A circle with center $S$ and radius $39 \, \text{mm}$ is given. We need to inscribe a triangle $ABC$ in the circle such that the length of side $AC$ is $72 \, \text{mm}$ and point $B$ lies in the half-plane determined by the line $AC$ and point $S$.

From the given data, calculate the height of triangle $ABC$ from vertex $B$ so that the problem has two solutions. Describe all possibilities.

(M. Krejčová)
The length of side $AC$ is indeed smaller than the diameter of the circle, so this side does not contain the center $S$. The height of triangle $ABC$ from vertex $B$ can be arbitrarily small, but it cannot be arbitrarily large: this height is the largest when it contains the center $S$. In such a case, the problem has a unique solution (the corresponding triangle is isosceles) and the height is equal to the length of segment $PB_{0}$ as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_04_17_614117edae892b473639g-4.jpg?height=743&width=765&top_left_y=1065&top_left_x=651)

The length of $PB_{0}$ is equal to the sum of the lengths of segments $PS$ and $SB_{0}$. The length of segment $SB_{0}$ is equal to the radius of the circle, i.e., $39 \text{ mm}$. The length of segment $PS$ is determined using the Pythagorean theorem in the right triangle $APS$:

$$
|PS| = \sqrt{|AS|^2 - |AP|^2} = \sqrt{39^2 - 36^2} = \sqrt{225} = 15 \text{ mm}.
$$

Thus, the maximum height is $39 + 15 = 54 \text{ mm}$. The problem has two solutions if the height from vertex $B$ is less than $54 \text{ mm}$ (and greater than $0 \text{ mm}$).

Evaluation. 3 points for expressing the limiting value; 3 points for analyzing the possibilities and the quality of the commentary.

Answers based on estimation or measurement from the drawn figure should be scored 0 points.