https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Aykw1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:55:58ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1973_USAMO_Problems/Problem_4&diff=946201973 USAMO Problems/Problem 42018-05-19T11:13:23Z<p>Aykw1: /* Solution 3 */</p>
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<div>==Problem==<br />
Determine all the [[root]]s, [[real]] or [[complex]], of the system of simultaneous [[equation]]s<br />
<center><math>x+y+z=3</math>,<br />
<math>x^2+y^2+z^2=3</math>,<br />
<math>x^3+y^3+z^3=3</math>.</center><br />
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==Solution==<br />
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.<br />
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==Solution 2==<br />
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.<br />
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==Solution 3==<br />
Let <math>a=x-1,</math> <math>b=y-1</math> and <math>c=z-1.</math> Then <br />
<cmath>a+b+c=0,</cmath><br />
<cmath>a^2+b^2+c^2=0,</cmath><br />
Since perfect squares are greater or equal to zero, a=0, b=0, c=0.<br />
Another method is<br />
<cmath>a^3+b^3+c^3=0.</cmath><br />
We have<br />
<cmath>\begin{align*}<br />
0&=(a+b+c)^3\\<br />
&=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\<br />
&=0-3a^3-3b^3-3c^3+6abc\\<br />
&=6abc.<br />
\end{align*}</cmath><br />
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.<br />
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J.Z.<br />
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{{alternate solutions}}<br />
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==See Also==<br />
[[Newton's Sums]]<br />
{{USAMO box|year=1973|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
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[[Category:Olympiad Algebra Problems]]</div>Aykw1