This is somehow embarrassing for me. So, I have been asked the following question (a similar one actually) by my friend who is currently an eight-grader:
Suppose $a$, $b$, and $c$ are known, find the length of $AI$.
Credit image: Wolfram MathWorld
I'm able to tackle this problem using the cosine rule and the cosine double-angle formula. I obtained this result: $$AI=r\sqrt{\frac{4bc}{2bc+a^2-b^2-c^2}}$$ but unfortunately, she hasn't been taught the cosine rule nor also trigonometry (sine, cosine, and tangent). I haven't figure it out using any 'simple methods'. Is it even possible? I guess I'm missing something obvious here. My question is how to deal with this problem using elementary ways preferably without using trigonometry? Any help would be greatly appreciated. Thank you.
We know that $AM_b = AM_c$ and so on, because lengths of tangents from a point to a circle are equal. So we let $AM_b=x, BM_c=y, CM_a=z$. Now,
$$AM_b + AM_c+ BM_a + BM_c + CM_b + CM_a = a+b+c$$ $$2(x+y+z)= a+b+c$$ $$x+y+z = s$$
We also know that $BC = y+z$. Thus $ x = s - BC = s-a$. Now since $AI$ is the hypotenuse of right triangle $AIM_b$, we have:
$$AI^2= r^2 + (s-a)^2$$
Now using Heron's formula and $rs =\Delta$, we can represent $AI$ in terms of $a,b,c$ as: $$AI = \sqrt{\frac{(s-a)(s-b)(s-c)}{s} + (s-a)^2}$$