| Solution by The Javelina Problem Solvers, Texas A&M University,
Kingsville, Texas.
Let alpha denote the smallest of the angles angle PA[i]A[i+1] and let
gamma[i]=angle A[i]PA[i+1]. For i=1, 2, ..., n, pick B[i+1] on PA[i+1]
so that angle PA[i]B[i+1]=alpha (see figure).
Then, using the law of sines,
PA[i] PA[i] sin(alpha+gamma[i])
------- <= ------- = -------------------.
PA[i+1] PB[i+1] sin(alpha)
P
/\ g=gamma[i]
/g \
/ \ /
\ / \ /
\ / ..*B[i+1]
\ /a ..... \ / a=alpha
A[i]\/...______ \/
---*A[i+1]
It follows that
n PA[i] sin(alpha+gamma[i])
1 = product ------- <= product ------------------- <=
PA[i+1] sin(alpha)
sin(sum i=1..n of (alpha+gamma[i])/n)
(-------------------------------------)^n =
sin(alpha)
sin(alpha+2pi/n)
(----------------)^n.
sin(alpha)
where, in the last inequality, we have used a special case of Jensen's
inequality (see, for example, _Maxima and Minima Without Calculus_,
Ivan Niven, MAA Dolciani Series, No. 6, 1981, pp. 92-95).
Thus, sin alpha <= sin(alpha+2pi/n). We conclude that alpha <= pi/2 -
pi/n; if alpha > pi/2-pi/n, then sin(alpha+2pi/n) < sin(pi/2+pi/n) <
sin alpha, a contradiction.
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