Eighteenth STARS OF MATHEMATICS Competition

Since the angles $\angle OXP$ and $\angle OYQ$ are both right, the circles $OXP$ and $OYQ$ cross again at a point $R$ on $PQ$, and $OR$ is perpendicular to $PQ$. Letting $A_1{C}_1$ and $A_2{B}_2$ cross at $S$, the conclusion then follows at once from the two facts below: (1) $S$ lies on $OR$; and (2) $OS$ is perpendicular to $A_1{A}_2$.

To prove (1), invert from $O$ with power $-O{B}_1\cdot O{B}_2$. Under this inversion, ${\gamma }_1$ and ${\gamma }_2$ correspond to one another, and so do the points in each of the pairs $(A_1,Y)$, $(B_1,B_2)$, $(C_1,C_2)$ and $(X,A_2)$.

Clearly, $B_1$ lies on the circle $OXP$, so this latter is mapped to $A_2{B}_2$. Similarly, the circle $OYQ$ (through $C_2$) is mapped to $A_1{C}_1$. Consequently, $R$ is mapped to $S$. This establishes (1).

To prove (2), let $A_1{A}_2$ cross $B_1{B}_2$ and $C_1{C}_2$ at $D_1$ and $D_2$, respectively; let $O_1{O}_2$ cross $A_1{C}_1$ and $A_2{B}_2$ at $X_1$ and $X_2$, respectively; and let $A_1{C}_1$ cross $O_1{D}_1$ at $Y_1$, and $A_2{B}_2$ cross $O_2{D}_2$ at $Y_2$. The argument hinges on the following two facts: (3) $X_1$ and $Y_1$ both lie on the circle ${\omega }_1$ on diameter $O{D}_1$; similarly, $X_2$ and $Y_2$ both lie on the circle ${\omega }_2$ on diameter $O{D}_2$; and (4) $X_1,X_2,Y_1,Y_2$ all lie on a circle $\omega$. Assume (3) and (4) to establish (2) as follows: $X_1{Y}_1$, i.e., $A_1{C}_1$, is the radical axis of $\omega$ and ${\omega }_1$, and $X_2{Y}_2$, i.e., $A_2{B}_2$, is the radical axis of $\omega$ and ${\omega }_2$. Hence $S$ is the radical centre of the three circles. As such, $S$ lies on the radical axis of ${\omega }_1$ and ${\omega }_2$. These two circles cross again at the orthogonal projection $O^{\prime }$ of $O$ on $D_1{D}_2$, i.e., on $A_1{A}_2$, and $O{O}^{\prime }$ is their radical axis. Consequently, $S$ lies on $O{O}^{\prime }$ and (2) follows.

We now turn to prove (3) and (4). To prove (3) only $X_1$ and $Y_1$ are dealt with. It is sufficient to show that the angles $\angle O_1{X}_1{D}_1$ and $\angle O{Y}_1{O}_1$ are both right. To prove that the angle $\angle O_1{X}_1{D}_1$ is right, we show that $X_1$ lies on the circle on diameter $O_1{D}_1$. Clearly, $A_1$ and $B_1$ both lie on this circle, so it is sufficient to show that $\angle O_1{A}_1{X}_1=\angle O_1{B}_1{X}_1$. Now, $\angle O_1{A}_1{X}_1=\angle O_1{A}_1{C}_1=\angle O_1{C}_1{A}_1=\angle O_1{C}_1{X}_1$, since $O_1{A}_1=O_1{C}_1$ (they both are radii of ${\gamma }_1$). On the other hand, $C_1$ and $B_1$ are reflexions of one another in $O{O}_1$, so $\angle O_1{C}_1{X}_1=\angle O_1{B}_1{X}_1$. Consequently, $\angle O_1{A}_1{X}_1=\angle O_1{B}_1{X}_1$, as desired. The proof that the angle $\angle O{Y}_1{O}_1$ is right is quite similar: This time we show that $Y_1$ lies on the circle on diameter $O{O}_1$. Clearly, $B_1$ and $C_1$ both lie on this circle, so it is sufficient to show that $\angle O_1{B}_1{Y}_1=\angle O_1{C}_1{Y}_1$. Notice that $B_1$ and $A_1$ are reflexions of one another in $O_1{D}_1$, to write $\angle O_1{B}_1{Y}_1=\angle O_1{A}_1{Y}_1$. Refer now again to $O_1{A}_1=O_1{C}_1$, to write $\angle O_1{A}_1{Y}_1=\angle O_1{A}_1{C}_1=\angle O_1{C}_1{A}_1=\angle O_1{C}_1{Y}_1$ and conclude that $O_1{B}_1{Y}_1=\angle O_1{C}_1{Y}_1$. This establishes (3).

