Problem 1245

a) Let the vertices of the quadrilateral $Q$ be denoted sequentially by $A$, $B$, $C$, $D$, and the centers of the circumcircles of triangles $BCD$, $CDA$, $DAB$, $ABC$ by $A'$, $B'$, $C'$, $D'$, respectively. The quadrilateral $Q$ is not cyclic by assumption; hence $|\measuredangle A| + |\measuredangle B| \neq |\measuredangle C| + |\measuredangle D|$. Without loss of generality, we can assume that

$$|\measuredangle A| + |\measuredangle B| < |\measuredangle C| + |\measuredangle D|$$

(if the opposite inequalities hold, we can perform a cyclic permutation of the labels).
From inequality (1), it follows that point $D$ lies inside the circumcircle of triangle $ABC$, while point $C$ lies outside the circumcircle of triangle $DAB$ (see Figure 12). Extend segment $BD$ to intersect the circle $\omega$ at a point denoted by $D_1$. Let the intersection of segment $AC$ with the circle $\sigma$ be denoted by $C_1$. Let $\lambda$ be the perpendicular bisector of segment $BD$, and $\chi$ be the perpendicular bisector of segment $AC$.
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The center of circle $\omega$, i.e., point $D'$, lies on the perpendicular bisector of segment $BD_1$, and thus on the same side of line $\lambda$ as point $D$.
The center of circle $\sigma$, i.e., point $C'$, lies on the perpendicular bisector of segment $AC_1$, and thus on the opposite side of line $\chi$ from point $C$.
By symmetry of the assumptions, we can interchange the roles of points $A$ and $C$ as well as $B$ and $D$ and conclude that:
- Point $B'$ lies on the same side of line $\lambda$ as point $B$;
- Point $A'$ lies on the opposite side of line $\chi$ from point $A$.
The center $A'$ of the circumcircle of triangle $BCD$ lies on the perpendicular bisector of side $BD$: $A'$; similarly for $C'$, $B'$, $D'$. Therefore, points $B'$ and $D'$ lie on opposite sides of line $A'$, and points $A'$ and $C'$ lie on opposite sides of line $B'$. This implies that points $A'$, $B'$, $C'$, $D'$ are the consecutive vertices of a convex quadrilateral. This is precisely the quadrilateral $\tau(Q)$.
b) We will show that the measures of the interior angles of quadrilaterals $\tau(Q)$ and $Q$ are related by the equations

$$\begin{aligned} |\measuredangle A'| &= 180^\circ - |\measuredangle A|, \\ |\measuredangle B'| &= 180^\circ - |\measuredangle B|, \\ |\measuredangle C'| &= 180^\circ - |\measuredangle C|, \\ |\measuredangle D'| &= 180^\circ - |\measuredangle D|. \end{aligned}$$

We will prove only the last of these equations, and from this, we can infer the other three (2) by abandoning the assumption that the angles of quadrilateral $Q$ satisfy condition (1).
Consider triangle $ABC$ and point $D'$, the center of the circumcircle. Let the midpoints of sides $AB$ and $BC$ be denoted by $M$ and $N$, respectively. As before, $\chi$ denotes the perpendicular bisector of segment $AC$.
We consider three cases:
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1. Angles $\measuredangle CAB$ and $\measuredangle BCA$ are acute (see Figure 13). The projection of point $B$ onto line $AC$ falls within the interior of segment $AC$, and thus the projections of points $M$ and $N$ fall on different halves of this segment. This means that points $M$ and $N$ lie on opposite sides of line $\chi$, and points $D'$, $M$, $B$, $N$ are the consecutive vertices of a convex quadrilateral with right angles at vertices $M$ and $N$. Therefore, the following equality holds:

$$|\measuredangle MD'B| = 90^\circ - |\measuredangle BCA|.$$

Thus,

$$|\measuredangle MD'B| = 90^\circ - |\measuredangle BCA|.$$

Point $A'$ lies on the perpendicular bisector of side $BC$, i.e., on line $ND'$; point $C'$ lies on the perpendicular bisector of side $AB$, i.e., on line $MD'$.
According to the reasoning in part a), line $\chi$ separates points $A'$ and $C'$, or they lie on the half-lines $D'$ and $D'$, or on their extensions. In any case, the angles $\measuredangle CD'$ and $\measuredangle MD'$ are either the same or vertical angles, and thus their measures are equal. Therefore, from (3),

