China Mathematical Competition (Extra Test)

(1) From the properties of an ellipse we know $$B_1{C}_0+C_0{B}_0=B_1{C}_1+C_1{B}_0.$$ Also, it is obvious that $$B_0{P}_0=B_0{Q}_0,C_1{B}_0+B_0{Q}_0=C_1{P}_1,$$ $$B_1{C}_1+C_1{P}_1=B_1{C}_0+C_0{Q}_1,C_0{Q}_1=C_0{B}_0+B_0{P}_0^{\prime }.$$ Adding these equations, we get $B_0{P}_0=B_0{P}_0^{\prime }$. Therefore $P_0^{\prime }$ and $P_0$ are coincident. Furthermore, as $P_0$, $C_0$ (the center of $\text{overarc}{Q}_1{P}_0$) and $B_0$ (the center of $\text{overarc}{P}_0{Q}_0$) are lying on the same line, we know that $\text{overarc}{Q}_1{P}_0$ and $\text{overarc}{P}_0{Q}_0$ are tangent at $P_0$.

(2) We have thus $\text{overarc}{Q}_1{P}_0$ and $\text{overarc}{P}_0{Q}_0$, $\text{overarc}{P}_0{Q}_0$ and $\text{overarc}{Q}_0{P}_1$, $\text{overarc}{Q}_0{P}_1$ and $\text{overarc}{P}_1{Q}_1$, $\text{overarc}{P}_1{Q}_1$ and $\text{overarc}{Q}_1{P}_0^{\prime }$ are tangent at points $P_0,Q_0,P_1,Q_1$ respectively. Now we draw common tangent lines $P_{0}T$ and $P_{1}T$ through $P_0$ and $P_1$ respectively, and suppose the two lines meet at point $T$. Also, we draw a common



tangent line $R_1{S}_1$ through $Q_1$, and suppose it intercepts $P_{0}T$ and $P_{1}T$ at point $R_1$ and $S_1$ respectively. Drawing segments $P_0{Q}_1$ and $P_1{Q}_1$, we get isosceles triangles $P_0{Q}_1{R}_1$ and $P_1{Q}_1{S}_1$ respectively. Then we have $$\begin{array}{rl}\angle P_0{Q}_1{P}_1& =\pi -\angle P_0{Q}_1{R}_1-\angle P_1{Q}_1{S}_1\\ & =\pi -(\angle P_1{P}_{0}T-\angle Q_1{P}_0{P}_1)\\ & -(\angle P_0{P}_{1}T-\angle Q_1{P}_1{P}_0).\end{array}$$ Since $$\pi -\angle P_0{Q}_1{P}_1=\angle Q_1{P}_0{P}_1+\angle Q_1{P}_1{P}_0,$$ we obtain $$\angle P_0{Q}_1{P}_1=\pi -\frac{1}{2}(\angle P_1{P}_{0}T+\angle P_0{P}_{1}T).$$ In the same way, we can prove that $$\angle P_0{Q}_0{P}_1=\pi -\frac{1}{2}(\angle P_1{P}_{0}T+\angle P_0{P}_{1}T).$$ It implies that points $P_0,Q_0,Q_1,P_1$ are concyclic.