\section*{Problem 2 - 071012}

Given is a triangle \(\triangle A B C\), whose sides have the following lengths (in cm): \(\overline{A B}=13, \overline{B C}=12\), \(\overline{A C}=5\). The side \(\overline{A B}\) is to be divided by a point \(T\) such that the perimeters of the triangles \(\triangle A T C\) and \(\triangle T B C\) are equal.

The area (in \(\mathrm{cm}^{2}\)) of the triangle \(\triangle T B C\) is to be calculated!
}

All numerical values are given in \(\mathrm{cm}\).

First, the converse of the Pythagorean theorem can be used to show that it is a right-angled triangle:

It holds that: \(\overline{A C}^{2}+\overline{B C}^{2}=5^{2}+12^{2}=25+144=169=13^{2}=\overline{A B}^{2}\), and thus, the triangle \(\triangle A B C\) is right-angled with the right angle at \(\angle A C B\).

For the perimeters of the triangles, we get:

(1) \(u_{\triangle A T C}=\overline{A C}+\overline{A T}+\overline{C T}\)

(2) \(u_{\triangle B T C}=\overline{B C}+\overline{B T}+\overline{C T}\)

Given that the perimeters of the sub-triangles are equal and \(\overline{A B}\) is composed of \(\overline{A T}\) and \(\overline{B T}\), we get for \(\overline{B T}\):

\[
\begin{aligned}
\overline{A C}+\overline{A T}+\overline{C T} & =\overline{B C}+\overline{B T}+\overline{C T} \\
\overline{A C}+\overline{A T} & =\overline{B C}+\overline{B T} \\
\overline{A C}+\overline{A B}-\overline{B T} & =\overline{B C}+\overline{B T} \\
\overline{B T} & =\frac{1}{2} \cdot(\overline{A B}+\overline{A C}-\overline{B C})
\end{aligned}
\]

The area of a right-angled triangle can be calculated in two different ways:

\[
\begin{aligned}
\frac{1}{2} \cdot g \cdot h & =\frac{1}{2} \cdot a \cdot b \\
\overline{A B} \cdot h & =\overline{B C} \cdot \overline{A C} \\
h & =\frac{\overline{B C} \cdot \overline{A C}}{\overline{A B}}
\end{aligned}
\]

Now, for the area of the sub-triangle \(\triangle T B C\) using (1) and (2):

\[
\begin{aligned}
A_{\triangle T B C} & =\frac{1}{2} \cdot \overline{B T} \cdot h=\frac{1}{2} \cdot \frac{1}{2} \cdot(\overline{A B}+\overline{A C}-\overline{B C}) \cdot \frac{\overline{B C} \cdot \overline{A C}}{\overline{A B}} \\
& =\frac{1}{4} \cdot(13+5-12) \cdot \frac{12 \cdot 5}{13} \approx 6.9
\end{aligned}
\]

Thus, the sub-triangle \(\triangle T B C\) has an area of approximately \(6.9 \mathrm{~cm}^{2}\).