Let's construct the triangle $ABC$, given the length of the altitude from point $A$ is $m_{a}$, the length of the median from point $A$ is $s_{a}$, and the length of the angle bisector from point $A$ is $f_{a}$. What is the solvability condition?
I. Solution: Imagine the problem is solved (Figure 1). The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_0d5523c100a3700533cbg-1.jpg?height=637&width=761&top_left_y=234&top_left_x=671)

Figure 1

Complete the diagram with the circumcircle of the triangle and let the perpendicular bisector through point $A_{1}$ intersect the arc $BC$ (not containing point $A$) at point $F$. Since the perpendicular bisector of a chord also bisects the corresponding arc, the arc $BF$ = arc $FC$, and thus, by the inscribed angle theorem, the angle bisector $f_{a}$ extended passes through point $F$. This also implies that $A_{3}$ always lies between $A_{1}$ and $A_{2}$ (unless $A_{1}$ coincides with $A_{2}$), so $m_{a} < f_{a} < s_{a}$. (If $A_{1}$ coincides with $A_{2}$, then the triangle is isosceles, $m_{a} = f_{a} = s_{a}$, and the problem is indeterminate.)

Accordingly, the construction procedure is as follows: Take the line segment $BC = a$ and a point $A_{2}$ on it. On the perpendicular to $a$ at $A_{2}$, measure the distance $A_{2}A = m_{a}$. From point $A$, with radii $f_{a}$ and $s_{a}$, draw arcs that intersect on the same side of $A_{2}$ at points $A_{3}$ and $A_{1}$. (Point $A_{3}$, due to $f_{a} < s_{a}$, lies between $A_{2}$ and $A_{1}$.) The perpendicular to $a$ at $A_{1}$ intersects the angle bisector of $\angle AA_{3}$ at point $F$. The perpendicular bisector of chord $AF$ intersects the line $A_{1}F$ at the center of the circumcircle of the triangle: $K$. This point $K$ with radius $KA = KF$ draws a circle that intersects the line containing $a$ at points $B$ and $C$.

Vera Kozma (Bp. V., 1st Textile Industry Technical School, 2nd grade)

II. Solution: After constructing points $A_{1}$ and $A_{3}$, points $B$ and $C$ can be constructed without using point $F$ and the circumcircle.

Draw the external angle bisector of $\angle A$ perpendicular to $f_{a}$, which intersects the line containing $a$ at point $A_{3}^{\prime}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_0d5523c100a3700533cbg-1.jpg?height=553&width=832&top_left_y=1713&top_left_x=641)

Figure 2

The circle drawn with $A_{3}A_{3}^{\prime}$ as the diameter (which, by Thales' theorem, passes through point $A$) is none other than the Apollonius circle of side $BC$ passing through $A$. We claim that the tangent from point $A_{1}$ to this circle, $A_{1}T$, is equal to $A_{1}B = A_{1}C = \frac{a}{2}$.

Proof: Let $A_{1}A_{3} = p$ and $A_{1}A_{3}^{\prime} = q$, then

$$
\frac{A_{3}C}{A_{3}B} = \frac{A_{3}^{\prime}C}{A_{3}^{\prime}B}
$$

or

$$
\frac{\frac{a}{2} - p}{\frac{a}{2} + p} = \frac{q - \frac{a}{2}}{q + \frac{a}{2}}
$$

Clearing the fractions and rearranging,

$$
2a^2 = 8pq,
$$

from which

$$
\frac{a}{2} = \sqrt{pq}.
$$

But the tangent from point $A_{1}$ to the circle, by a known theorem, is the geometric mean of the segments $p$ and $q$, so indeed $A_{1}T = \frac{a}{2}$.

Eligius Schmidt (Bp. I., Fürst S. g. III. o. t.)