# Locus of centroid, given the slopes of the sides

Find the locus of the centroid of a triangle if it is known that its orthocentre is at the origin and the slopes of the sides of the triangle are $$m_1, m_2, m_3$$ respectively.

I solved this problem by working with this diagram; Here, it's easy to see that the slopes of all the altitudes will be $$-1/m_1, -1/m_2, -1/m_3$$ and hence obtain relations in $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ by definition of the slope of altitude respectively (since the orthocentre is at (0,0)). Again by definition of slope, one may write also write the expressions; $$m_1=\frac{y_2-y_1}{x_2-x_1}$$ and so on, for $$m_2$$ and $$m_3$$. Then, from the 6 relations I obtained above, I expressed $$x_3$$ in terms of the slopes and $$x_1$$, and did the same for $$x_2$$ and $$x_1$$. Finally, now I assumed the centroid coordinates to be $$(h,k) =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$ and then divided the equations for h and k, and finally substituted $$x_2$$ and $$x_3$$ in terms of $$x_1$$ and the slopes so that $$x_1$$ canceled out. I then arrived at the following expression for the locus; $$y=\left(\frac{3+S_2}{S_1 +3S_3}\right)x$$ where $$S_i$$ are the symmetric sums of $$m_1, m_2, m_3$$ taken $$i$$ at a time. Since I eventually arrived at such a simple expression which I think I have seen in many places (the formula for $$\tan(A+B+C)$$ has a similar form, and yes I understand that the final answer ought to be symmetric in $$m_1,m_2,m_3$$ but I did not quite expect such a simple expression in symmetric sums), I am now looking for an alternate solution more geometric or trigonometric in flavor relying less heavily on coordinate geometry, perhaps using the Euler line (which has been brought to my attention by the comments), and possibly an interesting interpretation of the final equation. Does anyone have any ideas as to how else I can arrive at this conclusion with a more elegant method, without ploughing through so much messy algebra?

• "which I have seen in many places" : could you give a reference (it doesn't ring anything to me...) ? Jan 11 at 8:35
• Well, I just meant that the symmetric sums of 3 variables tend to pop up in lots of places, like for example the tan(A+B+C) formula and since tan sort of represents slope of the line, I was hoping to get a more elegant geometric answer rather than the brute force I've used. Jan 11 at 8:42
• IMHO, the locus of the centroid cannot be a line : it is the whole plane. Jan 11 at 8:54
• Wait what, how? Jan 11 at 8:55
• Not sure to bring an answer in the coming hour, but I will try. Jan 11 at 9:00

Proof that the locus is a straight line:

Since the slopes are constant, the angles between the sides are constant. So, all such triangles are similar. More importantly, their orthocentre remains constant and all such triangles can be formed by scaling one triangle about the origin.

So, homothety shows that the locus of the centroid is a straight line (unless the centroid coincides with the orthocentre) passing through the origin.

• I agree with the first part, but your "So, homothety shows that the locus of the centroid is a straight line" isn't based on any explanation. I think that reasoning on the "permanence" of Euler line would bring that explanation... Jan 11 at 10:15
• Since the tree points of the triangle follow homothety w.r.t the origin, then the centroid defined by them does too @JeanMarie
– D S
Jan 11 at 11:16
• Considering the Euler Line is also an alternative though..
– D S
Jan 11 at 11:18
• Could someone explain what the Euler line has to do with this? I've never really heard of it before Jan 11 at 13:16
• @Cognoscenti the Euler line is the line joining the circumcentre, centroid, nine-point centre, and orthocentre of a triangle
– D S
Jan 11 at 17:32