Let's construct a triangle, given the perimeter $k$, the angle $\alpha$, and the height from the vertex of the angle $\alpha$.
I. Solution: Imagine the problem is solved. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_1a66f95aaa35a5c611abg-1.jpg?height=403&width=759&top_left_y=205&top_left_x=672)

Figure 1

If we extend the side $BC = a$ beyond $B$ by the side $BA_1 = BA = c$ and beyond $C$ by the side $CA_2 = CA = b$, then in the isosceles triangles $A_1BA$ and $A_2CA$, the angles adjacent to the base are $\frac{\beta}{2}$ and $\frac{\gamma}{2}$, respectively, because the sum of two interior angles is equal to the exterior angle opposite them. Therefore, in the triangle $A_1AA_2$, the side $A_1A_2$ is the perimeter of the triangle, and the angle $\angle A = \alpha + \frac{\beta}{2} + \frac{\gamma}{2} = \frac{2\alpha + \beta + \gamma}{2} = \frac{180^\circ + \alpha}{2} = 90^\circ + \frac{\alpha}{2}$. Thus, the point $A$ lies on the arc of the circle drawn over $A_1A_2 = k$ such that the segment $A_1A_2$ is seen from the points of the arc at an angle of $90^\circ + \frac{\alpha}{2}$, and also lies on the line parallel to $A_1A_2$ and at a distance $m_a$ from it, on the same side of $A_1A_2$ as the arc.

The construction process is as follows: We take the distance $A_1A_2 = k$. We construct the two geometric places mentioned above. Their intersection point is $A$ (and $A^*$), one of the vertices of the sought triangle. The perpendicular bisectors of the segments $AA_1$ and $AA_2$ intersect the distance $A_1A_2$ to give the vertices $B$ and $C$ of the triangle. There is only one solution because $A^*$ provides the mirror image of the previous solution with respect to the perpendicular bisector of $A_1A_2$.

The solvability condition is that

$$
m_a \leq MN = ON - OM = OA_1 - OM = \frac{A_1M}{\cos \frac{\alpha}{2}} - A_1M \tan \frac{\alpha}{2} = \frac{k}{2 \cos \frac{\alpha}{2}} - \frac{k \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}} = \frac{k}{2} \cdot \frac{1 - \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}
$$

In the case of equality, we obtain an isosceles triangle.

Géza Aujeszky (Bp. II., Rákóczi g. II. o. t.)

II. Solution: It is known that the points of tangency of the incircle of a triangle on the extensions of the sides are at a distance of half the perimeter from the point of intersection of the tangents. Using this, the construction process is as follows: In the angle $\alpha$, we construct a circle that touches the sides at points $E_1$ and $E_2$, which are at a distance of $\frac{k}{2}$ from the vertex $A$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_1a66f95aaa35a5c611abg-1.jpg?height=447&width=525&top_left_y=1777&top_left_x=781)

Figure 2

The circle centered at $A$ with radius $m_a$ and the common internal tangent of the previous circle (tangency points $A_1$ and $E$) intersect the sides to give the vertices $B$ and $C$ of the sought triangle. Indeed, the triangle thus obtained satisfies our conditions: The perimeter of the triangle is $AB + BE + EC + CA = AB + BE_1 + E_2C + CA = AE_1 + AE_2 = \frac{k}{2} + \frac{k}{2} = k$ and the height to the side $BC = a$ is $AA_1 = m$.

The solvability condition is that the two circles do not intersect at two different points, i.e., $m_a \leq AO - OE_1 = \frac{k}{\alpha} - \frac{k}{2} \tan \frac{\alpha}{2}$, etc., as in the first solution.