Difference between revisions of "1973 USAMO Problems/Problem 4"
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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. | 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. | ||
− | ==See | + | |
+ | ==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. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>a=x-1,</math> <math>b=y-1</math> and <math>c=z-1.</math> Then | ||
+ | <cmath>a+b+c=0,</cmath> | ||
+ | <cmath>a^2+b^2+c^2=0,</cmath> | ||
+ | <cmath>a^3+b^3+c^3=0.</cmath> | ||
+ | We have | ||
+ | <cmath>\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*}</cmath> | ||
+ | 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. | ||
+ | |||
+ | J.Z. | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | ==See Also== | ||
[[Newton's Sums]] | [[Newton's Sums]] | ||
{{USAMO box|year=1973|num-b=3|num-a=5}} | {{USAMO box|year=1973|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 15:16, 21 May 2018
Problem
Determine all the roots, real or complex, of the system of simultaneous equations
,
.Solution
Let , , and be the roots of the cubic polynomial . 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.
Solution 2
Let have roots x, y, and z. Then using our system of equations, so . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let . Then we can use the system of equations to find that as well, and so is the only solution to the system of equations.
Solution 3
Let and Then We have Then one of and has to be 0, and easy to prove the other two are also 0. So 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
1973 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.