SELECTION and TRAINING SESSION

(Solution by V. Vityaz, A. Gaponenko.) Let without loss of generality $CX\ge CY$. Then $AP\le AN,BM\le BQ$ (see the Fig.) By the Menelaus theorem, we have: for the triangle $AQM$ and the line $XC$



$$\frac{AY}{YQ}\cdot \frac{QC}{CM}\cdot \frac{MX}{XA}=1,(1)$$

for the triangle $AQC$ and the line $NB$ $$\frac{AY}{YQ}\cdot \frac{QB}{BC}\cdot \frac{CN}{NA}=1,(2)$$

for the triangle $AMC$ and the line $PB$ $$\frac{AX}{XM}\cdot \frac{MB}{BC}\cdot \frac{CP}{PA}=1.(3)$$ Let $AC=b$, $AB=c$, $BC=a$, $p$ be the semiperimeter of the triangle $ABC$. Then (2) and (3) give $AY/YQ=a/(p-b)$ and $MX/XA=0.5(p-c)/(p-a)$, respectively. Substituting these equalities in (1) we get $$\frac{a}{p-b}\cdot \frac{2(p-c)}{a}\cdot \frac{p-c}{2(p-a)}=1⟺(p-c{)}^2=(p-a)(p-b)⟺$$ $$a^2+b^2=c(a+b)⟺a(p-c)=b(p-b)⟺\frac{YA}{YQ}=\frac{CA}{CQ},$$ the last proportion is equivalent to the fact that $CY$ bisects the angle $ACQ$.