Task B-1.3. (20 points)

The lengths of the legs of a right triangle are $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$. Inside the triangle, there is a point $T$ which is $1 \mathrm{~cm}$ away from the shorter leg and $2 \mathrm{~cm}$ away from the longer leg. How far is point $T$ from the hypotenuse? What is the length of the altitude to the hypotenuse?
Solution. If we connect point $T$ with the vertices $A, B, C$, we get triangles $A B T, B C T, C A T$. The desired distance $x$ is the length of the height $\overline{T D}$ of triangle $A B T$. Let $a=|B C|=8 \mathrm{~cm}$, $b=|C A|=6 \mathrm{~cm}$. The heights of triangles $B C T$ and $C A T$ are $2 \mathrm{~cm}$ and $1 \mathrm{~cm}$, respectively.

The length of the hypotenuse is $c=\sqrt{a^{2}+b^{2}}=\sqrt{8^{2}+6^{2}} \mathrm{~cm}=\sqrt{100} \mathrm{~cm}=10 \mathrm{~cm}$.

(1 point)

We will express the area of the triangle in two ways:

$$
\frac{|B C| \cdot|C A|}{2}=P_{A B C}=P_{A B T}+P_{B C T}+P_{C A T}
$$

(10 points)

Substituting the known values, we get,

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-03.jpg?height=359&width=585&top_left_y=226&top_left_x=744)

$$
\frac{8 \cdot 6}{2}=\frac{10 \cdot x}{2}+\frac{8 \cdot 2}{2}+\frac{6 \cdot 1}{2}
$$

from which we have $x=\frac{26}{10} \mathrm{~cm}=\frac{13}{5} \mathrm{~cm}=2.6 \mathrm{~cm}$.

(5 points)

We will also calculate the height to the hypotenuse using areas, i.e., $\frac{a b}{2}=\frac{c v_{c}}{2}$, from which we get $8 \cdot 6=10 \cdot v_{c}$, i.e., $v_{c}=\frac{48}{10} \mathrm{~cm}=\frac{24}{5} \mathrm{~cm}=4.8 \mathrm{~cm}$.

(4 points)