5. For triangle $ABC$, it is given that $|BC|=1$ and there exists exactly one point $D$ on side $BC$ such that $|DA|^2 = |DB| \cdot |DC|$. Determine all possible values of the perimeter of triangle $ABC$.

(Patrik Bak)
Solution. Let $E$ be the second intersection of line $A D$ with the circumcircle $k$ of triangle $A B C$. From the power of point $D$ with respect to circle $k$, it follows that $|D B| \cdot|D C|=|D A| \cdot|D E|$. Comparing this with the given condition $|D A|^{2}=|D B| \cdot|D C|$ yields $|D A|=|D E|$. Therefore, point $E$ lies on the image of line $B C$ under a homothety centered at $A$ with a coefficient of 2 (Fig. 3).

Conversely, for any intersection of line $p$ with circle $k$, we can reconstruct point $D$ on side $B C$ which will clearly satisfy the equality $|D A|^{2}=|D B| \cdot|D C|$. For such a point to be uniquely determined, line $p$ must be tangent to circle $k$ at point $E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ea51740f353cab5f511fg-7.jpg?height=517&width=574&top_left_y=1032&top_left_x=387)

Fig. 3

![](https://cdn.mathpix.com/cropped/2024_04_17_ea51740f353cab5f511fg-7.jpg?height=517&width=576&top_left_y=1032&top_left_x=1094)

Fig. 4

Let $S_{b}$ and $S_{c}$ be the midpoints of segments $A C$ and $A B$, respectively. Under a homothety centered at $A$ with a coefficient of $\frac{1}{2}$, the points $A, B, C, E$ lying on circle $k$ are mapped to points $A, S_{c}, S_{b}, D$ lying on circle $k^{\prime}$ (Fig. 4), and the image of line $p$ is the tangent to circle $k^{\prime}$ at point $D$. From the power of points $B, C$ with respect to this circle, we get $|B D|^{2}=|B A| \cdot|B S_{c}|=\frac{1}{2}|B A|^{2}$ and $|C D|^{2}=|C A| \cdot|C S_{b}|=\frac{1}{2}|C A|^{2}$. Therefore, for the perimeter of triangle $A B C$, we have

$$
|B C|+|A B|+|A C|=|B C|+\sqrt{2}(|B D|+|C D|)=|B C|+\sqrt{2} \cdot|B C|=1+\sqrt{2},
$$

which is the (only possible) size of the perimeter of triangle $A B C$.