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. :(



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

  • $\begingroup$ What do $k_{ab}$ and $k_{ac}$ represent? $\endgroup$ 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? $\endgroup$ May 22 '15 at 4:03
  • $\begingroup$ yes, you get it now. $\endgroup$
    – chenbai
    May 22 '15 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.