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

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==Solution 2==
 
==Solution 2==
 
Let <math>P(t)=t^3-at^2+bt-c</math> have roots x, y, and z. Then <cmath>0=P(x)+P(y)+P(z)=3-3a+3b-3c</cmath> using our system of equations, so <math>P(1)=0</math>. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let <math>x=1</math>. Then we can use the system of equations to find that <math>y=z=1</math> as well, and so <math>\boxed{(1,1,1)}</math> is the only solution to the system of equations.
 
Let <math>P(t)=t^3-at^2+bt-c</math> have roots x, y, and z. Then <cmath>0=P(x)+P(y)+P(z)=3-3a+3b-3c</cmath> using our system of equations, so <math>P(1)=0</math>. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let <math>x=1</math>. Then we can use the system of equations to find that <math>y=z=1</math> as well, and so <math>\boxed{(1,1,1)}</math> is the only solution to the system of equations.
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==Solution 3==
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Let <math>a=x-1,</math> <math>b=y-1</math> and <math>c=z-1.</math> Then
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<cmath>a+b+c=0,</cmath>
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<cmath>a^2+b^2+c^2=0,</cmath>
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<cmath>a^3+b^3+c^3=0.</cmath>
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We have
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<cmath>\begin{align*}
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0&=(a+b+c)^3\\
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&=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\
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&=0-3a^3-3b^3-3c^3+6abc\\
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&=6abc.
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\end{align*}</cmath>
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Then one of <math>a, b</math> and <math>c</math> has to be 0, and easy to prove the other two are also 0. So <math>\boxed{(1,1,1)}</math> is the only solution to the system of equations.
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J.Z.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 16:48, 18 May 2018

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.

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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