# INDEX OF TRIANGLES REFERENCED IN ETC

César E. Lozada - December 28, 2016

Last update: Aug 21, 2021.

# A:

## ABC

The reference triangle

A-vertex coordinates:

Trilinears    1 : 0 : 0

## AAOA (antiAOA)

Let L'A be the orthic axis of the A-anti-altimedial triangle, and define L'B, L'C cyclically. Let A" = L'B∩L'C, and define B", C" cyclically. The triangle A"B"C" is the antialtimedial orthic axes triangle, or AAOA triangle for short. (Reference: preamble just before X15015)

A-vertex coordinates:

Trilinears
(a^8+3*a^4*b^2*c^2-2*(b^2+c^2)*a^6+2*(b^4-c^4)*(b^2-c^2)*a^2-(b^4-c^4)^2)/(a*b*c) :
(-b^2*(a^2-b^2)-c^4+a^4)*c :
(-c^2*(a^2-c^2)-b^4+a^4)*b

## ABC-X3 reflections

This triangle is obtained by reflecting A, B, C about the circumcenter X3. Reference: TCCT 6.12, p. 159.

A-vertex coordinates:

Trilinears    -1/a : b/SC : c/SB

## AaBbCc and (AaBbCc)*

See 2nd and 4th Przybyłowski-Bollin triangle.

## AiBiCi and (AiBiCi)*

See 1st and 3rd Przybyłowski-Bollin triangle.

## altimedial

Let MA, MB, MC be the midpoints of sides BC, CA, AB, respectively, and HA, HB, HC the feet of the altitudes from A, B, C, respectively. The triangles HAMCMB, MCHBMA, MBMAHC are the A-, B- and C-altimedial triangles. (References: Hyacinthos 253 and preamble just before X15015)

A-altimedial triangle trilinear vertex coordinates:

HA  0 : (-c^2+a^2+b^2)*c : (-b^2+c^2+a^2)*b
MC  b : a : 0
MB  c : 0 : a

Note: Let P be a triangle center. Let PA be P-of-A-altimedial-triangle, and define PB, PC cyclically. Triangle PAPBPC is the P-of-altimedial-triangles triangle, or P-altimedial triangle for short.

## altimedial orthic axes (AOA)

Let LA be the orthic axis of the A-altimedial triangle, and define LB, LC cyclically. Let A' = LB∩LC, and define B', C' cyclically. The triangle A'B'C' is the altimedial orthic axes triangle, or AOA triangle for short. (Reference: preamble just before X15015)

A-vertex coordinates:

Trilinears
((b^2+c^2)*a^6-(b^4+c^4)*a^4-(b^4-c^4)*(b^2-c^2)*a^2+(b^4-c^4)^2)/a :
(b^8-3*b^6*a^2-(3*c^4-4*c^2*a^2-a^4)*b^4+(c^2-a^2)*(7*c^2-3*a^2)*b^2*a^2+2*(c^4-a^4)*(c^2-a^2)^2)/b :
(c^8-3*c^6*a^2-(3*b^4-4*b^2*a^2-a^4)*c^4+(b^2-a^2)*(7*b^2-3*a^2)*c^2*a^2+2*(b^4-a^4)*(b^2-a^2)^2)/c

## 1st and 2nd Altintas-isodynamic

Let ABC be a triangle. The parallel line to BC from P = X(15) (1st isodynamic point) intersect the circumcircle of ABC in A1, A2 and the Simson lines of A1, A2 cut at A'. Define B', C' cyclically. Then A'B'C' is equilateral and it is the 1st Altintas-isodynamic triangle. (Reference: Preamble just before X40933)

When P=X(16) (2nd isodynamic point), the parallel line to BC from P intersect the circumcircle of ABC in two imaginary points, but the Simson lines of these points still intersect in a real point A". The triangle A"B"C" built in this way is also equilateral and is the 2nd Altintas-isodynamic triangle.

A-vertex coordinates:

1st Altintas-isodynamic triangle:
Trilinears    -((S^2+SB*SC)*sqrt(3)*S+(SA+3*SW)*S^2-2*(S^2-SA*SW+SW^2)*SA)(a*(S^2-sqrt(3)*SA*S-2*SA*SW)) : SC/b : SB/c
2nd Altintas-isodynamic triangle:
Trilinears    -(-(S^2+SB*SC)*sqrt(3)*S+(SA+3*SW)*S^2-2*(S^2-SA*SW+SW^2)*SA)/(a*(S^2+sqrt(3)*SA*S-2*SA*SW)) : SC/b : SB/c

## Andromeda

Let A' be the center of the inverse-in-incircle of the A-excircle, and define B' and C' cyclically. The triangle with vertices A,' B, C' is the Andromeda triangle. (Reference: X5573)

A-vertex coordinates:

Trilinears    (a^2+3*(b-c)^2)/(3*a^2+(b-c)^2) : 1 : 1

Let AABACA, ABBBCB, ACBCCC be the A-, B- and C- anti-altimedial triangles of ABC. The triangles AAABAC, BABBBC, CACBCC are the A-, B- and C- adjunct anti-altimedial triangles, respectively. (Reference: preamble just before X15015)

A-adjunct anti-altimedial triangle trilinear vertex coordinates:

AA  -a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c
AB  1/a : 1/(-a^2+b^2+c^2)*b : -1/c
AC  1/a : -1/b : 1/(-a^2+b^2+c^2)*c

Note: Let P be a triangle center. Let P"A be P-of-A-adjunct-anti-altimedial-triangle, and define P"B, P"C cyclically. Triangle P"AP"BP"C is the P-of-adjunct-anti-altimedial-triangles triangle, or P-adjunct-anti-altimedial triangle for short.

## anticomplementary

The triangle A'B'C' whose medial triangle is ABC. It is also the anticevian-triangle of the centroid of ABC.

A-vertex coordinates:

Trilinears    -1/a : 1/b : 1/c

## anti-triangles

If T is a triangle of ABC then the anti-T triangle is the triangle whose T-triangle is ABC.

## anti-altimedial

The triangle AABACA whose A-altimedial triangle is ABC. (Reference: preamble just before X15015)

A-anti-altimedial triangle trilinear vertex coordinates:

AA  -a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c
BA  a/(a^2-b^2+c^2) : 1/b : -1/c
CA  a/(a^2+b^2-c^2) : -1/b : 1/c

Note: Let P be a triangle center. Let P'A be P-of-A-anti-altimedial-triangle, and define P'B, P'C cyclically. Triangle P'AP'BP'C is the P-of-anti-altimedial-triangles triangle, or P-anti-altimedial triangle for short.

## anti-Aquila

The triangle A'B'C' whose Aquila triangle is ABC.

Note: The anti-Aquila triangle is (the orthic triangle)-of-(the 2nd circumperp triangle).

A-vertex coordinates:

Trilinears    (2*a+b+c)/a : 1 : 1

## anti-Ara

The triangle A'B'C' whose Ara triangle is ABC.

Note: The anti-Ara triangle is (the intouch triangle)-of-(the orthic triangle.)

A-vertex coordinates:

Trilinears    (2*SA+SB+SC)/(a*SA) : b/SB : c/SC

## anti-Ascella

The triangle A'B'C' whose Ascella triangle is ABC.

Note: When ABC is acute, the anti-Ascella triangle is (the anticomplementary triangle)-of-((the anticomplementary triangle)-of-(the anti-Conway triangle)).

A-vertex coordinates:

Trilinears    -a : b*(SA+2*SB)/SB : c*(SA+2*SC)/SC

## anti-Atik

The triangle A'B'C' whose Atik triangle is ABC.

Notes: Only when ABC is acute. A' = AX(69) ∩ perpendicular(X(18909), BC)

A-vertex coordinates:

Trilinears    S^2*(4*R^2-SA)/(a*SA^2) : SB/b : SC/c

## anti-1st/2nd Auriga

Triangles A'B'C' whose 1st/2nd Auriga triangle is ABC..

A-vertex coordinates:

anti-1st Auriga:
Trilinears    -(D*a+(a+b+c)*(a*(b^2+c^2)-(b+c)*(b-c)^2))/a : b*(a+b+c)*(a-b+c)-D : c*(a+b+c)*(a+b-c)-D
anti-2nd Auriga:
Trilinears    -(-D*a+(a+b+c)*(a*(b^2+c^2)-(b+c)*(b-c)^2))/a : b*(a+b+c)*(a-b+c)+D : c*(a+b+c)*(a+b-c)+D
where where D=sqrt(r*R+4*R^2).

## 1st anti-Brocard

The triangle A'B'C' whose 1st Brocard triangle is ABC.

Note: For construction, see Bernard Gibert's anti-Brocard triangles and related topics.

A-vertex coordinates:

Trilinears    (a^4-b^2*c^2)/a : (c^4-a^2*b^2)/b : (b^4-a^2*c^2)/c

## 4th anti-Brocard

The triangle A'B'C' whose 4th Brocard triangle is ABC.

Note: For construction, see Bernard Gibert's anti-Brocard triangles and related topics.

A-vertex coordinates:

Trilinears    a*((a^2+b^2+c^2)^2-9*b^2*c^2)/(-5*a^2+b^2+c^2) : b*(a^2+b^2-2*c^2) : c*(a^2-2*b^2+c^2)

## 5th anti-Brocard

The triangle A'B'C' whose 5th Brocard triangle is ABC.

Note: For construction, see Bernard Gibert's anti-Brocard triangles and related topics.

A-vertex coordinates:

Trilinears    (a^2+c^2)*(a^2+b^2)/a : b^3 : c^3

## 6th anti-Brocard

The triangle A'B'C' whose 6th Brocard triangle is ABC.

Note: For construction, see Bernard Gibert's anti-Brocard triangles and related topics.

A-vertex coordinates:

Trilinears
a*(a^4-b^2*c^2) : (b^6-(a^2+c^2)*b^4-(a^2-c^2)*b^2*c^2+a^2*c^4)/b : (c^6-(a^2+b^2)*c^4-(a^2-b^2)*b^2*c^2+a^2*b^4)/c

## 1st anti-circumperp

The triangle A'B'C' whose 1st circumperp triangle is ABC.

Note: Only when ABC is acute. The 1st anti-circumperp triangle is (the orthic triangle)-of-(the anticomplementary triangle).

A-vertex coordinates:

Trilinears    -a : (b^2-c^2)/b : (c^2-b^2)/c

## 1st anti-circumperp-tangential

The triangle A'B'C' whose 1st circumperp-tangential triangle is ABC. It is the Anti-Mandart-incircle triangle.

## 2nd anti-circumperp-tangential

The triangle A'B'C' whose 2nd circumperp-tangential is ABC.

Notes: For any ABC. A' = AX(56) ∩ perpendicular(X(7354), BC)

A-vertex coordinates:

Trilinears    (b+c)^2/(a*(-a+b+c)) : b/(a-b+c) : c/(a+b-c)

## (1st) anti-Conway

The triangle A'B'C' whose (1st) Conway triangle is ABC.

Note: The A-vertex of the anti-Conway triangle is the A' = {perpendicular to BC through X(578)} ∩ AX(6).

A-vertex coordinates:

Trilinears    a^3*(-a^2+b^2+c^2)/((b^2+c^2)*a^2-(b^2-c^2)^2) : b : c

## 2nd anti-Conway

The triangle A'B'C' whose 2nd Conway triangle is ABC.

Note: The 2nd anti-Conway triangle is (the medial triangle)-of-(the orthic triangle).

A-vertex coordinates:

Trilinears    ((b^2+c^2)*a^2-(b^2-c^2)^2)/a/(-a^2+b^2+c^2) : b : c

## anti-Ehrmann-mid

The triangle A'B'C' whose Ehrmann-mid triangle is ABC.

A-vertex coordinates:

Trilinears    -3*a*SA : (S^2+3*SA*SC)/b : (S^2+3*SA*SB)/c

## anti-Euler

The triangle A'B'C' whose Euler triangle is ABC.

Note: A', B', C' are the reflections of the orthocenter of ABC in A, B, C, respectively.

A-vertex coordinates:

Trilinears    (3*a^4-4*(b^2+c^2)*a^2+(b^2-c^2)^2)/(a*(-a^2+b^2+c^2)) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 3rd anti-Euler

The triangle A'B'C' whose 3rd Euler triangle is ABC.

Note: The 3rd anti-Euler triangle is (the anti-Euler triangle)-of-(the circumorthic triangle).

A-vertex coordinates:

Trilinears    -((b^2+c^2)*a^2-b^4+b^2*c^2-c^4)*a : (a^4-(b^2-c^2)*(a^2-c^2))*b : (a^4-(c^2-b^2)*(a^2-b^2))*c

## 4th anti-Euler

The triangle A'B'C' whose 4th Euler triangle is ABC.

Note: The 4th anti-Euler triangle is (the anticomplementary triangle)-of-(the circumorthic triangle).

A-vertex coordinates:

Trilinears
-(-4*R^2*SA-S^2+SA^2-2*SB*SC)*a :
(S^2-2*SA*SC-SB^2+4*(R^2-SC)*SB)*b :
(S^2-2*SA*SB-SC^2+4*(R^2-SB)*SC)*c

## anti-excenters-incenter reflections

The triangle A'B'C' whose excenters-incenter reflections triangle is ABC.

Note: The anti-excenters-incenter reflections triangle is (the 5th mixtilinear triangle)-of-(the orthic triangle).

A-vertex coordinates:

Trilinears    2*a/(S^2-2*SB*SC) : 1/(b*SB) : 1/(c*SC)

## anti-excenters-reflections

See anti-excenters-incenter reflections.

## 2nd anti-extouch

The triangle A'B'C' whose 2nd extouch triangle is ABC.

Note: Only when ABC is acute. A' = AX(3) ∩ perpendicular(X(1181), BC)

A-vertex coordinates:

Trilinears    S^2*a/SA : SB*b : SC*c

## anti-inner-Garcia

The triangle A'B'C' whose inner-Garcia triangle is ABC.

A-vertex coordinates:

Trilinears    -(a^2-b^2+b*c-c^2)*a : a*(b*a-c^2)-(b^2-c^2)*(b-c) : a*(c*a-b^2)-(c^2-b^2)*(c-b)

## anti-inner-Grebe

The triangle A'B'C' whose inner-Grebe triangle is ABC.

Notes: For any ABC. A' = AX(6) ∩ perpendicular(X(1588), BC)

A-vertex coordinates:

Trilinears    (a^2-S)/a : b : c

## anti-outer-Grebe

The triangle A'B'C' whose outer-Grebe triangle is ABC.

Notes: For any ABC. A' = AX(6) ∩ perpendicular(X(1587), BC)

A-vertex coordinates:

Trilinears    (a^2+S)/a : b : c

## anti-Honsberger

The triangle A'B'C' whose Honsberger triangle is ABC.

Notes: Only for ABC acute. A' = AX(6) ∩ perpendicular(X(182), BC)

A-vertex coordinates:

Trilinears     -a^3/(b^2+c^2) : b : c

## anti-Hutson intouch

The triangle A'B'C' whose Hutson intouch triangle is ABC.

Note: Only when ABC is acute. The anti-Hutson intouch triangle is (the anti-excenters-reflections triangle)-of-(the Ara triangle).

A-vertex coordinates:

Trilinears    -(S^2+2*SA^2)*a : (S^2-2*SA*SB)*b : (S^2-2*SC*SA)*c

## anti-incircle-circles

The triangle A'B'C' whose incircle-circles triangle is ABC.

Note: The anti-incircle-circles triangle is (the anti-inverse-in-incircle triangle)-of-(the Johnson triangle).

A-vertex coordinates:

Trilinears    -a*(2*S^2+SA^2) : b*(2*S^2-SA*SB) : c*(2*S^2-SA*SC)

## anti-inverse-in-incircle

The triangle A'B'C' whose inverse-in-incircle triangle is ABC.

Note: The anti-inverse-in-incircle triangle is (the tangential triangle)-of-(the anticomplementary triangle).

A-vertex coordinates:

Trilinears    -(a^2+b^2+c^2)/a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## anti-1st/2nd Kenmotu-free-vertices

The triangle A'B'C' whose 1st/2nd Kenmotu-free-vertices triangle is ABC.

A-vertex coordinates:

anti-1st Kenmotu-free-vertices:
Trilinears    (SA+S)*(SB+SC+2*S)/a : (SB-S)*b : (SC-S)*c
anti-2nd Kenmotu-free-vertices:
Trilinears    (SA-S)*(SB+SC-2*S)/a : (SB+S)*b : (SC+S)*c

## anti-Mandart-incircle

The triangle A'B'C' whose Mandart-incircle triangle is ABC.

Note: The anti-Mandart-incircle triangle is (the tangential triangle)-of-(the 1st circumperp triangle).

