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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?

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  • $\begingroup$ Interesting question. The angles' ratio looks to be $atan(3x)/atan(x)$. Do you find the same thing ? $\endgroup$ – Jean Marie Mar 20 at 18:52
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No. For a convenient example, consider the following two cases:

  • $ABC$ 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 $B$.

There are some points that produce angles depending only on the one vertex angle - the circumcenter, orthocenter, and incenter all work, with different choices of exactly what angles you get. But the centroid? It's an affine object, not something that will play nice with angles.

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  • $\begingroup$ 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 $B$. $\endgroup$ – J. Fabian Meier Mar 20 at 19:10

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