$A B C$ triangle, perpendiculars are raised at $A$ and $B$ on side $AB$, which intersect $BC$ and $AC$ at $B_{1}$ and $A_{1}$, respectively. What is the area of the triangle if $AB=c$, $AB_{1}=m$, and $BA_{1}=n$?
Let the foot of the altitude from $C$ to side $AB$ be $D$ and $CD = x$. Since $\triangle B_{1}AB \sim \triangle CDB$ and $\triangle A_{1}BA \sim \triangle CDA$, we have:

$$
x : DB = m : c
$$

and

$$
c : AD = n : c
$$

from which:

$$
DB = \frac{xc}{m} \text{ and } AD = \frac{xc}{n}
$$

so

$$
DB + AD = c = \frac{xc}{m} + \frac{xc}{n}
$$

from which

$$
x = \frac{mn}{m+n}
$$

thus the area of the triangle is:

$$
t = \frac{mnc}{2(m+n)}
$$

Number of solutions: 52.