A-vertex coordinates:

Trilinears    -a^2+(b+c)*a-2*b*c : (a-b+c)*b : (a+b-c)*c

## anti-McCay

The triangle A'B'C' whose McCay triangle is ABC.

A-vertex coordinates:

Trilinears     -(5*a^4-2*(b^2+c^2)*a^2+(2*b^2-c^2)*(b^2-2*c^2))/a : ((2*a^2+2*b^2-c^2)^2-9*a^2*b^2)/b : ((2*a^2+2*c^2-b^2)^2-9*a^2*c^2)/c

## 6th anti-mixtilinear

The triangle A'B'C' whose 6th mixtilinear triangle is ABC.

Note: The 6th mixtilinear triangle is (the orthic triangle)-of-(the medial triangle).

A-vertex coordinates:

Trilinears    2*a : (a^2-b^2+c^2)/b : (a^2+b^2-c^2)/c

## anti-orthocentroidal

The triangle A'B'C' whose orthocentroidal triangle is ABC.

Note: The anti-orthocentroidal triangle is (the 4th anti-Brocard triangle)-of-(the circumsymmedial triangle) of ABC.

A-vertex coordinates:

Trilinears    a*((-a^2+b^2+c^2)^2-b^2*c^2) : b*(a^4-(2*b^2-c^2)*a^2+(b^2-c^2)*(b^2+2*c^2)) : c*(a^4+(b^2-2*c^2)*a^2-(b^2-c^2)*(2*b^2+c^2))

## 1st anti-orthosymmedial

The triangle A'B'C' whose 1st orthosymmedial triangle is ABC.

Notes: Only for ABC acute. A' = AX(1297) ∩ perpendicular(X(19158), BC)

A-vertex coordinates:

Trilinears
(S^4+3*SA^2*S^2+2*(SA^2-SB*SC-SW^2)*SA^2)*a/(2*SA+SB+SC) : ((SA-SB)*S^2-2*(SA*SC-SB^2)*SA)*b : ((SA-SC)*S^2-2*(SA*SB-SC^2)*SA)*c

## anti-1st Parry

The triangle A'B'C' whose 1st Parry triangle is ABC.

A-vertex coordinates:

Trilinears   -(a^4-2*(b^2+c^2)*a^2+3*(b^2-c^2)^2)/a : (a^2+b^2-5*c^2)*b/(a^2-c^2)*(b^2-c^2) : (a^2-5*b^2+c^2)*c*(c^2-b^2)/(a^2-b^2)

## anti-2nd Parry

The triangle A'B'C' whose 2nd Parry triangle is ABC.

A-vertex coordinates:

Trilinears   -(2*S^4+(5*SA^2-8*SB*SC+SW^2)*S^2+3*(SA^2-4*SB*SC-SW^2)*SA^2)/(a*(3*SA-SW)) : b*(S^2-3*SA*SB) : c*(S^2-3*SA*SC)

## anti-reflection

The triangle A'B'C' whose reflection triangle is ABC.

A-vertex coordinates:

## 1st anti-Sharygin

The triangle A'B'C' whose 1st Sharygin triangle is ABC.

Notes: Only for ABC acute. A' = AX(8795) ∩ perpendicular(X(8884), BC)

A-vertex coordinates:

Trilinears     -a*SB*SC/(S^2+SB*SC) : SC^2*(SA+SB)/(b*(S^2+SA*SC)) : SB^2*(SA+SC)/(c*(S^2+SA*SB))

## anti-tangential-midarc

The triangle A'B'C' whose tangential-midarc triangle is ABC.

Note: A' = AX(65) ∩ perpendicular(X(1), BC), when ABC is acute.

A-vertex coordinates:

Trilinears    -a/(-a+b+c) : (a+c)/(a-b+c) : (a+b)/(a+b-c)

## 3rd anti-tri-squares

The triangle A'B'C' whose 3rd tri-squares triangle is ABC.

Notes: Only for ABC acute. A' = AX(1328) ∩ perpendicular(X(486), BC)

A-vertex coordinates:

Trilinears     2*S*(2*S^2-(18*R^2+SA-4*SW)*S-3*SB*SC)/(a*(3*SA-S)*(SA-S)) : (3*SC-S)/b : (3*SB-S)/c

## 4th anti-tri-squares

The triangle A'B'C' whose 4th tri-squares triangle is ABC.

Notes: Only for ABC acute. A' = AX(1327) ∩ perpendicular(X(485), BC)

A-vertex coordinates:

Trilinears     -2*S*(2*S^2+(18*R^2+SA-4*SW)*S-3*SB*SC)/(a*(3*SA+S)*(SA+S)) : (3*SC+S)/b : (3*SB+S)/c

## anti-3rd/4th tri-squares-central

The triangle A'B'C' whose 3rd/4th tri-squares-central triangle is ABC.

A-vertex coordinates:

anti-3rd tri-squares-central:
Trilinears    -(b^2+c^2+3*S)/a : (b^2+S)/b : (c^2+S)/c
anti-4th tri-squares-central:
Trilinears    -(b^2+c^2-3*S)/a : (b^2-S)/b : (c^2-S)/c

## anti-Ursa minor

The triangle A'B'C' whose Ursa minor triangle is ABC.

Notes: Only when ABC is acute. The anti-Ursa minor triangle is (the anti-Hutson-intouch triangle)-of-(the Euler triangle).

Construction: Let A0B0C0 be the orthic triangle of ABC. Let A1 be the intersection of the parallel to A0B0 through B and the parallel to A0C0 through C. Let A2 be the midpoint of A and A1. Build B2, C2 cyclically. Then A2B2C2 is the anti-Ursa minor triangle of ABC. (Added Nov. 14, 2020)

A-vertex coordinates:

Trilinears    (b^2+c^2)/a : (c^2-a^2)/b : (b^2-a^2)/c

## anti-Wasat

The triangle A'B'C' whose Wasat triangle is ABC.

Notes: Only for ABC acute. The anti-Wasat triangle of ABC is (the 3rd anti-Euler triangle)-of-(the medial triangle).

A-vertex coordinates:

Trilinears    (S^2+SB*SC)*a : SB*(SA-SC)*b : SC*(SA-SB)*c

## anti-X3-ABC reflections

The triangle A'B'C' whose X3-ABC reflections triangle is ABC.

A-vertex coordinates:

Trilinears    -(2*a^4-3*(b^2+c^2)*a^2+(b^2-c^2)^2)/a : (a^2-b^2+c^2)*b : (a^2+b^2-c^2)*c

## anti-inner/outer Yff

The triangle A'B'C' whose inner-/outer- Yff triangle is ABC.

A-vertex coordinates:

anti-inner-Yff:
Trilinears    (a^4-2*(b^2+c^2)*a^2-2*(b+c)*b*c*a+(b^2-c^2)^2)/(2*a^2*b*c) : 1 : 1
anti-outer-Yff:
Trilinears    (a^4-2*(b^2+c^2)*a^2+2*(b+c)*b*c*a+(b^2-c^2)^2)/(2*a^2*b*c) : -1 : -1

## Antlia

Let A' be the center of the inverse-in-A-excircle of the incircle, and define B' and C' cyclically. The triangle A'B'C' is the Antlia triangle of ABC. (Reference: X5574)

A-vertex coordinates:

Trilinears    -(a^2+3*(b-c)^2)/(3*a^2+(b-c)^2) : 1 : 1

## AOA

See altimedial orthic axes triangle.

## Apollonius

The Apollonious circle is the circle internally tangent to all the excircles (Reference: MathWorld -- Apollonius Circle.). The respective touchpoints are the vertices of the Apollonius triangle.

A-vertex coordinates:

Trilinears    -((b+c)*a+b^2+c^2)^2*a/(a+b+c) : (a+c)^2*(a+b-c)*b : (a+b)^2*(a-b+c)*c

## Apus

Let A' be the insimilicenter of the circumcircle and A-excircle, and define B' and C' cyclically. The triangle with vertices A,' B, C' is the Apus triangle. (Reference: X5584)

A-vertex coordinates:

Trilinears    a/(a+b+c) : -b/(a+b-c) : -c/(a-b+c)

## Aquila

Let A' = reflection of the incenter in A, and define B' and C' cyclically. The triangle A'B'C' is the Aquila triangle. (Reference: X5586)
It is equivalent to the triangle T(1,2) in TCCT, p. 173.

A-vertex coordinates:

Trilinears    (a+2*b+2*c)/a : -1 : -1

## Ara

Let A'B'C' be the tangential triangle of triangle ABC. Let A" be the center of the A'-excircle of A'B'C', unless this is also the circumcircle of ABC, in which case let A" be the incenter of A'B'C'. Define B", C" cyclically. The triangle A"B"C" is the Ara triangle. (Reference: X5594)

A-vertex coordinates:

Trilinears    a*(a^2+b^2+c^2) : -b*(a^2+b^2-c^2) : -c*(a^2-b^2+c^2)

## Aries

Let A'B'C' be the tangential triangle of an acute triangle ABC. Let A" be the touchpoint of the A-excircle of A'B'C' and the line B'C'; define B" and C" cyclically. The triangle A"B"C" is the Aries triangle. (Reference: X5596)

A-vertex coordinates:

Trilinears    -(a^4+(b^2-c^2)^2)/(2*a) : b*(b^2-c^2) : c*(c^2-b^2)

## Artzt

The A-Artzt parabola of a triangle ABC is the parabola tangent at B and C to the sidelines AB and AC, respectively. The triangle A'B'C' bounded by the directrices of the Artzt parabolas is the Artzt triangle. (Reference: Preamble just before X9742)

A-vertex coordinates:

Trilinears    -(3*a^4+(b^2-c^2)^2)/(2*a) : ((2*b^2+c^2)*a^2+c^2*(b^2-c^2))/b : ((2*c^2+b^2)*a^2-(b^2-c^2)*b^2)/c

## Ascella

Let A' = incircle-inverse of A, and define B' and C' cyclically. Let OA be the circle {{B,C,B',C'}}, and define OB and OC cyclically. The three circles are othogonal to the incircle and their centers A",B",C" are the vertices of the Ascella triangle. (Reference: Preamble just before X8726)

A-vertex coordinates:

Trilinears    2*a : (a^2-2*(b+c)*a+c^2-b^2)/b : (a^2-2*(b+c)*a+b^2-c^2)/c

## Atik

Suppose that V is a point outside a circle (U,u). Let (V,v) be the circle with center V that is orthogonal to (U,u), so that v2 = |UV|2 - u2 = power of V with respect to (U,u). Let UV,A be the circle (V,w) obtained from (U,r) = A-excircle and V = incenter, and let LA be the radical axis of UV,A and the A-excircle. Define LB and LC cyclically. Let A' = LB∩LC, B' = LC∩LA, C' = LA∩LB. The triangle A'B'C' is the Atik triangle. (Reference: Preamble just before X8580))

The Atik triangle is also the triangle bounded by the polars of the incenter X(1) with respect to the excircles (Sept 10, 2017).

Let I, Ia be the incenter and A-excenter of ABC. Let a' be the radical axis of the A-excircle and the circle with diameter IIa and denote b', c' cyclically. The triangle bounded by these radical axes is the Atik triangle (July 7, 2020).

A-vertex coordinates:

Trilinears    -((b+c)*a^2-2*(b^2+c^2)*a+(b+c)^3)/a : a^2-2*(b-c)*a+(b+3*c)*(b-c) : a^2-2*(c-b)*a+(c+3*b)*(c-b)

## 1st Auriga

Let U be the inverter (see X5577) of the circumcircle and the incircle. There are two triangles that circumscribe U and are homothetic to triangle ABC, one of which has A-vertex on the same side of line BC as A. This triangle, A'B'C', is the 1st Auriga triangle, and the other, is the 2nd Auriga triangle. (Reference: X5597)

A-vertex coordinates:

Trilinears    (a^4-(b+c)^2*a^2-4*(b+c)*S*D)/a : (b^4-b^2*(a+c)^2+4*b*S*D)/b : (c^4-c^2*(a+b)^2+4*c*S*D)/c
where D=sqrt(r*R+4*R^2)

## 2nd Auriga

Let U be the inverter (see X5577) of the circumcircle and the incircle. There are two triangles that circumscribe U and are homothetic to triangle ABC, one of which has A-vertex on the opposite side of line BC as A. This triangle, A"B"C", is the 2nd Auriga triangle, and the other, is the 1st Auriga triangle. (Reference: X5597)

A-vertex coordinates:

Trilinears    (a^4-(b+c)^2*a^2+4*(b+c)*S*D)/a : (b^4-b^2*(a+c)^2-4*b*S*D)/b : (c^4-c^2*(a+b)^2-4*c*S*D)/c
where D=sqrt(r*R+4*R^2)

## Ayme

Let RA be the radical axis of the circumcircle and the A-excircle, and define RB and RC cyclically. Let TA = RB∩RC, and define TB and TC cyclically. (TA is also the radical center of the circumcircle and the B- and C- excircles.) The Ayme triangle is TATBTC. (Reference: X3610)

A-vertex coordinates:

Trilinears    (b+c)*(a^2+(b+c)^2)/a : -a^2-b^2+c^2 : -a^2-c^2+b^2

# B:

## Bankoff equilateral

Let ABC be a triangle with circumcenter O. Centering at O, let Ab be the rotation of A toward B by an angle |π/6| and let Ac be the rotation of A toward C by the same angle |π/6| (triangle OAbAc is an equilateral triangle). Build (Bc, Ba) and (Ca, Cb) cyclically and let Am, Bm, Cm be the midpoints of BcCb, CaAc, AbBa, respectively. The triangle AmBmCm is equilateral and is named the Bankoff equilateral triangle of ABC. (Reference: X34551)

A-vertex coordinates:

Trilinears    -((sqrt(3)-2)*SA+S)*a : (2*S^2+S*SC+(sqrt(3)-2)*SA*SC)/b : (2*S^2+S*SB+(sqrt(3)-2)*SA*SB)/c

## BCI

Let I be the incenter of ABC and A', B', C' the incenters of IBC, ICA and IAB, respectively. The triangle A'B'C' is the BCI triangle of ABC. (Reference: MathWorld -- BCI Triangle.)

A-vertex coordinates:

Trilinears    1 : 1+2*cos(C/2) : 1+2*cos(B/2)

## Bevan-antipodal

Let V be the Bevan point X(40) of ABC and Va, Vb, Vc the antipodes of V on the circumcircles of VBC, VCA and VAB, respectively. The triangle VaVbVc is the Bevan-antipodal triangle of ABC. (Reference: X223)

Note: The Bevan-antipodal triangle is the anticevian triangle of X(57).

A-vertex coordinates:

Trilinears    -1/(-a+b+c) : 1/(a-b+c) : 1/(a+b-c)

## 1st Brocard

Let Ω' and Ω" be the Brocard points of ABC and A* the intersection of BΩ' and CΩ"; similarly define B* and C*. A*B*C* is the 1st Brocard triangle of ABC. (Reference: MathWorld -- Brocard Triangles.)

A-vertex coordinates:

Trilinears    a*b*c : c^3 : b^3

## 1st Brocard-reflected

The triangles whose vertices are the reflections of the vertices of the 1st Brocard triangle in the sidelines of ABC is the 1st Brocard-reflected triangle of ABC. (Reference: CTC - Table 32)

A-vertex coordinates:

Trilinears    -a : (a^2+b^2)/b : (a^2+c^2)/c

## 2nd Brocard

Let Ω' and Ω" be the Brocard points of ABC. The circles ω'a and ω"a through {Ω',B,C} and {Ω",B,C}, respectively, meet again at A*. Define B* and C* similarly. A*B*C* is the 2nd Brocard triangle of ABC. (Reference: MathWorld -- Brocard Triangles.)

A-vertex coordinates:

Trilinears    b^2+c^2-a^2 : a*b : a*c

## 3rd Brocard

Let A'B'C' be the 1st Brocard triangle of ABC and A*, B*, C* the isogonal conjugates of A',B',C' with respect to ABC. A*B*C* is the 3rd Brocard triangle of ABC. (Reference: MathWorld -- Brocard Triangles.)

A-vertex coordinates:

Trilinears    b^2*c^2 : a*b^3 : a*c^3

## 4th Brocard

Let A"B"C" be the 2nd Brocard triangle of ABC and A*, B*, C* the isogonal conjugates of A",B",C" with respect to ABC. A*B*C* is the 4th Brocard triangle of ABC. (Reference: MathWorld -- Brocard Triangles.)

A-vertex coordinates:

Trilinears    a*b*c/(b^2+c^2-a^2) : c : b

## 5th Brocard

The 5th Brocard triangle is the vertex triangle of the circumcevian triangles of the 1st and 2nd Brocard points. (Reference: X32)

Note: The vertex triangle of A'B'C' and A"B"C" is the triangle A*B*C*, where A*=B'B"∩C'C" and similarly B* and C* (TCCT, p. 199).

