10.013. A triangle is given with sides of 12, 15, and 18 cm. A circle is drawn such that it is tangent to both of the shorter sides and has its center on the longest side. Find the segments into which the center of the circle divides the longest side of the triangle.
Solution.
Using the Law of Cosines, from Fig. 10.13 we have:
AB^2 = BC^2 + AC^2 - 2 \cdot BC \cdot AC \cdot \cos \alpha \Rightarrow
\Rightarrow 12^2 = 15^2 + 18^2 - 2 \cdot 15 \cdot 18 \cos \alpha, \cos \alpha = \frac{3}{4}, \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt{7}}{4};
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos \beta \Rightarrow
\Rightarrow 15^2 = 12^2 + 18^2 - 2 \cdot 12 \cdot 18 \cdot \cos \beta, \cos \beta = \frac{9}{16}, \sin \beta = \sqrt{1 - \cos^2 \beta} = \frac{5\sqrt{7}}{16}.
\sin \alpha = \frac{OM}{OC} = \frac{R}{18 - x}, \sin \beta = \frac{OK}{AO} = \frac{R}{x} and \frac{\sin \alpha}{\sin \beta} = \frac{x}{18 - x} = \frac{4}{5}.
5x = 72 - 4x, 9x = 72, x = 8, y = 10. AO = 8 cm, OC = 10 cm.
Answer: 8 and 10 cm.
