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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 $AI$. enter image description here

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.

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    Does she know Heron's formula? – 2017-01-26
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    @Sawarnik Yes, she knows. I've also used that formula but there are 2 sides that we haven't known their lengths yet – 2017-01-26
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    Well, if we know Heron's then we can do by finding the length of the angle bisector, for which we will need Stewart's theorem. Is she aware of that? I am thinking of alternative ways though. – 2017-01-26
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    @Sawarnik She doesn't know Stewart's theorem. – 2017-01-26
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    Ah, why aren't we using $AI^2 = r^2 + (s-a)^2$! – 2017-01-26
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    @Sawarnik How do you get $AM_b=s-a$? – 2017-01-26
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    That comes from $AM_b = AM_c$ and so on. Let $AM_b = x$, $BM_c = y$, $CM_a = z$, then we have that $x + y + z = s$, and also $ y + z = BC$. Do you get it? :) – 2017-01-26
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    @Sawarnik OMG!! How stupid I am. Could you post your comment as an answer so that I can accept it? Thank you so much... – 2017-01-26
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    Done :) .. btw Anastasiya, it was nice to talk to you after a long long while :) – 2017-01-26
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    @Sawarnik It's also nice to have you respond my question. Thank you so much... :) – 2017-01-26
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    You're welcome! Also, please do be active on MSE at times .. love to see your awesome integrals and series :) – 2017-01-26

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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: $$AI = \sqrt{\frac{(s-a)(s-b)(s-c)}{s} + (s-a)^2}$$