A-vertex coordinates:

Trilinears    ((b^2+c^2)*a^2+(b^2+c^2)^2-b^2*c^2)/a : -b^3 : -c^3

## 6th Brocard

Let Ω' and Ω" be the Brocard points of ABC. The line AΩ' cuts again the 2nd Brocard circle at A' and the line AΩ" cuts again the 2nd Brocard circle at A". Build B',B", C', C" similarly and define A*=B'B"∩C'C", B*=C'C"∩A'A" and C*=A'A"∩B'V". The triangle A*B*C* is the 6th Brocard triangle. (Reference: X384)

Note: The perpendicular bisector of the Brocard points of ABC passes through the circumcenter O of ABC. The circle with center at O and passing through both Brocard points is the 2nd Brocard circle.

A-vertex coordinates:

Trilinears    ((b^2+c^2)*a^2-b^2*c^2)/a : (c^4+c^2*b^2-b^4)/b : (b^4+b^2*c^2-c^4)/c

## 7th Brocard

In the plane of a triangle ABC, let O = X(3), the circumcenter. Let A' be the point, other than O, in which the line AO meets the Brocard circle, and define B' and C' cyclically. The triangle A'B'C', is the 7th Brocard triangle. (Reference: Dan Reznik and Peter Moses, preamble just before X39643)

A-vertex coordinates:

Trilinears    (a^4+(b^2-c^2)^2)/a : b*(a^2-b^2+c^2) : c*(a^2+b^2-c^2)

## 8th Brocard

Let A'B'C' be the 7th Brocard triangle of ABC and A", B", C" the circumcircle-inverses of A', B', C', respectively. The triangle A"B"C", is the 8th Brocard triangle. (Reference: Preamble just before X38643)

Note: A", B", C" are collinear on the Brocard axis of ABC.

A-vertex coordinates:

Trilinears    -2*a^3 : b*(a^2-b^2+c^2) : c*(a^2+b^2-c^2)

## 9th Brocard

Let A"B"C" be the 8th Brocard triangle of ABC and A*, B*, C* the isogonal conjugates of A", B", C", respectively. The triangle A*B*C*, is the 9th Brocard triangle. (Reference: Preamble just before X38643)

A-vertex coordinates:

Trilinears    -(a^2+b^2-c^2)*(a^2-b^2+c^2)/(2*a^3) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

# C:

## Caelum

Reflections of A,B,C on the incenter. For coordinates, see 5th mixtilinear triangle. (Reference: X5603)

## Carnot

See Johnson triangle.

## circumcircle-midarc

See 2nd circumperp triangle.

## circummedial

The circumcevian triangle of the centroid of ABC is its circummedial triangle.

A-vertex coordinates:

Trilinears    -a*b*c/(b^2+c^2) : c : b

## circumnormal

The circumnormal triangle is obtained by rotating the circumtangential triangle an angle π/6 around the circumcenter. It is an equilateral triangle. Reference: TCCT, p. 166.

A-vertex coordinates:

Trilinears    sec(C/3-B/3) : -sec(B/3+2*C/3) : -sec(C/3+2*B/3)

## circumorthic

The circumcevian triangle of the orthocenter of ABC is its circumorthic triangle.

A-vertex coordinates:

Trilinears    -a/((b^2+c^2)*a^2-(b^2-c^2)^2) : 1/(b*(a^2-b^2+c^2)) : 1/(c*(a^2+b^2-c^2))

## 1st circumperp

The perpendicular bisector of BC meets the circumcircle in two points A' and A" (A' in the same side of BC as A). Define B', B", C', C" cyclically. The triangle A'B'C' is the 1st circumperp triangle. (TCCT, p. 163).

A-vertex coordinates:

Trilinears    a : c-b : b-c

## 1st circumperp tangential

The tangential triangle of the 1st circumperp triangle. (TCCT, 6-23, p. 164).

See anti-Mandart-incircle triangle.

## 2nd circumperp

The perpendicular bisector of BC meets the circumcircle in two points A' and A" (A' in the same side of BC as A). Define B', B", C', C" cyclically. The triangle A"B"C" is the 2nd circumperp triangle. (TCCT, p. 163).

A-vertex coordinates:

Trilinears    -a : b+c : b+c

## 2nd circumperp tangential

The tangential triangle of the 1st circumperp triangle. (TCCT, 6-24, p. 165).

A-vertex coordinates:

Trilinears    b*c+(a+b)*(a+c) : -b*(a+b-c) : -c*(a-b+c)

## circumsymmedial

The circumcevian triangle of the symmedian point X(6) of ABC is its circumsymmedial triangle.

A-vertex coordinates:

Trilinears    -a/2 : b : c

## circumtangential

The circumtangential triangle has vertices three points X on the circumcircle of ABC at which the line XX-1 is tangent to the circumcircle, where X-1 means "the isogonal conjugate of X". It is an equilateral triangle. Reference: TCCT, p. 166 and 168-170.

A-vertex coordinates:

Trilinears    csc(C/3-B/3) : csc(B/3+2*C/3) : -csc(C/3+2*B/3)

## contact

See intouch triangle.

## Conway

Let a, b, c be the sidelengths BC, CA and AB, respectively, of ABC. Let Ab be the point on CA such that AAb = -(a/b)*AC and Ac the point on BA such that AAc = -(a/c)*AB. Define Bc, Ba, Ca, Cb similarly. The Conway triangle is the triangle bounded by the lines AbAc, BcBa and CaCb. (Reference: X7411)

Note: See this construction at MathWorld -- Conway Circle..

A-vertex coordinates:

Trilinears    -(a+b+c)/(b+c) : (a+b-c)/b : (a-b+c)/c

## 2nd Conway

Following the construction of the Conway triangle, let A2 = BaBc∩CaCb, and define B2 and C2 cyclically. The triangle A2B2C2 is the 2nd Conway triangle of ABC. (Reference: Preamble just before X9776)

A-vertex coordinates:

Trilinears    -(a+b+c)/a : (a+b-c)/b : (a-b+c)/c

## 3rd Conway

Following the construction of the Conway triangle, let A3 be the intersection of the tangents to the Conway circle at Ba and Ca, and define B3, C3 cyclically. The triangle A3B3C3 is the 3rd Conway triangle. (Reference: X10434)

Note: The six points Ab, Ac, Bc, Ba, Ca, Cb lie on a circle named the Conway circle.

A-vertex coordinates:

Trilinears
-a*(a+b+c)^2 : ((b+2*c)*a^3+2*a^2*b^2+(b-c)*(2*c^2+b*c+b^2)*a+2*(b^2-c^2)*b*c)/b : ((2*b+c)*a^3+2*a^2*c^2-(b-c)*(c^2+b*c+2*b^2)*a-2*(b^2-c^2)*b*c)/c

## 4th Conway

Following the construction of the Conway triangle, let A4 be the intersection of the tangents to the Conway circle at Ab and Ac, and define B4, C4 cyclically. The triangle A4B4C4 is the 4th Conway triangle. (Reference: X10434)

Note: The six points Ab, Ac, Bc, Ba, Ca, Cb lie on a circle named the Conway circle.

A-vertex coordinates:

Trilinears    (a^3-(b+c)*a^2-2*(b^2+b*c+c^2)*a-2*b*c*(b+c))/((a+b+c)*a^2) : 1 : 1

## 5th Conway

Following the construction of the Conway triangle, let A5 be the intersection of the tangents to the Conway circle at Bc and Cb, and define B5, C5 cyclically. Triangle A5B5C5 is named the 5th Conway triangle. (Reference: X10434)

Note: The six points Ab, Ac, Bc, Ba, Ca, Cb lie on a circle named the Conway circle.

A-vertex coordinates:

Trilinears    -((b+c)*a^2+(b^2+c^2)*a+2*b*c*(b+c))/((a+b+c)*(b+c)*a) : 1 : 1

## inner-Conway

Let a, b, c be the sidelengths BC, CA and AB, respectively, of ABC. Let Ab be the point on CA such that AAb = (a/b)*AC and Ac the point on BA such that AAc = (a/c)*AB. Define Bc, Ba, Ca, Cb similarly. The inner-Conway triangle is the triangle bounded by the lines AbAc, BcBa and CaCb. (Reference: Preamble just before X11677)

Notes: See this construction at MathWorld -- Conway Circle.
Points Ab, Ac, Bc, Ba, Ca and Cb do not lie on a conic.

A-vertex coordinates:

Trilinears    -1 : (b-c)/b : (c-b)/c

# D:

## D-triangle

See 4th Brocard triangle.

## de Villiers

See BCI triangle.

# E:

## 1st Ehrmann

In Hyacinthos #6098, December 2, 2002, Jean-Pierre Ehrmann defines a circle as follows. Let P be a point in the plane of ABC and not on the lines BC, CA, AB. Let AB the the point of intersection of the circle {{P,B,C}} and the line AB. Define AC symmetrically, and define BC, BA, CA, CB cyclically. These six points of intersection are on a circle if and only if P = X(6). The centers A1, B1, C1 of the circles {{X(6),B,C}}, {{X(6),C,A}}, {{X(6),A,B,}}, respectively, are the vertices of the 1st Ehrmann triangle. (Reference: Preamble just before X8537)

A-vertex coordinates:

Trilinears    -(a^4-b^4+4*b^2*c^2-c^4)*a : (a^2*(a^2+5*c^2)-b^4+3*b^2*c^2-2*c^4)*b : (a^2*(a^2+5*b^2)-2*b^4+3*b^2*c^2-c^4)*c

## 2nd Ehrmann

Following the construction of the 1st Ehrmann triangle, let A2 = BCBA∩CACB, and define B2 and C2 cyclically. The triangle A2B2C2 is the 2nd Ehrmann triangle. (Reference: Preamble just before X8537)

A-vertex coordinates:

Trilinears    a*(a^2-2*b^2-2*c^2)/(2*a^2-b^2-c^2) : b : c

## Ehrmann circumscribing

Let TC1=A1B1C1 be the triangle obtained by rotating ABC about the 1st Ehrmann pivot, P(5), by an angle of 2π/3, so that TC1 circumscribes ABC. Triangle TC1 is the 1st Ehrmann circumscribing triangle. Let TC2=A2B2C2 be the triangle obtained by rotating ABC about the 2nd Ehrmann pivot, U(5), by an angle of -2π/3, so that TC2 circumscribes ABC. Triangle TC2 is the 2nd Ehrmann circumscribing triangle. (Reference: Preamble just before X18300)

Note: Neither of both Ehrmann circumscribing triangles is a central triangle.

A-vertex coordinates:

1st Ehrmann circumscribing triangle:

Trilinears    -(S^2-sqrt(3)*(SB-SC)*S-3*SB*SC)/a : (S+sqrt(3)*SA)*(S+sqrt(3)*SC)/b : (S-sqrt(3)*SA)*(S-sqrt(3)*SB)/c

2nd Ehrmann circumscribing triangle:
Trilinears    -(S^2+sqrt(3)*(SB-SC)*S-3*SB*SC)/a : (S-sqrt(3)*SA)*(S-sqrt(3)*SC)/b : (S+sqrt(3)*SA)*(S+sqrt(3)*SB)/c

## Ehrmann cross

The cross-triangle of the 1st and 2nd Ehrmann inscribed triangles is the Ehrman cross-triangle. (Reference: Preamble just before X18300)

Note: if T1=A1B1C1 and T2=A2B2C2 are two distinct triangles, the cross-triangle T*=A*B*C* of T1 and T2 is the triangle having vertices A*=B1C2 ∩ B2C1, B*=C1A2 ∩ C2A1 and C*=A1B2 ∩ A2B1.

Note: The Ehrmann cross triangle is degenerate, its vertices lying on line X(3)X(523) (the trilinear polar of X(2986)).

A-vertex coordinates:

Trilinears    (S^2-3*SB*SC)/a : SB*(S^2-3*SC^2)/(b*(SB-SC)) : SC*(S^2-3*SB^2)/(c*(SC-SB))

## Ehrmann inscribed

Let TI1=A1B1C1 be the triangle obtained by rotating ABC about the 1st Ehrmann pivot, P(5), by an angle of -2π/3, so that TI1 is inscribed in ABC. Triangle TI1 is the 1st Ehrmann inscribed triangle. Let TI2=A2B2C2 be the triangle obtained by rotating ABC about the 2nd Ehrmann pivot, U(5), by an angle of 2π/3, so that TI2 is inscribed in ABC. Triangle TI2 is the 2nd Ehrmann inscribed triangle. (Reference: Preamble just before X18300)

Note: Neither of both Ehrmann inscribed triangles is a central triangle.

A-vertex coordinates:

1st Ehrmann inscribed triangle:

Trilinears    0 : (S-sqrt(3)*SA)*c : (S+sqrt(3)*SA)*b

2nd Ehrmann inscribed triangle:
Trilinears    0 : (S+sqrt(3)*SA)*c : (S-sqrt(3)*SA)*b

## Ehrmann mid

The mid-triangle of the 1st and 2nd Ehrmann circumscribing triangles is the Ehrman mid-triangle. (Reference: Preamble just before X18300)

Note: if T1=A1B1C1 and T2=A2B2C2 are two distinct triangles, the mid-triangle of T1 and T2 is the triangle having vertices the midpoints of {A1,A2}, {B1,B2} and {C1,C2}.

A-vertex coordinates:

Trilinears    -(S^2-3*SB*SC)/a : (S^2+3*SA*SC)/b : (S^2+3*SA*SB)/c

## Ehrmann side

The side-triangle of the 1st and 2nd Ehrmann circumscribing triangles is the Ehrman mid-triangle. (Reference: Preamble just before X18300)

Note: if T1=A1B1C1 and T2=A2B2C2 are two distinct triangles, the side-triangle T*=A*B*C* of T1 and T2 is the triangle having vertices A*=B1B2 ∩ C1C2, B*=C1C2 ∩ A1A2 and C*=A1A2 ∩ B1B2.

A-vertex coordinates:

Trilinears    a*(S^2+3*SB*SC) : SB*(S^2-3*SC^2)/b : SC*(S^2-3*SB^2)/c

## Ehrmann vertex

The vertex-triangle of the 1st and 2nd Ehrmann circumscribing triangles is the Ehrman vertex-triangle. (Reference: Preamble just before X18300)

Note: if T1=A1B1C1 and T2=A2B2C2 are two distinct triangles, the vertex-triangle T*=A*B*C* of T1 and T2 is the triangle having vertices A*=B1C1 ∩ B2C2, B*=C1A1 ∩ C2A2 and C*=A1B1 ∩ A2B2.

A-vertex coordinates:

Trilinears    -((4*SA+SB+SC)*S^2-(3*(SB+SC))*SA^2)/(a*(S^2-3*SA^2)) : SC/b : SB/c

## Euler

Let H be the orthocenter of ABC and A*, B*, C* the midpoints of AH, BH and CH, respectively. A*B*C* is the Euler triangle of ABC. (Reference: MathWorld -- Euler Triangle.)
A*, B*, C* are also the reflections of the vertices of the medial triangle in X(5).

A-vertex coordinates:

Trilinears    2((b^2+c^2)*a^2-(b^2-c^2)^2)/(a*(b^2+c^2-a^2)) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 2nd Euler

The reflections of the vertices of the orthic triangle in X(5) are the vertices of the 2nd Euler triangle. (Reference: X3758)

A-vertex coordinates:

Trilinears    -2*((b^2+c^2)*a^2-(b^2-c^2)^2)*a : (a^2-b^2+c^2)*(a^4-2*a^2*c^2+(b^2-c^2)^2)/b : (a^2+b^2-c^2)*(a^4-2*a^2*b^2+(b^2-c^2)^2)/c

## 3rd Euler

The perpendicular to BC through X(5) cuts the nine point circle in two points A', A", with A' farther from A than A". Build B', B", C', C" similarly. A'B'C' is the 3rd Euler triangle. (Reference: X3758)

A-vertex coordinates:

Trilinears    -(b-c)^2/a : (a^2+c*(b-c))/b : (a^2+b*(c-b))/c

## 4th Euler

The perpendicular to BC through X(5) cuts the nine-points-circle in two points A', A", with A" closer to A than A'. Build B', B", C', C" similarly. A"B"C" is the 4th Euler triangle. (Reference: X3758)

A-vertex coordinates:

Trilinears    -(b+c)^2/a : (a^2-c*(b+c))/b : (a^2-b*(b+c))/c

## 5th Euler

The nine-points-circle cuts again the median AA' in A", where A' is the midpoint of BC. If B" and C" are built similarly then A"B"C" is the 5th Euler triangle. (Reference: X3758)

A-vertex coordinates:

Trilinears    -2*(b^2+c^2)/(a*(a^2-b^2-c^2)) : 1/b : 1/c

## excenters-incenter reflections

The A-vertex is the reflection of the A-excenter in the incenter.

