Difference between revisions of "1980 USAMO Problems/Problem 3"
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== Problem == | == Problem == | ||
− | <math>A + B + C</math> is an integral multiple of <math>\pi</math>. <math>x, y, </math> and <math>z</math> are real numbers. If <math>x\sin(A) | + | <math>A + B + C</math> is an integral multiple of <math>\pi</math>. <math>x, y, </math> and <math>z</math> are real numbers. If <math>x\sin(A)+y\sin(B)+z\sin(C)=x^2\sin(2A)+y^2\sin(2B)+z^2\sin(2C)=0</math>, show that <math>x^n\sin(nA)+y^n \sin(nB) +z^n \sin(nC)=0</math> for any positive integer <math>n</math>. |
== Solution == | == Solution == | ||
− | {{ | + | Let <math>a=xe^{iA}</math>, <math>b=ye^{iB}</math>, <math>c=ze^{iC}</math> be numbers in the complex plane. |
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Newton Sums gives the following relationship between sums of the form <math>P_k=a^k+b^k+c^k</math> | ||
+ | <cmath>P_k-S_1P_{k-1}+S_2P_{k-2}-S_3P_{k-3}=0</cmath> | ||
+ | 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
is an integral multiple of . and are real numbers. If , show that for any positive integer .
Solution
Let , , be numbers in the complex plane.
Note that implies which is real. Also note that are the imaginary parts of and that are the imaginary parts of by de Moivre's Theorem. Therefore, and are real because their imaginary parts sum to zero.
Finally, note that is real as well.
It suffices to show that is real for all positive integer , which can be shown by induction.
Newton Sums gives the following relationship between sums of the form Where , , and . It is given that are real. Note that if are real, then clearly 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.