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Suppose $\vec{a}=\overrightarrow{\textrm{BC}}, \vec{b}=\overrightarrow{\textrm{CA}}, \vec{c}=\overrightarrow{\textrm{AB}}$ for a triangle $\triangle \rm{ABC}$ in a plane.

Let $\vec{p}=(\vec{a} \cdot \vec{b})\vec{c} + (\vec{b} \cdot \vec{c})\vec{a} + (\vec{c} \cdot \vec{a})\vec{b}$.

It can be shown that:

  • $\vec{p}=\vec{0}$ iff $\triangle \rm{ABC}$ is an equilateral traingle.
  • $\lvert \vec{p} \rvert = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \lvert \vec{c} \rvert$ iff $\triangle \rm{ABC}$ is a right traingle.

I met a problem where the objective is to prove above properties. But I can't quite know what does $\vec{p}$ mean geometrically. Arranging it gets:

$$\frac{\vec{p}}{\lvert\vec{a}\rvert\lvert\vec{b}\rvert\lvert\vec{c}\rvert}=-(\cos \textrm{C} \; \vec{c}+\cos \textrm{A} \; \vec{a}+\cos \textrm{B} \; \vec{b})$$

As the dimension make sense, this vector should have some kind of property if it does. But I don't see any specific property, nor a relationship to a particular point - like circumcenter, incenter, etc. Can anybody find an explanation of what this vector is?

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The following gives an interpretation of the magnitude $p = \left|\vec p\right|\,$. "Squaring" the definition of $\vec p\,$:

$$ \begin{align} p^2 = (\vec p \cdot \vec p) &= \left((\vec{a} \cdot \vec{b})\vec{c} + (\vec{b} \cdot \vec{c})\vec{a} + (\vec{c} \cdot \vec{a})\vec{b}\right) \cdot \left((\vec{a} \cdot \vec{b})\vec{c} + (\vec{b} \cdot \vec{c})\vec{a} + (\vec{c} \cdot \vec{a})\vec{b}\right) \\[5px] &= (\vec{a} \cdot \vec{b})^2 c^2 + (\vec{b} \cdot \vec{c})^2 a^2 + (\vec{c} \cdot \vec{a})^2 b^2 + 6 (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})(\vec{c} \cdot \vec{a}) \\[5px] &= a^2b^2c^2\left(\cos^2A+\cos^2B+\cos^2C - 6\cos A \cos B \cos C\right) \\[5px] &= a^2b^2c^2(1 - 8 \cos A \cos B \cos C) \end{align} $$

The last step used the identity $\cos^2A+\cos^2B+\cos^2C = 1 - 2 \cos A \cos B \cos C\,$, a proof of which can be found here for example.

It follows from the latter expression that:

  • $p = abc \iff \cos A \cos B \cos C = 0\;$ i.e. for right triangles

  • $p= 0 \iff \cos A \cos B \cos C = \frac{1}{8}\;$ which implies $A=B=C$ by AM-GM and the concavity of $\cos(x)$ for acute angles (full proof given at the same link).

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