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Difference between revisions of "1980 USAMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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Let <math>a=xe^{iA}</math>, <math>b=ye^{iB}</math>, <math>c=ze^{iC}</math> be numbers in the complex plane.
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Note that <math>A+B+C=k\pi</math> implies <math>abc=xyz(e^{ik\pi})=\pm xyz</math> which is real. Also note that <math>x\sin(A), y\sin(B), z\sin(C)</math> are the imaginary parts of <math>a, b, c</math> and that <math>x^2\sin(2A), y^2\sin(2B), z^2\sin(2C)</math> are the imaginary parts of <math>a^2, b^2, c^2</math> by de Moivre's Theorem. Therefore, <math>a+b+c</math> and <math>a^2+b^2+c^2</math> are real because their imaginary parts sum to zero.
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Finally, note that <math>\frac{1}{2}\left((a+b+c)^2-(a^2+b^2+c^2)\right)=ab+bc+ac</math> is real as well.
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It suffices to show that <math>P_n=a^n+b^n+c^n</math> is real for all positive integer <math>n</math>, which can be shown by induction.
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Newton Sums gives the following relationship between sums of the form <math>P_k=a^k+b^k+c^k</math>
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<cmath>P_k-S_1P_{k-1}+S_2P_{k-2}-S_3P_{k-3}=0</cmath>
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Where <math>S_1=a+b+c</math>, <math>S_2=ab+bc+ac</math>, and <math>S_3=abc</math>. It is given that <math>P_0, P_1, P_2</math> are real. Note that if  <math>P_{k-1}, P_{k-2}, P_{k-3}</math> are real, then clearly <math>P_k</math> is real because all other parts of the above equation are real, completing the induction.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:20, 6 May 2017

Problem

$A + B + C$ is an integral multiple of $\pi$. $x, y,$ and $z$ are real numbers. If $x\sin(A)+y\sin(B)+z\sin(C)=x^2\sin(2A)+y^2\sin(2B)+z^2\sin(2C)=0$, show that $x^n\sin(nA)+y^n \sin(nB) +z^n \sin(nC)=0$ for any positive integer $n$.

Solution

Let $a=xe^{iA}$, $b=ye^{iB}$, $c=ze^{iC}$ be numbers in the complex plane.

Note that $A+B+C=k\pi$ implies $abc=xyz(e^{ik\pi})=\pm xyz$ which is real. Also note that $x\sin(A), y\sin(B), z\sin(C)$ are the imaginary parts of $a, b, c$ and that $x^2\sin(2A), y^2\sin(2B), z^2\sin(2C)$ are the imaginary parts of $a^2, b^2, c^2$ by de Moivre's Theorem. Therefore, $a+b+c$ and $a^2+b^2+c^2$ are real because their imaginary parts sum to zero.

Finally, note that $\frac{1}{2}\left((a+b+c)^2-(a^2+b^2+c^2)\right)=ab+bc+ac$ is real as well.

It suffices to show that $P_n=a^n+b^n+c^n$ is real for all positive integer $n$, which can be shown by induction.

Newton Sums gives the following relationship between sums of the form $P_k=a^k+b^k+c^k$ \[P_k-S_1P_{k-1}+S_2P_{k-2}-S_3P_{k-3}=0\] Where $S_1=a+b+c$, $S_2=ab+bc+ac$, and $S_3=abc$. It is given that $P_0, P_1, P_2$ are real. Note that if $P_{k-1}, P_{k-2}, P_{k-3}$ are real, then clearly $P_k$ is real because all other parts of the above equation are real, completing the induction.

See Also

1980 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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