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 $\cot \left( \frac A2 \right) = 2$
- What is the locus of the centroid of $\Delta ABC$?
- What is the slope of side $BC$?
- 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. :(
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.