1980 USAMO Problems/Problem 3

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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
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. AMC logo.png