Problem 11994
(American Mathematical Monthly, Vol.124, August-September 2017) Proposed by M. Ochoa Sanchez (Peru) and L. Giugiuc (Romania).
Let ABC be a triangle with incenter I and circumcircle C. Let M , N , and P be the second points of intersection ofΓ with lines AI, BI, and CI, respectively. Let E and F be the points of intersection of N P with AB and AC, respectively. Similarly, let G and H be the points of intersection of M N with AC and BC, respectively, and let J and K be the points of intersection of M P with BC and AB, respectively. Prove
|EF | + |GH| + |JK| ≤ |KE| + |F G| + |HJ|.
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 the circumcircle Γ is the unit circle |z| = 1 of the complex plane. We will use the following know properties.
• There exist u, v, w ∈ C satisfying
A= u2, B= v2, C= w2, M = −vw, N = −wu, P = −uv.
(M , N and P are the midpoints of the arcs AB, BC, CA, which don’t contain C, A, B respectively.)
• If w1, w2, w3 and w4 lie on the unit circle then the intersection of the segments w1w2 and w3w4 is given by
f(w1, w2, w3, w4) = w1w2(w3+ w4) − w3w4(w1+ w2) w1w2− w3w4
.
We have that a := |BC| = |v2−w2|, b := |CA| = |w2−u2|, c := |AB| = |u2−v2|, m := |P N | = |v−w|, n:= |M P | = |w − u|, p := |M N | = |u − v|. It can be easily verified that
|EF | =
−(v + w)(w + u)(u + v) (v − w)
= abc
m2np and |F G| =
v(u2− w2)(u + w) (v − w)(u − v)
= b2
mnp where we used the above formula for E = f (A, B, P, N ), F = f (A, C, P, N ), and G = f (A, C, N, M ).
Similarly we obtain the other lengths, and then it follows that the desired inequality is equivalent to 1
m+1 n+1
p ≤ a bc+ b
ca + c ab.
Now, since I is the orthocenter of △M P N , we have that cM =B+ bb2C, and, by the Law of sines,
m= 2 sin Bb+ bC 2
!
, b= 2 sin( bB), c= 2 sin( bC) =⇒ m ≥b+ c 2 ≥ 2bc
b+ c. because the sine function is concave in [0, π]. In a similar way we have that n ≥c+a2ca, p ≥ a+b2ab. Hence
1 m+ 1
n+1
p≤ b+ c
2bc +c+ a
2ca +a+ b 2ab = 1
a+1 b +1
c ≤ a bc+ b
ca + c ab
where in the last step we used a2+ b2+ c2≥ ab + bc + ca.