Prove that the center of outer Napoleon triangle coincides with the centroid of the original triangle.

Question

Prove that the center of triangle $MLN$ coincides with the centroid of the triangle $ABC$. I have seen this from Wikipedia but I haven't been able to find the proof. So I hope you guys can help me.

Answer 1

Pure geometry solution.

Draw perpendicular bisector from $N$ to $AB$, intersecting $AB$ at $V$. Extend $NV$ to $W$ such that $NV=VW$. Connect $CW$ intersecting $ML$ at $U$. Connect other lines as shown in the picture.

First, we notice $\triangle AWN$ and $\triangle BWN$ are equilateral triangles because $WN=2VN=NB=NA$.

Next, we see $\triangle AMN \cong \triangle WLN$ because $AN=WN, MN=LN$ and $\angle ANM=\angle WNL$. Therefore $MA=LW$ and $MC=LW$ as well. Similarly we can show $CL=MW$ using $\triangle NLB \cong \triangle NMW$.

Therefore, $CMWL$ is a parallelogram. Hence $U$ is the midpoint of $CW$ and $ML$. Now because $V$ is already midpoint of $NW$, we know $UV$ is parallel to $CN$ and has length equal to half of $CN$.

If we connect $CV$ and $NU$, we know they will intersect at the point that divides both line segments into ratio $1:2$. What is that point? Firstly it is the centroid of $\triangle ABC$ because it is on $CV$ and divides the median into $1:2$. Secondly it is the centroid of $\triangle MNL$ because it is on $NU$ and divides the median into $1:2$.

Answer 2

Consider a vector-based approach: Set vectors A, B, C, X, Y, Z corresponding to their respective points in the plane.

Lemma: $\vec {AX}+\vec {BY}+\vec {CZ} = 0$

Proof: $LHS = \vec{AC}+\vec{CX}+\vec{BA}+\vec{AY}+\vec{CB}+\vec{BZ}$. Notice that $\vec{AC}+\vec{CB}+\vec{BA} = 0$. Notice again that after a 60 degrees clockwise rotation, $\vec{CX}+\vec{AY}+\vec{BZ} \to \vec{AC}+\vec{CB}+\vec{BA} = 0$, implying that $\vec{CX}+\vec{AY}+\vec{BZ}=0$. Thus, the total must also be 0.

Using our lemma, we know that $\vec A +\vec B +\vec C = \vec X+ \vec Y + \vec Z$. Notice that the center of the Napoleon triangle here is the centroid of several centroids, which we can express using the average of the vectors. We see here that: $\text{Center of Napoleon triangle}=\frac{\frac{a+b+z}{3}+\frac{b+c+x}{3}+\frac{c+a+y}{3}}{3} = \frac{2a+2b+2c+x+y+z}{9}=\frac{3a+3b+3c}{9} = \frac{a+b+c}{3}$, so it coincides with the centroid of $\triangle ABC$.