Let's construct a triangle, given the side $a$, the opposite angle $\alpha$, and the radius of the inscribed circle $\varrho$.
The task is identical to the 3rd problem of the 2nd round of the 1952 Arany Dániel poetry competition, with the simplification that instead of the radius of the circumscribed circle, the side opposite the given angle is directly provided. The solution can be found on pages 39-41 of the 2nd issue.

I. Solution: Imagine the problem is solved. Let the given angle be $\alpha$, the center of the circumscribed circle of the triangle be $K$ and its radius $r$, the center of the inscribed circle be $O$ and its radius $\varrho$. (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_dd851ca65831f2a4a6deg-1.jpg?height=680&width=678&top_left_y=397&top_left_x=707)

From $\triangle BOC$, $\angle BOC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = \frac{360^{\circ} - (\beta + \gamma)}{2} = \frac{180^{\circ} + [180^{\circ} - (\beta + \gamma)]}{2} = \frac{180^{\circ} + \alpha}{2} = 90^{\circ} + \frac{\alpha}{2}$.

According to the theorem of inscribed angles, in the circle with center $K$ and radius $r$, any intersection points of the sides of the inscribed angle with the circle determine the side $BC = a$ of the triangle.

The point $O$ - according to the above - lies on the arc of the circle where the distance $BC$ is seen at an angle of $90 + \frac{\alpha}{2}$ and which is on the same side of the side $BC$ as the vertex of the angle $\alpha$. Denoting the center of this angle-of-view circle by $P$, based on the relationship between the central and inscribed angles,

$$
\angle BPC = 360^{\circ} - 2\left(90 + \frac{\alpha}{2}\right) = 180^{\circ} - \alpha,
$$

which means that the quadrilateral $ABPC$ is a cyclic quadrilateral and thus the point $P$ is the midpoint of the arc $BC$.

Another geometric locus for $O$ is the line parallel to $BC$ at a distance $\varrho$ from $BC$, on the same side of the line $BC$ as the previous arc.

The intersection of these two geometric loci is the sought point $O$. The tangents from points $B$ and $C$ to the circle centered at $O$ with radius $\varrho$ intersect at point $A$ (which lies on the circumscribed circle) and is the third vertex of the sought triangle. (Generally, we get two points for $O$, but both lead to congruent triangles, so we are talking about one solution.)

Let's determine the solvability conditions. Denote the midpoint of side $a$ by $A_1$ and the intersection of the line $PA_1$ with the angle-of-view arc by $Q$. A solution clearly exists only if $\varrho \leq A_1Q$.

Since (1) implies that $\angle A_1BP = \frac{\alpha}{2}$, we have

$$
A_1P = \frac{a}{2} \tan \frac{\alpha}{2} = \frac{a \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}}
$$

and

$$
PQ = PB = \frac{a}{2 \cos \frac{\alpha}{2}}
$$

so

$$
\begin{gathered}
A_1Q = PQ - A_1P = \frac{a}{2 \cos \frac{\alpha}{2}} - \frac{a \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}} = \frac{a(1 - \sin \frac{\alpha}{2})}{2 \cos \frac{\alpha}{2}} \\
\text{But } a = 2r \sin \alpha = 4r \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}
\end{gathered}
$$

and thus

$$
A_1Q = 2r \sin \frac{\alpha}{2} (1 - \sin \frac{\alpha}{2})
$$

So a solution exists only if

$$
\varrho \leq 2r \sin \frac{\alpha}{2} (1 - \sin \frac{\alpha}{2})
$$

In the case of equality, $\varrho = A_1Q$ and the triangle is isosceles $(b = c)$.

For a constant $r$, the right-hand side is maximized when $\sin \frac{\alpha}{2} (1 - \sin \frac{\alpha}{2})$ is maximized. A product, where the sum of the factors is constant, takes its maximum value when the factors are equal, i.e., $\sin \frac{\alpha}{2} = \frac{1}{2}$, from which $\alpha = 60^{\circ}$ and $\varrho \leq \frac{r}{2} \cdot \varrho$ has a maximum value of $\frac{r}{2}$, which it attains when $\alpha = 60^{\circ}$ and additionally $\beta = \gamma = 60^{\circ}$, i.e., the triangle is equilateral. (L. K. M. L, 1st year, May 1948, problem 167.)

II. Solution: We use this theorem (1st grade textbook, 1950 edition, p. 285), which states that the distance between the two points of tangency of a common external tangent to the incircle and an excircle of a triangle is equal to the third side, which is the segment of the common internal tangent between the external tangents.

![](https://cdn.mathpix.com/cropped/2024_05_02_dd851ca65831f2a4a6deg-2.jpg?height=491&width=789&top_left_y=790&top_left_x=668)

2nd figure

According to this, the construction process is as follows: the side $BC = a$ of the triangle is constructed in the same way as in the first solution. We place the angle $\alpha$ and construct a circle with radius $\varrho$ that is tangent to both sides of the angle (2nd figure). On one of the sides of the angle, we measure the distance $a$ from the point of tangency, moving away from the vertex of the angle. The point thus obtained will be the point of tangency of the excircle, according to the aforementioned theorem. The common internal tangent of the incircle and the excircle intersects the sides of the angle $\alpha$ at points $B$ and $C$.

The solvability condition: Denoting the center of the excircle by $O_1$ and its radius by $\varrho_1$, the problem is solvable only if $\varrho + \varrho_1 \leq OO_1$. But

$$
\varrho_1 = \varrho + a \tan \frac{\alpha}{2}, \quad OO_1 = \frac{a}{\cos \frac{\alpha}{2}}
$$

and thus our condition is

$$
2\varrho + \frac{a \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \leq \frac{a}{\cos \frac{\alpha}{2}}
$$

from which $\varrho \leq \frac{a(1 - \sin \frac{\alpha}{2})}{\cos \frac{\alpha}{2}} = 2r \sin \frac{\alpha}{2} (1 - \sin \frac{\alpha}{2})$, which agrees with the condition derived in the first solution.