Given are two sides of a triangle ($a$ and $b$) and the bisector of the angle between them (v); calculate the third side of the triangle.
If the sides $a$ and $b$ enclose an angle $\gamma$, then the area of the triangle is:

$$
t=\frac{a b}{2} \sin \gamma
$$

On the other hand, the area of the triangle is equal to the sum of the areas of the two parts, that is,

$$
t=\frac{a v}{2} \sin \frac{\gamma}{2}+\frac{b v}{2} \sin \frac{\gamma}{2}
$$

thus

$$
\frac{a b}{2} \sin \gamma=\frac{v}{2}(a+b) \sin \frac{\gamma}{2}
$$

from which

$$
\cos \frac{\gamma}{2}=\frac{a+b}{2 a b} v
$$

It is known that $1+\cos \gamma=2 \cos ^{2} \frac{\gamma}{2}$, into which, substituting the value of $\cos \frac{\gamma}{2}$ from (4), we obtain:

$$
\cos \gamma=\frac{(a+b)^{2}}{2 a^{2} b^{2}} v^{2}-1
$$

On the other hand, according to the cosine rule,

$$
\cos \gamma=\frac{2 a^{2}+b^{2}-c^{2}}{2 a b}
$$

Thus, from (5) and (6),

$$
\begin{gathered}
\frac{v^{2}(a+b)^{2}}{2 a^{2} b^{2}}-1=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
\frac{v^{2}(a+b)^{2}}{a b}=(a+b)^{2}-c^{2}
\end{gathered}
$$

from which

$$
c^{2}=(a+b)^{2} \frac{a b-v^{2}}{a b}
$$

or

$$
c=(a+b) \sqrt{\frac{a b-v^{2}}{a b}}
$$

(Bernát Friedmann.)

The problem was also solved by: Barna D., Beck F., Bojedain F., Devecis M., Freibauer E., Goldziher K., Gross N., Kántor B., Kántor N., Kármán T., Kornis Ö., Krátky Gy., Petrogalli G., Roth M., Schiffer H., Schwartz E., Spitzer Ö., Szabó I., Szabó K., Weisz J.