Difference between revisions of "1973 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | Let <math>x</math>, <math>y</math>, and <math>z</math> be the roots of the cubic <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+ | + | Let <math>x</math>, <math>y</math>, and <math>z</math> be the roots of the cubic <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. |
==See also== | ==See also== |
Revision as of 12:01, 24 July 2009
Problem
Determine all the roots, real or complex, of the system of simultaneous equations
,
.Solution
Let , , and be the roots of the cubic . 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.
See also
1973 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |