Bulgarian Mathematical Olympiad

Solution:

Let $ABCD$ be a cyclic quadrilateral. Then the Simson theorem for $\triangle ABC$ gives $DB_1 \perp AC$. Hence $\angle B_1C_1D = \angle B_1AD = \angle CBD$, $\angle B_1DC_1 = \angle B_1AC_1 = \angle CDB$ and therefore $\triangle B_1C_1D \sim \triangle CBD$.

Analogously $\triangle B_1A_1D \sim \triangle ABD$, whence
$$DA : DB : DC = \frac{1}{DA_1} : \frac{1}{DB_1} : \frac{1}{DC_1}$$
This together with the Ptolemy's theorem for $ABCD$ gives
$$\frac{BC}{DA_1} + \frac{BA}{DC_1} = \frac{AC}{DB_1}$$

Conversely, suppose that the identity (1) is true. Set $x = \frac{DB_1}{DA_1}$ and $y = \frac{DB_1}{DC_1}$. Squaring (1)

and applying the Cosine theorem for $\triangle ABC$, we see that the ratio $\frac{BA}{BC}$ is a root of the equation
$$\left(y^2 - 1\right)t^2 + 2(xy + \cos \angle ABC)t + x^2 - 1 = 0$$
Since the point $B_1$ lies on the segment $A_1C_1$, the inequality $DB_1 \geq DA_1$ implies that $DB_1 < DC_1$. Hence $x \geq 1$ and $0 < y < 1$, which shows that (2) has at most one positive root.

On the other hand, it is easy to see that $C_1$ and $A_1$ lie on the open rays $BA \to$ and $BC \to$, and the line through $B_1$ perpendicular to $DB_1$ intersects these two rays. Denote these intersection points by $A'$ and $C'$. Then the converse Simson theorem implies that the convex quadrilateral $A'BC'D$ is cyclic. Hence the identity (1) for $A'BC'D$ is satisfied, i.e. $\frac{BA'}{BC'}$ is a root of the equation (2).

Therefore $\frac{BA}{BC} = \frac{BA'}{BC'}$, i.e. $AC \parallel A'C'$. But the lines $AC$ and $A'C'$ have a common point $B_1$ and this shows that $A = A'$ and $C = C'$.