In triangle $ABC$, $b+c=2l$ is given, and from point $A$, the height $m$ and the median $k$ can be drawn. Calculate the sides and angles of the triangle and construct the triangle.
Let $A A_{1}=m, A A_{2}=k$, and $A_{1} A_{2}=x$.

Then

$$
c^{2}=m^{2}+\left(\frac{a}{2}+x\right)^{2} \quad \text { and } \quad b^{2}=m^{2}+\left(\frac{a}{2}-x\right)^{2}
$$

therefore

$2 a x=c^{2}-b^{2}$

or

$$
2 a x=2 l(c-b)
$$

but

$$
x=\sqrt{k^{2}-m^{2}}
$$

so

$$
a=\frac{l(c-b)}{\sqrt{k^{2}-m^{2}}}, a^{2}=\frac{l^{2}(c-b)^{2}}{k^{2}-m^{2}}
$$

But

$$
\begin{gathered}
b+c=2 l \\
b^{2}+c^{2}=4 l^{2}-2 b c \\
(b-c)^{2}=4\left(l^{2}-b c\right)
\end{gathered}
$$

so

$$
a^{2}=\frac{4 l^{2}\left(l^{2}-b c\right)}{k^{2}-m^{2}}
$$

On the other hand

$$
4 k^{2}=2\left(b^{2}+c^{2}\right)-a^{2}
$$

or

$$
4 k^{2}=4\left(2 l^{2}-b c\right)-a^{2}
$$

from which

$$
a^{2}=4\left(2 l^{2}-b c\right)-4 k^{2}
$$

From (1) and (2):

$$
b c=\frac{2 l^{2} k^{2}-2 l^{2} m^{2}+k^{2} m^{2}-l^{4}-k^{4}}{k^{2}-m^{2}-l^{2}}
$$

Combining this equation with $b+c=2 l$, we get:

$$
\begin{aligned}
& b=l+\sqrt{\frac{\left(k^{2}-m^{2}\right)\left(k^{2}-l^{2}\right)}{k^{2}-m^{2}-l^{2}}} \\
& c=l-\sqrt{\frac{\left(k^{2}-m^{2}\right)\left(k^{2}-l^{2}\right)}{k^{2}-m^{2}-l^{2}}} \\
& \quad a=2 l \sqrt{\frac{k^{2}-l^{2}}{k^{2}-m^{2}-l^{2}}}
\end{aligned}
$$

The angles are given by the following equations

$$
\sin B=\frac{m}{c}, \sin C=\frac{m}{b}, A=180^{\circ}-(B+C)
$$

(Weisz Ármin.)

Construction. The vertex $A$ of the triangle is a point on an ellipse whose major axis is $b+c=2 l$, and whose foci are $B$ and $C$; if we can determine these points, the problem is solved. Based on this, we construct a right triangle whose one leg is $A A_{1}=m$, hypotenuse is $A A_{2}=k$, and the vertex $A_{2}$ is the midpoint of the line segment $b+c$; we draw a semicircle over $b+c$, which intersects the extension of $A A_{1}$ at $D$. We draw a perpendicular from $A$ to $A_{2} D$ to get $A E$ and measure $A_{2} E$ on the perpendicular raised at the midpoint $A_{2}$ of the major axis. Thus, we get one endpoint of the minor axis; from this point, we draw a circle with radius $\frac{b+c}{2}=l$, which intersects the major axis at the desired points $B$ and $C$.