5. Construct an isosceles triangle $ABC$ with base $BC$ of given length $a$, given the midpoint $P$ of side $AB$ and a point $Q (Q \neq P)$, which is the foot of the altitude $z$ from vertex $B$.
SOLUTION. The angle $B Q A$ is either a right angle or $Q=A$. Therefore, point $Q$ lies on the Thales' circle constructed over the diameter $B A$. (Fig. 2 shows the case of both an acute and an obtuse triangle $A B C$.) Since $P$ is the midpoint of segment $A B$, $|P Q|$ is the radius of this circle, so the diameter $|A B|$ of this circle is equal to $2|P Q|$. Triangle $A B C$ has a side length of $2|P Q|$, and since we know the length of the base, it is uniquely determined.
![](https://cdn.mathpix.com/cropped/2024_04_17_7fa1855ed15af0dc549ag-3.jpg?height=600&width=1590&top_left_y=1656&top_left_x=240)

Fig. 2

From this, the construction follows. First, we construct a triangle $A^{\prime} B^{\prime} C^{\prime}$ congruent to triangle $A B C$ with side lengths $\left|A^{\prime} B^{\prime}\right|=\left|A^{\prime} C^{\prime}\right|=2|P Q|$ and $\left|B^{\prime} C^{\prime}\right|=a$, which we then move so that the midpoint of side $A^{\prime} B^{\prime}$ is mapped to point $P$ and the foot of the altitude from vertex $B^{\prime}$ is mapped to point $Q$. This can be done uniquely up to axial symmetry with respect to the line $P Q$. Therefore, if triangle $A^{\prime} B^{\prime} C^{\prime}$ exists, the problem has two solutions symmetrically related with respect to the axis $P Q$.

The discussion is clear. Triangle $A B C$ can be constructed precisely when an isosceles triangle $A^{\prime} B^{\prime} C^{\prime}$ can be constructed, i.e., when $a<4|P Q|$ (triangle inequalities), in which case the problem has exactly two (congruent) solutions. Additionally, for $a<2 \sqrt{2}|P Q|$, triangle $A B C$ will be acute, for $a=2 \sqrt{2}|P Q|$ it will be right, and for $2 \sqrt{2}|P Q|<a<4|P Q|$ it will be obtuse. The proof of correctness follows from the analysis of the problem.