Problem 11900
(American Mathematical Monthly, Vol.123, March 2016) Proposed by G. Apostolopoulos (Greece).
LetABC be a triangle and let r be the radius of its incircle. The circle with radius rAis externally tangent to the incircle and internally tangent to sides AB and AC of ABC. Define rB and rC
similarly. Forn ≥ 1, prove that
r + rA
r − rA
n
+ r + rB
r − rB
n
+ r + rC
r − rC
n
≥ 3 · 2n.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let I be the center of the incircle and let IA the center of the circle with radius rA. Then IA is on the angle-bisector AI and |IAI| = rA+ r. Hence
rA
|AI| − (r + rA)= r
|AI| ⇒ rA
r = |AI| − r
|AI| + r ⇒ r + rA
r − rA
= |AI|
r . Since xn si convex, it follows that
r + rA
r − rA
n
+ r + rB
r − rB
n
+ r + rC
r − rC
n
≥ 3 |AI| + |BI| + |CI|
3r
n
and it suffices to show that (|AI| + |BI| + |CI|)/(3r) ≥ 2. Now r
|AI| = sin(A/2) =
r(s − b)(s − c) bc and similar formulas hold for r/|BI| and r/|CI|. Hence
|AI| + |BI| + |CI|
3r ≥ |AI||BI||CI|
r3
1/3
=
abc
(s − a)(s − b)(s − c)
1/3
= 4R r
1/3
≥ 2.
because abc = 4KR, (s − a)(s − b)(s − c) = K2/s, K = sr, and R ≥ 2r (Euler’s inequality).
