Difference between revisions of "1973 USAMO Problems/Problem 4"

 
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J.Z.
 
J.Z.
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== Solution 4 ==
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We are going to use Intermediate Algebra Techniques to solve this equation.
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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.
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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>.
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Squaring this equation: <math>x^2+y^2+z^2+2xy+2yz+2xz=9</math>
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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>.
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Now we try the same idea with the cubed terms. Cube the first equation:
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<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:
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<math>3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27</math>.
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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).
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Resubstituting in the second and third equation: <math>-3+3(3)=8-2xyz</math>. So <math>xyz=1</math>.
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So now we have three equations for the elementary symmetric sums of <math>x,y,z</math>:
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Equation 4: <math>x+y+z=3</math> (this is also equation 1)
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Equation 5: <math>xy+yz+xz=3</math>
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Equation 6: <math>xyz=1</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)
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Therefore, the only solution is <math>\boxed{(x,y,z)=(1,1,1)}</math>.
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~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

$x+y+z=3$,

$x^2+y^2+z^2=3$,

$x^3+y^3+z^3=3$.

Solution

Let $x$, $y$, and $z$ be the roots of the cubic polynomial $t^3+at^2+bt+c$. Let $S_1=x+y+z=3$, $S_2=x^2+y^2+z^2=3$, and $S_3=x^3+y^3+z^3=3$. From this, $S_1+a=0$, $S_2+aS_1+2b=0$, and $S_3+aS_2+bS_1+3c=0$. Solving each of these, $a=-3$, $b=3$, and $c=-1$. Thus $x$, $y$, and $z$ are the roots of the polynomial $t^3-3t^2+3t-1=(t-1)^3$. Thus $x=y=z=1$, and there are no other solutions.


Solution 2

Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then \[0=P(x)+P(y)+P(z)=3-3a+3b-3c\] using our system of equations, so $P(1)=0$. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$. Then we can use the system of equations to find that $y=z=1$ as well, and so $\boxed{(1,1,1)}$ is the only solution to the system of equations.

Solution 3

Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \[a+b+c=0,\] \[a^2+b^2+c^2=0,\] \[a^3+b^3+c^3=0.\] We have \begin{align*} 0&=(a+b+c)^3\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\ &=0-3a^3-3b^3-3c^3+6abc\\ &=6abc. \end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\boxed{(1,1,1)}$ 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: $x+y+z=3$. This will be referred as the FIRST equation.

We are going to use the first equation to relate to the SECOND one ($x^2+y^2+z^2=3$) and the THIRD one ($x^3+y^3+z^3=3)$.

Squaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$

Subtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$, so $xy+xz+yz=3$.

Now we try the same idea with the cubed terms. Cube the first equation:

$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$. Plug in $x^3+y^3+z^3=3$ and factor partially:

$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$.

Now here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$. So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).

Resubstituting in the second and third equation: $-3+3(3)=8-2xyz$. So $xyz=1$.

So now we have three equations for the elementary symmetric sums of $x,y,z$:

Equation 4: $x+y+z=3$ (this is also equation 1)

Equation 5: $xy+yz+xz=3$

Equation 6: $xyz=1$.

If we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$, then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$, which means that $t=1$. Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$. (This step uses Vieta's formulas)

Therefore, the only solution is $\boxed{(x,y,z)=(1,1,1)}$.

~hastapasta

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

Newton's Sums

1973 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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