# Angles at centroid of a triangle

If $$ABC$$ is a triangle with centroid $$P$$, I got the impression that the angle $$\angle BPA$$ at the centroid should only depend on the angle $$\angle BCA$$ (and not on the other angles).

Am I right? Is there a sensible formula?

• Interesting question. The angles' ratio looks to be $atan(3x)/atan(x)$. Do you find the same thing ? – Jean Marie Mar 20 at 18:52

• $$ABC$$ is equilateral. Then $$\angle BPA=120^\circ$$.
• $$\angle A = 90^\circ$$, $$\angle B = 30^\circ$$, $$\angle C = 60^\circ$$. For convenient coordinates, choose $$A=(0,0)$$, $$B=(3\sqrt{3},0)$$, $$C=(0,3)$$. Then $$P=(\sqrt{3},1)$$. The dot product $$(P-A)\cdot (P-B)$$ is $$(\sqrt{3},1)\cdot (-2\sqrt{3},1) = -5$$. Divide by the lengths, and we get $$\cos\angle BPA = \dfrac{-5}{\sqrt{4}\cdot\sqrt{13}}$$. That's not $$\cos(120^\circ)=-\frac12$$. Computing the arccosine, $$\angle BPA\approx 134^\circ$$.
Same angle at $$C$$, two different angles at $$B$$.
• Thanks. My argument was misguided. I found out that if $C$ moves on the circle for contant $\angle C$, then $P$ moves on a circle as well - but this circle is not a circle through $A$ and $B$. – J. Fabian Meier Mar 20 at 19:10