# Using side length of triangle, find radius of touching circle

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?

• If you can find the angle between the shortest side and the corresponding median, you can use en.wikipedia.org/wiki/Law_of_sines to determine the radius of the circle. Apr 11, 2017 at 2:05

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 $$R=\frac{MC}{2\sin \angle B} = \sqrt{\frac{149}{2}} \cdot \frac{80}{\sqrt{18711}}$$
• Hi, the circle I'm mentioning is the circle that passes through $B,C,M$, not $A,B,C$ :) $M$ is the midpoint of $AB$. Apr 12, 2017 at 0:22
Another proof of this that differs only slightly. I haven't seen the proof for the radius of an ex-circle, so while @Lazy Lee's method was fine, the last step just came out of the blue for me. Continuing from @Lazy Lee's answer, $$\cos{\angle B}=\frac{83}{160}$$ and $$E=(MC)^2=\frac{149}{2}$$. Let the ex-circle of $$\triangle BCM = O$$. It follows that $$BO=OC=MO=r$$. Then $$\angle MOC = 2\angle B$$ because the legs of $$\triangle MOC$$ and $$\triangle MBC$$ intercept the same arc $$\newcommand{\tarc}{\mbox{\frown}} \newcommand{\arc}[1]{\stackrel{\tarc}{#1}} \arc{MC}$$. Using the law of cosines on $$\triangle MOC$$ to find $$r$$, we see that $$E=(MC)^2=2r^2(1-\cos{\angle 2B})$$. Using the fundamental theorem of trigonometry, $$\sin^2{\angle B}+\cos^2{\angle B}=1$$, we see that $$\sin^2{\angle B}= 1-\big(\frac{83}{160}\big)^2$$. Then substituting $$\cos{\angle 2B}$$ with $$\cos^2{\angle B}-\sin^2{\angle B}$$, we arrive at the same conclusion when solving for r.