#isotomic conjugates

[squareallworthy]

Triangle Tuesday 11: The Gergonne point, circles, traces, and a new kind of conjugation

A couple of weeks ago, we found a point by drawing lines from the vertices to the points where the excircles tough the triangle (the extouch points). Those three points coincide, and that's the Nagel point, numbered as X(8) in the Encyclopedia of Triangle Centers.

This week we'll do the same, but different. Given a triangle, draw lines from its vertices to the points where the incircle touches the sides (the intouch points). These three lines coincide at the Gergonne point, X(7). Let's prove it exists.

Theorem: lines drawn from the vertices of a triangle to the intouch points of the opposite sides coincide.

Proof: let the intouch points be Ia, Ib, and Ic. The segments AIb and AIc (in red) are tangent to the incircle drawn from the same point, and therefore equal. Similarly BIa = BIc (blue) and Cia = CIb (green).

Let these three lengths be d, e, and f respectively. Then

d/e * e/f * f/d = 1

and therefore by Ceva's theorem, the lines AIa, BIb, and CIc coincide.

Very simple, and similar to the proof of regarding splitters that we looked at with the Nagel point. Too simple! The incircle and excircles are closely related, so these two points that are based on intouch and extouch points should be closely related as well, right?

Here they are together. The Nagel point is defined by the orange lines, and the Gergonne point by the green lines. The lines are in turn defined by the extouch and intouch points. What can we say about their arrangement?

Let's define a certain relationship here. Take a point P and draw its three cevians (blue lines). Call the places that the cevians intercept the sidelines the traces of P, in this case P1, P2, and P3. (This is not exactly standard terminology for these points, but I have seen at least a couple authors use it.) Reflect the traces of P across the midpoints of the sides M1, M2, M3 to produce Q1, Q2, Q3. Then the lines AQ1, BQ2, CQ3 intersect at Q. We say that P and Q are isotomic conjugates.

We can also express the idea of reflecting, for instance P3 across M3 to produce Q3 by saying the segments M3P3 and M3Q3 are equal, or that AQ3 and BP3 are equal.

Will this procedure always produce three lines that coincide at Q? Yes, and we can prove this with Ceva's theorem. Since P1, P2, and P3 are defined by cevians of P, they cut the sides of triangle ABC in ratios that multiply to 1:

AP3/P3B * BP1/P1C * CP2/P2A = 1.

If we go the same way around the triangle but look at the ratios made by the points Q1, Q2, Q3, we get the reciprocals of the three ratios for point P, which must also multiply to 1. Therefore Q1, Q2, and Q3, serving as endpoints of cevians, define the point Q.

And, just as we do with isogonal conjugates, we have a formula in trinlinear coordinates for converting a point to its isotomic conjugate. if a point P has trilinears coordinates

d : e : f

then its isotomic conjugate has trilinears

1/(d * a^2) : 1/(e * b^2) : 1/(f * c^2).

Now we can get back to looking for a connection between the Nagel and Gergonne points.

Theorem: the Nagel and Gergonne points are isotomic conjugates.

Recall from our discussion of the Nagel point that the orange lines are splitters, meaning they cut the triangle's perimeter in half. If we let a, b, and c be the lengths of the sides opposite vertices A, B, and C, then the length AC + ATc is half the perimeter, or (a + b + c)/2. Then the blue segment on the left is

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

Referring to the drawing above where we defined segments d, e, and f, we have

a = e + f -> e = a - f

b = d + f

c = d + e -> e = c - d.

Adding the two equations for e, we have

2 * e = a + c - d - f

and then substituting from the second equation, we have

2 * e = a + c - b

e = (a + c - b)/2

for the length of the blue segment on the right, and we have proved that the two blue segments are equal. Thus, Tc is the reflection of Ic across the midpoint of side c, and the same for the intouch and extouch points on the other sides.

Isotomic conjugates (from Greek iso meaning "same" and tomos meaning "cut" -- same root as in tomography) don't come up as often as isogonal conjugates. The isotomic conjugate of a well-known point is not that often interesting in itself. But I thought this relationship between the Nagel and Gergonne points was worth pointing out because they are defined similarly from the extouch and intouch points, which makes them my favorite isotomic conjugate pair.

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.

An index of all posts in this series is available here.

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Isotomic

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