If point O lies inside triangle ABC then $\left\|\frac{\vec{OA}}{|OA|}+\frac{\vec{OB}}{|OB|}+\frac{\vec{OC}}{|OC|}\right\| \leq 1$

Question

Given triangle ABC and a point O lies inside the triangle. Prove that $$\left\|\frac{\vec{OA}}{|OA|}+\frac{\vec{OB}}{|OB|}+\frac{\vec{OC}}{|OC|}\right\| \leq 1$$

I am undergraduate student and I couldn't remember this elementary property. Can anyone give me some detail for the proof? Thank you in advance !

Answer 1

Let $\alpha = \angle AOB, \beta = \angle BOC, \gamma = \angle COA, \vec{u} = \dfrac{\vec{OA}}{||\vec{OA}||}, \vec{v} = \dfrac{\vec{OB}}{||\vec{OB}||}, \vec{w} = \dfrac{\vec{OC}}{||\vec{OC}||}\Rightarrow ||\vec{u}||=||\vec{v}||=||\vec{w}||=1, \alpha+\beta+\gamma = 2\pi \Rightarrow ||\vec{u}+\vec{v}+\vec{w}|| \leq 1\iff ||\vec{u}+\vec{v}+\vec{w}||^2 \leq 1\iff <\vec{u}+\vec{v}+\vec{w},\vec{u}+\vec{v}+\vec{w}> \leq 1\iff ||\vec{u}||^2+||\vec{v}||^2+||\vec{w}||^2+2\left(<\vec{u},\vec{v}>+<\vec{v},\vec{w}>+<\vec{w},\vec{u}>\right)\leq 1\iff <\vec{u},\vec{v}>+<\vec{v},\vec{w}>+<\vec{w},\vec{u}> \leq -1\iff \cos \alpha+\cos \beta+\cos \gamma\leq -1\iff 2\cos^2\left(\dfrac{\alpha+\beta}{2}\right)-1+2\cos\left(\dfrac{\alpha+\beta}{2}\right)\cos \left(\dfrac{\alpha-\beta}{2}\right)\leq -1\iff \cos \left(\dfrac{\alpha+\beta}{2}\right)\left(\cos\left(\dfrac{\alpha+\beta}{2}\right)+\cos\left(\dfrac{\alpha-\beta}{2}\right)\right)\leq 0\iff-2\cos \left(\dfrac{\alpha}{2}\right)\cos \left(\dfrac{\beta}{2}\right)\cos \left(\dfrac{\gamma}{2}\right)\leq 0$. This last inequality is true since each factor is positive as $0 < \alpha, \beta, \gamma < \pi$, proving the claim.

Answer 2

Remind that: $$\forall\vec{u},\|\vec{u}\|^2=\vec{u}\cdot\vec{u},$$ $$\forall\vec{u},\vec{v},\vec{u}\cdot\vec{v}=\|\vec{u}\|\|\vec{v}\|\cos\left(\widehat{\vec{u},\vec{v}}\right).$$ From there, notice that: $$\left\|\frac{\vec{OA}}{|OA|}+\frac{\vec{OB}}{|OB|}+\frac{\vec{OC}}{|OC|}\right\|^2=3+2\left[\cos\left(\widehat{AOB}\right)+\cos\left(\widehat{AOC}\right)+\cos\left(\widehat{BOC}\right)\right].$$ One has now to show that: $$\cos\left(\widehat{AOB}\right)+\cos\left(\widehat{AOC}\right)+\cos\left(\widehat{BOC}\right)\leqslant -1.$$ Since $O$ is inside $ABC$, one has: $$\widehat{AOB}+\widehat{AOC}+\widehat{BOC}=2\pi.$$ Perhaps you or someone else will see how to proceed from there.

N.B. If this doesn't help let me know, I'll delete my answer. :)