Triangle center

In some cases these functions are not defined on the whole of ${\displaystyle \mathbb {R} ^{3}.}$ For example, the trilinears of X₃₆₅ which is the 365th entry in the Encyclopedia of Triangle Centers, are ${\displaystyle a^{1/2}:b^{1/2}:c^{1/2}}$ so a, b, c cannot be negative. Furthermore, in order to represent the sides of a triangle they must satisfy the triangle inequality. So, in practice, every function's domain is restricted to the region of ${\displaystyle \mathbb {R} ^{3}}$ where

$${\displaystyle a\leq b+c,\quad b\leq c+a,\quad c\leq a+b.}$$

This region T is the domain of all triangles, and it is the default domain for all triangle-based functions.

There are various instances where it may be desirable to restrict the analysis to a smaller domain than T. For example:

• The centers X₃, X₄, X₂₂, X₂₄, X₄₀ make specific reference to acute triangles, namely that region of T where
  $${\displaystyle a^{2}\leq b^{2}+c^{2},\quad b^{2}\leq c^{2}+a^{2},\quad c^{2}\leq a^{2}+b^{2}.}$$

  
• When differentiating between the Fermat point and X₁₃ the domain of triangles with an angle exceeding 2π/3 is important; in other words, triangles for which any of the following is true:

$${\displaystyle a^{2}>b^{2}+bc+c^{2};\quad b^{2}>c^{2}+ca+a^{2};\quad c^{2}>a^{2}+ab+b^{2}.}$$

• A domain of much practical value since it is dense in T yet excludes all trivial triangles (i.e. points) and degenerate triangles (i.e. lines) is the set of all scalene triangles. It is obtained by removing the planes b = c, c = a, a = b from T.

Not every subset D ⊆ T is a viable domain. In order to support the bisymmetry test D must be symmetric about the planes b = c, c = a, a = b. To support cyclicity it must also be invariant under 2π/3 rotations about the line a = b = c. The simplest domain of all is the line (t, t, t) which corresponds to the set of all equilateral triangles.

The point of concurrence of the perpendicular bisectors of the sides of triangle △ABC is the circumcenter. The trilinear coordinates of the circumcenter are

$${\displaystyle a(b^{2}+c^{2}-a^{2}):b(c^{2}+a^{2}-b^{2}):c(a^{2}+b^{2}-c^{2}).}$$

Let ${\displaystyle f(a,b,c)=a(b^{2}+c^{2}-a^{2}).}$ It can be shown that f is homogeneous:

$${\displaystyle {\begin{aligned}f(ta,tb,tc)&=ta{\Bigl [}(tb)^{2}+(tc)^{2}-(ta)^{2}{\Bigr ]}\\[2pt]&=t^{3}{\Bigl [}a(b^{2}+c^{2}-a^{2}){\Bigr ]}\\[2pt]&=t^{3}f(a,b,c)\end{aligned}}}$$

as well as bisymmetric:

$${\displaystyle {\begin{aligned}f(a,c,b)&=a(c^{2}+b^{2}-a^{2})\\[2pt]&=a(b^{2}+c^{2}-a^{2})\\[2pt]&=f(a,b,c)\end{aligned}}}$$

so f is a triangle center function. Since the corresponding triangle center has the same trilinears as the circumcenter, it follows that the circumcenter is a triangle center.

Let △A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let △AB'C and △ABC' be similarly constructed equilateral triangles based on the other two sides of triangle △ABC. Then the lines AA', BB', CC' are concurrent and the point of concurrence is the 1st isogonal center. Its trilinear coordinates are

${\displaystyle \csc \left(A+{\frac {\pi }{3}}\right):\csc \left(B+{\frac {\pi }{3}}\right):\csc \left(C+{\frac {\pi }{3}}\right).}$

Expressing these coordinates in terms of a, b, c, one can verify that they indeed satisfy the defining properties of the coordinates of a triangle center. Hence the 1st isogonic center is also a triangle center.

Let

$${\displaystyle f(a,b,c)={\begin{cases}1&\quad {\text{if }}a^{2}>b^{2}+bc+c^{2}&\iff {\text{if }}A>2\pi /3\\[8pt]0&\quad \!\!\displaystyle {{{\text{if }}b^{2}>c^{2}+ca+a^{2}} \atop {{\text{ or }}c^{2}>a^{2}+ab+b^{2}}}&\iff \!\!\displaystyle {{{\text{if }}B>2\pi /3} \atop {{\text{ or }}C>2\pi /3}}\\[8pt]\csc(A+{\frac {\pi }{3}})&\quad {\text{otherwise }}&\iff A,B,C>2\pi /3\end{cases}}}$$

Then f is bisymmetric and homogeneous so it is a triangle center function. Moreover, the corresponding triangle center coincides with the obtuse angled vertex whenever any vertex angle exceeds 2π/3, and with the 1st isogonic center otherwise. Therefore, this triangle center is none other than the Fermat point.

