What property does a triangle have if the following relationship holds between two of its sides and the angle bisector of the third side:

$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{f_{c}}
$$
I. Solution: Write that the area of the triangle is equal to the sum of the areas of the two triangles into which the angle bisector $f_{c}$ divides the triangle (see the figure):

$$
a f_{c} \sin \frac{\gamma}{2} + b f_{c} \sin \frac{\gamma}{2} = a b \sin \gamma
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_4baf5fbe5316b8fa6972g-1.jpg?height=496&width=555&top_left_y=343&top_left_x=774)

Dividing both sides by $a b f_{c} \sin \frac{\gamma}{2}$ ( $\sin \frac{\gamma}{2}$ is clearly not 0 )

$$
\frac{1}{a} + \frac{1}{b} = \frac{1}{f_{c}} \cdot \frac{\sin \gamma}{\sin \frac{\gamma}{2}} = \frac{1}{f_{c}} \cdot 2 \cos \frac{\gamma}{2}
$$

According to this, (1) holds if and only if

$$
\cos \frac{\gamma}{2} = \frac{1}{2}
$$

Since $\frac{\gamma}{2} < 90^{\circ}$, it follows that $\frac{\gamma}{2} = 60^{\circ}$, and thus

$$
\gamma = 120^{\circ}
$$

Therefore, the necessary and sufficient condition for (1) to hold is that the angle bisected by $f_{c}$ is $120^{\circ}$. Mária Jáky (Pécs, Bányaip. t. II. o. t.)

II. Solution: The relationship under (1) can also be written as

$$
f_{c} = \frac{a b}{a + b}
$$

Draw a line through vertex $B$ parallel to $f_{c}$, intersecting the extension of side $AC$ at point $E$ (see the figure). In the resulting $\triangle BCE$, the angles at vertices $B$ and $E$, being alternate and corresponding angles, are equal to $\frac{\gamma}{2}$, and thus $CE = CB = a$. Clearly, $\triangle ADC \sim \triangle ABE$, and thus

$$
f_{c} : b = x : (a + b)
$$

from which

$$
f_{c} = \frac{b x}{a + b}
$$

Comparing (3) and (4), it follows that $x = a$, so $\triangle BCE$ is equilateral, and thus every angle, including $\frac{\gamma}{2}$, is $60^{\circ}$.