China Mathematical Competition (Complementary Test)

(1) As shown in Fig. 2, join $NI$, $MI$. Since $PC\parallel MN$ and $P,C,M,N$ are concyclic, $PCMN$ is an isosceles trapezoid. Therefore, $NP=MC$, $PM=NC$. Join $AM$, $CI$. Then $AM$ intercepts $CI$ at $I$. We have $$\begin{array}{rl}\angle MIC& =\angle MAC+\angle ACI\\ & =\angle MCB+\angle BCI\\ & =\angle MCI.\end{array}$$ Fig. 2

Therefore, $MC=MI$. In the same way, $NC=NI$. Then $NP=MI$, $PM=NI$. This means that $MPNI$ is a parallelogram. Therefore, $S_{△PMT}=S_{△PNT}$, for the two triangles having the same base and height. On the other hand, $\angle TNP+\angle PMT=180^{\circ }$, as $P,N,T,M$ are concyclic. Then we have $$\begin{array}{rl}{S}_{△PMT}& =\frac{1}{2}PM\times MT\sin \angle PMT\\ & =S_{△PNT}=\frac{1}{2}PN\times NT\sin \angle PNT\\ & =\frac{1}{2}PN\times NT\sin \angle PMT.\end{array}$$ Therefore, $MP\times MT=NP\times NT$.

(2) As shown in Fig. 3, we have $$\angle NC{I}_1=\angle NCA+\angle AC{I}_1=\angle NQC+\angle QC{I}_1=\angle C{I}_{1}N.$$ Therefore, $NC=N{I}_1$. In the same way, $MC=M{I}_2$. From $MP\times MT=NP\times NT$ we get $$\frac{NT}{MP}=\frac{MT}{NP}.$$ From (1) we know that $MP=NC$, $NP=MC$. Then Fig. 3 $$\frac{NT}{N{I}_1}=\frac{MT}{M{I}_2}.$$ Furthermore, $$\angle I_{1}NT=\angle QNT=\angle QMT=\angle I_{2}MT.$$ Therefore, $△I_{1}NT\sim △I_{2}MT$. Consequently, $\angle NT{I}_1=\angle MT{I}_2$. Then we have $$\angle I_{1}Q{I}_2=\angle NQM=\angle NTM=\angle I_{1}T{I}_2.$$ This means that $Q,I_1,I_2,T$ are concyclic.