$$|\measuredangle CD'| = 180^\circ - |\measuredangle CBA|,$$

which is what we needed to prove.
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2. One of the angles $\measuredangle CAB$ and $\measuredangle BCA$ is a right angle; let, for example, $|\measuredangle BCA| = 90^\circ$ (see Figure 14). Point $D'$ then coincides with $M$. As in the previous case, points $A'$ and $C'$ must lie on the perpendicular bisector of side $BC$ (i.e., on line $ND'$) and on the perpendicular bisector of side $AB$; and they lie on opposite sides of line $\chi$, which in this case is equivalent to stating that they lie on opposite sides of line $AB$. Therefore, angle $\measuredangle C'$ is obtuse, with perpendicular sides to the sides of angle $\measuredangle ABC$; thus, the equality (4) holds.
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3. One of the angles $\measuredangle CAB$ and $\measuredangle BCA$ is obtuse; say, $|\measuredangle BCA| > 90^\circ$ (see Figure 15). Each of points $M$ and $N$ is then closer to point $C$ than to point $A$, which means that points $M$ and $N$ lie on the same side of $\chi$. Angle $\measuredangle MD'$ is then acute, and its sides are perpendicular to the sides of angle $\measuredangle ABC$. Therefore,

$$|\measuredangle MD'B| = 90^\circ - |\measuredangle BCA|.$$

As in case 1, points $A'$ and $C'$ lie on lines $ND'$ and $MD'$, respectively, and are separated by line $\chi$. This time, it means that they cannot both lie on the half-lines $D'$ and $D'$, nor can they both lie on their extensions. Therefore, angle $\measuredangle C'$ is the complement of angle $\measuredangle MD'$ to a straight angle, and from (5) we obtain the desired equality (4).
All equalities (2) can be considered proven. It immediately follows from them that

$$|\measuredangle A'| + |\measuredangle B'| \neq |\measuredangle C'| + |\measuredangle D'|$$

(by the analogous inequality for quadrilateral $Q$), and thus quadrilateral $\tau(Q)$ is not cyclic.
c) According to the conclusion of part b), quadrilateral $\tau(\tau(Q))$ is well-defined. Let its vertices be denoted by $A''$, $B''$, $C''$, $D''$ ($A''$ is the center of the circumcircle of triangle $B''C''D''$ and so on). The sides and diagonals of quadrilaterals $Q$ and $\tau(Q)$ are related by the equations:

$$\begin{aligned} A''B'' &= \frac{1}{2} AC, \\ B''C'' &= \frac{1}{2} BD, \\ C''D'' &= \frac{1}{2} CA, \\ D''A'' &= \frac{1}{2} DB, \\ A''C'' &= \frac{1}{2} AD, \\ B''D'' &= \frac{1}{2} BC. \end{aligned}$$

These relations arise from the positions of the centers of the circumcircles of triangles $BCD$, $CDA$, $DAB$, $ABC$ on the perpendicular bisectors of the corresponding segments.
Analogous relations hold between the sides and diagonals of quadrilaterals $\tau(Q)$ and $\tau(\tau(Q))$. In other words, if we replace points $A$, $B$, $C$, $D$ with $A'$, $B'$, $C'$, $D'$, respectively, the relations (6) remain valid. Therefore,

$$\begin{aligned} A''B'' &= \frac{1}{2} A'C', \\ B''C'' &= \frac{1}{2} B'D', \\ C''D'' &= \frac{1}{2} C'A', \\ D''A'' &= \frac{1}{2} D'B', \\ A''C'' &= \frac{1}{2} A'D', \\ B''D'' &= \frac{1}{2} B'C'. \end{aligned}$$

We infer from this that triangle $A''B''C''$ is similar to triangle $ABC$, and triangle $C''D''A''$ is similar to triangle $CDA$. Consequently, quadrilateral $\tau(\tau(Q))$ is similar to quadrilateral $Q$.
Note 1. It is not difficult to see that the main difficulty of the problem lies in part b), in issues concerning the orientation of certain angles. The reasoning presented in part c) is almost independent of the preceding text; it only relies on the fact that the symbol $\tau(\tau(Q))$ makes sense.
The essence of part b) is the relations (2). In each of them, there is a pair of angles with perpendicular sides (which immediately follows from the fact that the centers of the considered circumcircles lie on the perpendicular bisectors of the corresponding segments). The measures of such angles either add up to $180^\circ$ or are equal; the goal is to exclude the latter possibility. This is the mentioned issue of orientation; for this, we needed the considerations regarding the positioning of the vertices of $\tau(Q)$ relative to lines $\chi$ and $\lambda$.