An Unexpected Pair of Similar Triangles Which Are Equal

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12 July 2013, Created with GeoGebra

Problem

Given two similar triangles $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2}.$

  1. The centroids of the triangles $A_{1}B_{2}C_{2},$ $A_{2}B_{1}C_{2},$ and $A_{2}B_{2}C_{1}$ form a similar triangle $ABC$.

  2. The centroids of the triangles $A_{2}B_{1}C_{1},$ $A_{1}B_{2}C_{1},$ and $A_{1}B_{1}C_{2}$ form a similar triangle $A'B'C'$.

  3. Triangles $ABC$ and $A'B'C'$ are centrally symmetric in the centroid $(\alpha_1+\beta_1+\gamma_1+\alpha_2+\beta_2+\gamma_2)/6$ of the six given vertices.

  4. Two similar triangles - surprise in itself - that happen to be equal, even centrally symmetric

Hint

As in several similar problems, complex numbers show the clearest way to achieve the goal.

Solution

Two similar triangles - surprise in itself - that happen to be equal, even centrally symmetric

I associate indexed complex numbers $\alpha$ with corresponding $A$ vertices, $\beta$ with $B,$ and $\gamma$ with $C$. Let $d$ be a complex number that defines the shape of the given triangles $A_1B_1C_1$ and $A_2B_2C_2$ in the following sense:

$\gamma_1=\alpha_1+d(\beta_1-\alpha_1)$ and
$\gamma_2=\alpha_2+d(\beta_2-\alpha_2).$

We also have

$\alpha=(\alpha_1+\beta_2+\gamma_2)/3,$
$\beta=(\alpha_2+\beta_1+\gamma_2)/3,$
$\gamma=(\alpha_2+\beta_2+\gamma_1)/3,$
${\alpha}'=(\alpha_2+\beta_1+\gamma_1)/3,$
${\beta}'=(\alpha_1+\beta_2+\gamma_1)/3,$
${\gamma}'=(\alpha_1+\beta_1+\gamma_2)/3.$

First, we prove Dao Thanh Oai's unexpected observation that $\gamma=\alpha+d(\beta-\alpha).$ Check that

\( \begin{align} \gamma &= (\alpha_2+\beta_2+\gamma_1)/3 \\ &= (\alpha_2+\beta_2+\alpha_1+d(\beta_1-\alpha_1))/3 \end{align} \)

while

\( \begin{align} \alpha+d(\beta-\alpha) &= (\alpha_1+\beta_2+\gamma_2)/3 \\ &\space\space\space+ d[(\alpha_2+\beta_1+\gamma_2)/3-(\alpha_1+\beta_2+\gamma_2)/3] \\ &= (\alpha_1+\beta_2+\alpha_2+d(\beta_2-\alpha_2+\alpha_2+\beta_1-\alpha_1-\beta_2))/3 \\ &= (\alpha_1+\beta_2+\alpha_2+d(\beta_1-\alpha_1))/3. \end{align} \)

Hence, $\gamma=\alpha+d(\beta-\alpha),$ making $\Delta ABC$ similar to the given two triangles. Obviously, in the same manner we can show that $\Delta A'B'C'$ is also similar to the given two.

Now, let's find the midpoints of segments $AA',$ $BB',$ and $CC'.$ Say,

$ \begin{align} (\alpha+{\alpha}')/2&=[(\alpha_1+\beta_2+\gamma_2)/3+(\alpha_2+\beta_1+\gamma_1)/3]/2 \\ &=(\alpha_1+\beta_1+\gamma_1+\alpha_2+\beta_2+\gamma_2)/6. \end{align} $

Similarly,

$(\beta+{\beta}')/2=(\alpha_1+\beta_1+\gamma_1+\alpha_2+\beta_2+\gamma_2)/6$ and
$(\gamma+{\gamma}')/2=(\alpha_1+\beta_1+\gamma_1+\alpha_2+\beta_2+\gamma_2)/6,$

making triangles $ABC$ and $A'B'C'$ centrally symmetric and, hence, equal.

Acknowledgment

Dao Thanh Oai has defined and proved in a post at the CutTheKnotMath facebook page that $\Delta ABC$ defined above is similar to triangles $A_1B_1C_1$ and $A_2B_2C_2.$ It was just one extra step to reverse the indices $1\leftrightarrow 2\space.$

P.S.

Dao Thanh Oai has observed that the centroids of the four triangles and the six vertices are collinear. For example,

$(\alpha+\beta+\gamma)/3=[(\alpha_1+\beta_1+\gamma_1)+2(\alpha_2+\beta_2+\gamma_2)]/3.$

Similarly,

$({\alpha}'+{\beta}'+{\gamma}')/3=[2(\alpha_1+\beta_1+\gamma_1)+(\alpha_2+\beta_2+\gamma_2)]/3.$

Thus, if we denote the centroids of $A_1B_1C_1,$ $A'B'C',$ $ABC,$ and $A_2B_2C_2$ as, respectively, $K_1,$ $K',$ $K,$ and $K_2,$ the points are indeed collinear and are always located in the following sequence:

Two similar triangles - five collinear points

Points $K_1,$ $K',$ $K,$ and $K_2,$ are equally spaced, while $G$ is the midpoint of two pairs of points: $K_1,$ $K_2,$ and $K',$ $K.$

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