Question:
The side lengths of a triangle are equal to lengths $8, 9$ and $10$. Find the exact value of the radius of the circle passing through the endpoints of the longest side and the midpoint of the shortest side.
I am very confused as to where to begin to solve the problem. If I solve the angles of the triangle, how will that help me to find the radius of the circles?
Let the triangle be $\triangle ABC$ with $AB=8$, $AC=9$, $BC = 10$, and let the midpoint of $AB$ be $M$. Then, our goal is to find the radius of the excircle of $\triangle BCM$. By the cosine law, we know that $$\cos\angle B=\frac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\frac{83}{160}$$ and similarly by the cosine law, $$\cos \angle B = \frac{BM^2+BC^2-MC^2}{2\cdot BM\cdot BC}=\frac{116-MC^2}{80}\implies MC = \sqrt{\frac{149}{2}}$$ Hence, by the sine law, the radius of the excircle of $\triangle BCM$ is $$R=\frac{MC}{2\sin \angle B} = \sqrt{\frac{149}{2}} \cdot \frac{80}{\sqrt{18711}}$$