A triangle has one angle $\alpha$, and the lengths of the internal and external angle bisectors from its vertex are $f$ and $g$, respectively. Calculate the sides of the triangle.

Numerical data: $\alpha=30^{\circ}, f=106, g=239$ units.
I. Solution. I. Let the triangle be $ABC$, the angle bisectors $f=AF, g=AG$, and choose the labeling so that $AB > AC$ (since the finiteness of $g$ implies $AB \neq AC$), thus the order of our points on the line $BC$ is $B, F, C, G$.

![](https://cdn.mathpix.com/cropped/2024_05_02_feac7ff23498fd3e421ag-1.jpg?height=274&width=574&top_left_y=302&top_left_x=770)

The internal and external angle bisectors are perpendicular to each other, so from the right triangle $FGA$, the angle $\angle AFG = \angle AFC = \varphi$ can be determined, and thus the angles of our triangle are:

$$
\beta = \varphi - \frac{\alpha}{2}, \quad \gamma = 180^\circ - \left(\varphi + \frac{\alpha}{2}\right)
$$

Using the Law of Sines and the addition formula, first in triangles $ABF$ and $ACF$:

$$
c = AB = f \frac{\sin (180^\circ - \varphi)}{\sin \left(\varphi - \frac{\alpha}{2}\right)} = \frac{f \cdot FG \sin \varphi}{FG \sin \varphi \cos \frac{\alpha}{2} - FG \cos \varphi \sin \frac{\alpha}{2}} = \frac{f g}{g \cos \frac{\alpha}{2} - f \sin \frac{\alpha}{2}}
$$

Similarly,

$$
b = AC = f \frac{\sin \varphi}{\sin \gamma} = \frac{f g}{g \cos \frac{\alpha}{2} + f \sin \frac{\alpha}{2}}
$$

Finally, from the desired triangle:

$$
a = \frac{b \sin \alpha}{\sin \beta} = \frac{f g \sqrt{f^2 + g^2} \sin \alpha}{g^2 \cos^2 \frac{\alpha}{2} - f^2 \sin^2 \frac{\alpha}{2}}
$$

This gives the expressions for the sides in terms of the given data.

The calculation yields a real triangle if the denominator of $c$ is positive (since then the denominator of $a$ is also positive):

$$
g \cos \frac{\alpha}{2} - f \sin \frac{\alpha}{2} > 0
$$

which implies

$$
\tan \frac{\alpha}{2} < \frac{g}{f} = \tan \varphi, \quad \frac{\alpha}{2} < \varphi
$$

since $f, g$ are positive, and $\alpha / 2$ and $\varphi$ are acute angles.

The same condition is derived from the construction solution of the triangle, where the point $B$ on the ray $AB$ must lie beyond the extension of segment $FG$ at $F$, hence the angle $\angle FAB$ must be smaller than the alternate interior angle of $\angle AFG$ at $A$.

II. With the given data, $\varphi = 66^\circ 5', \beta = 51^\circ 5', \gamma = 98^\circ 55', b = 98.08, c = 124.5, a = 63.03$ units. (The numerical calculation is simpler by calculating $\varphi, \beta$, and $\gamma$ from $\tan \varphi = g / f$ rather than from (2)-(4).)

András Garay (Sopron, Széchenyi I. Gymnasium, 3rd grade)

II. Solution. Using area considerations, we derive the above expressions. The area of the right triangle $FGA$ is equal to the sum of the areas of triangles $ACF$ and $ACG$, and also to the difference of the areas of triangles $ABG$ and $ABF$. Each area, doubled, can be written as the product of two sides and the sine of the included angle:

$$
b f \sin \frac{\alpha}{2} + b g \cos \frac{\alpha}{2} = f g, \quad c g \sin \left(90^\circ + \frac{\alpha}{2}\right) - c f \sin \frac{\alpha}{2} = f g
$$

From these, by factoring and dividing, we get (3) and (2).

Moreover, the altitude from $A$ in triangle $FGA$ is the same as the altitude from $A$ in triangle $ABC$, so the ratio of their areas is equal to the ratio of their bases:

$$
\frac{a}{\sqrt{f^2 + g^2}} = \frac{b c \sin \alpha}{f g}
$$

Substituting the expressions for $b$ and $c$, we get (4).