Saudi Arabian Mathematical Competitions

Let $K$ be a point on the ray $BN$ such that $\widehat{BMA} = \widehat{BCK}$. It is clear that $K$ is outside of triangle because $\widehat{BMA} > \widehat{ACB}$. We have $\triangle ABM \sim \triangle KBC$, hence
$$\frac{AB}{BK} = \frac{BM}{BC} = \frac{AM}{CK} . \tag{1}$$
Now, we have $\triangle ABK \sim \triangle MBC$ (since $\widehat{ABK} = \widehat{MBC}$ and $\frac{AB}{BK} = \frac{BM}{BC}$), hence
$$\frac{AB}{BM} = \frac{BK}{BC} = \frac{AK}{CM} . \tag{2}$$
We obtained $\widehat{CKN} = \widehat{MAB} = \widehat{NAC}$. It follows that quadrilateral $ANCK$ is cyclic, and by using Ptolemy's theorem we get
$$AC \cdot (BK - BN) = AN \cdot CK + CN \cdot AK . \tag{3}$$
From (1) and (2) it follows
$$CK = \frac{AM \cdot BC}{BM}, \quad AK = \frac{AB \cdot CM}{BM}, \quad BK = \frac{AB \cdot BC}{BM} .$$
Replacing these relations in (3) we get
$$AC \cdot \left( \frac{AB \cdot BC}{BM} - BN \right) = \frac{AM \cdot AN \cdot BC}{BM} + \frac{CM \cdot CN \cdot AB}{BM},$$
hence
$$$$
$\text{AM}$ AN}{AB $\cdot$ AC} + $\text{BM}$ BN}{BA $\cdot$ BC} + $\text{CM}$ CN}{CA $\cdot$ CB} = 1 .

Using Ceva's theorem in trigonometric form for points $M$ and $N$ we get
$$\frac{\sin \widehat{MAB}}{\sin \widehat{MAC}} \cdot \frac{\sin \widehat{MBC}}{\sin \widehat{MBA}} \cdot \frac{\sin \widehat{MCA}}{\sin \widehat{MCB}} = 1,$$
and
$$\frac{\sin \widehat{NAB}}{\sin \widehat{NAC}} \cdot \frac{\sin \widehat{NBC}}{\sin \widehat{NBA}} \cdot \frac{\sin \widehat{NCA}}{\sin \widehat{NCB}} = 1$$
Multiplying these relations, it follows
$$\frac{\sin \widehat{MCA}}{\sin \widehat{MBC}} \cdot \frac{\sin \widehat{NCA}}{\sin \widehat{NCB}} = 1,$$
hence
$$\frac{\sin \widehat{MCA}}{\sin \widehat{MCB}} = \frac{\sin \widehat{NCB}}{\sin \widehat{NCA}} . \tag{2}$$
Denote $\widehat{MCB} = \alpha$ and $\widehat{MCA} = \beta$ and get
$$\frac{\sin (C - \alpha)}{\sin \alpha} = \frac{\sin (C - \beta)}{\sin \beta} . \tag{3}$$
or
$$\sin C \cdot \cot \alpha - \cos C = \sin C \cdot \cot \beta - \cos C$$
hence $\alpha = \beta$.

Using complex coordinates, relation $\widehat{MCB} = \widehat{ACN}$ shows that
$$\arg \frac{m-c}{b-c} = \arg \frac{a-c}{n-c},$$
hence the number $\frac{m-c}{b-c} : \frac{a-c}{n-c}$ is real. We have
$$\frac{m-c}{b-c} \cdot \frac{n-c}{a-c} = \left| \frac{m-c}{b-c} \cdot \frac{n-c}{a-c} \right| = \frac{CM \cdot CN}{CB \cdot CA} .$$
Finally, we get
$$\sum \frac{AM \cdot AN}{AB \cdot AC} = \sum \frac{(m-a)(n-a)}{(b-a)(c-a)} \\ = \frac{\sum (m-a)(n-a)(b-c)}{(a-b)(b-c)(c-a)} = 1$$