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I got this problem on a test yesterday

Consider $\Delta ABC$ with incenter $I(1,0)$. Equations of the straight lines $AI$, $BI$, and $CI$ are $x=1$, $y+1=x$ and $x+3y=1$ respectively and $\cot \left( \frac A2 \right) = 2$

  1. What is the locus of the centroid of $\Delta ABC$?
  2. What is the slope of side $BC$?
  3. If $A$ lies above the X axis and the area of $\Delta ABC$ is 30 square units then what is the inradius of $\Delta ABC$?

I assumed $A=(1,\alpha)$, $B=(\beta,\beta-1)$ and $C=(1-3\gamma , \gamma)$ and $BC=a$, $CA=b$, and $AB=c$.

I'm not sure how to proceed from here. I have a feeling I might end up using the relation $\cot \left(\frac A2\right)=\frac{s(s-a)}{\Delta}=\frac{(b+c)^2-a^2}{4\Delta} = 2$ at some point.

Using the formula for the incentre of a triangle given the vertices and the fact that $I=(1,0)$ I obtained the ratios of the sides in terms of $\alpha$, $\beta$, and $\gamma$ as $\frac ac=\frac{-4\gamma}\alpha$, $\frac bc=\frac{3\gamma}{\beta-1}$, and $\frac ba=\frac{3\alpha}{-4(\beta-1)}$

I'm stuck here. :(

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hint:

$\cot \left( \frac A2 \right) = 2 \implies k_{ab}=2,k_{ac}=-2,$ if you set $A(1,a)$, then you can find $B,C$ with $a$,rest is simple.

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  • $\begingroup$ What do $k_{ab}$ and $k_{ac}$ represent? $\endgroup$ – Danish Mohammed May 22 '15 at 3:07
  • $\begingroup$ Oh those are the slopes right? I get it. The slopes and the point $A$ uniquely determine $AB$ and $AC$ and $B$ and $C$ can be found from the intersections with the equations of the angle bisectors. Is that right? $\endgroup$ – Danish Mohammed May 22 '15 at 4:03
  • $\begingroup$ yes, you get it now. $\endgroup$ – chenbai May 22 '15 at 5:12

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