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?
[Math] Angles at centroid of a triangle
geometrytriangles
Related Solutions
Remember the theorems "In a right triangle, the length of the median to the hypotenuse equals half the hypotenuse", and also "The centroid in a triangle is a point which divides each median in a proportion 2:1, where the longest side is always between the centroid and its vertex.
So we have
Call the centroid $\;M\;$ and let $\;K\;$ be the midpoint of $\;AB\;$, and let $\;x=|AB|=$ the hypotenuse's length . Thus, we have:
$$\cos\alpha=\frac{AC}x\implies AC=x\cos\alpha\implies S=\frac{AC\cdot AB\cdot\sin\alpha}2=\frac{x^2\cos\alpha\sin\alpha}2=\frac14x^2\sin2\alpha$$
Now, observe that summing angles of triangle $\;\Delta BCK\;$ , we have that $$\;\angle BKC=2\alpha\implies\angle MKA=180^\circ-2\alpha\;$$
Also
$$|CK|=\frac12x\implies |MK|=\frac13|CK|=\frac x6$$
so we can know form a little right triangle $\;\Delta MKL\;$ , with $\;L\;$ a point on the hypotenuse $\;AB\;$ s.t. $\;ML\perp AB\;$ , and then in this right triangle:
$$\sin(180-2\alpha)=\frac{ML}{MK}\implies ML=MK\sin2\alpha=\frac {x\sin2\alpha}6$$
Finally, observe that
$$\frac12\sqrt{S\sin2\alpha}=\frac12\sqrt{\frac14x^2\sin^22\alpha}=\frac14 x\sin2\alpha$$
You can see I get a different result, so either I am wrong (check the above ) or else it should be $\;\frac13\;$ instead of $\;\frac12\;$ in your formula...
$BA\cdot AM\cdot\sin\widehat{BAM} = AM\cdot AC\sin\widehat{CAM}$ (a median splits a triangle in two parts with the same area) hence $$\frac{\sin\widehat{BAM}}{\sin\widehat{CAM}}=\frac{b}{c}=\frac{\sin\widehat{ABC}}{\sin\widehat{ACB}}.$$
Best Answer
No. For a convenient example, consider the following two cases:
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.