United States — Team Selection Test, 2010

Solution:

Solution 1. Extend $BP$ through $P$ to $P_1$ so that $\angle PAP_1 = \angle PCD = C - z$. By (1), $\angle DPC = 180^{\circ} - \angle APB = \angle APP_1$. Hence triangles $PAP_1$ and $PCD$ are similar. In particular, there is a spiral similarity $S_1$ centered at $P$ sending $PAP_1$ to $PCD$. Thus, there is another spiral similarity $S_2$ centered at $P$ sending $PAC$ to $PP_1D$, from which it follows that $\angle PP_1D = \angle PAC = A - x = \angle BAQ$. Combining the last equation with $\angle ABQ = B - y = \angle PBQ = \angle P_1BQ$, we deduce that triangle $QAB$ is similar to triangle $DP_1B$; that is, there is a spiral similarity $S_3$ centered at $B$ sending $BAQ$
to $BP_1D$. Thus, there is another spiral similarity $S_4$ centered at $B$ sending $BAP_1$ to $BQD$. It therefore follows that
$$\angle BQD = \angle BAP_1 = \angle BAP + \angle PAP_1 = \angle QAC + \angle PCD = \angle QAC + \angle QCA = 180^\circ - \angle AQC,$$
as desired.

Solution 2. Because $\angle BQC > \angle BAC > \angle PAC$, there is a point $Q_1$ on segment $BC$ such that $\angle Q_1QC = \angle PAC = A-x$. Because $\angle Q_1QC = \angle PAC$ and $\angle Q_1CQ = \angle PCA = z$, we conclude that triangle $Q_1QC$ is similar to triangle $PAC$. In particular, there is a spiral similarity $T_1$ centered at $C$ sending $Q_1QC$ to $PAC$. It follows that there is another similarity $T_2$ centered at $C$ sending $Q_1PC$ to $QAC$, implying that $\angle Q_1PC = \angle QAC = x$. By (1), we have $\angle DPQ_1 = \angle CPD - \angle CPQ_1 = B-y = \angle QBC = \angle QBQ_1$.
Extend $PD$ to a point $Q_2$ such that $\angle Q_2BQ_1 = \angle QBQ_1$. We have that $\angle DPQ_1 = \angle QBQ_1 = \angle QBQ_1$, implying that $BPQ_1Q_2$ is cyclic, from which it follows that $\angle BQ_1Q_2 = \angle BPQ_2$. By (1) and because $\angle APC = \angle QQ_1C$ (from the similar triangles $APC$ and $QQ_1C$), we have
$$\angle BQ_1Q_2 = \angle BPQ_2 = 180^\circ - \angle APC = 180^\circ - \angle QQ_1C = \angle BQQ_1.$$
Combining the last equality with the fact that $\angle QBQ_1 = \angle Q_2BQ_1$, we see that $BQQ_1Q_2$ is a kite with line $BQ_1$ as its symmetry axis. In particular, by symmetry, cyclic quadrilateral $BPQ_1Q_2$, and similar triangles $AQC$ and $PQ_1C$, we obtain
$$\angle BQD = \angle BQ_2D = \angle BQ_2P = \angle BQ_1P = 180^\circ - \angle CQ_1P = 180^\circ - AQC,$$
completing the proof.

Solution 3. There is a unique point $D_1$ on side $BC$ such that $\angle CPD_1 + \angle APB = 180^\circ$ and a unique point $D_2$ on side $BC$ such that $\angle BQD_2 + \angle AQC = 180^\circ$. It suffices to show that $D_1 = D_2$ or
$$\frac{BD_1}{D_1C} = \frac{BD_2}{D_2C} \quad \text{or} \quad \frac{BD_1}{D_1C} \cdot \frac{CD_2}{D_2B} = 1. \qquad (2)$$
Applying the law of sines to triangles $BPD_1$, $CPD_1$, and $BCP$ gives
$$\frac{BD_1}{D_1C} = \frac{BP \sin \angle BPD_1}{CP \sin CPD_1} = \frac{\sin(C-z) \sin \angle BPD_1}{\sin y \sin \angle CPD_1} = \frac{\sin(C-z)}{\sin y} \cdot \frac{\sin \angle BPD_1}{\sin \angle CPD_1}.$$
By the definition of $D_1$, (1) becomes
$$\angle APB + \angle CPD = 180^{\circ} \quad \text{if and only if} \quad \angle APC + \angle BPD = 180^{\circ}.$$
In particular, $\sin \angle BPD_1 = \sin \angle APC$ and $\sin \angle CPD_1 = \sin \angle APB$. Applying the law of sines in triangles $APB$ and $APC$, we have
$$\begin{aligned} \frac{BD_1}{D_1C} &= \frac{\sin(C - z)}{\sin y} \cdot \frac{\sin \angle BPD_1}{\sin \angle CPD_1} \\ &= \frac{\sin(C - z)}{\sin y} \cdot \frac{\sin \angle APC}{\sin \angle APB} \\ &= \frac{BP}{CP} \cdot \frac{\sin \angle APC \cdot AP}{\sin \angle APB \cdot AP} \\ &= \frac{\sin(C - z)}{\sin y} \cdot \frac{\sin \angle ACP \cdot AC}{\sin \angle ABP \cdot AB}, \end{aligned}$$
which we may rewrite as
$$\frac{BD_1}{D_1C} = \frac{\sin(C - z)}{\sin y} \cdot \frac{\sin z}{\sin(B - y)} \cdot \frac{AC}{AB}.$$
In exactly the same way, we can show that
$$\frac{CD_2}{D_2B} = \frac{\sin(B - y)}{\sin z} \cdot \frac{\sin(C - z)}{\sin y} \cdot \frac{AB}{AC}.$$
Multiplying the last two equations together yields (2), completing the proof.