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I already know that, for a triangle $\Delta ABC$

  • $G$ is the triangle centroid, we have $$\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}.$$
  • $I$ is the incenter, we have $$a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}.$$
  • $H$ is the the orthocenter, we have $$(a\cos B\cos C)\vec{HA} + b(\cos C\cos A)\vec{HB} + (c\cos A\cos B) \vec{HC} = \vec{0}.$$

But I don't know how to do the same with the circumcenter. Anyone can give me the hint?

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1 Answer 1

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Let us assume we want to find $\alpha_{A,B,C}$ such that $$\alpha_A \vec{OA} + \alpha_B \vec{OB} + \alpha_C \vec{OC} = \vec{0},$$ where $O$ is a given point (orthocenter, etc).

You can vector-multiply the above equality as in:

(1) multiply to the right by $\vec{OB}$ gives: $$\alpha_A \vec{OA} \times \vec{OB} + \alpha_C \vec{OC} \times \vec{OB} = \vec{0}.$$

(2) multiply to the right by $\vec{OA}$ gives: $$\alpha_B \vec{OB} \times \vec{OA} + \alpha_C \vec{OC} \times \vec{OA} = \vec{0}.$$

You notice that $\alpha_A \propto \vec{OB} \times \vec{OC}$, $\alpha_C \propto \vec{OA} \times \vec{OB}$, etc. BUT, the vector products are signed areas, eg, $\vec{OA} \times \vec{OC}$ is $S_{OAC}$ and so on (taking all ares as positive). After you compute the area, last thing you must take into account is to normalize $\alpha's$ such that their sum is $1$: $\alpha_A + \alpha_B + \alpha_C=1$ (the latter normalization is not required, since it will simplify out in the sum).

Final answer is: $\sin(2A) \vec{OA} + \dots = 0$ (since the area $OBC = R^2/2 \sin2A$). https://en.wikipedia.org/wiki/Circumscribed_circle#Barycentric_coordinates

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  • $\begingroup$ @GAVD: you got it wrong. I quote from the link: "In terms of the triangle's angles α , β , γ , the barycentric coordinates of the circumcenter are[2]: sin ⁡ 2 α : sin ⁡ 2 β : sin ⁡ 2 γ" $\endgroup$
    – Chip
    Jul 4, 2016 at 10:01
  • $\begingroup$ Ah, thank you! After some computation, I got the result. I delete the previous comment. $\endgroup$
    – GAVD
    Jul 4, 2016 at 23:50

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