Points collinear with the orthocenter and incenter

Question

Let $ABC$ be a triangle, $H$ be its orthocenter and $I$ its incenter.

1. What properties does the line $HI$ have? What known points in the triangle belong to it?
2. What is the barycentric equation of the line $HI$?

(I tried to write it by using the formulas $H = ((a^2 +b^2 −c^2)(c^2 +a^2 −b^2) : (b^2 +c^2 −a^2)(a^2 +b^2 −c^2) : (c^2 +a^2 −b^2)(b^2 +c^2 −a^2))$ and $I=(a:b:c)$, but the calculations seems terrible, and I think that there is a smarter way to do this)

Answer 1

Let us use the ETC as a reference. Then for a triangle $\Delta ABC$ with sides $a,b,c$ the "remarkable" triangle centers are humanly labeled as $X(j)$, and $j$ is in a finite list of indices $1,2,3,4...$ - for instance $X(1)$ is the incenter, denoted by $I$ in the question, and $X(4)$ is the orthocenter.

Question $(1)$: The line $X(1)X(4)$ - or $HI$ in the question - is determined by going through these two points. Which properties it has? (This is rather a soft question...) Maybe it is enough to see some other points on this line. Searching for 1,4 in ETC gives some points on it, the points $X(j)$ for $j$ among... $$ 33\ ,\ 34\ ,\ 73\ ,\ 225\ ,\ 226\ ,\ 243\ ,\ 278\ ,\ 388\ ,\ 497\ ,\ 515\ ,\ 518\ ,\ 944\ ,\ 946\ ,\ 948\ ,\ 950\ ,\ 1056\ ,\ \dots $$ and so on. (I may have missed some points. Some point $X(j)$ above is defined by the intersection of $X(1)X(4)$ with some other specific line, so this is not really giving a striking property of $X(j)$ by definition, but other properties of $X(j)$ lead to geometric results.)

Question $(2)$ The equation is $$ 0= \begin{vmatrix} x&y&z\\ a&b&c\\ \displaystyle\frac 1{b^2+c^2-a^2}& \displaystyle\frac 1{c^2+a^2-b^2}& \displaystyle\frac 1{a^2+b^2-c^2} \end{vmatrix} $$ The coefficient in $x$ on the R.H.S. is $$ \begin{aligned} & \begin{vmatrix} b&c\\ \displaystyle\frac 1{c^2+a^2-b^2}& \displaystyle\frac 1{a^2+b^2-c^2} \end{vmatrix} \\ &\qquad= \frac {b(a^2-b^2+c^2)-c(a^2+b^2-c^2)} {(a^2-b^2+c^2)(a^2+b^2-c^2)} \\ &\qquad= \frac {(b-c)(a^2+b^2+c^2)-2(b^3-c^3)} {(a^2-b^2+c^2)(a^2+b^2-c^2)} \\ &\qquad= (b-c)\cdot \frac {(a^2+b^2+c^2)-2(b^2+bc+c^2)} {(a^2-b^2+c^2)(a^2+b^2-c^2)} \\ &\qquad= (b-c)\cdot \frac {a^2-(b+c)^2} {(a^2-b^2+c^2)(a^2+b^2-c^2)} \\ &\qquad= - \underbrace{ \frac{a+b+c}{(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)} }_{\text{symmetric part}} \cdot (b-c) (b+c-a)(b^2+c^2-a^2) \ . \end{aligned} $$ So the equation of the line $IH$ in barycentric coordinates $(x,y,z)$ is: $$ 0 = \sum x\cdot(b-c)(b+c-a)(b^2+c^2-a^2)\ . $$

---

Computer check, here using sage:

var('a,b,c,x,y,z');
def f(x, a, b, c):
    return x*(b - c)*(b + c - a)*(b^2 + c^2 -a^2)

def eq_HI(x, y, z):
    return bool(f(x, a, b, c) + f(y, b, c, a) + f(z, c, a, b) == 0)


