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?


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.