Problem 12086
(American Mathematical Monthly, Vol.126, January 2019) Proposed by M. O. Sanchez (Peru) and L. Giugiuc (Romania).
LetABC be a triangle with right angle at A, and let H be the foot of the altitude from A. Let M , N , and P be the incenters of triangles ABH, ABC, and ACH, respectively. Prove that the ratio of the area of triangleM N P to the area of triangle ABC is at most (√
2 − 1)3/2, and determine when equality holds.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Without loss of generality, we may assume that B = (−1, 0), C = (1, 0) and A = (cos(θ), sin(θ)) with θ ∈ [0, π]. Then H = (cos(θ), 0) and
M = (cos(θ) − rB, rB), N = (cos(θ) − (sin(θ) − r) tan π 4 −θ
2
, r), P = (cos(θ) + rC, rC) where rB, r and rC are the inradii of the triangles ABH, ABC, and ACH respectively,
rB= sin(θ) cos2(θ2)
cos(θ2) + cos2(θ2) +12sin(θ), r = sin(θ)
sin(θ2) + cos(θ2) + 1, rC= sin(θ) sin2(θ2) sin(θ2) + sin2(θ2) +12sin(θ). Hence
Area(M N P )
Area(ABC) = |MN × P N|
2 sin(θ) = (1 − sin(θ2))(1 − cos(θ2))
sin(θ2) + cos(θ2) + 1 = f (t) ≤ f(√
2 + 1) = (√ 2 − 1)3
2 where t = sin(θ2) + cos(θ2) + 1 =√
2 sin π4+θ2 + 1 ∈ [2,√
2 + 1] and
f (t) := (t − 2)2 2t = 1
2
t +4
t
− 2
is strictly increasing in [2,√
2 + 1]. It follows that equality holds if and only if t =√
2 + 1, i.e. θ = π2
or ABC is an isosceles right triangle.