Finally, proving (4) requires more work. We will show that $\angle X_1{Y}_1{Y}_2+\angle X_1{X}_2{Y}_2=180^{\circ }$. Clearly, $\angle X_1{Y}_1{Y}_2=180^{\circ }-(\angle O_1{Y}_1{X}_1+\angle D_1{Y}_1{Y}_2)$, so it is sufficient to prove that $$\angle X_1{X}_2{Y}_2=\angle O_1{Y}_1{X}_1+\angle D_1{Y}_1{Y}_2.(\ast )$$ Deal with each angle in the right-hand member separately. By (3), $\angle O_1{Y}_1{X}_1=\angle O_{1}O{D}_1$, and an obvious angle chase yields $\angle O_{1}O{D}_1=\angle O_{1}O{B}_1=\angle O_{2}O{B}_2=\angle O_{2}O{C}_2=\angle O_{2}O{D}_2$. Hence $\angle O_1{Y}_1{X}_1=\angle O_{2}O{D}_2$. To deal with $\angle D_1{Y}_1{Y}_2$, let $O_1{D}_1$ and $O_2{D}_2$ cross at $Z$. By (3), the angles $\angle O{Y}_1{D}_1$ and $\angle O{Y}_2{D}_2$ are both right, so $O{Y}_{1}Z{Y}_2$ is cyclic, and hence $\angle D_1{Y}_1{Y}_2=\angle Z{Y}_1{Y}_2=\angle ZO{Y}_2$. Thus, () now reads $$\angle X_1{X}_2{Y}_2=\angle O_{2}O{D}_2+\angle ZO{Y}_2.(\ast \ast )$$ Our purpose now is to replace $\angle ZO{Y}_2$ by a more tractable angle. To this end, notice that $O_1{D}_1$ is the internal angle bisectrix of $\angle A_1{D}_1{B}_1$, and $O_2{D}_1$ is the internal angle bisectrix of $\angle A_2{D}_1{B}_2$, so $O_1{D}_1$ and $O_2{D}_1$ are perpendicular; that is, $O_2{D}_1$ is the $O_2$-altitude of the triangle $Z{O}_1{O}_2$. Similarly, $O_1{D}_2$ is the $O_1$-altitude of the triangle $Z{O}_1{O}_2$, so it crosses $O_2{D}_1$ at the orthocentre $H$ of this triangle. On the other hand, $O_2{D}_1$ and $O_1{D}_2$ are clearly the internal angle bisectrices of the triangle $O{D}_1{D}_2$ at $D_1$ and $D_2$, respectively. Hence $OH$ is the internal angle bisectrix of $\angle D_{1}O{D}_2$. As such, it is perpendicular to external bisectrix $O_1{O}_2$ of this angle. Recalling that $H$ is the orthocentre of the triangle $Z{O}_1{O}_2$, it follows that $OH$ is the $Z$-altitude of this triangle. Hence $\angle ZO{Y}_2=\angle HO{Y}_2$, and (*) now reads $$\angle X_1{X}_2{Y}_2=\angle O_{2}O{D}_2+\angle HO{Y}_2.$$ To prove this, split $\angle X_1{X}_2{Y}_2=\angle X_1{X}_2{D}_2+\angle D_2{X}_2{Y}_2$. The angle $\angle X_1{X}_2{D}_2=\angle O{X}_2{D}_2$ is right, by (3). As such, $\angle X_1{X}_2{D}_2=\angle HO{X}_2$, since $OH$ is perpendicular to $O_1{O}_2$. Split further $\angle HO{X}_2=\angle HO{Y}_2+\angle Y_{2}O{X}_2$, to write $\angle X_1{X}_2{D}_2=\angle HO{Y}_2+\angle Y_{2}O{X}_2$. By (3), $\angle D_2{X}_2{Y}_2=\angle D_{2}O{Y}_2$, so $\angle X_1{X}_2{Y}_2=\angle HO{Y}_2+\angle Y_{2}O{X}_2+\angle D_{2}O{Y}_2=\angle HO{Y}_2+\angle D_{2}O{X}_2$, as desired. This establishes (4) and completes the proof.