In an isosceles triangle $A B C$, the base $A B$ is $6 \mathrm{~cm}$ long and the angle $B C A$ has a size of $45^{\circ}$.

Calculate the radius of the circle circumscribed around this triangle. (L. Růžičková)
The center $S$ of the circumcircle of triangle $ABC$ is denoted as $S$. The distance from the center $S$ to each of the points $A, B, C$ is equal to the sought radius, which we denote as $r$. Triangles $ACS$, $BCS$, and $ABS$ are therefore isosceles. First, we will show that triangle $ABS$ is a right triangle with a right angle at vertex $S$.

The center $S$ lies on the perpendicular bisector of the base $AB$, which is also the axis of symmetry of triangle $ABC$. The isosceles triangles $ACS$ and $BCS$ are thus congruent. Therefore, the angles $SAC$, $ACS$, $SCB$, and $CBS$ are all congruent with measures

$$
\alpha = \frac{1}{2} \cdot 45^{\circ} = 22^{\circ} 30'
$$

The sum of the measures of the interior angles $SAC$ and $ACS$ in triangle $ACS$ is $45^{\circ}$, which is also the measure of the exterior angle at vertex $S$ (denoted as $\omega$ in the figure). This angle is half of the angle $ASB$, so angle $ASB$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_04_17_71346d6af5a10a586345g-6.jpg?height=825&width=740&top_left_y=176&top_left_x=658)

In the right triangle $ABS$, both legs have length $r$ and the hypotenuse $AB$ is $6 \text{ cm}$. According to the Pythagorean theorem,

$$
r^2 + r^2 = 6^2,
$$

thus $r = \sqrt{18} = 3\sqrt{2} \text{ cm}$. The circumcircle of triangle $ABC$ has a radius of $3\sqrt{2} \text{ cm}$.

Scoring suggestions. 1 point for recognizing the congruence of triangles $ACS$ and $BCS$, or the angles $SAC$, $ACS$, $SCB$, and $CBS$; 3 points for deriving that angle $ASB$ is a right angle; 2 points for calculating the radius (consider any of the above values as correct).

Award the first 4 points even if the measure of angle $ASB$ is derived by referring to the theorem about central and inscribed angles.

Award no more than 1 point for using the Pythagorean theorem in triangle $ABS$ without justification for why this triangle is a right triangle.

Notes. After calculating (1), the measure of angle $ASB$ can be determined in various ways. For example, by determining the measures of the interior angles at the base in the isosceles triangle $ABC$,

$$
\frac{1}{2} \left(180^{\circ} - 45^{\circ}\right) = 67^{\circ} 30'
$$

then expressing the measures of the interior angles at the base in the isosceles triangle $ABS$,

$$
\beta = 67^{\circ} 30' - 22^{\circ} 30' = 45^{\circ},
$$

from which the measure of angle $ASB$ is calculated as

$$
\delta = 180^{\circ} - 2 \cdot 45^{\circ} = 90^{\circ}.
$$

Another derivation can be based on determining the measures of the remaining interior angles in the congruent isosceles triangles $ACS$ and $BCS$,

$$
\gamma = 180^{\circ} - 2 \cdot 22^{\circ} 30' = 135^{\circ},
$$

from which the measure of angle $ASB$ is calculated as

$$
\delta = 360^{\circ} - 2 \cdot 135^{\circ} = 90^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_71346d6af5a10a586345g-7.jpg?height=774&width=737&top_left_y=407&top_left_x=662)