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The triangle $\triangle ABC$ is an isosceles triangle where $AB = 4\sqrt{2}$ and $\angle B$ is a right angle. If $I$ is the incenter of $\triangle ABC,$ then what is $BI$?

Express your answer in the form $a + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect squares integers other than $1.$

Below is a picture of what I have done. I have found the answer, which I believe is correct, but i did not find it in the square root version. Could someone please help me find the square root version and how to get to the answer. Thanks a lot! Edit: the property of an incenter is that it is equidistant from all of the sides. enter image description here

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  • $\begingroup$ Note you can find the value of sin(45/n) with respect to $\sqrt{2}$. $\endgroup$ – Sam Palmer Jun 21 '18 at 9:03
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enter image description here

\begin{align} |AC|=|BB'|&=\sqrt2|AB|=8 ,\\ |BI|=x&=\sqrt2 r ,\\ |BB'|&=2x+2r=2x+\sqrt2\sqrt2 r =x(2+\sqrt2) ,\\ x&=\frac{8}{2+\sqrt2} \\ &=\frac{8(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)} \\ &=\frac{8(2-\sqrt2)}{(2^2-(\sqrt2)^2)} \\ &=8-4\sqrt2 . \end{align}

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Note that the inradius of the triangle is $r=\dfrac{(2-\sqrt2)}{2}(4\sqrt2).$ The proof is simple: use the fact that $$\text{the area of the whole triangle}=\text{sum of 3 individual triangles}.$$

Then the line from point $I$ to $AB$ is equal in length to the line from point $I$ to $BC$. This is a square. So by Pythagorean theorem, $$r^2+r^2=(BI)^2\implies BI=\sqrt{2}\left[\dfrac{(2-\sqrt2)}{2}(4\sqrt2)\right]=8-4\sqrt2.$$

Addendum: I add the proof of the inradius here. Let the inradius be $r$. Observe $$\begin{align}\text{the area of the whole triangle}&=\text{sum of 3 individual triangles}\\\dfrac{(4\sqrt2)^2}{2}&=\dfrac{1}{2}(4\sqrt2r+4\sqrt2r+8r) \\r&=\dfrac{(4\sqrt2)^2}{2(4+4\sqrt2)}=4(\sqrt2-1)=\dfrac{(2-\sqrt2)}{2}(4\sqrt2).\end{align}$$

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