Note: The original name of this triangle is T(-1,3). Reference: TCCT, p. 173.

A-vertex coordinates:

Trilinears    (-a+3*b+3*c)/(-3*a+b+c) : 1 : 1

## excenters-midpoints

The A-vertex is the midpoint of A and the A-excenter.

Note: The original name of this triangle is T(-2,1). Reference: TCCT, p. 173.

A-vertex coordinates:

Trilinears    (-2*a+b+c)/a : 1 : 1

## excenters-reflections

See excenters-incenter reflections.

## excentral

The triangle whose vertices are the excenters of ABC.

Note: The excentral triangle is the anticevian and the antipedal triangle of X(1).

A-vertex coordinates:

Trilinears    -1 : 1 : 1

## 1st and 2nd excosine

Let Ab, Ac be the points where the A-excosine circle cuts again the sidelines AC and AB, respectively, and build {Bc, Ba}, {Ca, Cb} cyclically. The triangle A1B1C1 bounded by the lines AbAc, BcBa and CaCb is the 1st excosine triangle of ABC and the triangle A2B2C2 bounded by the lines BcCb, CaAc and AbBa is the 2nd excosine triangle of ABC. (Reference: X17807)

Note: If the tangents at B and C to the circumcircle of a triangle ABC intersect at A', then the circle with center A' passing through B and C is called A-excosine circle of ABC. This circle cuts AB and AC again at two points which are the extremities of a diameter of it. (Reference: MathWorld -- Excosine Circle.)

A-vertex coordinates:

1st excosine triangle:

Trilinears    -S^2*a : (S^2-2*SA*SB)*b : (S^2-2*SA*SC)*c

2nd excosine triangle:
Trilinears    -S^2*SB*SC/(SA*a) : (S^2-2*SA*SB)*b : (S^2-2*SA*SC)*c

## extangents

The triangle A'B'C' formed by the points of pairwise intersection of the three exterior tangents to the excircles.

A-vertex coordinates:

Trilinears    -a*(a+b+c) : (a+c)*(a+b-c) : (a+b)*(a-b+c)

## extouch

Let A' be the point where the A-excircle touches the side BC and define B', C' similarly. A'B'C' is the extouch triangle of ABC.

A-vertex coordinates:

Trilinears    0 : 1/(b*(a+b-c)) : 1/(c*(a-b+c))

## 2nd extouch

Let AA,AB, AC be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define BB, BC, BA and CC, CA, CB cyclically.
Let A2 = BCBA∩CACB, and define B2 and C2 cyclically. The 2nd extouch triangle is A2B2C2. (Reference: X5927)

A-vertex coordinates:

Trilinears    -2*(b+c) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 3rd extouch

Let AA,AB, AC be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define BB, BC, BA and CC, CA, CB cyclically.
Let A3 = CAAC∩ABBA, and define B3 and C3 cyclically. The 3rd extouch triangle is A3B3C3. (Reference: X5927)

A-vertex coordinates:

Trilinears    -2*(b+c)*(a+b-c)*(a-b+c)/((-a+b+c)*(a+b+c)) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 4th extouch

Let AA,AB, AC be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define BB, BC, BA and CC, CA, CB cyclically.
Let A4 = BCCA∩BACB, and define B4 and C4 cyclically. The 4th extouch triangle is A4B4C4. (Reference: X5927)

A-vertex coordinates:

Trilinears    -2*(b+c)*(a+b+c)/(-a+b+c) : (a^2-b^2+c^2)/b : (a^2+b^2-c^2)/c

## 5th extouch

Let AA,AB, AC be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define BB, BC, BA and CC, CA, CB cyclically.
Let D1 = BBBC∩CCCA, and define D2 and D3 cyclically. Let E1 = BBBA∩CCCB, and define E2 and E3 cyclically. Define A5 = D2E2∩D3E3, and define B5 and C5 cyclically. The 5th extouch triangle is A5B5C5. (Reference: X5927)

A-vertex coordinates:

Trilinears    -2*(b+c)/(-a+b+c) : (b^2+c^2+a^2+2*c*a)/(b*(a-b+c)) : (b^2+2*a*b+a^2+c^2)/(c*(a+b-c))

# F:

## inner-Fermat

In a triangle ABC build equilateral BCA, CAB, ABC inwards ABC. Then ABC is the inner-Fermat triangle of ABC.

A-vertex coordinates:

Trilinears    -sqrt(3)*a : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c

## outer-Fermat

In a triangle ABC build equilateral BCA, CAB, ABC outwards ABC. Then ABC is the outer-Fermat triangle of ABC.

A-vertex coordinates:

Trilinears    -sqrt(3)*a : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

## 1st and 2nd Fermat-Dao equilateral

Let F=X(13) be the 1st Fermat point of ABC and La the parallel to BC through F. Let Ab and Ac be the points, others than F, where La cuts the circumcircles of FAC and FAB, respectively. Construct Bc, Ba and Ca, Cb cyclically. Let A'=AbCb∩AcBc and similarly B' and C'. The triangle A'B'C' is equilateral and is named the 1st Fermat-Dao equilateral triangle. (Reference: X16247)

If the above construction is done from F=X(14) (the 2nd Fermat point of ABC), the obtained triangle A'B'C' is also equilateral and is called the 2nd Fermat-Dao equilateral triangle. (Reference: X16247)

You can see a sketch of 1st and 2nd Fermat-Dao equilateral triangles here.

A-vertex coordinates:

1st Fermat-Dao equilateral triangle:

Trilinears
-a*(sqrt(3)*S*((-6*R^2+3*SA+12*SW)*S^2+(7*SA^2+SB*SC+SW^2)*SW+18*R^2*SB*SC)+(20*S^2-6*R^2*(3*SA-SW)+19*SA^2-SB*SC+11*SW^2)*S^2+3*(2*SA-SW)*SA*SW^2)/(sqrt(3)*a^2+2*S) : (S+sqrt(3)*SB)*(S^2+sqrt(3)*(6*R^2-SW+SC)*S+SW*SC)*b : (S+sqrt(3)*SC)*(S^2+sqrt(3)*(6*R^2-SW+SB)*S+SB*SW)*c

2nd Fermat-Dao equilateral triangle:
Trilinears
-a*(-sqrt(3)*S*((-6*R^2+3*SA+12*SW)*S^2+(7*SA^2+SB*SC+SW^2)*SW+18*R^2*SB*SC)+(20*S^2-6*R^2*(3*SA-SW)+19*SA^2-SB*SC+11*SW^2)*S^2+3*(2*SA-SW)*SA*SW^2)/(sqrt(3)*a^2-2*S) : (-S+sqrt(3)*SB)*(S^2-sqrt(3)*(6*R^2-SW+SC)*S+SW*SC)*b : (-S+sqrt(3)*SC)*(S^2-sqrt(3)*(6*R^2-SW+SB)*S+SB*SW)*c

## 3rd and 4th Fermat-Dao equilateral

Let F=X(13) be the 1st Fermat point of ABC. Let Ab and Ac be points on AC and AB, respectively, such that FAbAc is an equilateral triangle with the same orientation as ABC. Denote the centroid of FAbAc as A' and similarly build B' and C'. The triangle A'B'C' is equilateral and is named the 3rd Fermat-Dao equilateral triangle. (Reference: X16267)

Let F=X(14) be the 2nd Fermat point of ABC. Let Ab and Ac be points on AC and AB, respectively, such that FAbAc is an equilateral triangle with opposite orientation than ABC. Denote the centroid of FAbAc as A" and similarly build B" and C". The triangle A"B"C" is equilateral and is named the 4th Fermat-Dao equilateral triangle. (Reference: X16268)

You can see a sketch of 3rd and 4th Fermat-Dao equilateral triangles here.

A-vertex coordinates:

3rd Fermat-Dao equilateral triangle:

Trilinears
(sqrt(3)*(3*S^2+SB*SC)+5*(SB+SC)*S)/(a*(sqrt(3)*SA+S)) : (SC+sqrt(3)*S)/b : (SB+sqrt(3)*S)/c

4th Fermat-Dao equilateral triangle:
Trilinears
(sqrt(3)*(3*S^2+SB*SC)-5*(SB+SC)*S)/(a*(sqrt(3)*SA-S)) : (SC-sqrt(3)*S)/b : (SB-sqrt(3)*S)/c

## 5th and 6th Fermat-Dao equilateral

Let F=X(13) be the 1st Fermat point of ABC. Let Ba and Ca be points on BC such that the triangle FBaCa is equilateral. Define the pairs {Cb, Ab} and {Ab, Bc}cyclically. Let A' be the point other than F that lies on the circles (FAcBc) and (FCbAb), and define B' and C' cyclically. Then the triangle A'B'C' is equilateral and is the 5th Fermat Dao equilateral triangle of ABC. (Reference: X16459)

When F=X(14), i.e, the 2nd Fermat point of ABC, the obtained triangle A'B'C' is also equilateral and is named the 6th Fermat-Dao equilateral triangle. (Reference: X16459)

You can see a sketch of 5th and 6th Fermat-Dao equilateral triangles here.

A-vertex coordinates:

5th Fermat-Dao equilateral triangle:

Trilinears
((7*SA+4*SW)*S^2+3*SA^3+5*sqrt(3)*(S^2+SA^2)*S)/(a*(sqrt(3)*SA+S)^2) : b*(sqrt(3)*c^2+2*S)/(sqrt(3)*SB+S) : c*(sqrt(3)*b^2+2*S)/(sqrt(3)*SC+S)

6th Fermat-Dao equilateral triangle:
Trilinears
((7*SA+4*SW)*S^2+3*SA^3-5*sqrt(3)*(S^2+SA^2)*S)/(a*(sqrt(3)*SA-S)^2) : b*(sqrt(3)*c^2-2*S)/(sqrt(3)*SB-S) : c*(sqrt(3)*b^2-2*S)/(sqrt(3)*SC-S)

## 7th and 8th Fermat-Dao equilateral

Let F=X(13) be the 1st Fermat point of ABC. Let Ab and Ac be points on AC and AB, respectively, such that FAbAc is an equilateral triangle with the same orientation as ABC. Let A' be the midpoint of Ab and Ac and define B' and C' cyclically. The triangle A'B'C' is equilateral and is named the 7th Fermat-Dao equilateral triangle. (Reference: X396)

Let F=X(14) be the 2nd Fermat point of ABC. Let Ab and Ac be points on AC and AB, respectively, such that FAbAc is an equilateral triangle with opposite orientation than ABC. Let A" be the midpoint of Ab and Ac and define B" and C" cyclically. The triangle A"B"C" is equilateral and is named the 8th Fermat-Dao equilateral triangle. (Reference: X395)

You can see a sketch of 7th and 8th Fermat-Dao equilateral triangles here.

A-vertex coordinates:

7th Fermat-Dao equilateral triangle:

Trilinears
2*(sqrt(3)*(SB+SC)+2*S)/(a*(sqrt(3)*SA+S)) : 1/b : 1/c

8th Fermat-Dao equilateral triangle:
Trilinears
2*(sqrt(3)*(SB+SC)-2*S)/(a*(sqrt(3)*SA-S)) : 1/b : 1/c

## 9th and 10th Fermat-Dao equilateral

Let F = X(13) be the 1st Fermat point of ABC. Let Ba, Ca be the points on BC such that FBaCa is an equilateral triangle. Define Cb, Ab and Ac, Bc cyclically. Let A', B', C' be the midpoints of AbAc, BcBa, CaCb, respectively. The triangle A'B'C' is equilateral and is named here the 9th Dao-Fermat equilateral triangle. (Reference: X16536)

When F=X(14), i.e., the 2nd Fermat point of ABC, the obtained triangle A'B'C' is also equilateral and is called the 10th Fermat-Dao equilateral triangle. (Reference: X16536)

You can see a sketch of 9th and 10th Fermat-Dao equilateral triangles here.

A-vertex coordinates:

9th Fermat-Dao equilateral triangle:

Trilinears
2*S*((2*S^2+SA^2+SB*SC)*sqrt(3)+(6*R^2+SA+SW)*S)/(a*(sqrt(3)*SA+S)^2) : b : c

10th Fermat-Dao equilateral triangle:
Trilinears
-2*S*((2*S^2+SA^2+SB*SC)*sqrt(3)-(6*R^2+SA+SW)*S)/(a*(sqrt(3)*SA-S)^2) : b : c

## 11th and 12th Fermat-Dao equilateral

Let F = X(13) be the 1st Fermat point of ABC. Let Ab and Ac be the points where the circle FBC cuts again AC and AB, respectively. Define Bc, Ba and Ca, Cb cyclically. Denote A', B', C' the circumcenters of ABcCb, BCaAc and CBaAb, respectively. Then the triangle A'B'C' is equilateral and is here named the 11th Dao-Fermat equilateral triangle.

When F=X(14), i.e., the 2nd Fermat point of ABC, the resulting triangle is also equilateral and is here named the 12th Fermat-Dao equilateral triangle. (Reference: X16626)

A-vertex coordinates:

11th Fermat-Dao equilateral triangle:

Trilinears
(sqrt(3)*(S^2+SB*SC)+(SB+SC)*S)/(a*(sqrt(3)*SA-S)) : SC/b : SB/c

12th Fermat-Dao equilateral triangle:
Trilinears
(sqrt(3)*(S^2+SB*SC)-(SB+SC)*S)/(a*(sqrt(3)*SA+S)) : SC/b : SB/c

## 13th and 14th Fermat-Dao equilateral

Let F = X(13) be the 1st Fermat point of ABC. Let Ab and Ac be the points where the circle FBC cuts again AC and AB, respectively. Define Bc, Ba and Ca, Cb cyclically. Let A', B', C' be centroids of ABcCb, BCaAc, CAbBa respectively. Then the triangle A'B'C' is equilateral and is the 13th Fermat-Dao equilateral triangle.

When F=X(14), i.e., the 2nd Fermat point of ABC, the resulting triangle is also equilateral and is the 14th Fermat-Dao equilateral triangle. (Reference: X16636)

A-vertex coordinates:

13th Fermat-Dao equilateral triangle:

Trilinears
(sqrt(3)*(2*S^2+SA^2+SB*SC)+(SA+SW)*S)/(a*(sqrt(3)*SA-S)) : b : c

14th Fermat-Dao equilateral triangle:
Trilinears
(sqrt(3)*(2*S^2+SA^2+SB*SC)-(SA+SW)*S)/(a*(sqrt(3)*SA+S)) : b : c

## 15th and 16th Fermat-Dao equilateral

Let AoBoCo be the outer-Fermat triangle of ABC. Denote p'a the polar of A with respect to the circle with center Ao and radius a=BC. Define similarly p'b and p'c. Then the triangle A'B'C' bounded by these polars is equilateral and is the 15th Fermat-Dao equilateral triangle.

Let AiBiCi be the inner-Fermat triangle of ABC. Denote p"a the polar of A with respect to the circle with center Ai and radius a=BC. Define similarly p"b and p"c. Then the triangle A"B"C" bounded by these polars is equilateral and is the 16th Fermat-Dao equilateral triangle. (Reference: X16960)

A-vertex coordinates:

15th Fermat-Dao equilateral triangle:

Trilinears
(sqrt(3)*(3*S^2+SB*SC)+5*(SB+SC)*S)/(a*(SA-sqrt(3)*S)) : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

16th Fermat-Dao equilateral triangle:
Trilinears
(sqrt(3)*(3*S^2+SB*SC)-5*(SB+SC)*S)/(a*(SA+sqrt(3)*S)) : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c

## 1st to 4th inner- and outer- Fermat-Dao-Nhi

Let ABC be a triangle with centroid G and inner-Fermat (or outer-Fermat) triangle AfBfCf.

1. Let A* be the reflection of G in Af and A1 the reflection of A* in A; build B1 and C1 cyclically.
2. Let A* be the reflection of G in A and A2 the reflection of A* in Af; build B2 and C2 cyclically.
3. Let A* be the reflection of A in G and A3 the reflection of A* in Af; build B3 and C3 cyclically.
4. Let A* be the reflection of Af in G and A4 the reflection of A* in A; build B4 and C4 cyclically.
Then triangles A1B1C1, A2B2C2, A3B3C3, A4B4C4 are equilateral and perspective to ABC. (Reference: Preamble just before X33602.)

These triangles built from the inner-Fermat triangle are the 1st, 2nd, 3rd and 4th inner-Fermat-Dao-Nhi triangles, respectively, and their correspondents built from the outer-Fermat triangle are the 1st, 2nd, 3rd and 4th outer-Fermat-Dao-Nhi triangles, respectively.

