In a $P T S$ triangle, the angle at $P$ is $60^{\circ}$. Express the radius of the circle that touches the line $P T$ at point $T$ and passes through point $S$ in terms of $P T$ and $P S$.
I. Solution. Let the radius of the circle be $r$, the projection of $S$ on $PT$ be $M$, which is different from $T$, and the diameter of the circle starting from $T$ be $TN$. Then, $\triangle STM$ and $\triangle TNS$ are similar right triangles because the angles at $S$ and $T$ are alternate interior angles. Furthermore, $SM$ is the height of the equilateral triangle with side $PS$, so

$$
\frac{NT}{ST}=\frac{2r}{ST}=\frac{ST}{MS}; \quad r=\frac{ST^2}{2MS}=\frac{PT^2+PS^2-PT \cdot PS}{\sqrt{3} \cdot PS}
$$

We applied the cosine theorem to $\triangle PTS$ (Figure 1).
![](https://cdn.mathpix.com/cropped/2024_05_02_694fb358f7ed707148eeg-1.jpg?height=462&width=1132&top_left_y=510&top_left_x=477)

Figure 1

Our result is also valid if $S$ coincides with $N$, i.e., $ST \perp PT$, and thus $\triangle PTS$ is half of an equilateral triangle cut by its height, $PS=2PT$, so $NT=\sqrt{3} \cdot PT, r=PT \sqrt{3} / 2=PS \cdot \sqrt{3} / 4$; however, we can reach the goal more quickly from the observation $PS=2PT$. Similarly, we can easily obtain the result in the special case where $PT=2PS$, i.e., $\angle PST=90^{\circ}$.

László Hárs (Budapest, Berzsenyi D. Gymnasium III. grade)

II. Solution. Let the projection of the center $O$ of the circle on the chord $TS$ be $U$, a point different from $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_694fb358f7ed707148eeg-1.jpg?height=402&width=705&top_left_y=1390&top_left_x=691)

Thus, the sides of $\angle TOU$ are perpendicular to the sides of $\angle PTS$, so from the right triangle $\triangle TOU$ and applying the sine theorem to $\triangle PTS$,

$$
\begin{gathered}
\sin \angle TOU=\frac{TU}{r}=\frac{TS}{2r}= \\
=\sin \angle PTS=\frac{PS \sin \angle TPS}{TS}=\frac{PS \sqrt{3}}{2TS}.
\end{gathered}
$$

Comparing the two results and applying the cosine theorem:

$$
r=\frac{TS^2}{\sqrt{3} \cdot PS}=\frac{PT^2+PS^2-PT \cdot PS}{\sqrt{3} \cdot PS}
$$

If $U$ coincides with $O$, then the special case $ST \perp PT$ holds, which we have already seen in the first solution.