Distance between triangle incenter and vertices

Question

after many researches on the subject, I can't find any convincing argument anywhere, so I come to you about this problem which has been brought by some of my high school students.

Let $ABC$ be a random triangle and $I$ his incenter. It is known that $AB=c$, $AC=b$ and $BC=a$. I'm looking for a clean way to express the distance AI only from $a$ $b$ and $c$ parameters (no angles). When I searched on the internet, I found the formula : $$ AI^{2}=\frac{p-a}{p}bc$$ Where $p=\frac{a+b+c}{2}$ is the semi-perimeter. The formula is quite nice, but I can't find a proof. I tried with law of cosines, heron's formula, but I can't quite catch the idea which will bring me this particular formula. Any idea ?

Answer 1

I'm using $s$ for semiperimeter (that's more common).

First, drop perpendiculars from $I$ onto $a,b,c$ and call them $ID,IE,IF$ respectively. All are equal to the inradius $r$.

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Then, it is well known that $AE = AF = s-a$. I provide a short proof below:

Since the two tangents drawn from a point to a circle are equal, we have $AE=AF$. Similarly, $BD=BF$ and $CD=CE$. However, the perimeter $$2s = AE+CE+CD+BD+BF+AF = 2(AE+CD+BD) = 2(AE +a)$$ which gives $AE = s-a$. Analogously, $BD=BF = s-b, CD=CE=s-c$.

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It is also well known that if $\Delta$ is the area of $\Delta ABC$ (by a slight abuse of notation, but it is commonly used), then $\Delta = rs$. Here's a proof:

$$\begin{align}\Delta &= \mathrm{Ar}(AIC)+\mathrm{Ar}(BIC)+\mathrm{Ar}(AIB) \\ &= 0.5(rb+ra+rc) = rs \end{align}$$

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Then, by Pythagoras' theorem in $\Delta AIE$ (since $\angle AEI = 90^\circ$): $$\begin{align}AI^2 &= r^2+(s-a)^2 \\ &= \Delta^2/s^2 +(s-a)^2 \\ &= (s-a)\left(\frac{(s-b)(s-c)}{s} + (s-a)\right) \tag{Heron's Formula} \\ &= (s-a)\left(\frac{2s^2 - sb - sc -sa +bc}{s}\right) \\ &= \frac{(s-a)bc}{s} \tag{as $s(a+b+c) = 2s^2$} \end{align}$$ QED

Answer 2

We can use angle relations and convert the formulas to one in terms of a, b and c.We can easily see that:

$AI=\frac r{sin\frac{\alpha}2}\space\space\space \space\space\space \space\space\space (1)$

$BI=\frac r{sin\frac{\beta}2}\space\space\space \space\space\space \space\space\space (2)$

$CI=\frac r{sin\frac{\gamma}2}\space\space\space \space\space\space \space\space\space (3)$

We also know:

$r=(p-a)tan\frac{\alpha}2\space\space\space \space\space\space \space\space\space (4)$

Where $p=\frac{a+b+c}2$

(1)&(4) gives:

$AI=\frac{p-a}{cos\frac{\alpha}2}\space\space\space \space\space\space \space\space\space (5)$

We also know:

$cos\frac{\alpha}2=\sqrt{\frac{p(p-a)}{bc}}\space\space\space \space\space\space\space\space\space \space\space\space(6)$

(5)&(6) give:

$AI=\frac{\sqrt{bc(p-a)}}{\sqrt p}\Rightarrow AI^2=\frac{p-a}p bc\space\space\space \space\space\space\space\space\space \space\space\space(7)$

If you want to prove for exapmple formula (6) , we have:

$a^2=b^2+c^2-2(bc) cos\alpha\Rightarrow cos\alpha=\frac{b^2+c^2-a^2}{2bc}\space\space\space \space\space\space\space\space\space \space\space\space(8)$

$cos^2\frac{\alpha}2=\frac{1+cos\alpha}2\space\space\space \space\space\space\space\space\space \space\space\space(9)$

Finally (8) and (9) give:

$cos \frac{\alpha}2=\sqrt{\frac{p(p-a)}{bc}}\space\space\space \space\space\space\space\space\space \space\space\space(10)$