, and <math>z</math> be the [[root]]s of the [[cubic polynomial]] <math>t^3+at^2+bt+c</math>. Let <math>S_1=x+y+z=3</math>, <math>S_2=x^2+y^2+z^2=3</math>, and <math>S_3=x^3+y^3+z^3=3</math>. From this, <math>S_1+a=0</math>, <math>S_2+aS_1+2b=0</math>, and <math>S_3+aS_2+bS_1+3c=0</math>. Solving each of these, <math>a=-3</math>, <math>b=3</math>, and <math>c=-1</math>. Thus <math>x</math>, <math>y</math>, and <math>z</math> are the roots of the polynomial <math>t^3-3t^2+3t-1=(t-1)^3</math>. Thus <math>x=y=z=1</math>, and there are no other solutions. | Let <math>x</math>, <math>y</math>, and <math>z</math> be the [[root]]s of the [[cubic polynomial]] <math>t^3+at^2+bt+c</math>. Let <math>S_1=x+y+z=3</math>, <math>S_2=x^2+y^2+z^2=3</math>, and <math>S_3=x^3+y^3+z^3=3</math>. From this, <math>S_1+a=0</math>, <math>S_2+aS_1+2b=0</math>, and <math>S_3+aS_2+bS_1+3c=0</math>. Solving each of these, <math>a=-3</math>, <math>b=3</math>, and <math>c=-1</math>. Thus <math>x</math>, <math>y</math>, and <math>z</math> are the roots of the polynomial <math>t^3-3t^2+3t-1=(t-1)^3</math>. Thus <math>x=y=z=1</math>, and there are no other solutions. | ||
− | |||
==Solution 2== | ==Solution 2== | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
0&=(a+b+c)^3\\ | 0&=(a+b+c)^3\\ | ||
− | &=(a^3+b^3+c^3)+3a^2(b+c)+3b^ | + | &=(a^3+b^3+c^3)+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc\\ |
&=0-3a^3-3b^3-3c^3+6abc\\ | &=0-3a^3-3b^3-3c^3+6abc\\ | ||
&=6abc. | &=6abc. | ||
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J.Z. | J.Z. | ||
+ | |||
== Solution 4 == | == Solution 4 == | ||
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Let's start with the first one: <math>x+y+z=3</math>. This will be referred as the FIRST 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>. | + | 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> | Squaring this equation: <math>x^2+y^2+z^2+2xy+2yz+2xz=9</math> | ||
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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) | 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>. | + | Therefore, the only solution is <math>\boxed{(x,y,z)=(1,1,1)}</math>. |
~hastapasta | ~hastapasta | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | From the question | ||
+ | |||
+ | Equation 1 : <math>x+y+z=3</math> | ||
+ | |||
+ | Equation 2 : <math>x^2+y^2+z^2=3</math> | ||
+ | |||
+ | Equation 3 : <math>x^3+y^3+z^3=3</math> | ||
+ | |||
+ | From Eq(1) and Eq(2), | ||
+ | |||
+ | Equation 4 : <math>(x+y+z)^2=9 \implies x^2+y^2+z^2+2\sum xy = 9 \implies \sum xy = 3</math> | ||
+ | |||
+ | From Eq(1), Eq(3), and the fact that <math>(x+y+z)^3=x^3+y^3+z^3+3\prod(x+y)</math>, | ||
+ | |||
+ | Equation 5 : <math>(x+y+z)^3=x^3+y^3+z^3+3\prod(x+y)= 27 \implies \prod(x+y)=8</math> | ||
+ | |||
+ | As from Eq(1), <math>\ x+y=3-z</math>,<math>\ y+z=3-x</math>, <math>\ z+x = 3- y</math> | ||
+ | |||
+ | Equation 6 : <math>\prod(3-x) =8</math> | ||
+ | |||
+ | Now let there be a cubic function <math>f(t)</math> with the roots <math>x,y,z</math> so from the relation between roots and coefficients, | ||
+ | |||
+ | Equation 7 : <math>f(t)= t^3-3t^2+3t-d</math> (As we don't know the product let the constant be <math>d</math>) | ||
+ | |||
+ | As <math>f(t) = (t-x)(t-y)(t-z)</math> then at <math>t=3</math>, | ||
+ | |||
+ | Equation 8 : <math>f(3)=9-d=(3-x)(3-y)(3-z)</math> | ||
+ | |||
+ | And from Eq(6), | ||
+ | |||
+ | Equation 9 : <math>d= -1</math> | ||
+ | |||
+ | So finally, | ||
+ | |||
+ | Equation 10 : <math>f(t)=t^3-3t^2+3t-1=(t-1)^{3}</math> | ||
+ | |||
+ | So, <math>\ (x,y,z)=(1,1,1)</math> | ||
+ | |||
+ | ~Sparrow_1827 | ||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 08:18, 22 April 2024
Problem
Determine all the roots, real or complex, of the system of simultaneous equations
![$x+y+z=3$](http://latex.artofproblemsolving.com/a/d/5/ad5ff13ebc979a03c144ba96c88e6281c5556894.png)
,
![$x^3+y^3+z^3=3$](http://latex.artofproblemsolving.com/e/4/7/e47fe3685e8d2812abd631eb34e54bcffd0d2643.png)
Solution 1
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
Solution 5
From the question
Equation 1 :
Equation 2 :
Equation 3 :
From Eq(1) and Eq(2),
Equation 4 :
From Eq(1), Eq(3), and the fact that ,
Equation 5 :
As from Eq(1), ,
,
Equation 6 :
Now let there be a cubic function with the roots
so from the relation between roots and coefficients,
Equation 7 : (As we don't know the product let the constant be
)
As then at
,
Equation 8 :
And from Eq(6),
Equation 9 :
So finally,
Equation 10 :
So,
~Sparrow_1827
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.