Task B-1.3. (20 points) The length of the base of isosceles triangle $\mathrm{ABC}$ is $12 \mathrm{~cm}$, and the length of the leg is $10 \mathrm{~cm}$. Points $P$ and $Q$ are the midpoints of legs $\overline{A C}$ and $\overline{B C}$, and $S$ and $R$ are their orthogonal projections on $\overline{A B}$. Determine the distance between the feet of the perpendiculars from points $P$ and $R$ to the line segment $\overline{S Q}$.
Solution. The length of $\overline{P Q}$ is the midline of triangle $A B C$, so $|P Q|=\frac{1}{2}|A B|=6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-07.jpg?height=412&width=524&top_left_y=473&top_left_x=651)

Point $D$ is the midpoint of side $\overline{A B}$. By applying the Pythagorean theorem, we have

$$
|C D|=\sqrt{|B C|^{2}-|B D|^{2}}=\sqrt{100-36}=8 \mathrm{~cm}
$$

Point $Q$ is the midpoint of side $\overline{B C}$, so $|B Q|=|C Q|=5 \mathrm{~cm}$.

From the similarity of triangles $B D C$ and $B R Q$, we have $\frac{|C D|}{|Q R|}=\frac{|B D|}{|B R|}=\frac{2}{1}$, hence $|Q R|=\frac{|C D|}{2}=4$ $\mathrm{cm}$. Quadrilateral $S R Q P$ is a rectangle and $|S R|=|P Q|=6 \mathrm{~cm}$. The length of its diagonal is

$$
|S Q|=\sqrt{|S R|^{2}+|Q R|^{2}}=\sqrt{36+16}=2 \sqrt{13}
$$

From $P_{S R Q}=\frac{1}{2} P_{S R Q P}$, we have $|R N|=|P M|=\frac{P_{S R Q P}}{|S Q|}=\frac{|S R| \cdot|Q R|}{|S Q|}=\frac{12}{\sqrt{13}}=\frac{12 \sqrt{13}}{13}$.

Furthermore, by the Pythagorean theorem, $|N Q|=\sqrt{|Q R|^{2}-|R N|^{2}}=\sqrt{16-\frac{144}{13}}=\frac{8 \sqrt{13}}{13}$.

By symmetry, $|S M|=|N Q|$, so we have $|M N|=|S Q|-2|N Q|=\frac{10 \sqrt{13}}{13}$.

The desired distance between the feet is $|M N|=\frac{10 \sqrt{13}}{13} \mathrm{~cm}$.