A-vertices coordinates:

1th to 4th inner-Fermat-Dao-Nhi triangles
Trilinears    A1 = (7*S-3*sqrt(3)*a^2)/a : (-2*S+3*sqrt(3)*SC)/b : (-2*S+3*sqrt(3)*SB)/c
Trilinears    A2 = (5*S-3*sqrt(3)*a^2)/a : (-4*S+3*sqrt(3)*SC)/b : (-4*S+3*sqrt(3)*SB)/c
Trilinears    A3 = (-4*S-3*sqrt(3)*a^2)/a : ( 5*S+3*sqrt(3)*SC)/b : ( 5*S+3*sqrt(3)*SB)/c
Trilinears    A4 = (-8*S-3*sqrt(3)*a^2)/a : ( S+3*sqrt(3)*SC)/b : ( S+3*sqrt(3)*SB)/c
1th to 4th outer-Fermat-Dao-Nhi triangles
Trilinears    A1 = (-7*S-3*sqrt(3)*a^2)/a : ( 2*S+3*sqrt(3)*SC)/b : ( 2*S+3*sqrt(3)*SB)/c
Trilinears    A2 = (-5*S-3*sqrt(3)*a^2)/a : ( 4*S+3*sqrt(3)*SC)/b : ( 4*S+3*sqrt(3)*SB)/c
Trilinears    A3 = ( 4*S-3*sqrt(3)*a^2)/a : (-5*S+3*sqrt(3)*SC)/b : (-5*S+3*sqrt(3)*SB)/c
Trilinears    A4 = ( 8*S-3*sqrt(3)*a^2)/a : ( -S+3*sqrt(3)*SC)/b : ( -S+3*sqrt(3)*SB)/c

## Feuerbach

The Feuerbach triangle is the triangle formed by the three points of tangency of the nine-point circle with the excircles. (Reference: MathWorld -- Feuerbach Triangle.)

A-vertex coordinates:

Trilinears    -(b-c)^2*(a+b+c)/a : (a+c)^2*(a+b-c)/b : (a+b)^2*(a-b+c)/c

## Fuhrmann

The Fuhrmann triangle has vertices the reflections of the vertices of 2nd circumperp triangle on the sidelines of ABC.

A-vertex coordinates:

Trilinears    a : -(a^2-c*(b+c))/b : -(a^2-b*(b+c))/c

## 2nd Fuhrmann

The 2nd Fuhrmann triangle has vertices the reflections of the vertices of 1st circumperp triangle on the sidelines of ABC.

A-vertex coordinates:

Trilinears     -a : (b*c+a^2-c^2)/b : (b*c+a^2-b^2)/c

# G:

## inner- and outer- Gallatly-Kiepert

inner-Gallatly-Kiepert: See 1st Brocard

outer-Gallatly-Kiepert: See 1st Brocard reflected

## inner-Garcia

Let TATBTC be the extouch triangle of a triangle ABC, and let LA be the line perpendicular to line BC at TA. Of the two points on LA at distance r=inradius-of-ABC from TA, let A' be the one farther from A and let A" be the closer. Define B', C' and B", C" cyclically. A'B'C' is the outer Garcia triangle and A"B"C" is the inner Garcia triangle. (Reference: X5587)

A-vertex coordinates:

Trilinears    a : (c*a-b^2+c^2)/b : (a*b+b^2-c^2)/c

## outer-Garcia

Let TATBTC be the extouch triangle of a triangle ABC, and let LA be the line perpendicular to line BC at TA. Of the two points on LA at distance r=inradius-of-ABC from TA, let A' be the one farther from A and let A" be the closer. Define B', C' and B", C" cyclically. A'B'C' is the outer Garcia triangle and A"B"C" is the inner Garcia triangle. (Reference: X5587)

A-vertex coordinates:

Trilinears    -1 : (a+c)/b : (a+b)/c

## Garcia-reflection

The Garcia-reflection triangle has vertices the reflections of the excenters on the midpoints of their corresponding sides. (Reference: X15995)

A-vertex coordinates:

Trilinears    1 : (c - a)/b : (b - a)/c

See here.

## Gossard

The Euler line of ABC cuts BC, CA, AB at A',B',C', respectively. The Euler lines of AB'C', BC'A' and CA'B' enclose the Gossard triangle. (Reference: X402)

A-vertex coordinates:

Trilinears
SA^2*(SB-SC)^2*(S^2-3*SB*SC)/(a*S^2) :
(S^2-3*SA*SC)*(4*R^2*(-6*SB+SW)+5*S^2-SW^2+6*SB^2-4*SA*SC)/b :
(S^2-3*SA*SB)*(4*R^2*(-6*SC+SW)+5*S^2-SW^2+6*SC^2-4*SA*SB)/c

## inner-Grebe

Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of triangle ABC (inwards). Then, the triangle A'B'C' enclosed by the lines BaCa, CbAb, AcBc is the inner-Grebe triangle of ABC.

A-vertex coordinates:

Trilinears    -(b^2+c^2-S)/a : b : c

## outer-Grebe

Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the lines BaCa, CbAb, AcBc is the outer-Grebe triangle of ABC. (Reference: Hyacinthos #6445)

A-vertex coordinates:

Trilinears    -(b^2+c^2+S)/a : b : c

# H:

## half-altitude

See midheight triangle.

## 1st and 2nd half-diamonds

Build equilateral triangles AB'C', BC'A' and CA'B'. There are two triads of such triangles leading to two distinct triangles A'B'C', called the 1st and 2nd half-diamonds triangles of ABC. (Reference: Preamble just before of X33338.)

A construction: Let P=X(17) or P=X(18) (Napoleon points). Let Bc the reflection of B in the C-cevian of P and Cb the reflection of C in the B-cevian-of P. Denote as a' the circle {{A,Bc,Cb}} and build b', c', cyclically, naming their centers as A', B', C', respectively. Then, for P=X(18), A'B'C' is the 1st half-diamonds triangle, and, for P=X(17), A'B'C' is the 2nd half-diamonds triangle.

A-vertex coordinates:

1st half-diamonds triangle:

Trilinears     -(sqrt(3)*a^2-2*S)/a : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

2nd half-diamonds triangle:
Trilinears     -(sqrt(3)*a^2+2*S)/a : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c

## 1st and 2nd half-diamonds-central

The centers of the triangles AB'C', BC'A' and CA'B' in the construction of the 1st and 2nd half-diamonds triangles are vertices of equilateral triangles, here named the 1st and 2nd half-diamonds-central triangles of ABC. (Reference: Preamble just before of X33338.)

A-vertex coordinates:

1st half-diamonds-central equilateral triangle:

Trilinears      -(a^2+2*sqrt(3)*S)/a : (SC-sqrt(3)*S)/b : (SB-sqrt(3)*S)/c

2nd half-diamonds-central equilateral triangle:
Trilinears      -(a^2-2*sqrt(3)*S)/a : (SC+sqrt(3)*S)/b : (SB+sqrt(3)*S)/c

## 1st and 2nd half-squares

Build isosceles rectangle triangles AB'C', BC'A' and CA'B', having vertical angles at A, B, C, respectively. There are two triads of such triangles leading to two distinct triangles A'B'C', called the 1st and 2nd half-squares triangles of ABC. (Reference: Preamble just before of X33338.)

A-vertex coordinates:

1st half-squares triangle:

Trilinears      -(a^2+S)/a : SC/b : SB/c

2nd half-squares triangle:
Trilinears      -(a^2-S)/a : SC/b : SB/c

## 1st Hatzipolakis

Let A'B'C' be the intouch triangle of ABC. Let CA be the point other than C' in which the perpendicular to BC from C' meets the incircle, let BA be the point other than B' in which the perpendicular to BC from B' meets the incircle, and let A0 be the point of intersection of lines BCA and CBA. Define B0 and C0 cyclically. A0B0C0 is the 1st Hatzipolakis triangle. (Reference: X1118)

A-vertex coordinates:

Trilinears    (-a+b+c)*a : (a^2+b^2-c^2)^2/(b*(a-b+c)) : (a^2-b^2+c^2)^2/(c*(a+b-c))

## 2nd Hatzipolakis

Let A0B0C0 be the 1st Hatzipolakis triangle of ABC. Let A1 be the orthogonal projection of A0 onto line BC, and define B1 and C1 cyclically. The triangle A1B1C1 is the 2nd Hatzipolakis triangle. (Reference: X1119)

A-vertex coordinates:

Trilinears    0 : 1/(b*(a^2-b^2+c^2)*(a-b+c)^2) : 1/(c*(a^2+b^2-c^2)*(a+b-c)^2)

## 3rd Hatzipolakis

Let A'B'C' be the orthic triangle of ABC. Let Ab and Ac be the orthogonal projections of A' on AC and AB, respectively, and denote Na the nine-point-center of AAbAc; build Nb and Nc cyclically. NaNbNc is the 3rd Hatzipolakis triangle. (Reference: X17817)

A-vertex coordinates:

Trilinears    2*S^2*(10*R^2-2*SA-SB-SC)/a : b*(S^2+SA*SB) : c*(S^2+SA*SC)

## Hatzipolakis-Moses

Let A'B'C' be the orthic triangle of ABC. Let AB be the orthogonal projection of A' onto line AB, and define BC and CA cyclically. Let AC be the orthogonal projection of A' onto line AC, and define BA and CB cyclically. Let A'B be the reflection of A in AB, and let A'C be the reflection of A in AC . Let NA be the nine-point center of the triangle AA'BA'C, and define NB and NC cyclically. The triangle NANBNC, is the Hatzipolakis-Moses triangle. (Reference: X6145)

A-vertex coordinates:

Trilinears    2*S^2*(6*R^2-2*SA-SB-SC)/a : b*(S^2+SA*SB) : c*(S^2+SC*SA)

## hexyl

Given a triangle ABC with excenters Ja, Jb, Jc, define A* as the point in which the perpendicular to AB through the excenter Jb meets the perpendicular to AC through the excenter Jc, and similarly define B* and C*. Then A*B*C* is the hexyl triangle of ABC. (Reference: MathWorld -- Hexyl Triangle.)

A-vertex coordinates:

Trilinears    a^3+(-b-c)*a^2-(b+c)^2*a+(b^2-c^2)*(b-c) : a^3+(c-b)*a^2-(b-c)^2*a+(b^2-c^2)*(b+c) : a^3+(b-c)*a^2-(b-c)^2*a-(b^2-c^2)*(b+c)

## Honsberger

Let A'B'C' be the intouch triangle of ABC. Let LA be the line through X(7) parallel to B'C', and let AB = AB∩LA and AC = AC∩LA. Define BC and CA cyclically, and define BA and CB cyclically. (The six points AB, BC, CA, AC, BA, CB lie on the Adams circle). Let A* = ABCB∩ACBC, and define B* and C* cyclically. The triangle A*B*C* is the Honsberger triangle. (Reference: X7670)

A-vertex coordinates:

Trilinears    1/((b+c)*a-(b-c)^2) : -1/(b*(a-b+c)) : -1/(c*(a+b-c))

## Hung-Feuerbach

Let FaFbFc be the Feuerbach triangle of ABC and (Ka) the circle, other than the nine-point-circle, passing through Fb and Fc and tangent to the A-excircle (Ia). Let A' be the touchpoint of (Ka) and (Ia) and name as ta the tangent to both circles in A'. Build tb and tc cyclically. The triangle bounded by ta, tb and tc is named here the Hung-Feuerbach triangle of ABC. (Reference: X5973)

A-vertex coordinates:

Trilinears
-(b+c)^2*(a+b-c)*(a-b+c)*(a^2+(b+c)^2)*(a^2+c*a+b*(b-c))*(a^2+b*a+3*(c-b))/a :
(a+c)*(a^5+c*a^4-(b^2-c^2)^2*a+c*(b^2-c^2)*(3*b^2+c^2))*((a+c)*b+a^2+c^2)*((-a+b)*c+a^2+b^2)/b :
(a+b)*(a^5+b*a^4-(c^2-b^2)^2*a+b*(c^2-b^2)*(3*c^2+b^2))*((a+b)*c+a^2+b^2)*((-a+c)*b+a^2+c^2)/c

## inner-Hutson

The internal bisector of angle A meets the A-excircle in two points. Let PA be the point closer to line BC and let QA be the other point. Define PB and PC cyclically, and define QB and QC cyclically. Let LA be the line tangent to the A-excircle at PA, and define LB and LC cyclically. The inner-Hutson triangle is the triangle bounded by the lines LA, LB and LC. (Reference: X363)

A-vertex coordinates:

Trilinears
-(2*δ2*sin(A/2)-y2*sin(B/2)-z2*sin(C/2)) :
(2*(c*z-δ2)*b*sin(A/2)-(a-c)*y2*sin(B/2)-b*z2*sin(C/2))/b :
(2*(b*y-δ2)*c*sin(A/2)-(a-b)*z2*sin(C/2)-c*y2*sin(B/2))/c

where x = (-a+b+c)/2, y = (a-b+c)/2, z = (a+b-c)/2 and δ2 = (4*R*r+r2)/2

## outer-Hutson

Continuing the construction of the inner-Hutson triangle, let MA be the line tangent to the A-excircle at QA, and define MB and MC cyclically. The outer-Hutson triangle is the triangle bounded by the lines MA, MB and MC. (Reference: X363)

A-vertex coordinates:

Trilinears
-(2*δ2*sin(-A/2)-y2*sin(B/2)-z2*sin(C/2)) :
(2*(c*z-δ2)*b*sin(-A/2)-(a-c)*y2*sin(B/2)-b*z2*sin(C/2))/b :
(2*(b*y-δ2)*c*sin(-A/2)-(a-b)*z2*sin(C/2)-c*y2*sin(B/2))/c

where x = (-a+b+c)/2, y = (a-b+c)/2, z = (a+b-c)/2 and δ2 = (4*R*r+r2)/2

## Hutson extouch

Let A' be the antipode of the A-extouch point in the A-excircle, and define B' and C' cyclically. A'B'C' is the Hutson-extouch triangle. (Reference: X5731)

A-vertex coordinates:

Trilinears    -4*a/(a+b+c) : (a+b-c)/b : (a-b+c)/c

## Hutson intouch

Let A' be the antipode of the A-intouch point in the incircle, and define B' and C' cyclically. A'B'C' is the Hutson-intouch triangle. (Reference: X5731)

A-vertex coordinates:

Trilinears    4*a/(-a+b+c) : (a-b+c)/b : (a+b-c)/c

## 1st Hyacinth

Let A'B'C' be the orthic triangle of a triangle ABC, and
O = circumcenter of ABC
Ab = reflection of A' in BO, and define Bc and Ca cyclically
Ac = reflection of A' in CO, and define Ba and Cb cyclically
(Na) = nine-point circle of A'AbAc, and define (Nb) and (Nc) cyclically
Na = nine-point center of A'AbAc, and define Nb and Nc cyclically
The triangle NaNbNc is the 1st Hyacinth triangle. (Reference: X10111)

A-vertex coordinates:

Trilinears
-a^3*SA*(SA^2-3*S^2)/(2*S^2) :
((12*R^2-2*SW)*S^2-(2*R^2*(SA+SC)-SC^2+S^2-(SA+SC)^2)*SB)/b :
((12*R^2-2*SW)*S^2-(2*R^2*(SA+SB)-SB^2+S^2-(SA+SB)^2)*SC)/c

## 2nd Hyacinth

Let A'B'C' be the orthic triangle of ABC. Define A* as the orthogonal projection of A on B'C', and B* and C* similarly. A*B*C* es the 2nd Hyacinth triangle of ABC.
(Reference: Hyacinthos #24020)

A-vertex coordinates:

Trilinears    -((b^2+c^2)*a^4-2*(b^2-c^2)^2*a^2+(b^2+c^2)*(b^2-c^2)^2)/(a*(a^2-b^2-c^2)) : (a^2-b^2+c^2)*b : (a^2+b^2-c^2)*c

# I:

## incentral

Cevian triangle of the incenter.

A-vertex coordinates:

Trilinears    0 : 1 : 1

## incircle-circles

Let a', b', c' be the incircle-inverses of the sidelines BC, CA and AB, respectively of ABC and A', B', C' their centers. A'B'C' is the incircle-circles triangle of ABC. (Reference: X11034)

A-vertex coordinates:

Trilinears    2*a : (a^2+4*a*b+b^2-c^2)/b : (a^2+4*a*c+c^2-b^2)/c

## infinite altitude

Let A', B', C' be the intersections of the line at infinity and the altitudes AX(4), BX(4), CX(4) of ABC, respectively. The triangle A'B'C' is the infinite altitude triangle of ABC. (Reference: Preamble just before X36436)

A-vertex coordinates:

Trilinears    -2*a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## intangents

In ABC, let ta be the other than BC interior common tangent of the incircle and the A-excircle; define tb and tc cyclically. The triangle A'B'C' bounded by these lines is the intangents triangle.

