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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?

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  • $\begingroup$ 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. $\endgroup$
    – stewbasic
    Apr 11, 2017 at 2:05

2 Answers 2

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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 $$R=\frac{MC}{2\sin \angle B} = \sqrt{\frac{149}{2}} \cdot \frac{80}{\sqrt{18711}}$$

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  • $\begingroup$ By ex-circle, do you mean circum-circle? $\endgroup$
    – Mick
    Apr 11, 2017 at 4:12
  • $\begingroup$ yes, that's what I mean :) $\endgroup$
    – Lazy Lee
    Apr 11, 2017 at 4:17
  • $\begingroup$ Hi, thank you for your answer, but I'm not sure if the circle in this case would be the circumcircle. As the circle goes through the endpoint of the longest side and the midpoint of the shortest side, I'm thinking that part of the circle may be inside the triangle as well. $\endgroup$ Apr 11, 2017 at 23:24
  • $\begingroup$ Hi, the circle I'm mentioning is the circle that passes through $B,C,M$, not $A,B,C$ :) $M$ is the midpoint of $AB$. $\endgroup$
    – Lazy Lee
    Apr 12, 2017 at 0:22
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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.

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