XIV APMO

We are going to show that this can only happen when $$\angle CBP=\angle BCQ=15^{\circ }.$$

Lemma. If $\angle CBP>\angle BCQ$, then $RT>ST$.

Proof. Let $AD$, $BE$ and $CF$ be the altitudes of triangle $ABC$ concurrent at its centre $G$. Then $P$ lies on $CE$, $Q$ lies on $BF$, and thus $T$ lies in triangle $BDG$.



Note that - $$\angle FAS=\angle FCQ=30^{\circ }-\angle BCQ>30^{\circ }-\angle CBP=\angle EBP=\angle EAR.$$ Since $AF=AE$, we have $FS>ER$ so that $$GS=GF-FS<GE-ER=GR.$$ Let $T_x$ be the projection of $T$ onto $BC$ and $T_y$ be the projection of $T$ onto $AD$, and similarly for $R$ and $S$. We have $$R_x{T}_x=D{R}_x+D{T}_x>|D{S}_x-D{T}_x|=S_x{T}_x$$ and $$R_y{T}_y=G{R}_y+G{T}_y>G{S}_y+G{T}_y=S_y{T}_y.$$ It follows that $RT>ST$.

[1 mark for stating the Lemma, 3 marks for proving it.]

Thus, if $△TRS$ is equilateral, we must have $\angle CBP=\angle BCQ$.



It is clear from the symmetry of the figure that $TR=TS$, so $△TRS$ is equilateral if and only if $\angle RTA=30^{\circ }$. Now, as $BR$ is an altitude of the triangle $ABC$, $\angle RBA=30^{\circ }$. So $△TRS$ is equilateral if and only if $RTBA$ is a cyclic quadrilateral. Therefore, $△TRS$ is equilateral if and only if $\angle TBR=\angle TAR$. But $$\begin{array}{rl}90^{\circ }& =\angle TBA+\angle BAR\\ & =(\angle TBR+\angle RBA)+(\angle BAT+\angle TAR)\\ & =(\angle TBR+30^{\circ })+(30^{\circ }+\angle TAR)\end{array}$$ and so $$30^{\circ }=\angle TAR+\angle TBR.$$ But these angles must be equal, so $\angle TAR=\angle TBR=15^{\circ }$. Therefore $\angle CBP=\angle BCQ=15^{\circ }$.

[3 marks for finishing the proof with the assumption that $\angle CBP=\angle BCQ$.]