, 2006

We will work with oriented angles between lines. For two straight lines $\ell, m$ in the plane, $\angle(\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\pi$ are identified, so that
$$\angle(\ell, m) = -\angle(m, \ell), \quad \angle(\ell, m) + \angle(m, n) = \angle(\ell, n).$$
If $\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\angle(KL, MN)$ for $\angle(\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.
The counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then
$$K, L, M, N \text{ are concyclic if and only if } \angle(KM, LM) = \angle(KN, LN). \tag{1}$$
Passing to the solution proper, we first show that the three circles $(AB_{1}C_{1})$, $(BC_{1}A_{1})$, $(CA_{1}B_{1})$ have a common point. So, let $(AB_{1}C_{1})$ and $(BC_{1}A_{1})$ intersect at the points $C_{1}$ and $P$. Then by (1)
$$\begin{aligned} & \angle(PA_{1}, CA_{1}) = \angle(PA_{1}, BA_{1}) = \angle(PC_{1}, BC_{1}) \\ = & \angle(PC_{1}, AC_{1}) = \angle(PB_{1}, AB_{1}) = \angle(PB_{1}, CB_{1}) \end{aligned}$$
Denote this angle by $\varphi$.
The equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.
From now on the basic property (1) will be used without explicit reference. We have
$$\varphi = \angle(PA_{1}, BC) = \angle(PB_{1}, CA) = \angle(PC_{1}, AB). \tag{2}$$


Let lines $A_{2}P$, $B_{2}P$, $C_{2}P$ meet the circle $(ABC)$ again at $A_{4}$, $B_{4}$, $C_{4}$, respectively. As
$$\angle(A_{4}A_{2}, AA_{2}) = \angle(PA_{2}, AA_{2}) = \angle(PC_{1}, AC_{1}) = \angle(PC_{1}, AB) = \varphi,$$
we see that line $A_{2}A$ is the image of line $A_{2}A_{4}$ under rotation about $A_{2}$ by the angle $\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2\varphi$ about $O$, the centre of $(ABC)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $ABC$ is the image of $A_{4}B_{4}C_{4}$ in this map. Thus
$$\angle(A_{4}B_{4}, AB) = \angle(B_{4}C_{4}, BC) = \angle(C_{4}A_{4}, CA) = 2\varphi. \tag{3}$$
Since the rotation by $2\varphi$ about $O$ takes $B_{4}$ to $B$, we have $\angle(AB_{4}, AB) = \varphi$. Hence by (2)
$$\angle(AB_{4}, PC_{1}) = \angle(AB_{4}, AB) + \angle(AB, PC_{1}) = \varphi + (-\varphi) = 0,$$
which means that $AB_{4} \parallel PC_{1}$.


Let $C_{5}$ be the intersection of lines $PC_{1}$ and $A_{4}B_{4}$; define $A_{5}, B_{5}$ analogously. So $AB_{4} \parallel C_{1}C_{5}$ and, by (3) and (2),
$$\angle(A_{4}B_{4}, PC_{1}) = \angle(A_{4}B_{4}, AB) + \angle(AB, PC_{1}) = 2\varphi + (-\varphi) = \varphi; \tag{4}$$
i.e., $\angle(B_{4}C_{5}, C_{5}C_{1}) = \varphi$. This combined with $\angle(C_{5}C_{1}, C_{1}A) = \angle(PC_{1}, AB) = \varphi$ (see (2)) proves that the quadrilateral $AB_{4}C_{5}C_{1}$ is an isosceles trapezoid with $AC_{1} = B_{4}C_{5}$.
Interchanging the roles of $A$ and $B$ we infer that also $BC_{1} = A_{4}C_{5}$. And since $AC_{1} + BC_{1} = AB = A_{4}B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4}B_{4}$ and partitions it into segments $A_{4}C_{5}$, $B_{4}C_{5}$ of lengths $BC_{1} (= AC_{3})$ and $AC_{1} (= BC_{3})$. In other words, the rotation which maps triangle $A_{4}B_{4}C_{4}$ onto $ABC$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3}B_{3}C_{3}$ and $A_{5}B_{5}C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2}B_{2}C_{2}$.
Lines $B_{4}C_{5}$ and $PC_{5}$ coincide respectively with $A_{4}B_{4}$ and $PC_{1}$. Thus by (4)
$$\angle(B_{4}C_{5}, PC_{5}) = \varphi.$$
Analogously (by cyclic shift) $\varphi = \angle(C_{4}A_{5}, PA_{5})$, which rewrites as
$$\varphi = \angle(B_{4}A_{5}, PA_{5}).$$
These relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore
$$\angle(A_{5}B_{5}, C_{5}B_{5}) = \angle(A_{5}B_{5}, PB_{5}) + \angle(PB_{5}, C_{5}B_{5}) = \angle(A_{5}C_{4}, PC_{4}) + \angle(PA_{4}, C_{5}A_{4}). \tag{5}$$
On the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(ABC)$, we have
$$\angle(A_{2}B_{2}, C_{2}B_{2}) = \angle(A_{2}B_{2}, B_{4}B_{2}) + \angle(B_{4}B_{2}, C_{2}B_{2}) = \angle(A_{2}A_{4}, B_{4}A_{4}) + \angle(B_{4}C_{4}, C_{2}C_{4}). \tag{6}$$
But the lines $A_{2}A_{4}, B_{4}A_{4}, B_{4}C_{4}, C_{2}C_{4}$ coincide respectively with $PA_{4}, C_{5}A_{4}, A_{5}C_{4}, PC_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\angle(A_{5}B_{5}, C_{5}B_{5}) = \angle(A_{2}B_{2}, C_{2}B_{2})$. Hence (by cyclic shift, once more) also $\angle(B_{5}C_{5}, A_{5}C_{5}) = \angle(B_{2}C_{2}, A_{2}C_{2})$ and $\angle(C_{5}A_{5}, B_{5}A_{5}) = \angle(C_{2}A_{2}, B_{2}A_{2})$. This means that the triangles $A_{5}B_{5}C_{5}$ and $A_{2}B_{2}C_{2}$ have their corresponding angles equal, and consequently they are similar.