The trilinear coordinates of the first Brocard point are:

$${\displaystyle {\frac {c}{b}}\$$ :\ {\frac {a}{c}}\ :\ {\frac {b}{a}}}

These coordinates satisfy the properties of homogeneity and cyclicity but not bisymmetry. So the first Brocard point is not (in general) a triangle center. The second Brocard point has trilinear coordinates:

$${\displaystyle {\frac {b}{c}}\$$ :\ {\frac {c}{a}}\ :\ {\frac {a}{b}}}

and similar remarks apply.

The first and second Brocard points are one of many bicentric pairs of points,[6] pairs of points defined from a triangle with the property that the pair (but not each individual point) is preserved under similarities of the triangle. Several binary operations, such as midpoint and trilinear product, when applied to the two Brocard points, as well as other bicentric pairs, produce triangle centers.

In the following table of more recent triangle centers, no specific notations are mentioned for the various points. Also for each center only the first trilinear coordinate f(a,b,c) is specified. The other coordinates can be easily derived using the cyclicity property of trilinear coordinates.

ETC reference; Name		Center function ${\displaystyle f(a,b,c)}$	Year described
X21	Schiffler point	${\displaystyle {\frac {1}{\cos B+\cos C}}}$	1985
X22	Exeter point	${\displaystyle a(b^{4}+c^{4}-a^{4})}$	1986
X111	Parry point	${\displaystyle {\frac {a}{2a^{2}-b^{2}-c^{2}}}}$	early 1990s
X173	Congruent isoscelizers point	${\displaystyle \tan {\tfrac {A}{2}}+\sec {\tfrac {A}{2}}}$	1989
X174	Yff center of congruence	${\displaystyle \sec {\tfrac {A}{2}}}$	1987
X175	Isoperimetric point	${\displaystyle \sec {\tfrac {A}{2}}\cos {\tfrac {B}{2}}\cos {\tfrac {C}{2}}-1}$	1985
X179	First Ajima-Malfatti point	${\displaystyle \sec ^{4}{\tfrac {A}{4}}}$
X181	Apollonius point	${\displaystyle {\frac {a(b+c)^{2}}{b+c-a}}}$	1987
X192	Equal parallelians point	${\displaystyle bc(ca+ab-bc)}$	1961
X356	Morley center	${\displaystyle \cos {\tfrac {A}{3}}+2\cos {\tfrac {B}{3}}\cos {\tfrac {C}{3}}}$	1978[7]
X360	Hofstadter zero point	${\displaystyle {\frac {A}{a}}}$	1992

In honor of Clark Kimberling who created the online encyclopedia of more than 32,000 triangle centers, the triangle centers listed in the encyclopedia are collectively called Kimberling centers.[8]

A triangle center P is called a polynomial triangle center if the trilinear coordinates of P can be expressed as polynomials in a, b, c.

A triangle center P is called a regular triangle point if the trilinear coordinates of P can be expressed as polynomials in △, a, b, c, where △ is the area of the triangle.

A triangle center P is said to be a major triangle center if the trilinear coordinates of P can be expressed in the form ${\displaystyle f(A):f(B):f(C)}$ where ${\displaystyle f(X)}$ is a function of the angle X alone and does not depend on the other angles or on the side lengths.[9]

A triangle center P is called a transcendental triangle center if P has no trilinear representation using only algebraic functions of a, b, c.

Let f be a triangle center function. If two sides of a triangle are equal (say a = b) then

$${\displaystyle {\begin{aligned}f(a,b,c)&=f(b,a,c)&({\text{since }}a=b)\\&=f(b,c,a)&{\text{(by bisymmetry)}}\end{aligned}}}$$

so two components of the associated triangle center are always equal. Therefore, all triangle centers of an isosceles triangle must lie on its line of symmetry. For an equilateral triangle all three components are equal so all centers coincide with the centroid. So, like a circle, an equilateral triangle has a unique center.

Let

$${\displaystyle f(a,b,c)={\begin{cases}-1&\quad {\text{if }}a\geq b{\text{ and }}a\geq c,\\\;\;\;1&\quad {\text{otherwise}}.\end{cases}}}$$

This is readily seen to be a triangle center function and (provided the triangle is scalene) the corresponding triangle center is the excenter opposite to the largest vertex angle. The other two excenters can be picked out by similar functions. However, as indicated above only one of the excenters of an isosceles triangle and none of the excenters of an equilateral triangle can ever be a triangle center.

