# A Triangle’s Tridpoint-Hexagon Any triangle may be named triangle ABC. Each of its sides will contain exactly two points — called “tridpoints” — which divide that side into three segments of equal length. In triangle ABC above, the tridpoints are named in such a way that two of them, E and F, are encountered, in that order, if one moves from A to B. On the way from B to C, two more tridpoints are encountered: first F, and then G. Finally, going from C back to A, the last two tridpoints are found: first H, and then J. If a polygon is formed using those six tridpoints in alphabetical order (matching the order of their placement), that polygon is a convex hexagon, DEFGHJ. Another name for it is hexagon DJHGFE, which I mention only because Geometer’s Sketchpad called it that, in the picture above, when I asked it for the area of this hexagon, shown in green. The original triangle, ABC, includes both the yellow and green regions, and I asked Sketchpad for the area of this triangle, also, as well as the hexagon-to-triangle area ratio, which is shown above as the familiar “decimalized” version of the fraction 2/3.

A nice feature of Sketchpad is that you can do things like this — and then move points around, to see what effect that has on measured and calculated values. When I move points A, B, and/or C, the triangle and hexagon areas, of course, change. Their area ratio, however, remains at a decimal which is a rounded-off version of 2/3. It doesn’t change at all, no matter where A, B, and C are placed. Any triangle’s tridpoint-hexagon has an area exactly 2/3 that of the original triangle.

This is not yet a theorem — because what is written above is an explanation, not a proof. I’ve started working on a proof for this conjecture in my head, and will post it on this blog when/if I successfully complete it.

[Later edit — one of my readers provided a proof, so now it’s a theorem. For his proof, see the first comment on this post.]

## 2 thoughts on “A Triangle’s Tridpoint-Hexagon”

1. Okay, you got me interested. Hope you don’t mind if I take a crack.

Put any side at the base, such as BC. The top tridpoint connecting line, in this case DJ, is parallel to the base. It forms the triangle ADJ, which is similar to the full triangle because its base, DJ, is parallel to the big one, BC. The height and base are both 1/3 that of the original triangle, as they are similar triangles and the tridpoints are 1/3 along.

Therefore, triangle ADJ’s area is 1/9 of the total area (1/2 b/3 h/3 = A/9).

You can do the same for each base, so each corner triangle has 1/9 the overall area. Add these three corner triangles up and they make 1/3 the total area, leaving 2/3 for the hexagon.

(If you locate the centroid, you can further divide that hexagon into 6 more triangles, then geometrically you can see the ratio 1:2.)

Thanks for piquing my interest. 🙂

Liked by 1 person

• Nicely done — NOW it’s a theorem!

Like