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Let E and F be the points of intersection of N P with AB and AC, respectively

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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. 

Riferimenti

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