A function f is biantisymmetric if

$${\displaystyle f(a,b,c)=-f(a,c,b)\quad {\text{for all}}\quad a,b,c.}$$

If such a function is also non-zero and homogeneous it is easily seen that the mapping

$${\displaystyle (a,b,c)\to f(a,b,c)^{2}\,f(b,c,a)\,f(c,a,b)}$$

is a triangle center function. The corresponding triangle center is

$${\displaystyle f(a,b,c):f(b,c,a):f(c,a,b).}$$

On account of this the definition of triangle center function is sometimes taken to include non-zero homogeneous biantisymmetric functions.

Any triangle center function f can be normalized by multiplying it by a symmetric function of a, b, c so that n = 0. A normalized triangle center function has the same triangle center as the original, and also the stronger property that

$${\displaystyle f(ta,tb,tc)=f(a,b,c)\quad {\text{for all}}\quad t>0,\ (a,b,c).}$$

Together with the zero function, normalized triangle center functions form an algebra under addition, subtraction, and multiplication. This gives an easy way to create new triangle centers. However distinct normalized triangle center functions will often define the same triangle center, for example f and ${\displaystyle (abc)^{-1}(a+b+c)^{3}f.}$

Assume a, b, c are real variables and let α, β, γ be any three real constants. Let

$${\displaystyle f(a,b,c)={\begin{cases}\alpha &\quad {\text{if }}a<b{\text{ and }}a<c&(a{\text{ is least}}),\\[2pt]\gamma &\quad {\text{if }}a>b{\text{ and }}a>c&(a{\text{ is greatest}}),\\[2pt]\beta &\quad {\text{otherwise}}&(a{\text{ is in the middle}}).\end{cases}}}$$

Then f is a triangle center function and α : β : γ is the corresponding triangle center whenever the sides of the reference triangle are labelled so that a < b < c. Thus every point is potentially a triangle center. However the vast majority of triangle centers are of little interest, just as most continuous functions are of little interest.

If f is a triangle center function then so is af and the corresponding triangle center is

$${\displaystyle a\,f(a,b,c):b\,f(b,c,a):c\,f(c,a,b).}$$

Since these are precisely the barycentric coordinates of the triangle center corresponding to f it follows that triangle centers could equally well have been defined in terms of barycentrics instead of trilinears. In practice it isn't difficult to switch from one coordinate system to the other.

There are other center pairs besides the Fermat point and the 1st isogonic center. Another system is formed by X₃ and the incenter of the tangential triangle. Consider the triangle center function given by:

$${\displaystyle f(a,b,c)={\begin{cases}\cos A&{\text{if }}\triangle {\text{ is acute}},\\[2pt]\cos A+\sec B\sec C&{\text{if }}\measuredangle A{\text{ is obtuse}},\\[2pt]\cos A-\sec A&{\text{if either}}\measuredangle B{\text{ or }}\measuredangle C{\text{ is obtuse}}.\end{cases}}}$$

For the corresponding triangle center there are four distinct possibilities:

$${\displaystyle {\begin{aligned}&{\text{if reference }}\triangle {\text{ is acute:}}\quad \cos A\$$ :\,\cos B\ :\,\cos C\\[6pt]&{\begin{array}{rcccc}{\text{if }}\measuredangle A{\text{ is obtuse:}}&\cos A+\sec B\sec C&:&\cos B-\sec B&:&\cos C-\sec C\\[4pt]{\text{if }}\measuredangle B{\text{ is obtuse:}}&\cos A-\sec A&:&\cos B+\sec C\sec A&:&\cos C-\sec C\\[4pt]{\text{if }}\measuredangle C{\text{ is obtuse:}}&\cos A-\sec A&:&\cos B-\sec B&:&\cos C+\sec A\sec B\end{array}}\end{aligned}}}

Note that the first is also the circumcenter.

Routine calculation shows that in every case these trilinears represent the incenter of the tangential triangle. So this point is a triangle center that is a close companion of the circumcenter.

Reflecting a triangle reverses the order of its sides. In the image the coordinates refer to the (c, b, a) triangle and (using "|" as the separator) the reflection of an arbitrary point ${\displaystyle \gamma$ :\beta :\alpha } is ${\displaystyle \gamma \ |\ \beta \ |\ \alpha .}$ If f is a triangle center function the reflection of its triangle center is ${\displaystyle f(c,a,b)\ |\ f(b,c,a)\ |\ f(a,b,c),}$ which, by bisymmetry, is the same as ${\displaystyle f(c,b,a)\ |\ f(b,a,c)\ |\ f(a,c,b).}$ As this is also the triangle center corresponding to f relative to the (c, b, a) triangle, bisymmetry ensures that all triangle centers are invariant under reflection. Since rotations and translations may be regarded as double reflections they too must preserve triangle centers. These invariance properties provide justification for the definition.

Some other names for dilation are uniform scaling, isotropic scaling, homothety, and homothecy.