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Let $A A_1, BB_1, CC_1$ be the angle bisectors of triangle $ABC$ and $\vec{AA_1}+\vec{BB_1}+\vec{CC_1}=0.$ Prove that $ABC$ is equilateral.

I know that the vectors $\vec{AA_1}$ and $\frac{\vec{AB}}{|AB|}+\frac{\vec{AC}}{|AC|}$ ( and so on) are collinear but not find how it can help.

Any ideas?

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    By "bisector", do you mean *angle* bisector or *side* bisector? – 2017-01-29
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    angle bisector, bisectrix – 2017-01-29

3 Answers 3

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Using trilinear or barycentric coordinates is a sensible choice, but we may go through "pure" geometry as well. Let $B''$ be such that $A'B''$ is parallel to $BB'$ and has the same length. We have $\vec{AA'}+\vec{BB'}+\vec{CC'}=0$ iff $CC'AB''$ is a parallelogram. In such a case $B'$ has to be the midpoint of $AC$, hence $BA=BC$ by the bisector theorem. Since $C',B',B''$ have to be collinear, by Thales' theorem $C'$ is the midpoint of $AB$ and $CB=CA$. Done.

enter image description here

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    [+1] Short and interesting proof. – 2017-01-30
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    Why $CC'AB''$ must be a parallelogram? – 2017-01-30
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    @Leox: $BA'B''B'$ is a parallelogram by construction, and we want that $AA'+BB'+CC'=0$, i.e. $\vec{B''A}=\vec{CC'}$. – 2017-01-30
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We have, for suitable non-zero constants $u,v,w$, $$ u\left(\frac{\vec{AB}}{|AB|}+\frac{\vec{AC}}{|AC|}\right)+ v\left(\frac{\vec{BC}}{|BC|}+\frac{\vec{BA}}{|AB|}\right)+ w\left(\frac{\vec{CA}}{|AC|}+\frac{\vec{CB}}{|BC|}\right)=0, $$ or $$\frac {u-v}{|AB|}\vec{AB}+\frac {v-w}{|BC|}\vec{BC}+\frac {w-u}{|AC|}\vec{CA}=0,$$ or, using $\vec{CA}=-\vec{AB}-\vec{BC}$, $$ \left(\frac {u-v}{|AB|}-\frac {w-u}{|AC|}\right)\vec{AB}+\left(\frac {v-w}{|BC|}-\frac {w-u}{|AC|}\right)\vec{BC}=0.$$ As $\vec{AB}$, $\vec{BC}$ are linearly independent, this menas that both coefficients are $0$.

if $u>v$, then form the coefficient of $\vec{AB}$, we find $w>u$, and then from the other coefficient $v>w$, contradiction. Likewise, $u

Can you take it from here?

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    not yet. Why $u=v=w$ implies that $ABC$ is equilateral? – 2017-01-29
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    [+1] Your solution and mine have some common points... – 2017-01-29
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enter image description here Using the so-called angle bisector theorem, one can write (with usual notations $a,b,c$ for the side lengths $BC,CA,AB$ resp.):

$$\tag{1}\vec{AA_1}=\dfrac{b}{b+c}\vec{AC}+\dfrac{c}{b+c}\vec{AB}$$

In the same way, by cyclic permutation:

$$\vec{BB_1}=\dfrac{c}{c+a}\vec{BA}+\dfrac{a}{c+a}\vec{BC}$$

which can be written:

$$\tag{2}\vec{BB_1}=-\dfrac{c}{c+a}\vec{AB}+\dfrac{a}{c+a}\vec{AC}-\dfrac{a}{c+a}\vec{AB}$$

and finally:

$$\vec{CC_1}=\dfrac{a}{a+b}\vec{CB}+\dfrac{b}{a+b}\vec{CA}$$

which can be written:

$$\tag{3}\vec{CC_1}=\dfrac{a}{a+b}\vec{AB}-\dfrac{a}{a+b}\vec{AC}-\dfrac{b}{a+b}\vec{AC}$$

adding (1)+(2)+(3) gives

$$\tag{4}\vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\alpha \vec{AB}+\beta \vec{AC}$$

with $\alpha=\frac{c}{b+c}-\frac{c}{c+a}-\frac{a}{c+a}+\frac{a}{a+b}=\frac{ac-b^2}{(a+b)(a+c)}$ and $\beta=\frac{b}{b+c}+\frac{a}{a+c}-\frac{a}{a+b}-\frac{b}{a+b}=\frac{ab-c^2}{(b+c)(a+c)}.$

$\{\vec{AB},\vec{AC} \}$ being a basis of the plane, we will have a zero sum in relationship (4) if and only if the two numerators in the expressions of $\alpha$ and $\beta$ are zero, i.e.,

$ac=b^2$ and $ab=c^2$, whence $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}$ from which, denoting the common value of these ratios by $k$ we obtain $a=kb, b=kc,c=ka$. Thus $k^3=1$ which is equivalent to $k=1$. Consequence: $a=b=c$.

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    Thank you. please, explain more the coefficients in (1). – 2017-01-29
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    Do you know the definition of $G$ as barycenter of $P$ and $Q$ with resp. weights $p$ and $q$ as being such that, for any origin $O, \ $ $(p+q)\vec{OG}=p\vec{OP}+q\vec{OQ}$, or equivalently $\vec{OG}=\frac{p}{p+q}\vec{OP}+\frac{q}{p+q}\vec{OQ}$ ? – 2017-01-29
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    yes, I see now.. – 2017-01-29