Recently, I became aware that there are many more triangle centers than the four I learned about in school. This reminded me of a thought I had when I first learned about the incenter: what point would you get if you took a triangle of the three tangent points of the inscribed circle (that I now know is called the contact triangle) an infinite number of times?

Where is it located (e.g. what are its barycentric and/or trilinear coordinates), and does it have a name/number in the Encyclopedia of Triangle Centers? Is it transcendental?


1 Answer 1


Where is it located?

Unfortunately I couldn't find a closed formula for the trilinears yet. Here is what I have so far. Let the outer triangle have edge lengths proportional to $a_1:b_1:c_1$. The trilinear coordinates of the recursive contact center with respect to that triangle would be called $x_1:y_1:z_1$. Now the edge lengths of the contact triangle would be proportional to

$$ a_2 = \sqrt{a_1(b_1+c_1-a_1)} \quad:\quad b_2 = \sqrt{b_1(c_1+a_1-b_1)} \quad:\quad c_2 = \sqrt{c_1(a_1+b_1-c_1)} $$

Now you can use these to recursively compute the trilinear coordinates of the recursive contact center in that contact triangle. (If you truncate the recursion, it makes sense to return the incenter $1:1:1$ as a simple approximation of the recursive contact center.) Once you have the trilinears from the recursive evaluation, which I'll call $x_2:y_2:z_2$, then you'll have to convert them back to the coordinate system of the original triangle, which you can do like this:

$$ x_1 = b_2c_2(c_2y_2 + b_2z_2) \quad:\quad y_1 = c_2a_2(c_2x_2 + a_2z_2) \quad:\quad z_1 = b_2a_2(b_2x_2 + a_2y_2) $$

Note that if you perform the computations as above, without divisions, the numbers will soon become huge. So if you are working with machine floating point numbers, please normalize your homogeneous coordinates from time to time to avoid overflow. This holds for both the edge lengths and the center point.

This is the formula I used in my own computations, and the desired closed formulation would have to be a fixed point of this recursion. Can anyone find a closed formula matching these requirements?

Is it in the Encyclopedia of Triangle Centers?

I've written some code to compute this triangle center recursively. A student of mine has also translated most of the encyclopedia of triangle centers into executable code. Doing numeric comparisons for a number of random test triangles, I found no close match.

Is it transcendental?

Numerical evidence suggests that it is. Using the triangle with $a=3,b=5,c=7$, I found trilinears $x:y:z$ satisfying $x/z\approx0.29$. Computing that ratio with a lot of precision, I found the value to be approximately


I believe all of these digits to be reliable, since they are backed up by two more computation, each at twice the number of bits and twice the recursion depth as the one before. The two verifying computations agreed with one another in approximately twice as many digits as I gave above, so the ones above really should be stable under increased precision or depth.

Approximating that number using PARI's algdep function, and using about half the digits as input, will yield a polynomial (tested up to degree 64) which explains these digits nicely, but completely fails to reproduce the remaining digits. So the number appears not to be algebraic.

If someone has a different approach to test whether a number given by a truncated decimal representation is likely algebraic, feel free to feed my number above to your tools and report your findings in a comment or edit.


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