A-vertex coordinates:

Trilinears    -a*(-a+b+c) : (a-c)*(a-b+c) : (a-b)*(a+b-c)

## intouch

The intouch triangle of ABC, also called the contact triangle, has as vertices the touchpoints of the incircle and the sidelines of ABC.

A-vertex coordinates:

Trilinears    0 : 1/(b*(a-b+c)) : 1/(c*(a+b-c))

## inverse-in-excircles

The inverse-in-excircles triangle of ABC has as vertices the inverses of A,B,C in the A-, B-, C- excircle of ABC, respectively. (Reference: X8055)

A-vertex coordinates:

Trilinears    -((b+c)*a+(b-c)^2)/(a*(a+b+c)) : 1 : 1

## inverse-in-incircle

The inverse-in-incircle triangle of ABC has as vertices the inverses of A,B,C in the incircle of ABC. (Reference: X11018)

A-vertex coordinates:

Trilinears    ((b+c)*a-(b-c)^2)/(a*(-a+b+c)) : 1 : 1

## 1st and 2nd isodynamic-Dao

Let P1 = X(15) be the 1st isodynamic point of ABC. Circles (A, P1), (B, P1) and (C, P1) intersect by pairs at P1, A', B', C'. Then the triangle A'B'C' is equilateral and is named the 1st isodynamic-Dao equilateral triangle.

Let P2 = X(16) be the 2nd isodynamic point of ABC. Circles (A, P2), (B, P2) and (C, P2) intersect by pairs at P2, A", B", C". Then the triangle A"B"C" is equilateral and is called the 2nd isodynamic-Dao equilateral triangle. (Reference: X16802)

A-vertex coordinates:

1st isodynamic-Dao equilateral triangle:

Trilinears
-(SB+SC)*(sqrt(3)*SA+S)/a : (sqrt(3)*(S^2+SA*SC)+(SA+3*SC)*S)/b : (sqrt(3)*(S^2+SA*SB)+(SA+3*SB)*S)/c

2nd isodynamic-Dao equilateral triangle:
Trilinears
-(SB+SC)*(sqrt(3)*SA-S)/a : (sqrt(3)*(S^2+SA*SC)-(SA+3*SC)*S)/b : (sqrt(3)*(S^2+SA*SB)-(SA+3*SB)*S)/c

## 3rd and 4th isodynamic-Dao

Let AoBoCo be the orthic triangle of ABC. Denote by A', B', C' the 1st isodynamic points X(15) of ABoCo, BCoAo and CAoBo, respectively, and A", B", C" the 2nd isodynamic points X(16) of ABoCo, BCoAo and CAoBo, respectively. Then the triangles A'B'C' and A"B"C" are equilateral and are the 3rd and 4th isodynamic-Dao of ABC, respectively. (Reference: preamble just before X31683)

A-vertex coordinates:

3rd isodynamic-Dao equilateral triangle:

Trilinears
(sqrt(3)*(S^2+SB*SC)+2*(SB+SC)*S)/(a*SA) : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

4th isodynamic-Dao equilateral triangle:
Trilinears
(sqrt(3)*(S^2+SB*SC)-2*(SB+SC)*S)/(a*SA) : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c

# J:

## 1st, 2nd and 3rd Jenkins

Let {a'} be the circle internally tangent to the A-excircle and externally tangent to the B- and C- excircles and build {b'}, {c'} cyclically. Let A', B', C' be the centers of these just defined circles. The triangle A'B'C' is the 1st Jenkins triangle of ABC. (Reference: "Hechos Geométricos en el Triángulo", by Angel Montesdeoca).

Notes: Circles {a'}, {b'}, {c'} are named the A-, B- and C- Jenkins circles. (Reference: Jenkins circles at "The Triangle Web", by Quim Castellsaguer). These circles concur at X(10).

Let A" be the intersection, other than X(10), of the B- and C- Jenkins circles and denote B", C" cyclically. The triangle A"B"C" is the 2nd Jenkins triangle of ABC. (Reference: "Hechos Geométricos en el Triángulo", by Angel Montesdeoca).

Let Tab and Tac be the touchpoints of the A-Jenkins circle with the B- and C- excircles, respectively. Denote (Tbc, Tba) and (Tca, Tcb) cyclically. The triangle bounded by the lines TabTac, TbcTba and TcaTcb is named the 3rd Jenkins triangle of ABC.

A-vertex coordinates:

1st Jenkins triangle:
Trilinears
-(2*(b+c)*a^4+(3*b^2+4*b*c+3*c^2)*a^3-(b+c)*(b^2+c^2)*a^2-(b^2-c^2)^2*a+(b^2-c^2)^2*(b+c))/a :
((2*b^2+2*b*c+c^2)*a^3+(b+c)*(2*b^2+c^2)*a^2+(b^2-c^2)*(2*b+c)*c*a+(b^2-c^2)*(b+c)*c^2)/b :
((2*c^2+2*c*b+b^2)*a^3+(c+b)*(2*c^2+b^2)*a^2+(c^2-b^2)*(2*c+b)*b*a+(c^2-b^2)*(c+b)*b^2)/c

2nd Jenkins triangle:
Trilinears    -(a+b+c)*((b+c)*a+b^2+c^2)/(a*(a-b+c)*(a+b-c)) : (a+c)/b : (a+b)/c
3rd Jenkins triangle:
Trilinears
(a+b+c)*(2*(b+c)*a^4+(3*b^2+4*b*c+3*c^2)*a^3+(b+c)*(b^2+5*b*c+c^2)*a^2+3*(b^2+b*c+c^2)*b*c*a+(b+c)*b^2*c^2)/a :
((b+c)*a^3+(2*b^2+b*c+c^2)*a^2+(b^2+b*c+3*c^2)*b*a+b^2*c*(b+c))*(a-b+c)*(a+c)/b :
((c+b)*a^3+(2*c^2+c*b+b^2)*a^2+(c^2+c*b+3*b^2)*c*a+c^2*b*(c+b))*(a-c+b)*(a+b)/c :

## Jenkins-contact

Let Ta be the touchpoint of the A-Jenkins circle and the A-excircle and denote Tb, Tc cyclically. The triangle TaTbTc is the Jenkins-contact triangle of ABC. (See Jenkins triangle). (Reference: "Hechos Geométricos en el Triángulo", by Angel Montesdeoca).

A-vertex coordinates:

Trilinears    -(b+c)^2*(a+b-c)*(a-b+c)/(a*(a+b+c)) : c^2*(a-b+c)/b : b^2*(a-c+b)/c

## Jenkins-tangential

Let τa be the tangent to the A-Jenkins circle at Ta (see Jenkins-contact triangle) and define τb and τc cyclically. The triangle bounded by these tangents is the Jenkins-tangential triangle of ABC. (References: "Hechos Geométricos en el Triángulo", by Angel Montesdeoca and X(37865)).

A-vertex coordinates:

Trilinears
-((3*(b+c))*a^3+(2*b^2+3*b*c+2*c^2)*a^2-(b+c)^2*b*c-(b+c)*(b^2+c^2)*a)*b*c/a^2 :
((b+c)*a^3+(2*b-c)*b*a^2+(b^3-c^3-b*c*(b+c))*a+(b^2-c^2)*b*c)*(a+c)/b :
((c+b)*a^3+(2*c-b)*c*a^2+(c^3-b^3-c*b*(c+b))*a+(c^2-b^2)*c*b)*(a+b)/c

## Johnson

The vertices of the Johnson triangle are the centers of the Johnson circles Ja, Jb and Jc. (Reference: MathWorld -- Johnson Circles.)
Explanation: Johnson's theorem states that if three equal circles mutually intersect one another in a single point, then the circle J passing through their other three pairwise points of intersection is congruent to the original three circles. If Ja is the circle {{B,C,X(4)}} and Jb and Jc are defined cyclically, then each one has radius=circumradius and J is the circumcircircle of ABC.

Note: The Johnson triangle is also known as Carnot triangle.

A-vertex coordinates:

Trilinears    -a*(a^2-b^2-c^2) : (a^4+(-b^2-2*c^2)*a^2-c^2*(b^2-c^2))/b : (a^4+(-2*b^2-c^2)*a^2+(b^2-c^2)*b^2)/c

## inner-Johnson

Of the two tangents to the Jb and Jc Johnson circles of ABC (see Johnson triangle), let LA be the one on the opposite side of BC from A, and let MA be the other. Define LB, MB, LC, MC cyclically. The inner-Johnson triangle is the triangle enclosed by the lines LA, LB and LC. (Reference: X10522)

A-vertex coordinates:

Trilinears    -a^2+(b+c)*a-2*b*c : (a-b+c)*(a-c)^2/b : (a+b-c)*(a-b)^2/c

## outer-Johnson

Of the two tangents to the Jb and Jc Johnson circles of ABC (see Johnson triangle), let LA be the one on the opposite side of BC from A, and let MA be the other. Define LB, MB, LC, MC cyclically. The outer-Johnson triangle is the triangle enclosed by the lines MA, MB and MC. (Reference: X10522)

A-vertex coordinates:

Trilinears    -a^2-(b+c)*a-2*b*c : (a+c)^2*(a+b-c)/b : (a+b)^2*(a-b+c)/c

## 1st Johnson-Yff

The Yff circles are the two triplets of congruent circles in which each circle is tangent to two sides of a reference triangle (reference: MathWorld -- Yff Circles.). The circles Y'a, Y'b and Y'c of the first triplet concur at X(55) and the circles Y"a, Y"b and Y"c of the second triplet concur at X(56). The vertices of the 1st Johnson-Yff triangle are the other points of intersection of Y'a, Y'b and Y'c. (Reference: X10523).

Note: By Johnson theorem, if three congruent circles concur then the circle through their other points of intersection is congruent to the original circles. This is the reason for the name of the triangle.

A-vertex coordinates:

Trilinears    a/(-a+b+c) : (a+c)^2/(b*(a-b+c)) : (a+b)^2/(c*(a+b-c))

## 2nd Johnson-Yff

The Yff circles are the two triplets of congruent circles in which each circle is tangent to two sides of a reference triangle (reference: MathWorld -- Yff Circles.). The circles Y'a, Y'b and Y'c of the first triplet concur at X(55) and the circles Y"a, Y"b and Y"c of the second triplet concur at X(56). The vertices of the 2nd Johnson-Yff triangle are the other points of intersection of Y"a, Y"b and Y"c. (Reference: X10523).

Note: By Johnson theorem, if three congruent circles concur then the circle through their other points of intersection is congruent to the original circles. This is the reason for the name of the triangle.

A-vertex coordinates:

Trilinears    -a*(-b-c+a) : (a-b+c)*(a-c)^2/b : (a+b-c)*(a-b)^2/c

# K:

## K798e and K798i

Let Ae, Ai be the points where the perpendicular bisector of BC is cut by the circle through the nine-point-center of ABC and centered at the midpoint of BC, with Ai closer to A than Ae. Build Bi, Be, Ci, Ce cyclically. AeBeCe and AiBiCi are the K798e & K798i triangles, respectively. (References: CTC K798 and X(11230) & X(11231))

A-vertex coordinates of K798e:

Trilinears    -a*b*c : (b*c*SC+2*S^2)/b : (b*c*SB+2*S^2)/c

A-vertex coordinates of K798i:

Trilinears    -a*b*c : (b*c*SC-2*S^2)/b : (b*c*SB-2*S^2)/c

## 1st Kenmotu diagonals

Kenmotu squares of ABC are three equal squares such that each square touches two sides of ABC and all three squares touch at a single common point (reference: MathWorld -- Kenmotu Point.). There are two triplets of such squares: in the first of them the squares have X(371) as common vertex, in the second triplet the common vertex is X(372). The diagonals not through the common point of the squares in first triplet surround the 1st Kenmotu diagonal triangle.

A-vertex coordinates:

Trilinears    a*(b^2+c^2-a^2-2*S)/(b^2+c^2-a^2+2*S) : b : c

## 2nd Kenmotu diagonals

Kenmotu squares of ABC are three equal squares such that each square touches two sides of ABC and all three squares touch at a single common point (reference: MathWorld -- Kenmotu Point.). There are two triplets of such squares: in the first of them the squares have X(371) as common vertex, in the second triplet the common vertex is X(372). The diagonals not through the common point of the squares in second triplet surround the 2nd Kenmotu diagonal triangle.

A-vertex coordinates:

Trilinears    a*(b^2+c^2-a^2+2*S)/(b^2+c^2-a^2-2*S) : b : c

## 1st & 2nd Kenmotu-free-vertices

Let Ka, Kb, Kc be the free vertices of the 1st(2nd) Kenmotu squares. Triangle KaKbKc is the 1st(2nd) Kenmotu free vertices triangle. (Reference: X12375)

A-vertex coordinates:

1st Kenmotu free vertices:
Trilinears    -(4*S^2-(SB+SC)*(SA-S))/a : (SB-S)*b : (SC-S)*c

2nd Kenmotu free vertices:
Trilinears    -(4*S^2-(SB+SC)*(SA+S))/a : (SB+S)*b : (SC+S)*c

## Kosnita

The vertices of the Kosnita triangle are the circumcenters of the triangles BOC, COA, AOB, where O is the circumcenter of ABC. (Reference: X1658)

A-vertex coordinates:

Trilinears    (a^4+(-2*b^2-2*c^2)*a^2+b^4+c^4)*a : -b*(a^4+(-2*b^2-c^2)*a^2+(b^2-c^2)*b^2) : -(a^4+(-b^2-2*c^2)*a^2-c^2*(b^2-c^2))*c

# L:

## largest-circumscribed-equilateral

Considere all equilateral triangles AeBeCe circumscribing ABC and such that A lies between Be and Ce, B lies between Ce and Ae and C lies between Ae and Be. The triangle defined here is the AeBeCe with largest area. (Reference: Preamble just before X36761.)

The A-vertex of the largest-circumscribed-equilateral is the antipode of X(13) in the circle {{X(13), B, C}}, and the other two vertices are found cyclically.

A-vertex coordinates:

Trilinears
(-6*sqrt(3)*S*a^2-3*(a^2+b^2+c^2)*a^2+2*(b^2-c^2)^2)/a :
((7*b^2+2*c^2)*a^2-3*b^4+5*b^2*c^2-2*c^4+2*sqrt(3)*S*(2*a^2+b^2))/b :
((2*b^2+7*c^2)*a^2-3*c^4+5*b^2*c^2-2*b^4+2*sqrt(3)*S*(2*a^2+c^2))/c

## inner- & outer- Le Viet An

Let ABC be a triangle and BCA', CAB', ABC' equilateral triangles erected in/out - wardly of ABC. Let Bc, Cb be the circumcenters of CC'B and BB'C, respectively, and build Ca, Ac, Ab and Ba cyclically. Denote the circumcenters of BcCbA, CaAcB, AbBaC as Oa, Ob, Oc, respectively. Then, in each case, the triangle OaObOc is equilateral. These triangles are named the inner- and outer- Le Viet An triangles. (Reference: X14169)

Note: Both triangles are equilateral.

A-vertex coordinates:

inner-Le Viet An triangle:
Trilinears     (SW-sqrt(3)*S)*a : (SB-SC)*b : (SC-SB)*c

outer-Le Viet An triangle:
Trilinears     (SW+sqrt(3)*S)*a : (SB-SC)*b : (SC-SB)*c

## Lemoine

The Lemoine triangle is the cevian triangle of X(598).

A-vertex coordinates:

Trilinears    0 : a*c/(2*a^2+2*c^2-b^2) : a*b/(2*a^2+2*b^2-c^2)

## 1st and 2nd Lemoine-Dao

Like in the construction of the 2nd Lemoine circle of a triangle ABC, let Ab, Ac be the points where the antiparallel to BC through X(6) cuts the sides AC and AB, respectively. Define Bc, Ba, Ca, Cb cyclically. Build equilateral triangles BcCbA', CaAcB' and AbBaC' all with the same orientation as ABC and build equilateral triangles BcCbA", CaAcB" and AbBaC", all with opposite orientation than ABC. Then A'B'C' and A"B"C" are both equilateral triangles and are named the 1st Lemoine-Dao equilateral triangle and the 2nd Lemoine-Dao equilateral triangle, respectively. (Reference: X16940)

A-vertex coordinates:

1st Lemoine-Dao equilateral triangle:

Trilinears
-(sqrt(3)*SA+2*S)*(SB+SC)/(a*SA) : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c

2nd Lemoine-Dao equilateral triangle:
Trilinears
-(sqrt(3)*SA-2*S)*(SB+SC)/(a*SA) : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

## Lucas triangles

Build outwards ABC the square BCCABA (see figure). Lines ABA and ACA cut BC at A'B and A'C, respectively, and the perpendiculars through these points cut AB and AC at A"C and A"B, respectively. The quadrilateral A'BA'CA"BA"C is a square named the A-Lucas square (and also the A-inner-inscribed-square). The B- and C- Lucas squares, and the points B"C, B"A, C"A, C"B can be built similarly.

