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Geometry Difficulty 8.8 Shortlist Prove it Bulgaria

Problem:

Let ABCABC be a non-equilateral triangle and let MM and NN be interior points of it such that BAM = CAN\text{BAM = CAN}, ABM = CBN\text{ABM = CBN} and
AM⋅AN⋅BC=BM⋅BN⋅CA=CM⋅CN⋅AB=k AM \cdot AN \cdot BC = BM \cdot BN \cdot CA = CM \cdot CN \cdot AB = k
Prove that:

a) 3k=AB⋅BC⋅CA3k = AB \cdot BC \cdot CA;

b) the midpoint of the segment MNMN is the centroid of △ABC\triangle ABC.

Solution

Solution:

The angle equality implies that MM and NN are isogonal conjugate points in △ABC\triangle ABC. Therefore BCM = ACN\text{BCM = ACN}. Denote by small letters the affixes of the corresponding points in the complex plane. We have that
arg⁡b−am−a=arg⁡n−ac−a \arg \frac{b-a}{m-a} = \arg \frac{n-a}{c-a}
and
∣(m−a)(n−a)(b−c)∣=k |(m-a)(n-a)(b-c)| = k
Thus,
(b−a)(c−a)(m−a)(n−a)=AB⋅BC⋅CAk=:K \frac{(b-a)(c-a)}{(m-a)(n-a)} = \frac{AB \cdot BC \cdot CA}{k} =: K
Analogously (c−b)(a−b)=K(m−b)(n−b)(c-b)(a-b) = K(m-b)(n-b) and (a−c)(b−c)=K(m−c)(n−c)(a-c)(b-c) = K(m-c)(n-c). After subtraction we obtain
(b−c)(K(m+n)−(K−1)(b+c)−2a)=0 (b-c)(K(m+n) - (K-1)(b+c) - 2a) = 0
Hence
m+n=(K−1)(b+c)+2aK m+n = \frac{(K-1)(b+c) + 2a}{K}
and analogously m+n=(K−1)(c+a)+2bKm+n = \frac{(K-1)(c+a) + 2b}{K}. Therefore K=3K = 3 and m+n2=a+b+c3\frac{m+n}{2} = \frac{a+b+c}{3}, which completes the proof.

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