18th Turkish Mathematical Olympiad

Since $PS\parallel AC$ and $\angle CAP=\angle BCP$ we have $\angle QPC=\angle ACB=\angle AQB=\angle PQB$. Therefore $BQ\parallel PC$.

We will first show that $SQ=R{B}^{\prime }$ if and only if $AB\parallel B^{\prime }Q$. Since $PS$ and $A{B}^{\prime }$ are parallel, we have $\frac{PQ}{PA}=\frac{SQ}{S{B}^{\prime }}$. If $SQ=R{B}^{\prime }$, then $\frac{PQ}{PA}=\frac{R{B}^{\prime }}{RQ}=\frac{P{B}^{\prime }}{PB}$ as $BQ\parallel PC$, and hence $AB\parallel B^{\prime }Q$. On the other hand, if $AB\parallel B^{\prime }Q$, then using $BQ\parallel PC$ we obtain $\frac{R{B}^{\prime }}{RQ}=\frac{P{B}^{\prime }}{PB}=\frac{PQ}{PA}=\frac{SQ}{S{B}^{\prime }}$, and hence $SQ=R{B}^{\prime }$.

Now we will show that $AB\parallel B^{\prime }Q$ if and only if $\angle BAT=\angle B{B}^{\prime }Q$. Let $AP\cap B^{\prime }{C}^{\prime }=\{U\}$ and let $D$ be the second point of intersection of the line $AP$ and the circumcircle of the triangle $C^{\prime }P{B}^{\prime }$. Using $P{C}^{\prime }\parallel QT$ we obtain $\frac{UD}{U{B}^{\prime }}=\frac{U{C}^{\prime }}{UP}=\frac{UT}{UQ}$

and conclude that the triangles $TD{C}^{\prime }$ and $Q{B}^{\prime }P$ are similar. In particular, $\angle P{B}^{\prime }Q=\angle C^{\prime }DT$ and $\angle DT{C}^{\prime }=\angle B^{\prime }QP$.

We will show that $D$ cannot coincide with $A$. Suppose it does. Then $A,C^{\prime },P,B^{\prime }$ are concyclic and hence $C^{\prime }{B}^{\prime }$ and $BC$ are parallel. Let $A^{\prime }$ be the point of intersection of $AP$ and $BC$. It follows that the triangles $PB{A}^{\prime }$ and $BA{A}^{\prime }$ are similar, and therefore $A^{\prime }{B}^2=A^{\prime }P\cdot A^{\prime }A$. Similarly we obtain $A^{\prime }{C}^2=A^{\prime }P\cdot A^{\prime }A$ and conclude that $A^{\prime }B=A^{\prime }C$, which is a contradiction as $P$ is not on the median belonging to $A$.

Let us assume that $D$ is between $A$ and $U$. A similar argument works if $A$ is between $D$ and $U$. If $\angle BAT=\angle B{B}^{\prime }Q$, then $\angle C^{\prime }AT=\angle BAT=\angle B{B}^{\prime }Q=\angle P{B}^{\prime }Q=\angle C^{\prime }DT$, and $T,A,D,C^{\prime }$ are concyclic. Therefore $\angle PAB=\angle DA{C}^{\prime }=\angle DT{C}^{\prime }=\angle B^{\prime }QP$ and hence $AB\parallel B^{\prime }Q$. On the other hand if $AB\parallel B^{\prime }Q$, then $\angle DT{C}^{\prime }=\angle B^{\prime }QP=\angle DA{C}^{\prime }$, and $T,A,D,C^{\prime }$ are concyclic. Therefore $\angle BAT=\angle C^{\prime }AT=\angle C^{\prime }DT=\angle P{B}^{\prime }Q=\angle B{B}^{\prime }Q$.