#splitters

Today we're going to use one of the most ancient methods of mathematical exploration: alteration of an established theme. Or in other words, we're going to say "Hey, I liked that thing you did. How about if we do that, but different. Is that anything?"

Earlier, we looked at the medians, which are cevians (lines drawn through a vertex) that divide the triangle's area in half. This week, we're going to alter that to look for cevians that divide the triangle's perimeter in half and see what comes of that.

These lines are called splitters, and finding where to draw them turns out to be a surprisingly simply problem. We just use the excircles. The places where they are tangent to the sides of the reference triangle are called extouch points, and those are our targets.

Theorem: a line drawn from a vertex to the extouch point on the opposite side is a splitter of the triangle.

Proof: Let Ta be the extouch point opposite A. Let U and V be points of tangency with the extensions of sides AB and AC respectively. Then because the segments AU and AV are both tangent to the same circle, they are equal. Furthermore we may divide the red segments at B and C, and then

AB + BU = AC + VC.

Tangent segments BU and BTa are also equal, and the same for segments CV and CTa. Then

AB + BTa = AC + CTa

which shows that ATa splits the perimeter in half.

And of course what we love is when three lines all cross at one point. Here are all three excircles and extouch points. If we draw in the splitters, will they all cross together?

Theorem: the splitters of a triangle coincide at a point.

To prove this, we will use Ceva's theorem. We look at the ratios that the sides are divided into by the extouch points, that is, ATc / TcB, BTa / TaC and CTb / Tba. If the product of these three ratios is 1, then the lines ATa, BTb, and CTc all cross at the same point.

Then what are those ratios? The semiperimeter of the triangle is s = (a + b + c)/2, which is the length of either the red or the blue path above. Taking the red path and removing the side c, we get

BTa = s - c

BTa = (a + b - c) /2.

Similarly for the blue path we find

TaC = (a + c - b) /2

and the ratio of the two segments of side a is

(a + b - c) / (a + c - b).

Working our way around the triangle, we get

(a + b - c) / (a + c - b) * (b + c - a) / (a + b - c) * (a + c - b) / (b + c -a)

which cancels out to 1.

Here then is the Nagel point, the intersection of the vertices joined to the extouch points, which is also the intersection of the splitters, and given the symbol X(8).

Oh yeah, I can't keep using single letters as symbols for points indefinitely. There's only so many available. From now on, I'll be referring to triangle centers by their Kimberling numbers, which are named after (and assigned by) Clark Kimberling, a professor of mathematics at the University of Evansville. He is the keeper of the Encyclopedia of Triangle Centers, the sacred texts of our cult of triangles. There are currently 65,385 listed, which I realize is quite a few. (Don't try to memorize them all at once.) Most of them are technical and not of general interest, but many of them are a lot of fun to look at. I'll be back next week with a different alteration on the idea of cutting a triangle in half.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

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