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On the outside of triangle ABC construct equilateral triangles $ABC_1,BCA_1, CAB_1$, and inside of ABC, construct equilateral triangles $ABC_2,BCA_2, CAB_2$. Let $G_1,G_2,G_3$ $G_3,G_4,G_6$be respectively the centroids of triangles $ABC_1,BCA_1, CAB_1$, $ABC_2,BCA_2, CAB_2$.

Prove that the centroids of triangle $G_1G_2G_3$ and of triangle $G_4G_5G_6$ coincide

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Not only do the centroids of $\triangle G_1 G_2 G_3$ and $\triangle G_4 G_5 G_6$ coincide with each other, they coincide with the centroid of $\triangle ABC$. In fact, the "$G$" triangles are themselves equilateral, and they have a name: (inner and outer) Napoleon Triangles.

Note that we can "construct" $G_1$ by rotating $\overrightarrow{AB}$ by $30^\circ$ about $A$ and then scaling by $\frac{\sqrt{3}}{2}\cdot \frac{2}{3} = \frac{\sqrt{3}}{3}$. Writing $R$ for the rotation operator, and $k$ for the scale factor, we have $$G_1 = A + k\;R\;(B-A) \qquad G_2 = B + k\;R\;(C-B) \qquad G_3 = C + k\;R\;(A-C)$$ (In $\mathbb{R}^2$, take $R$ to be the appropriate rotation matrix; in the complex plane, take $R = e^{i\pi/6}$ (or $e^{-i\pi/6}$, if we should be rotating clockwise).) Then, answering the question is easy: $$\begin{align} \text{centroid of } \triangle G_1 G_2 G_3 &= \frac{1}{3}\left( G_1 + G_2 + G_3 \right) \\[6pt] &=\frac{1}{3}\left( A + B + C + k R\left(\; B - A + C - B + A - C \;\right)\;\right) \\[6pt] &= \frac{1}{3}\left( A + B + C \right) \\[4pt] &= \text{ centroid of } \triangle ABC \end{align}$$

The same goes for $\triangle G_4 G_5 G_6$, except that we rotate the vectors in the opposite direction.

Proving that the "$G$" triangles are equilateral is left as an exercise for the reader. (One might notice that the specific angle for $R$, and the specific value of $k$, are actually irrelevant to the proof of the centroid relation, since the rotated terms cancel-out. They are, however, key to guaranteeing the equilateral property.)

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The easiest way to get at the result is to apply an affine transformation to the original triangle so as to make it equilateral. The constructed "interior" triangles will all then coincide with the original triangle, and the conclusion can be easily drawn using basic geometry.

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  • $\begingroup$ it's hard for me to understand the solution which all in English , can you use math formular? $\endgroup$ Commented Oct 19, 2012 at 14:18
  • $\begingroup$ Can you solve the case when ABC is equilateral? $\endgroup$
    – NoClue
    Commented Oct 19, 2012 at 16:39
  • $\begingroup$ Better yet, do you see that $G_1G_2G_3$ and $G_4G_5G_6$ are equilateral? $\endgroup$
    – NoClue
    Commented Oct 19, 2012 at 16:42
  • $\begingroup$ ok thanks you :D $\endgroup$ Commented Oct 20, 2012 at 14:04
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    $\begingroup$ @NoClue: The triangles that were equilateral before the affine transformation are probably not equilateral after the affine transformation. I think that this messes up the symmetry of the transformed picture. $\endgroup$
    – robjohn
    Commented Jul 14, 2013 at 19:20

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