Problem 11690
(American Mathematical Monthly, Vol.120, May 2013) Proposed by Proposed by Pal Peter Dalyay (Hungary). .
Let M be a point in the interior of a convex polygon with vertices A1, . . . , Anin order. For 1 ≤ i ≤ n, let ribe the distance from M to Ai, and let Ribe the radius of the circumcircle of triangle M AiAi+1, where An+1= A1. Show that
n
X
i=1
Ri
ri+ ri+1
≥ n
4 cos(π/n).
Solution proposed by Radouan Boukharfane, Polytechnique de Montreal, Canada, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.
We first note that, by Mollweide’s formula, for any non-degenerate triangle ∆ABC we have that c
a + b= sin(C/2) cos((A − B)/2).
Hence, since |A − B| < π, it follows that 0 < cos((A − B)/2) ≤ 1 and, by the law of sines, R
a + b = c
2 sin(C)(a + b) = sin(C/2)
2 sin(C) cos((A − B)/2) ≥ 1 4 cos(C/2) where R is the radius of the circumcircle of ∆ABC.
Now, let us consider the triangle ∆M AiAi+1 for i = 1, . . . , n, then by the above inequality
n
X
i=1
Ri
ri+ ri+1 ≥ 1 4
n
X
i=1
1
cos(AiM Ai+1/2) ≥ n 4 cos (Pn
i=1AiM Ai+1/2n)= n 4 cos(π/n) where we used the following facts: 1/ cos(x) is convex in [0, π/2), and
n
X
i=1
AiM Ai+1 = 2π
because M is an interior point of a convex polygon.
