shortlistBMO 2011

Let $\gamma$ be the circle through the $A_i$, centred at $O$, and let tangents to $\gamma$ at $A_i$ and $A_{i+1}$ meet at $B_i^{\prime }$. Notice that the line $B_i^{\prime }{B}_{i+3}^{\prime }$ is the image of the circle $B_{i}O{B}_{i+3}$ under the inversion of pole $O$ and power $r^2$, where $r$ is the radius of $\gamma$. By Brianchon's theorem, the three lines $B_i^{\prime }{B}_{i+3}^{\prime }$ are concurrent at a point $Q$. Notice further that $Q$ is different from $O$, for no opposite sides of the hexagon $A_0{A}_1\cdots A_5$ are parallel. Consequently, the three circles $B_{i}O{B}_{i+3}$ share a second point $P$, different from $O$: the image of the point $Q$ under the inversion.

The lemma below shows that the points $P,C_i,C_j$ and $D_k$ are concyclic (not necessarily in this order), and the lines $PO$ and $P{D}_k$ are isogonal with respect to the lines $P{C}_i$ and $P{C}_j$, where $\{i,j,k\}=\{0,1,2\}$. Finally, a standard angle-chase argument shows that $P$ and the three points $D_i$ are concyclic: with reference to the figure below, write successively



$$\begin{array}{rl}\angle D_{i}P{D}_j& =\angle D_{i}P{C}_i+\angle C_{i}P{D}_j\\ & =\angle D_{i}P{C}_i+\angle C_{k}PO(\text{for }P{D}_j\text{ and }PO\text{ are isogonal relative to }P{C}_k\text{ and }P{C}_i)\\ & =\angle D_{i}P{C}_i+\angle C_{j}P{D}_i(\text{for }P{C}_k\text{ and }P{C}_j\text{ are isogonal relative to }P{D}_i\text{ and }PO)\\ & =\angle C_{i}P{C}_j\\ & =\angle C_i{D}_k{C}_j(\text{for }P,C_i,C_j,D_k\text{ are concyclic})\\ & =\angle D_j{D}_k{D}_i,\end{array}$$

to conclude that the points $P,D_0,D_1$ and $D_2$ are indeed concyclic.

Lemma. Two circles, ${\gamma }_1$ and ${\gamma }_2$, meet at the points $X$ and $Y$. A line through $Y$ meets again ${\gamma }_1$ at the point $Y_1$, and ${\gamma }_2$ at the point $Y_2$. The tangent to ${\gamma }_1$ at $Y_1$ meets the tangent to ${\gamma }_2$ at $Y_2$ at the point $Z$. Then the points $X,Z,Y_1$ and $Y_2$ are concyclic, and the lines $XY$ and $XZ$ are isogonal with respect to the lines $X{Y}_1$ and $X{Y}_2$.



Proof. If $X$ and $Z$ lie on opposite sides of the line through $Y$, then the angle $Y_{1}X{Y}_2$ is the sum of the angles $X{Y}_{1}Y$ and $X{Y}_{2}Y$ which are respectively equal to the angles $Y_{1}Z{Y}_2$ and $Y_{2}Z{Y}_1$, whose sum is supplementary to the angle $Y_{1}Z{Y}_2$. Consequently, the quadrangle $X{Y}_{1}Z{Y}_2$ is cyclic. It then follows that the angles $ZX{Y}_2$ and $Z{Y}_1{Y}_2$ are equal, and since the latter is equal to the angle $X{Y}_{1}Y$, we conclude that the lines $XY$ and $XZ$ are indeed isogonal with respect to the lines $X{Y}_1$ and $X{Y}_2$. If $X$ and $Z$ lie on the same side of the line through $Y$, then the angle $Y_{1}X{Y}_2$ is the difference of the angles $X{Y}_{1}Y$ and $X{Y}_{2}Y$ in some order, depending on which side of the line $XY$ the line $Y_1{Y}_2$ is situated. The later angles are respectively supplementary to the angles $Y_{1}ZY$ and $Y_{2}ZY$ whose difference in the corresponding order is equal to the angle $Y_{1}Z{Y}_2$. Consequently, the quadrangle $X{Y}_1{Y}_{2}Z$ or $X{Y}_2{Y}_{1}Z$ is cyclic. Isogonality is proved by adapting the argument in the former case.