The circles {{A,A"B,A"C}}, {{B,B"C,B"A}} and {{C,C"A,C"B}} are the A-, B-, C- Lucas circles and result to be pairwise tangents and simultaneously tangent internally to the circumcircle of ABC. The circle externally tangent to the Lucas circles is named the Lucas inner circle.

The other circle tangent to the circumcircle and to the B- and C- Lucas circles is named the A-Lucas secondary central circle, and the B- and C-Lucas secondary central circles are defined in the same way.

Denote:
A0 = the center of the A-Lucas circle, and similarly B0, C0;
AT = the touchpoint of the B- and C- Lucas circles, and similarly BT, CT;
Ap = the antipode of A in the A-Lucas circle, and similarly Bp, Cp;
Ai = the touchpoint of the A- Lucas circle and the Lucas inner circle, and similarly Bi, Ci;
A2 = the center of the A-Lucas secundary central circle, and similarly B2, C2;
A2O = the touchpoints of the A-Lucas secundary central circle and the circumcircle, and similarly B2O, C2O;
A2B, A2C = the touchpoints of the A-Lucas secundary central circle and the B- and C- Lucas circles, respectively, and similarly B2C, B2A, C2A, C2B. These six point lie on the Lucas secondary tangents circle.

If the construction is made with squares drawn inwards ABC then all the above terms can be reused just by replacing the name Lucas with Lucas(-1); examples: A-Lucas(-1) square and A-Lucas(-1) circle instead of A-Lucas square and A-Lucas circle. Moreover, if the construction is generalized and BCCABA is a rectangle whose sides are in the proportion length:width=t:1 (i.e, BBA = CCA = t*BC), then all the terms described can be reused just by replacing Lucas by Lucas(t), assuming t>0 for rectangles built outwards ABC and t<0 otherwise. In this case we must be awared that Lucas(t) circles are pairwise tangent if, and only if, t=1 or t=-1.

## Lucas antipodal and Lucas(-1) antipodal

Vertices: Ap, Bp and Cp. See Lucas triangles. (Reference: X6457)

A-vertex coordinates:

Lucas antipodal:

Trilinears    cos(B)*cos(C)-sin(A) : -cos(B) : -cos(C)

Lucas(-1) antipodal:
Trilinears    cos(B)*cos(C)+sin(A) : -cos(B) : -cos(C)

## Lucas antipodal tangents and Lucas(-1) antipodal tangents

The tangents at Ap, Bp and Cp to the A-, B- and C-Lucas circles, respectively, bound the Lucas antipodal tangents triangle.

A-vertex coordinates:

Lucas antipodal tangents:

Trilinears    (2*SA+SB+SC+S)*SA*a : (S^2+SB*SC+(2*SB+SA)*S)*b : (S^2+SB*SC+(2*SC+SA)*S)*c

Lucas(-1) antipodal tangents:
Trilinears    (2*SA+SB+SC-S)*SA*a : (S^2+SB*SC-(2*SB+SA)*S)*b : (S^2+SB*SC-(2*SC+SA)*S)*c

## Lucas Brocard and Lucas(-1) Brocard

The Lucas Brocard triangle is bounded by the radical axis of the Brocard circle and the Lucas circles. See Lucas triangles. (Reference: X6421)

A-vertex coordinates:

Lucas Brocard:

Trilinears    -a*(a^2-2*S) : b*(a^2+c^2-S) : c*(a^2+b^2-S)

Lucas(-1) Brocard:
Trilinears    -a*(a^2+2*S) : b*(a^2+c^2+S) : c*(a^2+b^2+S)

## Lucas central and Lucas(-1) central

Vertices: A0, B0 and C0. See Lucas triangles.

A-vertex coordinates:

Lucas central:

Trilinears    cos(A)+2*sin(A) : cos(B) : cos(C)

Lucas(-1) central:
Trilinears    cos(A)-2*sin(A) : cos(B) : cos(C)

## Lucas homothetic and Lucas(-1) homothetic

The Lucas homothetic triangle is enclosed by the lines A"BA"C, B"CB"A and C"BC"B. See Lucas triangles. (Reference: X493)

A-vertex coordinates:

Lucas homothetic:

Trilinears    -(b^2+c^2-a^2)^2/(4*a) : b*(c^2+S) : c*(b^2+S)

Lucas(-1) homothetic:
Trilinears    -(b^2+c^2-a^2)^2/(4*a) : b*(c^2-S) : c*(b^2-S)

## Lucas inner and Lucas(-1) inner

Vertices: Ai, Bi and Ci. See Lucas triangles.

A-vertex coordinates:

Lucas inner:

Trilinears    2*cos(A)+3*sin(A)/2 : 2*cos(B)+sin(B) : 2*cos(C)+sin(C)

Lucas(-1) inner:
Trilinears    2*cos(A)-3*sin(A)/2 : 2*cos(B)-sin(B) : 2*cos(C)-sin(C)

## Lucas inner tangential and Lucas(-1) inner tangential

This triangle is bounded by the tangents to the Lucas inner circle at Ai, Bi and Ci. See Lucas triangles. (Reference: Preamble just before X6395)

A-vertex coordinates:

Lucas inner tangential:

Trilinears    4*cos(A)+sin(A) : 4*cos(B)+3*sin(B) : 4*cos(C)+3*sin(C)

Lucas(-1) inner tangential:
Trilinears    4*cos(A)-sin(A) : 4*cos(B)-3*sin(B) : 4*cos(C)-3*sin(C)

## Lucas reflection and Lucas(-1) reflection

Let La be the reflection of the line B0C0 in BC and define Lb, Lc cyclically (see Lucas triangles). The Lucas reflection triangle is the triangle limited by La, Lb, Lc. (Reference: X6401)

A-vertex coordinates:

Lucas reflection:

Trilinears
-cos(B)*cos(C)*sin(A)^2+(cos(B)+(1+sin(2*B))*cos(A-C)+sin(B))*(cos(C)+(1+sin(2*C))*cos(A-B)+sin(C)) :
cos(B)*(cos(C)*sin(A)*sin(B)-(cos(C)+(1+sin(2*C))*cos(A-B)+sin(C))*sin(C)) :
cos(C)*(cos(B)*sin(A)*sin(C)-(cos(B)+(1+sin(2*B))*cos(A-C)+sin(B))*sin(B))

Lucas(-1) reflection:
Trilinears
-cos(B)*cos(C)*sin(A)^2+(cos(B)+(1-sin(2*B))*cos(A-C)-sin(B))*(cos(C)+(1-sin(2*C))*cos(A-B)-sin(C)) :
cos(B)*(cos(C)*sin(A)*sin(B)+(cos(C)+(1-sin(2*C))*cos(A-B)-sin(C))*sin(C)) :
cos(C)*(cos(B)*sin(A)*sin(C)+(cos(B)+(1-sin(2*B))*cos(A-C)-sin(B))*sin(B))

## Lucas secondary central and Lucas(-1) secondary central

Vertices: A2, B2 and C2. See Lucas triangles. (Reference: X6199)

A-vertex coordinates:

Lucas secondary central:

Trilinears    cos(A)-2*sin(A) : cos(B)+4*sin(B) : cos(C)+4*sin(C)

Lucas(-1) secondary central:
Trilinears    cos(A)+2*sin(A) : cos(B)-4*sin(B) : cos(C)-4*sin(C)

## Lucas 1st secondary tangents and Lucas(-1) 1st secondary tangents

Triangle enclosed by the lines B2AC2A, C2BA2B and A2CB2C (see Lucas triangles). (Reference: X6199)

A-vertex coordinates:

Lucas 1st secondary tangents:

Trilinears    cos(A)-2*sin(A) : cos(B)+3*sin(B) : cos(C)+3*sin(C)

Lucas(-1) 1st secondary tangents:
Trilinears    cos(A)+2*sin(A) : cos(B)-3*sin(B) : cos(C)-3*sin(C)

## Lucas 2nd secondary tangents and Lucas(-1) 2nd secondary tangents

Triangle enclosed by the lines A2BA2C, B2CB2A and C2AC2B (see Lucas triangles). (Reference: X6199)

A-vertex coordinates:

Lucas 2nd secondary tangents

Trilinears    cos(A)+6*sin(A) : cos(B)-sin(B) : cos(C)-sin(C)

Lucas(-1) 2nd secondary tangents
Trilinears    cos(A)-6*sin(A) : cos(B)+sin(B) : cos(C)+sin(C)

## Lucas tangents and Lucas(-1) tangents

Vertices: AT, BT and CT. See Lucas triangles.

A-vertex coordinates:

Lucas tangents:

Trilinears    cos(A) : cos(B)+sin(B) : cos(C)+sin(C)

Lucas(-1) tangents:
Trilinears    cos(A) : cos(B)-sin(B) : cos(C)-sin(C)

# M:

## MacBeath

The MacBeath triangle is the triangle whose vertices are the contact points of the MacBeath inconic with the reference triangle ABC. (MathWorld -- MacBeath Triangle.).

Note: The MacBeath inconic of a triangle is the inconic having foci the orthocenter and the circumcenter, giving the center as the nine-point center.

A-vertex coordinates:

Trilinears    0 : 1/(b^3*(a^2-b^2+c^2)) : 1/(c^3*(a^2+b^2-c^2))

## Malfatti

The Malfatti triangle has as vertices the centers of the Malfatti circles.

Note: The Malfatti circles are three circles packed inside a triangle such that each is tangent to the other two and to two sides of the triangle. (Reference: MathWorld -- Malfatti Circles.)

A-vertex coordinates:

Trilinears    (2*cos(B/2)*cos(C/2)+2*cos(B/2)+2*cos(C/2)-cos(A/2)+1)/(cos(A/2)+1) : 1 : 1

## Mandart-excircles

Let A'B'C' be the intangents triangle of ABC and A* the point where B'C' touchs the A-excircle of ABC. Build B* and C* cyclically. The triangle A*B*C* is the Mandart-excircles triangle of ABC. (Reference: X10974)

A-vertex coordinates:

Trilinears    -(b-c)^2*(a+b+c)/a : b*(a+b-c) : c*(a-b+c)

## Mandart-incircle

Let A'B'C' be the intangents triangle of ABC and A* the point where B'C' touchs the incircle of ABC. Build B* and C* cyclically. The triangle A*B*C* is the Mandart-incircle triangle of ABC.

A-vertex coordinates:

Trilinears    (b+c-a)*(b-c)^2/a, b*(a-b+c), c*(a+b-c)

## McCay

The McCay triangle is the triangle whose vertices are the centers of the McCay circles. (Reference: X7606)

Note: Let G be the centroid and A'B'C' the 2nd Brocard triangle of ABC. The A-, B- and C- McCay circles of ABC are the circles {{G,B',C'}}, {{G,C',A'}} and {{G,A',B'}}, respectively. (Reference: MathWorld -- McCay Circles.)

A-vertex coordinates:

Trilinears    a*(a^2+b^2+c^2) : -(2*a^4+(-3*c^2-2*b^2)*a^2+(b^2-c^2)*(-c^2+2*b^2))/b : -(2*a^4+(-2*c^2-3*b^2)*a^2+(b^2-c^2)*(-2*c^2+b^2))/c

## medial

The medial triangle is the cevian triangle of the centroid.

A-vertex coordinates:

Trilinears    0 : 1/b : 1/c

## midarc

The A-angle bisector cuts the incircle at two points A' and A", with A closer from A than A". If B', B", C' C" are defined similarly then A'B'C' is the midarc triangle of ABC. (Reference: MathWorld -- Mid-Arc Triangle.)

A-vertex coordinates:

Trilinears    (cos(B/2)+cos(C/2))^2 : cos(A/2)^2 : cos(A/2)^2

## 2nd midarc

The A-angle bisector cuts the incircle at two points A' and A", with A closer from A than A". If B', B", C' C" are defined similarly then A"B"C" is the 2nd midarc triangle of ABC.

A-vertex coordinates:

Trilinears    (cos(B/2)-cos(C/2))^2 : cos(A/2)^2 : cos(A/2)^2

## midheight

Let A'B'C' be the orthic triangle of ABC and A*, B*, C* the midpoints of AA', BB', CC', respectively. A*B*C* is the midheight triangle of ABC. (It is also known as the half-altitude triangle)

A-vertex coordinates:

Trilinears    2*a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## mixtilinear

The mixtilinear triangle is the triangle connecting the centers of the mixtilinear incircles. It is alson known as the inner-mixtilinear triangle. (Reference: MathWorld -- Mixtilinear Triangle.)

Note: A circle that is internally tangent to two sides of a triangle and to the circumcircle is called a mixtilinear incircle. There are three mixtilinear incircles, one corresponding to each angle of the triangle (reference: MathWorld -- Mixtilinear Incircles.). Also, there are three circles that are externally tangent to two sides of a triangle and to the circumcircle and are called a mixtilinear excircles.

A-vertex coordinates:

Trilinears    -(a^3+(b+c)*a^2-(b+c)^2*a-(b^2-c^2)*(b-c))/(4*a*b*c) : 1 : 1

## 2nd mixtilinear

The triangle whose vertices are the centers of the mixtilinear excircles is the 2nd mixtilinear triangle. It is alson known as the outer-mixtilinear triangle. (See note in mixtilinear triangle). (Reference: Preamble just before X7955)

A-vertex coordinates:

Trilinears    (a^3-(b+c)*a^2-(b+c)^2*a+(b^2-c^2)*(b-c))/(4*a*b*c) : 1 : 1

## 3rd mixtilinear

The 3rd mixtilinear triangle has as vertices the touchpoints of the circumcircle and the mixtilinear incircles. (See note in mixtilinear triangle). (Reference: Preamble just before X7955)

A-vertex coordinates:

Trilinears    -1 : 2*b/(a-b+c) : 2*c/(a+b-c)

## 4th mixtilinear

The 4th mixtilinear triangle has as vertices the touchpoints of the circumcircle and the mixtilinear excircles. (See note in mixtilinear triangle). (Reference: Preamble just before X7955)

Note: The 4th mixtilinear triangle is the circumcevian triangle of X(55).

A-vertex coordinates:

Trilinears    -1 : 2*b/(a+b-c) : 2*c/(a-b+c)

## 5th mixtilinear

Let Iab and Iac be the touchpoints of the A-mixtilinear incircle and the sidelines AC and AB, respectively, and define Ibc, Iba, Ica, Icb similarly. Denote A5=IabIba∩IacIca, B5=IbcIcb∩IbaIab, C5=IcaIac∩IcbIbc. A5B5C5 is the 5th mixtilinear triangle. (See note in mixtilinear triangle).

Note: This triangle is also named Caelum triangle. (Reference: Preamble just before X7955)

A-vertex coordinates:

Trilinears    -(b+c-a)/(2*a) : 1 : 1

## 6th mixtilinear

Let Eab and Eac be the touchpoints of the A-mixtilinear excircle and the sidelines AC and AB, respectively, and define Ebc, Eba, Eca, Ecb similarly. The 6th mixtilinear triangle is the triangle bounded by the lines EabEac, EbbEba and EcaEcb. (See note in mixtilinear triangle). (Reference: Preamble just before X7955)

A-vertex coordinates:

Trilinears    a^2-2*(b+c)*a+(b-c)^2 : a^2-(b-c)*(2*a-b-3*c) : a^2-(c-b)*(2*a-c-3*b)

## 7th mixtilinear

Let Eab and Eac be the touchpoints of the A-mixtilinear excircle and the sidelines AC and AB, respectively, and define Ebc, Eba, Eca, Ecb similarly. Denote A7=EabEba∩EacEca, B7=EbcEcb∩EbaEab, C7=EcaEac∩EcbEbc. A7B7C7 is the 7th mixtilinear triangle. (See note in mixtilinear triangle). (Reference: X8916)

A-vertex coordinates:

Trilinears     (a-b+c)*(a+b-c)*(a^2-2*(b+c)*a+(b-c)^2)/(2*(-a+b+c)*a) : a^2-2*(b-c)*a+(b+3*c)*(b-c) : a^2+2*(b-c)*a+(c+3*b)*(c-b)

## inner-mixtilinear

See mixtilinear triangle.

## outer-mixtilinear

See 2nd mixtilinear triangle.

## Montesdeoca-Hung

Let (Ap) be the Apollonius circle and (Ka), (Kb), (Kc) the circles defined in the Hung-Feuerbach triangle. Let La be the radical axis of (Ap) and (Ka) and define Lb and Lc similarly. The triangle enclosed by these lines is the Montesdeoca-Hung triangle. (Reference: X6042)

Note: The Apollonious circle is the circle internally tangent to all the excircles (MathWorld -- Apollonius Circle.)

