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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

1977.0. "Crux Mathematicorum 2031" by RUSURE::EDP (Always mount a scratch monkey.) Thu Jun 15 1995 10:37

    Proposed by Toshio Seimiya, Kawasaki, Japan.
    
    Suppose that a, b, c are acute angles such that
    
    	sin(a-b)   sin(b-c)   sin(c-a)
    	-------- + -------- + -------- = 0.
    	sin(a+b)   sin(b+c)   sin(c+a)
    
    Prove that at least two of a, b, c are equal.

T.R	Title	User	Personal Name	Date	Lines
1977.1		HANNAH::OSMAN	see HANNAH::IGLOO$:[OSMAN]ERIC.VT240	`Thu Jun 15 1995 12:05`	30
	I assume acute means quadrant one ? (less than 90 degress ?) o.k. Suppose a,b,c are unequal. By symmetric nature of the formula, assume a < b < c. (We'll attempt to reach a contradiction). Here's the original formula: sin(a-b) sin(b-c) sin(c-a) -------- + -------- + -------- = 0. sin(a+b) sin(b+c) sin(c+a) Since a,b,c are in q1, a-b is in q4 (lower right), so sin is negative. b-c is q4, c-a is q1. a+b, b+c, c+a are all in q2 (upper left) so sin is positive. Let's look at the signs of the sins: minus/plus + minus/plus + plus/plus = 0. minus + minus + plus = 0. Hence minus + minus = - plus Is this getting somewhere ? I assume the next step is to start applying some trigy identities. /Eric
1977.2		AUSSIE::GARSON	achtentachtig kacheltjes	`Fri Jun 16 1995 00:00`	7
	re .1 Yes, acute means quadrant one (less than 90 degrees) as distinct from obtuse meaning quadrant two. I haven't looked at the problem yet myself. So little time. So many problems.
1977.3		HERON::BUCHANAN	Et tout sera bien et	`Mon Jun 19 1995 09:57`	17
	> Suppose that a, b, c are acute angles such that > > sin(a-b) sin(b-c) sin(c-a) > -------- + -------- + -------- = 0. > sin(a+b) sin(b+c) sin(c+a) > > Prove that at least two of a, b, c are equal. Expand by sin(P�Q) = sinPcosQ � sinQcosP Place under a common denominator It's the numerator we must look at Divide by sinasinbsinccosacosb*cosc (which we know to be non-zero) Substitute: A = tana/tanb, B = tanb/tanc (A-1)(B-1)(AB-1) = 0 whence the result.