Dynamic Triangle Geometry: Families of Lines with Equal Intercepts

Appendices

Appendix

We record the coordinates of points and equations of lines and conics necessary to justify the results stated in the present paper, which were obtained by experimentation with the Geometer’s Sketchpad$^{\circledR}.$ We use homogeneous barycentric coordinates with respect to triangle ABC. For every point P in the plane of triangle ABC, its homogeneous barycentric coordinates are proportional to the signed areas $(\triangle PBC:\triangle PCA:\triangle PAB).$ Some elementary examples on the use of barycentric coordinates can be found in Yiu (2000).

Classical Triangle Centers

Triangle center	Notation	Homogeneous barycentric coordinates
Centroid	G	(1:1:1)
Circumcenter	O	(a 2(b 2 + c 2 − a 2):b 2(c 2 + a 2 − b 2):c 2(a 2 + b 2 − c 2))
Orthocenter	H	$\left(\frac{1}{b^2\,+\,c^2\,-\,a^2}:\frac{1}{c^2\,+\,a^2\,-\,b^2}:\frac{1} {a^2\,+\,b^2\,-\,c^2}\right)$
Incenter	I	(a:b:c)
Gergonne point	G e	$\left(\frac{1}{b\,+\,c\,-\,a}:\frac{1}{c\,+\,a\,-\,b}: \frac{1}{a\,+\,b\,-\,c}\right)$
Nagel point	N a	(b + c − a : c + a − b : a + b − c)

The Lines ${\mathcal{L}}_a(t)$ etc and Their Equations

$$ \begin{array}{llll} {\mathcal{L}}_a(t): & & Y_a = (t:0:b-t), & Z_a = (t:c-t:0); \\ {\mathcal{L}}_b(t): & X_b = (0:a-t:t), & \qquad & Z_b = (c-t:t:0);\\ {\mathcal{L}}_c(t): & X_c = (0:t:a-t), & Y_c = (b-t:0:t).& \end{array} $$

$$ \begin{array}{ll} {\mathcal{L}}_a(t): & -(b-t)(c-t)x + t(b-t)y + t(c-t)z = 0\\ {\mathcal{L}}_b(t): & t(a-t)x -(c-t)(a-t)y + t(c-t)z = 0\\ {\mathcal{L}}_c(t): & t(a-t)x + t(b-t)y-(a-t)(b-t)z = 0\\ \end{array} $$

The Envelope of ${\mathcal{L}}_a(t)$

The envelope of ${\mathcal{L}}_t$ has equation

$$ (b-c)^2x^2+b^2y^2+c^2z^2+2bcyz-2c(b-c)zx+2b(b-c)xy =0. $$

Its focus is the point F = (−a ² : b(b − c):c(c − b)) and its axis is the line cy + bz = 0, and directrix

$$ -(b-c)(b^2+c^2-a^2)x + b(c^2+a^2-b^2)y - c(a^2+b^2-c^2)z = 0. $$

The line ${\mathcal{L}}_a(t)$ intersects the external bisector of angle A at

$$ Q'= ((b-c)t(b+c-t):b(b-t)(c-t):-c(b-t)(c-t)). $$

The reflection of Q′ in the segment Y _a Z _a is the point

$$ Q = ((b-c)t^2:-b(c-t)^2:c(b-t)^2), $$

which is the point of tangency of ${\mathcal{L}}_a(t)$ with the envelope.

The Triangle Γ

The vertices of the triangle Γ(t) are

$$ \begin{aligned} X_t=\,& (-a(a-2t)(b-t)(c-t) : t(a-t)((a+b-c)t-ab) : t(a-t)((c+a-b)t-ca)), \\ Y_t=\,& (t(b-t)((a+b-c)t-ab):-b(a-t)(b-2t)(c-t):t(b-t)((b+c-a)t-bc)),\\ Z_t=\,& (t(c-t)((c+a-b)t-ca):t(c-t)((b+c-a)t-bc):-c(a-t)(b-t)(c-2t)). \end{aligned} $$

This triangle is perspective with ABC at the point

$$ P(t)=\left(\frac{1}{bc - (b+c-a)t} : \frac{1}{ca-(c+a-b)t} : \frac{1}{ab-(a+b-c)t} \right). $$

The Three Lines

$$ \begin{array}{ll} {\mathcal{L}}_a(a): & (a-b)(c-a)x - a(a-b)y + a(c-a)z = 0\\ {\mathcal{L}}_b(b): & b(a-b)x +(a-b)(b-c)y - b(b-c)z = 0\\ {\mathcal{L}}_c(c): & -c(c-a)x + c(b-c)y+(b-c)(c-a)z = 0\\ \end{array} $$

all contain the infinite point (a(b − c):b(c − a):c(a − b)) and are perpendicular to the line OI:

$$ bc(b-c)(b+c-a)x + ca(c-a)(c+a-b)y + ab(a-b)(a+b-c)z = 0. $$

The Three Lines

$$ \begin{array}{ll} {\mathcal{L}}_a(-a): & (a+b)(c+a)x + a(a+b)y + a(c+a)z = 0\\ {\mathcal{L}}_b(-b): & b(a+b)x +(a+b)(b+c)y + b(b+c)z = 0\\ {\mathcal{L}}_c(-c): & c(c+a)x + c(b+c)y+(b+c)(c+a)z = 0\\ \end{array} $$

bound the triangle with vertices

$$ \begin{aligned} X=\,& (-a(b+c)(a+b+c) : b(c+a)(a+b-c) : c(a+b)(c+a-b)), \\ Y=\,& (a(b+c)(a+b-c) : -b(b+c)(a+b+c) : c(a+b)(b+c-a)),\\ Z=\,& (a(b+c)(a+b-c) : b(c+a)(b+c-a) : -c(a+b)(a+b+c)). \end{aligned} $$

This is perspective with ABC at the point

$$ \left(\frac{a(b+c)}{b+c-a}:\frac{b(c+a)}{c+a-b}:\frac{c(a+b)}{a+b-c}\right), $$

which is the orthocenter of the intouch triangle.