I = (a, b, c)
H = (1/(b^2 + c^2 - a^2), 1/(c^2 + a^2 - b^2), 1/(a^2 + b^2 - c^2))
print(f'Is I satisfying the equation? {eq_HI(*I)}')
print(f'Is H satisfying the equation? {eq_HI(*H)}')

Results:

Is I satisfying the equation? True
Is H satisfying the equation? True

Answer 2

The $u:v:w$ barycentric equation of a line between points $P$ and $Q$ is "simply" $$\left|\begin{array}{ccc}u & v & w \\ P_u & P_v & P_w \\ Q_u & Q_v & Q_w\end{array}\right|=0$$ For $H = \dfrac{1}{-a^2+b^2+c^2}:\dfrac{1}{-b^2+c^2+a^2}:\dfrac{1}{-c^2+a^2+b^2}$ and $I= a:b:c$, that equation gets a bit messy, but reduces to

$$\begin{align} u(b-c)(-a+b+c)(-a^2+b^2+c^2) &\\ +\,v(c-a)(-b+c+a)(-b^2+c^2+a^2) &\\ +w(a-b)(-c+a+b)(-c^2+a^2+b^2) &= 0 \end{align}$$

I don't know any particular properties of line $HI$ off-hand.

One way to find additional points on the line is to visit Clark Kimberling's Encyclopedia of Triangle Centers. (A "triangle center" is a point with a fully-symmetric definition in terms of elements of the triangle. Not every notable point is a center, but arguably "most" of them are. Incenter $I$ and orthocenter $H$ have Kimberling identifiers $X(1)$ and $X(4)$.)

As @jjagmath points out in a comment, lists of collinear triangle centers appear in the "Central Lines" entry from the ETC's Tables page. As of 6 December, 2020, line $X(1)X(4)$ is documented as containing $199$ referenced centers with the following Kimberling indices:

1, 4, 33, 34, 73, 223, 225, 226, 243, 278, 388, 497, 515, 581, 944, 946, 948, 950, 1056, 1058, 1064, 1066, 1068, 1070, 1072, 1079, 1457, 1478, 1479, 1490, 1519, 1528, 1541, 1543, 1558, 1659, 1699, 1745, 1750, 1785, 1838, 1848, 1870, 1877, 1891, 2356, 2635, 2647, 2654, 2655, 2947, 3465, 3468, 3475, 3476, 3485, 3486, 3487, 3488, 3583, 3585, 3586, 3944, 4857, 5225, 5229, 5236, 5270, 5290, 5307, 5603, 5658, 5691, 5712, 5713, 5714, 5715, 5716, 5717, 5882, 5930, 6198, 6256, 6260, 6261, 6637, 7009, 7718, 7952, 7967, 8555, 8763, 8862, 8923, 9118, 9121, 9577, 9612, 9613, 9614, 10106, 10368, 10382, 10393, 10454, 10478, 10531, 10532, 10571, 10572, 10595, 10596, 10597, 10629, 10805, 10806, 10888, 11254, 11392, 11393, 11522, 12047, 12053, 12115, 12116, 12608, 12650, 12667, 13161, 13390, 13407, 13464, 13607, 14547, 15237, 15500, 15836, 15942, 16869, 16870, 17985, 18283, 18340, 18393, 18446, 18483, 18513, 18514, 20130, 20264, 20277, 21147, 21620, 21740, 23706, 23710, 23987, 24042, 24210, 24268, 26098, 26332, 26333, 26389, 26413, 30384, 31516, 31528, 31529, 31532, 31533, 31673, 31866, 32056, 32082, 32083, 33106, 33144, 33152, 34029, 34036, 34039, 34045, 34231, 34500, 34913, 34936, 34937, 35015, 35635, 36844, 36985, 38295, 38945, 39579, 39599, 40257, 40264, 40270

I don't believe the ETC is necessarily exhaustive (or infallible) in documenting such collinearities. Lists of collinear points seem to be auto-generated (numerically?) when a new center is added, which happens quite often; the Central Lines list seems to be updated with far less regularity.