Problem 11905
(American Mathematical Monthly, Vol.123, April 2016) Proposed by C. Lupu (USA).
From a point P inside a triangle ABC, the perpendiculars P PA, P PB, andP PC are drawn to its sides. LetR be the circumradius and r the inradius of the triangle. Prove that
R
2r ≤ |P A||P B||P C|
(|P PB| + |P PC|)(|P PA| + |P PC|)(|P PA| + |P PB|).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We have that
|P PB| + |P PC|
|P A| = sin α1+ sin α2 where α1= |P bAC| and α2= |P bAB|.
Hence, by using similar notations for the other factors, the inquality is equivalent to
(sin α1+ sin α2) (sin β1+ sin β2) (sin γ1+ sin γ2) ≤ 2r
R = 8 sin
α1+ α2 2
sin
β1+ β2 2
sin
γ1+ γ2 2
which holds because the function sin x is positive and concave in (0, π) and
(sin x1+ sin x2) ≤ 2 sin
x1+ x2 2
for all x1, x2∈ (0, π).
