Prove that the triangle $IJK$ is equilateral

Question

$ABC$ triangle , $M,N\in (AB)$ , $P,Q\in (BC)$ and $R,S\in (AC)$ such that

(divided side by $3$)

$\vec{AB}=3\vec{AM}$ , $\vec{BA}=3\vec{BN}$,

$\vec{BC}=3\vec{BP}$ , $\vec{CB}=3\vec{CQ}$,

$\vec{CA}=3\vec{CR}$ , $\vec{AC}=3\vec{AS}$

Let $I,J,K$ outside triangle $ABC$ such that

$MNI,PQJ,RSK$ Equilateral triangle

Prove that the triangle $IJK$ is equilateral

I need see simple method

My try :

I'm going to solve it by axes $(O,\vec{ON},\vec{j})$

such that $2\vec{AO}=\vec{AB}$ , $j=ON$

and try to prove that

$IJ=IK=JK$

So let searching coordinate of all point $A=(-3,0)$ and $B=(3,0)$ but how I find other coordinate

Please see if I'm correct in chose axes and I like to see other esay method

Answer 1

Vector notation doesn't easily allow for rotations, so we use complex numbers instead:

Let $ \omega$ be the rotation by $60^\circ$. $\omega^6 = 1$, $|\omega| = 1$, $\omega+ \omega^5 = 1$.
$\vec{I} = i = a + (m-a) + (i-m) = a + \frac{1}{3} (b-a) + \frac{1}{3} (b-a) \omega = \frac{(2 - \omega)a + (1+\omega) b}{3}$
$\vec{J} = j = \frac{(2 - \omega)b + (1+\omega) c}{3} $
$\vec{IJ} = j-i = \frac{-(2 - \omega)a + (1 - 2\omega)b + (1+\omega) c}{3}$.
Likewise, $\vec{IK} = - \vec{KI} = -\frac{-(2 - \omega)c + (1 - 2\omega)a + (1+\omega) b}{3} $

Now, observe that $ (1+\omega)\omega^{2} = -(2-\omega) $, $-(2-\omega)\times \omega^2 = (1-2\omega) $, $ ( 1 - 2 \omega) \times \omega^2 = (1+ \omega)$, hence the sides of $IJK$ have equal length and we are done.

Note: We have actually shown that $ \vec{IJ} \times \omega^2 \times (-1) = \vec{IK}$, so these vectors are indeed obtained by a rotation of $60^\circ$.

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Note: If you really have to use vector notation, then you can replace $ \omega \times k$ by $\frac{k}{2} + \frac{\sqrt{3}k^{\perp}}{2}$. It makes the calculations less obvious and more tedious, but essentially identical.

Answer 2

This answer is rather a discussion connecting the OP with the well known

Napoleon's Theorem

and with a synthetic picture to have a better orientation while computing vectorialy in the sequel. So let us enrich the picture with the mid points $A',B',C'$ of the sides of $\Delta ABC$, let $G=AA'\cap BB'\cap CC'$ be its centroid, and we also draw the equilateral triangles $\Delta ABI'$, $\Delta BCJ'$, and $\Delta CAK''$ in the exterior of the given triangle. The picture collects now the following many points:

It is clear that $I, J,K$ are the centers of the three triangles constructed on the sides $AB,BC,CA$.

(Since we have for instance $C'I:C'I'=C'M:C'A=1:3$, so the point $I$ is the centroid of $\Delta ABI'$, so also its center (of symmetry), all first important centers becoming this point.)

(Note also that the three medians $AA',BB',CC'$ and the parallels to the sides $NR,PS,QM$ all go through $G$. And that $II'A'\cap JJ'B'\cap KK'C'$ is $O$, the circumcenter of $\Delta ABC$.)

Let us now show in a vectorial manner:

$\Delta IJK$ is equilateral with center $G$, the same as the centroid of $\Delta BC$.

(There are of course "better" synthetic proofs.)

Proof: We first note the equality: $$ \begin{aligned} &\overline{GI} + \overline{GJ} + \overline{GK} \\ &\qquad = ( \overline{GA'} + \overline{A'J} ) + ( \overline{GB'} + \overline{B'K} ) + ( \overline{GC'} + \overline{C'I} ) \\ &\qquad = ( \overline{GA'} + \overline{GB'} + \overline{GC'} ) + ( \overline{A'J} + \overline{B'K} + \overline{C'I} ) \\ &\qquad= 0+0 \\ &\qquad=0\ . \end{aligned} $$ Here, the equality $\overline{GA'} + \overline{GB'} + \overline{GC'}= -\frac 12(\overline{GA} + \overline{GB} + \overline{GC})=0$ is obvious. We need to waste some words for the second vanishing. For it we observe that the vector $\overline{C'I}$ differs from the side vector $\overline{AB}$

• in magnitude by the factor $\sqrt 3/6$,
• in angle by rotation by $90^\circ$.

Similar properties hold for the other two pairs of vectors, $\overline{A'J}$ and the side vector $\overline{BC}$, respectively $\overline{B'K}$ and the side vector $\overline{CA}$. Then the equality $\overline{AB}+\overline{BC}+\overline{CA}=0$ leads after rotation and rescaling to $\overline{C'I}+\overline{A'J}+\overline{B'K}=0$.

We know so far that $G$ is the centroid of $\Delta IJK$.

So it is enough to show also $GI=GJ=GK$. We then compute $GI$ in terms of the sides $a,b,c$ of $\Delta ABC$, and in terms of other quantities associated with it, till we get a symmetric expression. So we blindly compute: $$ \begin{aligned} {GC'}^2 &=\left(\frac 13 CC'\right)^2 =\frac 19\cdot\frac 14(2a^2+2b^2-c^2)\ , \\ {C'I}^2 & =\left(C'N\sqrt 3\right)^2 =\left(\frac 16 AB\sqrt 3\right)^2 =\frac 3{6^2}c^2\ , \\ \overline{GC'}\cdot\overline{C'I} &=\frac 13 \overline{CC'}\cdot\overline{C'I} \\ &=\frac 16 (\overline{CA}+\overline{CB})\cdot\overline{C'I} \\ &=\frac 16 (\overline{CB}+\overline{BA}+\overline{CB})\cdot\overline{C'I} \\ &=\frac 13 \overline{CB}\cdot\overline{C'I} \\ &=\frac 13 CB\cdot C'I\cdot \cos\widehat{B'C'I} \\ &=\frac 13 a\cdot \frac {\sqrt 3}6c\cdot \cos(B+90^\circ) \\ &=\frac {\sqrt 3}9\cdot \underbrace{\frac 12 ac\sin B}_{\operatorname{Area}(ABC)}\ , \\ GI^2 & = \overline{GI}^2 \\ & = (\overline{GC'}+\overline{C'I})^2 \\ & = \overline{GC'}^2+\overline{C'I}^2 +2\overline{GC'}\cdot\overline{C'I} \\ &= \frac 1{36}(2a^2+2b^2-c^2) + \frac 1{36}\cdot 3c^2 + 2\cdot \frac {\sqrt 3}9\cdot \operatorname{Area}(ABC) \\ &= \frac 1{18}(a^2+b^2+c^2) + \frac {2\sqrt 3}9\cdot \operatorname{Area}(ABC) \ . \end{aligned} $$ The last expression is symmetric w.r.t. the permutation of the vertices of $\Delta ABC$, so $GI=GJ=GK$.

$\square$