I came across the following geometry problem.

In the exterior of a triangle $ABC$ three equilateral triangles $ABC' , BCA'$ and $CAB'$ are constructed. Prove that the centroids of these triangles are the vertices of an equilateral triangle.

The book presents a proof using complex numbers which is fairly elementary.

What intrigues me is that the author leaves a note: "The reader who wants a 'proof without words' should examine the diagram"

Diagram attatched below, can someone show me how this proves anything at all?

enter image description here

  • $\begingroup$ This figure is very interesting. Do you remember the book where you found it ? $\endgroup$ – Jean Marie Dec 6 '17 at 20:13

I've annotated the image below. $K$ is the triangle we start with, K1,K2,K3 are the three equilateral triangles constructed around it, and the solid green triangle is the triangle having the three centroids of K1,K2,K3 as vertices

Now the idea is that the green line connecting the centroid of K1 and the centroid of K2 has the same length as the dotten green line connecting the centroids of K2 and K1'. This is the case because the two lines are constructed in exactly the same way, only that $K1'$ takes the place of $K1$ and K3' takes the place of K3.

The same reasoning works for all the other dotted green lines, which shows that these lines form an equilateral hexagon. Call the remaining vertices of the hexagon $K2', K1'',$ and $K2''$, so that the hexagon's vertices read $K1K2K1'K2'K1''K2''$ in clockwise order.

Now consider the angles of this hexagon. The angles at $K1$, $K1'$, and $K1''$ are equal by construction. The angles at $K2$, $K2'$, and $K2''$ are also equal by construction. Then notice that this hexagon, when translated (but not rotated!), tiles the plane in the way shown in the diagram. Therefore, looking at vertex $K1'$ where three hexagons meet, we have that angles $K1$, $K1'$, and $K1''$ must add up to $360^\circ$. Therefore each is equal to $120^\circ$. It follows that $K2$, $K2'$, and $K2''$ are also equal to $120^\circ$, so the hexagon is regular.

Since the hexagon is regular, it lies on a circle. Since $K1K3 = K1'K3 = K1''K3$ (by construction), the center of the circle is $K3$. Finally, for a regular hexagon, the sides have the same length as the radius, so we conclude $$ K1K3 = K2K3 = K1K2 $$ which is the desired result.

Annotated Image

Hardly a full proof without words when it requires so much justification, but perhaps the "proof without words" was more of a visualization. Or perhaps there is a simpler way to reason this through using the fact that the hexagon tiles the plane directly.

  • $\begingroup$ You've shown that the $6$-sided polygon is equilateral, but how does that imply it is regular? And what do you mean by "second radius"? $\endgroup$ – 6005 Apr 27 '14 at 18:23
  • $\begingroup$ @Goos With radius I mean one of the lines connecting a vertex on the circumference of the polygon to it's (presumed) center, the centroid of K3. There are two types of radii (or whatever the plural of radius us) - those starting at the centroid of K2 or one of it's copies, and those starting at the centroid of K1 or one of it's copies. To show that the centroid of K3 is the center of the polygon, it suffices (I think) to know that every second radius has the same length, which is the case because they're constructed the same way. $\endgroup$ – fgp Apr 27 '14 at 19:10
  • $\begingroup$ @Goos If a 6-sided polygon is equilateral, it's regular, no? $\endgroup$ – fgp Apr 27 '14 at 19:11
  • $\begingroup$ Not at all @fgp... $\endgroup$ – 6005 Apr 27 '14 at 19:11
  • 1
    $\begingroup$ @Goos Very nice! $\endgroup$ – fgp Apr 27 '14 at 21:12

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