65th Czech and Slovak Mathematical Olympiad

Let us denote by $E$ the second intersection of $AD$ with the circumcircle $k$. The power of $D$ with respect to $k$ gives $|DB|\cdot |DC|=|DA|\cdot |DE|$, which together with the given condition $|DA{|}^2=|DB|\cdot |DC|$ yields $|DA|=|DE|$. That is $E$ lies on the image $p$ of the line $BC$ in the homothety with center $A$ and a coefficient $2$ (Fig. 1).

Vice versa, to any intersection of a line $p$ with the circle $k$ we reconstruct the point $D$ on $BC$, which fulfills $|DA{|}^2=|DB|\cdot |DC|$. If the reconstruction has to be unique, the line $p$ has to touch $p$ in $E$.

Fig. 2 Fig. 3

Let us denote $S_b$ and $S_c$ in order the centers of $AC$ and $AB$. The homothety with the center $A$ and a coefficient $\frac{1}{2}$ sends $A,B,C,E$ (lying on the circle $k$) to $A,S_c,S_b,D$ which lie on the circle $k^{\prime }$ (Fig. 2), while the image of $p$ is the tangent $BC$ of $k^{\prime }$ in $D$. The powers of $B$ and $C$ with respect to $k^{\prime }$ give $|BD{|}^2=|BA|\cdot |B{S}_c|=\frac{1}{2}|BA{|}^2$ and $|CD{|}^2=|CA|\cdot |C{S}_b|=\frac{1}{2}|CA{|}^2$. All together for the perimeter of $ABC$:

$$|BC|+|AB|+|AC|=|BC|+\sqrt{2}(|BD|+|CD|)=|BC|+\sqrt{2}\cdot |BC|=1+\sqrt{2},$$

which is the only possible value.