Problem 1645

1. Coordinate System Setup:
- Place the coordinate origin at point $A$ and align the positive x-axis along the ray $AK$.
- Let $K = (k, 0)$ and $\kappa = \tan \measuredangle MKL$. The equation of the line $MK$ is $y = \kappa(x - k)$.

2. **Point $B$ on Line $MK$**:
- Let $B$ be an arbitrary point on $MK$, so $B = (x_B, y_B)$ where $y_B = \kappa(x_B - k)$.

3. Circle Center and Radius:
- The perpendicular bisector of segment $[AB]$ intersects $AK$ at the center $O$ of the circle with radius $[OA] = [OB]$.
- The line $BO$ intersects the circle again at point $D = (x_D, y_D)$, where $y_D = -y_B = -\kappa(x_B - k)$.

4. **Power of Point $B_0$**:
- Let $B_0$ be the perpendicular projection of $B$ on $AK$. The power of $B_0$ to the circle is given by:
$$-x_B \cdot x_D = y_B \cdot y_D = -y_D^2$$

5. **Equation of Locus $\mathcal{D}$**:
- Eliminate the arbitrary parameter $x_B$ to find the equation of the locus $\mathcal{D}$ of $D$:
$$\frac{y_D^2}{x_D} = x_B = -\frac{y_D}{\kappa} + k \implies h(x, y) = y^2 + \frac{xy}{\kappa} - kx = 0$$
- This is the general equation of a conic section $ax^2 + bxy + cy^2 + dx + ey + f = 0$ with $a = e = f = 0$. Since $b^2 - 4ac = \frac{1}{\kappa^2} > 0$, it is a hyperbola.

6. Asymptotes of the Hyperbola:
- The slopes of the asymptotes $u$ and $v$ are obtained by factoring the first three terms:
$$ax^2 + bxy + cy^2 = y^2 + \frac{xy}{\kappa} = y \left(y + \frac{x}{\kappa}\right) = 0$$
- The slopes are $0$ and $-\frac{1}{\kappa}$. Thus, one asymptote $u \parallel AK$ and the other $v \perp MK$.

7. Equations of Asymptotes:
- Let the equations of the asymptotes be $y = p$ and $y = -\frac{x}{\kappa} + q$. For any $(x, y) \in u \cup v$:
$$0 = (y - p) \left(y + \frac{x}{\kappa} - q\right) = h(x, y) - \left(\frac{p}{\kappa} - k\right)x - (p + q)y + pq$$
- Since $h(x, y) = \text{const}$, we have $h(x, y) = -pq$ and $p = \kappa k$, $q = -p = -\kappa k$.

8. Intersection Points and Tangents:
- Let $MK$ cut the y-axis at $Q$ and let $P$ be the reflection of $Q$ in $A$. The asymptote $u \parallel AK$ passes through $P$ and $v \perp MK$ passes through $Q$.
- Since the absolute term of the hyperbola equation is $0$, $A \in \mathcal{D}$. Since $A$ is the midpoint of $[PQ]$, $PQ$ is tangent to $\mathcal{D}$ at $A$.

9. Hyperbola Center and Major Axis:
- The intersection $X \equiv u \cap v$ is the hyperbola center. The internal bisector $w$ of $\measuredangle PXQ$ is the hyperbola major axis line.
- Let $E$ and $F$ be the hyperbola foci, which need to be constructed.

10. Construction of Foci:
- The hyperbola tangent $PAQ$, bisecting $\measuredangle FAE$, intersects the perpendicular bisector of $[EF]$ through $X$ at $S$.
- The circumcircle $(Y)$ of $\triangle AEF$, centered on $XS$ and passing through $A$ and $S$, intersects the hyperbola major axis line $w$ at the hyperbola foci $E$ and $F$.

11. Hyperbola Vertices:
- The perpendicular projection $E_0$ of the focus $E$ on the hyperbola tangent $PQ$ lies on the hyperbola pedal circle $(X)$, which intersects the hyperbola major axis line $w$ at the hyperbola vertices $U$ and $V$.

12. Circle Construction:
- Let $(E)$ be a circle centered at $E$ with radius $[UV]$. Let $F'$ be the reflection of $F$ in $LM$.
- The two circles $(D_1)$ and $(D_2)$ through $F$ and $F'$, and externally tangent to $(E)$ at $T_1$ and $T_2$, are centered on the line $LM$.

13. Intersection Points:
- Since $[D_1E] - [D_1F] = [D_1E] - [D_1T_1] = [T_1E] = [UV]$ and similarly, $[D_2E] - [D_2F] = [UV]$, $D_1$ and $D_2 \in \mathcal{D}$ are intersections of the line $LM$ and the hyperbola $\mathcal{D}$.

14. Radical Axes and Tangents:
- Let $(G)$ be an arbitrary circle through $F$ and $F'$, preferably intersecting $(E)$ at $J$ and $J'$. The radical axes $FF'$ and $JJ'$ of circle pairs $(D_1), (G)$ and $(E), (G)$ meet at the radical center $R$ of $(D_1), (E), (G)$.
- Tangents $RT_1$ and $RT_2$ of $(E)$ from $R$ are radical axes of externally tangent circle pairs $(D_1), (E)$ and $(D_2), (E)$. Thus, $D_1 \equiv ET_1 \cap LM$ and $D_2 \equiv ET_2 \cap LM$.

15. Circumcenters of Rectangles:
- The perpendicular bisectors of $[AD_1]$ and $[AD_2]$ intersect the line $AK$ at the circumcenters $O_1$ and $O_2$ of rectangles $AB_1C_1D_1$ and $AB_2C_2D_2$, respectively, providing the two problem solutions.

The final answer is the construction of the rectangles $\boxed{ AB_1C_1D_1 }$ and $AB_2C_2D_2$.