4.1. For triangle $A B C$, the following is known: $A B=12, B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R-$ is the radius of the smallest circle in which this triangle can be placed.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bdc1b7f380f62058621g-04.jpg?height=397&width=589&top_left_y=201&top_left_x=745)
Answer: 91

![](https://cdn.mathpix.com/cropped/2024_05_06_6bdc1b7f380f62058621g-04.jpg?height=426&width=422&top_left_y=935&top_left_x=820)

Solution. Since segment $AC$ fits inside the circle, $2R \geqslant AC$. On the other hand, the circle constructed with $AC$ as its diameter covers the triangle $ABC$, since $\angle ABC > 90^{\circ}$. Therefore, the desired value of $R$ is $AC / 2$, and $R^2 = AC^2 / 4$.

Next, by the cosine rule, we have $AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos 120^{\circ} = 12^2 + 10^2 + 2 \cdot 12 \cdot 10 / 2 = 4(6^2 + 5^2 + 6 \cdot 5) = 4 \cdot 91$