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The problem came from Needham's Visual Complex analysis, so I have been using complex numbers. The set-up is equilateral triangles drawn on the sides of an arbitrary triangle. I have shown that the centroid of the triangle formed by the centroids of the equilateral triangles is the centroid of the original triangle, whether the equilateral triangles are drawn inside or outside the arbitrary triangle. I used a method similar to that of the answer to LevanDokite's question Oct 18'12.
If the centroids of the equilateral triangle are $G_1,G_2,G_3$ I have tried rotating $G_1$ about $G_3$, hoping it would end up as $G_2$. It didn't. I tried several different ways of using complex numbers to describe the set-up, with one vertex of the original triangle as origin or not. No help. I am working with a very basic knowledge of maths and wonder if the proof I need involves theorems I haven't met yet. I don't know how to show lengths are the same.

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  • $\begingroup$ I think a better choice of origin is the centrold of the original triangle. It works out (I verified it in Maple) that with that choice of origin, a rotation of $120^\circ$ about origin maps each of $G_1,G_2,G_3$ to the next counterclockwise one. By hand, it's cumbersome, but with Maple, it's easy. $\endgroup$ – quasi Dec 31 '17 at 14:06
  • $\begingroup$ I did try that. Maple-less I must have made a mistake... $\endgroup$ – Kang Dec 31 '17 at 15:32
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Let the original triangle be $ABC$ with equilateral triangle $ABC_1$, $CAB_1$, and $BCA_1$ built on its sides. Think of all the vertices involved as complex numbers.We apply a classical criterion to the three equilateral triangles. Let $j$ be a suitable rotation through $120°$. Then the fact that triangles $ABC_1$, $CAB_1$, and $BCA_1$ are equilateral may be expressed as

$$A+jB+j^2C_1=0$$ $$C+jA+j^2B_1=0$$ $$B+jC+j^2A_1=0$$

The center of $ΔABC_1$ is given by $$ P=\frac{A+B+C_1}{3}$$, and similarly for centers $Q$ and $R$ of triangles $CAB_1$ and $$BCA_1: Q=\frac{C+A+B_1}{3}$$ and $$R=\frac{B+C+A_1}{3}$$. We want to show that $P+jQ+j^2R=0$. Hence

$$3(P+jQ+j^2R)=A+B+C_1+j(C+A+B_1)+j_2(B+C+A1)=(B+jC+j^2A_1)+j(A+jB+j^2C_1)+j2(C+jA+j^2B_1)=0.$$

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