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

Revision as of 15:16, 21 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.

@@ Line 16: / Line 16: @@
 <cmath>a+b+c=0,</cmath>
 <cmath>a^2+b^2+c^2=0,</cmath>
-Since perfect squares are greater or equal to zero, a=0, b=0, c=0.
-Another method is
 <cmath>a^3+b^3+c^3=0.</cmath>
 We have

1973 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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All USAMO Problems and Solutions

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