Let's construct a triangle, given the perimeter of the triangle, $\alpha$ and $m_{a}$ (the altitude from vertex $A$).
Let $ABC \Delta$ be the triangle that meets the conditions. Extend the sides $AB$ and $AC$, and construct the circle that lies between the rays of $\alpha$ and is externally tangent to the triangle. The center $I_{1}$ of this circle lies on the angle bisector of $\alpha$; if the points of tangency on the sides are $A_{1}, B_{1}, C_{1}$, then

$$
BA_{1} = BC_{1} \quad \text{and} \quad CA_{1} = CB_{1}
$$

Since $AB_{1} = AC_{1}$ and

$$
AB_{1} + AC_{1} = AC + CB_{1} + AB + BC_{1} = AC + AB + BC = 2s
$$

it is clear that

$$
AB_{1} = AC_{1} = s
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_41e9dcb5289a8f86a77bg-1.jpg?height=542&width=420&top_left_y=640&top_left_x=842)

The side $BC$ is a tangent to the circle centered at $A$ with radius $AH = m_{a}$, so it is a common internal tangent to this circle and the circle with center $I_{1}$.

The construction based on this is as follows. On the rays of $\alpha$, measure the semiperimeter: $AB_{1} = AC_{1} = s$. At points $B_{1}$ and $C_{1}$, draw perpendiculars to $AB_{1}$ and $AC_{1}$, respectively. These perpendiculars intersect the angle bisector of $\alpha$ at point $I_{1}$. Draw the circle with center $I_{1}$ and radius $I_{1}B_{1} = I_{1}C_{1}$, and the circle with center $A$ and radius $AH$. The common internal tangent of these two circles will be the line containing the side $BC$. (There are two solutions, symmetric with respect to the angle bisector of $\alpha$.)

Let's examine the condition for the possibility of the construction. In the limiting case - i.e., when there is still a common internal tangent to the two circles defined earlier - these two circles are externally tangent to each other at the intersection point $K$ of the angle bisector $AI_{1}$ and the circle with center $I_{1}$. Therefore, it is necessary that $AH = m_{a} < AK$.

Now,

$$
AK = AI_{1} - KI_{1} = \frac{AB_{1}}{\cos \frac{\alpha}{2}} - I_{1}B_{1} = \frac{s}{\cos \frac{\alpha}{2}} - s \tan \frac{\alpha}{2} = \frac{s}{\cos \frac{\alpha}{2}} \left(1 - \sin \frac{\alpha}{2}\right)
$$

Accordingly, the condition for the possibility of the construction is:

$$
m_{a} < \frac{s \left(1 - \sin \frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}}
$$

**Second Solution.** Extend the side $BC$ of the triangle $ABC$ in both directions and measure the distances $BD = BA$ and $CE = CA$. Thus, we obtain a triangle $ADE$ that can be constructed. Indeed,

$$
DE = DB + BC + CE = AB + BC + CA = 2s
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_41e9dcb5289a8f86a77bg-1.jpg?height=415&width=634&top_left_y=2237&top_left_x=735)

The altitude to side $DE$ is $AH = m_{a}$. The triangle $ABD$ is isosceles; the angles at the base $AD$ sum to $\beta$, so each is $\frac{\beta}{2}$. Similarly, in triangle $ACE$, the angles at the base $AE$ are each $\frac{\gamma}{2}$. Therefore,

$$
\angle DAE = \alpha + \frac{\beta}{2} + \frac{\gamma}{2} = \alpha + \frac{\beta + \gamma}{2} = \alpha + 90^\circ - \frac{\alpha}{2} = 90^\circ + \frac{\alpha}{2}
$$

The side $DE$ of triangle $ADE$ is a chord of a circle, and the larger inscribed angle corresponding to this chord is $90^\circ + \frac{\alpha}{2}$. The corresponding central angle is $180^\circ + \alpha$, so $\angle DOE = 180^\circ - \alpha$ and in the isosceles triangle $DOE$, each base angle is $\frac{\alpha}{2}$.

The construction process is as follows: on an arbitrary line, measure the distance $DE = 2s$ (the perimeter of the triangle to be constructed). At the endpoints, measure angles of $\frac{\alpha}{2}$ such that the isosceles triangle $DOE$ is formed. Draw a circle with radius $OD = OE$; the point $A$ must lie on the smaller arc of this circle, at a distance $m_{a}$ from $DE$. Draw a perpendicular to $DE$ at any point, measure the distance $m_{a}$ on this perpendicular, and draw a line $g$ parallel to $DE$ at this distance. The point where $g$ intersects the circle $O$ will be the vertex $A$ of the triangle. The perpendicular bisectors of the chords $AD$ and $AE$ intersect $DE$ at points $B$ and $C$, respectively.

The line $g$ intersects the circle $O$ at two points ($A$ and $A'$); thus, we get two solutions, symmetric with respect to the line $OF \perp DE$: the two solutions provide triangles with the same side lengths. If $g$ is tangent to the circle, the two solutions coincide. Points $A$ and $A'$ coincide at $F$, and triangle $ABC$ becomes isosceles.

For a solution to exist, it is necessary and sufficient that $m_{a} \leq FI$, where $I$ is the midpoint of $DE$. In the right triangle $FIE$, $IE = s$, and

$$
\angle IFE = \frac{1}{2} \angle DAE = \frac{1}{2} \left(90^\circ + \frac{\alpha}{2}\right) = 45^\circ + \frac{\alpha}{4}
$$

Thus, $FI = s \cot \left(45^\circ + \frac{\alpha}{4}\right)$.

The condition for the possibility of the construction is: $m_{a} < s \cot \left(45^\circ + \frac{\alpha}{4}\right)$.

Pál Holzer (Faludi Ferenc g. VIII. o. Szombathely.)

Note. From the two conditions for the solution, we must conclude that

$$
\frac{1 - \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} = \cot \left(45^\circ + \frac{\alpha}{4}\right) = \tan \left(45^\circ - \frac{\alpha}{4}\right)
$$

The proof of this is given in problem 1382.