## Task B-3.4.

What is the area of a triangle whose sum of the squares of the side lengths is 2016, and the sum of the cotangents of the angles is 18?
## Solution.

We have

$$
a^{2}+b^{2}+c^{2}=2016 \text { and } \operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma=18
$$

By the cosine rule,

$$
\begin{aligned}
& a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \\
& b^{2}=a^{2}+c^{2}-2 a c \cos \beta \\
& c^{2}=a^{2}+b^{2}-2 a b \cos \gamma
\end{aligned}
$$

it follows that

$$
a^{2}+b^{2}+c^{2}=2 b c \cos \alpha+2 a c \cos \beta+2 a b \cos \gamma
$$

From the area formulas

$$
P=\frac{a b \sin \gamma}{2}=\frac{b c \sin \alpha}{2}=\frac{a c \sin \beta}{2}
$$

we get

$$
a b=\frac{2 P}{\sin \gamma}, \quad a c=\frac{2 P}{\sin \beta}, \quad b c=\frac{2 P}{\sin \alpha}
$$

Thus,

$$
a^{2}+b^{2}+c^{2}=4 P \frac{\cos \alpha}{\sin \alpha}+4 P \frac{\cos \beta}{\sin \beta}+4 P \frac{\cos \gamma}{\sin \gamma}=4 P(\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma)
$$

so

$$
P=\frac{a^{2}+b^{2}+c^{2}}{4(\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma)}=\frac{2016}{4 \cdot 18}=28 \text { square units. }
$$