Let $ABC$ be a triangle and $I$ its incenter. It is given that $BC=a$, $CA=b$, $AB=c$, $\angle A=\alpha$, $\angle B=\beta$ and $\angle C=\gamma$. Find the distance $AI$ in terms of these given values.

I've tried everything. I managed to find $AD=\frac{2bc\cos (\alpha/2)}{b+c}$, where $D$ is the point of intersection of line $AI$ with side $BC$. However this doesn't seem to help unless I can also find $ID$, because then $AI$ is given by $AD-ID$.

  • $\begingroup$ Its simple, consider the triangle $AIE$, where $E$ is a point on $AB$ such that $IE\perp AB$. You have a triangle with all angles known and a side as well, apply some trig and you are done. Should I write that as an elaborated answer? $\endgroup$ – Sawarnik Feb 2 '15 at 19:23

let us look at the triangle $AIB.$ the $\angle IAB = \alpha/2, \angle IBA = \beta/2, AB = c.$ applying the rule of sines you get $$\dfrac{AI}{\sin (\beta/2)} = \dfrac{c}{\sin(\alpha+\beta)/2)} = \dfrac{c}{\cos (\gamma/2)}$$

that is $$AI = \dfrac{c\sin(\beta/2)}{\cos(\gamma/2)}. $$

  • $\begingroup$ Although the question didn't ask for it, it's worth noting that, using the Law of Sines on $\triangle ABC$ with $d$ the circumdiameter, you can write $c = d\;\sin\gamma = 2d\;\sin(\gamma/2)\;\cos(\gamma/2)$, so that $$AI = 2 d\;\sin(\beta/2)\;\sin(\gamma/2)$$ This representation has better "balance". $\endgroup$ – Blue Feb 2 '15 at 20:00
  • $\begingroup$ @Blue, i know that i could replace $\sin c$ by the diameter. i did not want to make it more complicated than what the op asked for. $\endgroup$ – abel Feb 2 '15 at 20:02
  • $\begingroup$ I know that you know. :) Still, I thought my comment might be helpful to someone out there. $\endgroup$ – Blue Feb 2 '15 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.