A-vertex coordinates:

Trilinears
2*a^5*(a+2*b+2*c)+4*a^3*(b^2+3*b*c+c^2)*(a+b+c)+(3*b^4+3*c^4+4*b*c*(3*b^2+5*b*c+3*c^2))*a^2+(b+c)^2*(2*(b+c)*(b^2+c^2)*a+b^4+c^4) :
-(a+c)^2*(a^2+b*a+c*(b+c))^2 :
-(a+b)^2*(a^2+c*a+b*(c+b))^2

## 1st Morley

Let ba and bc be the two angle trisectors of B, and ca and cb the two angle trisectors of C, with ba and ca closer to BC than the others. Build ab and ac similarly. Define A1 = ba∩ca and similarly B1 and C1. The triangle A1B1C1 is equilateral and is named the 1st Morley triangle. (MathWorld -- First Morley Triangle.)

A-vertex coordinates:

Trilinears    1 : 2*cos(C/3) : 2*cos(B/3)

## 2nd Morley

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate ba around B an angle Pi/3 (counter-clockwise) and let b'a be the resulting line. Next rotate ca around C an angle Pi/3 (clockwise) and denote c'a the resulting line. Define A2= b'a∩c'a and similarly B2 and C2. The triangle A2B2C2 is equilateral and is named the 2nd Morley triangle. (MathWorld -- Second Morley Triangle.)

Note: Invert the senses of rotation if ABC is labeled clockwise.

A-vertex coordinates:

Trilinears    -1 : 2*cos(C/3+π/3) : 2*cos(B/3+π/3)

## 3rd Morley

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate ba around B an angle Pi/3 (clockwise) and let b"a be the resulting line. Next rotate ca around C an angle Pi/3 (counter-clockwise) and denote c"a the resulting line. Define A3= b"a∩c"a and similarly B3 and C3. The triangle A3B3C3 is equilateral and is named the 3rd Morley triangle. (MathWorld -- Third Morley Triangle.)

Note: Invert the senses of rotation if ABC is labeled clockwise.

A-vertex coordinates:

Trilinears    -1 : 2*cos(C/3-π/3) : 2*cos(B/3-π/3)

With the same notation used in 1st Morley triangle, denote A'1 = bc∩cb and similarly B'1 and C'1. The triangle A'1B'1C'1 is the 1st Morley-adjunct triangle. (MathWorld -- First Morley Adjunct Triangle.)

Note: The 1st Morley-adjunct triangle is not equilateral.

A-vertex coordinates:

Trilinears    2 : sec(C/3) : sec(B/3)

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate bc around B an angle Pi/3 (counter-clockwise) and let b'c be the resulting line. Next rotate cb around C an angle Pi/3 (clockwise) and denote c'b the resulting line. Define A'2= b'c∩c'b and similarly B'2 and C'2. The triangle A'2B'2C'2 is equilateral is the 2nd Morley-adjunct triangle. (MathWorld -- Second Morley Adjunct Triangle.)

Note: Invert the senses of rotation if ABC is labeled clockwise. This triangle is not equilateral.

A-vertex coordinates:

Trilinears    -2 : sec(C/3+π/3) : sec(B/3+π/3)

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate bc around B an angle Pi/3 (clockwise) and let b"a be the resulting line. Next rotate cb around C an angle Pi/3 (counter-clockwise) and denote c"a the resulting line. Define A'3= b"c∩c"b and similarly B'3 and C'3. The triangle A'3B'3C'3 is the 3rd Morley-adjunct triangle. (MathWorld -- Third Morley Adjunct Triangle.)

Note: Invert the senses of rotation if ABC is labeled clockwise. This triangle is not equilateral.

A-vertex coordinates:

Trilinears    -2 : sec(C/3-π/3) : sec(B/3-π/3)

## 1st, 2nd and 3rd Morley-adjunct-midpoint

Let AnBnCn be the nth Morley-adjunct triangle (n=1,2,3). Let Ba and Ca be points on BC such that AnBaCa is an equilateral triangle having the same orientation as ABC; define Cb and Ab cyclically, and define Ac and Bc cyclically. Let A"n be the midpoint of Ab and Ac, and define B"n and C"n cyclically. Then A"nB"nC"n is named the nth Morley-adjunct-midpoint triangle. (Reference: X16839-to-X16841)

A-vertex coordinates:

Trilinears
(3*a*b*c*(2*a*y*z+b*x*z+c*x*y)+2*(2*sqrt(3)*S+3*SB+3*SC)*b*c*x+2*(3*SA+6*SC+sqrt(3)*S)*a*c*y+2*(6*SB+3*SA+sqrt(3)*S)*a*b*z+12*S^2+4*sqrt(3)*(SA+SW)*S+12*SB*SC)/(a*(3*b*c*x+6*SA-2*sqrt(3)*S)) : 2*b+x*c+z*a : 2*c+y*a+x*b

where x=sec((A-2*(n-1)*π)/3), y=sec((B-2*(n-1)*π)/3), z=sec((C-2*(n-1)*π)/3)

## 1st, 2nd and 3rd Morley-midpoint equilaterals

Let AnBnCn be the nth Morley triangle (n=1,2,3). Let Ba and Ca be points on BC such that AnBaCa is an equilateral triangle having the same orientation as ABC; define Cb and Ab cyclically, and define Ac and Bc cyclically. Let A"n be the midpoint of Ab and Ac, and define B"n and C"n cyclically. Then A"nB"nC"n is an equilateral triangle and is named the nth Morley-midpoint equilateral triangle. (Reference: X3602-to-X3604)

A-vertex coordinates:

Trilinears    (12*a*b*c*(x*z*b+x*y*c+2*z*y*a)+2*(3*a^2+2*sqrt(3)*S)*b*c*x+2*(3*SA+6*SC+sqrt(3)*S)*a*c*y+2*(3*SA+6*SB+sqrt(3)*S)*a*b*z+3*S^2+sqrt(3)*(SA+SW)*S+3*SB*SC)/(a*(3*SA+6*x*b*c-sqrt(3)*S)) : 2*z*a+b+2*x*c : 2*y*a+2*x*b+c

where x=cos((A-2*(n-1)*π)/3), y=cos((B-2*(n-1)*π)/3), z=cos((C-2*(n-1)*π)/3)

## Moses-Hung

Let HaHbHc be the orthic triangle of ABC and (Ja) the circle, other than the nine-point-circle, passing through Hb and Hc and tangent to the A-excircle (Ia). Let A' be the touchpoint of (Ja) and (Ia) and define B' and C' cyclically. The triangle A'B'C' is named here the Moses-Hung triangle of ABC. (Reference: X6044)

A-vertex coordinates:

Trilinears    -(2*a^3+(b+c)*a^2+(b^2-c^2)*(b-c))^2/(a*(a+b+c)) : (a+c)^2*(a+b-c)^3/b : (a+b)^2*(a-b+c)^3/c

## Moses-Soddy

Let A' be the pole of Soddy line in the Soddy A-circle, and define B' and C' cyclically. The triangle A'B'C' is the Moses-Soddy triangle. (Reference: X44311)

A-vertex coordinates:

Trilinears    1/a : (a-c)/(b*(b-c)) : (a-b)/(c*(c-b))

## Moses-Steiner osculatory

A' is the A-center of curvature of the Steiner circumellipse, and similarly B' and C'. A'B'C' is the Moses-Steiner osculatory triangle of ABC. (Reference: X34505)

A-vertex coordinates:

Trilinears    (-3*a^4+2*(b^2+c^2)*a^2-(b^2-c^2)^2)/a^3 : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## Moses-Steiner reflection

Let A', B', C' be the reflections of X(99) in the sides of Moses-Steiner osculatory triangle of ABC. A'B'C' is the Moses-Steiner reflection triangle of ABC. (Reference: X34505)

A-vertex coordinates:

Trilinears    (-a^2+b^2+c^2)/a : (a^2-b^2-2*c^2)/b : (a^2-2*b^2-c^2)/c

# N:

## inner-Napoleon

The inner Napoleon triangle is the triangle A"B"C" formed by the centers of internally erected equilateral triangles BCA', CAB' and ABC' on the sides of a given triangle ABC. It is an equilateral triangle. (MathWorld -- Inner Napoleon Triangle.)

A-vertex coordinates:

Trilinears    -a : (SC-sqrt(3)*S)/b : (SB-sqrt(3)*S)/c

## outer-Napoleon

The outer Napoleon triangle is the triangle A"B"C" formed by the centers of externally erected equilateral triangles BCA', CAB' and ABC' on the sides of a given triangle ABC. It is an equilateral triangle. (MathWorld -- Outer Napoleon Triangle.)

A-vertex coordinates:

Trilinears    -a : (SC+sqrt(3)*S)/b : (SB+sqrt(3)*S)/c

## 1st Neuberg

The triangle A1B1C1 formed by joining the centers of the Neuberg circles is the 1st Neuberg triangle. (Weisstein, Eric W. "First Neuberg Triangle." From MathWorld)

Note: The Neuberg A-circle is the locus of A of a triangle on a given base BC and with a given Brocard angle ω. (MathWorld -- Neuberg Circles.)

A-vertex coordinates:

Trilinears    -a*SW : (SC^2-SA*SB)/b : (SB^2-SA*SC)/c

## 2nd Neuberg

The triangle A2B2C2 formed by joining the centers of the reflections of the Neuberg circles on the opposite sides of ABC is the 2nd Neuberg triangle. (MathWorld -- Second Neuberg Triangle.)

A-vertex coordinates:

Trilinears    -a*SW : (SC^2-SA*SB+2*S^2)/b : (SB^2-SA*SC+2*S^2)/c

# O:

## orthic

Cevian and pedal triangle of X(4).

A-vertex coordinates:

Trilinears    0 : 1/(b*(a^2-b^2+c^2)) : 1/(c*(a^2+b^2-c^2))

## orthic axes

Let LA be the orthic axis of triangle BCX(4), and define LB, LC cyclically. The triangle bounded by these lines is the orthic axes triangle. (Reference: X2501)

The orthic axes triangle is also the cevian triangle of X(4) with respect to the orthic triangle.

A-vertex coordinates:

Trilinears    2*SB*SC/a : SA*SC/b : SA*SB/c

## orthocentroidal

Let A' be the intersection, other than X(4), of the A-altitude and the orthocentroidal circle, and define B' and C' cyclically. The triangle A'B'C'is the orthocentroidal triangle. (Reference: X5476)

Note: The orthocentroidal circle has diameter X(2)X(4).

A-vertex coordinates:

Trilinears    a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 1st orthosymmedial

Let A' be the intersection, other than the orthocenter, of the A-altitude and the circle with diameter X(4)X(6). Define B' and C' similarly. The triangle A'B'C' is the 1st orthosymmedial triangle.

A-vertex coordinates:

Trilinears    2*a^3/(b^2+c^2) : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## 2nd orthosymmedial

Let A" be the intersection, other than the symmedian center, of the A-symmedian and the circle with diameter X(4)X(6). Define B" and C" similarly. The triangle A"B"C" is the 2nd orthosymmedial triangle.

A-vertex coordinates:

Trilinears    -((b^4+c^4)*a^2+(b^2-c^2)^2*(-b^2-c^2))/(a*(b^2+c^2)*(a^2-b^2-c^2)) : b : c

# P:

## Paasche-Hutson

Let pa be the parabola with focus A and directrix BC. Let Ba be the point, closer to C, at which pa cuts AC and let Ca be the point, closer to B, at which pa cuts AB. Build Cb, Ab and Ac, Bc cyclically. Points Ab, Ac, Bc, Ba, Ca, Cb lie on an ellipse named the Paasche ellipse.

Let AH = AbCb∩AcBc, BH = BcAc∩BaCa, CH = CaBa∩CbAb. AHBHCH is the Paasche-Hutson triangle. (Reference: X(37994))

A-vertex coordinates:

Trilinears    S*(S-2*a*R)/a : (S+a*b)*c : (S+a*c)*b

## 1st Pamfilos-Zhou

The Pamfilos-Zhou A-rectangle RA = AABAAAC is the rectangle of maximal area such that A is a vertex of RA, B lies on the line AAAC, and C lies on the line AAAB. The Pamfilos-Zhou B- and C-rectangles, RB and RC, are defined cyclically.

Let A' = BCBA∩CACB, and define B' and C' cyclically. The 1st Pamfilos-Zhou triangle is A'B'C'. (Reference: X7594)

A-vertex coordinates:

Trilinears
-a^4*(a^2+b*c)+(b^2+c^2)*(b+c)^2*a^2+b*c*(b+c)*(4*S*a-(b^2-c^2)*(b-c)) :
(a^2-b^2+c^2)*((a^2+b^2+c^2)*a*b+2*c^2*S) :
(a^2+b^2-c^2)*((a^2+b^2+c^2)*a*c+2*b^2*S)

## 2nd Pamfilos-Zhou

The Pamfilos-Zhou A-rectangle RA = AABAAAC is the rectangle of maximal area such that A is a vertex of RA, B lies on the line AAAC, and C lies on the line AAAB. The Pamfilos-Zhou B- and C-rectangles, RB and RC, are defined cyclically.

Let A" = CAAC∩ABBA, and define B" and C" cyclically. The 2nd Pamfilos-Zhou triangle is A"B"C". (Reference: X7594)

A-vertex coordinates:

Trilinears
-2*(b+c)*S-(-a+b+c)*((b+c)*a+(b-c)^2) :
(2*(a-c)*a*S+(-a+b+c)*(a^2*c+(b-c)*(b^2+a*b+c^2)))/b :
(2*(a-b)*a*S+(-a+b+c)*(a^2*b+(c-b)*(c^2+a*c+b^2)))/c

## 1st Parry

Let A1 be the intersection, other than X(111), of the Parry circle and the line AX(111), and define B1 and C1 cyclically. The triangle A1B1C1 is the 1st Parry triangle. (Reference: Preamble just before X9123)

Note: The Parry circle is the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.

A-vertex coordinates:

Trilinears    3*a^4+(-2*b^2-2*c^2)*a^2-b^2*c^2+c^4+b^4 : -b*(a^2+b^2-2*c^2)*a : -c*(a^2-2*b^2+c^2)*a

## 2nd Parry

Let A2 be the intersection, other than X(110), of the Parry circle and line AX(110), and define B2 and C2 cyclically. The triangle A2B2C2 is the 2nd Parry triangle. (Reference: Preamble just before X9123)

Note: The Parry circle is the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.

A-vertex coordinates:

Trilinears    (a^4+b^2*c^2-b^4-c^4)*(b^2-c^2) : (a^2-b^2)*b*a*(2*a^2-b^2-c^2) : -(a^2-c^2)*c*a*(2*a^2-b^2-c^2)

## 3rd Parry

Let A3 be the intersection, other than X(2), of the Parry circle and the A-median, and define B3 and C3 cyclically. The triangle A3B3C3, is the 3rd Parry triangle. (Reference: Preamble just before X9123)

Note: The Parry circle is the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.

A-vertex coordinates:

Trilinears    a*(a^4+(-3*b^2-3*c^2)*a^2+2*b^4+b^2*c^2+2*c^4) : b*c^2*(2*a^2-b^2-c^2) : b^2*c*(2*a^2-b^2-c^2)

## Pelletier

The Pelletier triangle is the vertex triangle of ABC and the intangets triangle. Reference: TCCT, p. 162.

Note: The vertex triangle A*B*C* of triangles T1=A1B1C1 and T2=A2B2C2 has vertex A*=B1B2∩C1C2, and B*, C* are defined cyclically.

Note: The Pelletier triangle is the anticevian triangle of X(650)

A-vertex coordinates:

Trilinears    -(b-c)*(-a+b+c), (c-a)*(a-b+c), (a-b)*(a-c+b)

## 1st Przybyłowski-Bollin

Let D1 be the 1st isodynamic center X(15) of ABC and I'a, I'b and I'c the incenters of BCD1, CAD1 and ABD1, respectively. The triangle I'aI'bI'c is the 1st Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as AiBiCi in ETC.

A-vertex coordinates:

Trilinears    a*(sqrt(3)*SA+S) : sqrt(3)*SB*b+(b+U)*S : sqrt(3)*SC*c+(c+U)*S
where U = 2*sqrt(SW+sqrt(3)*S)

## 2nd Przybyłowski-Bollin

Let D1 be the 1st isodynamic center X(15) of ABC and J'a the excenter of BCD1 opposed to D1. Define J'b and J'c similarly. The triangle J'aJ'bJ'c is the 2nd Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as AaBbCb in ETC.

A-vertex coordinates:

Trilinears    a*(sqrt(3)*SA+S) : sqrt(3)*b*SB+(b-U)*S : sqrt(3)*c*SC+(c-U)*S
where U = 2*sqrt(SW+sqrt(3)*S)

## 3rd Przybyłowski-Bollin

Let D2 be the 2nd isodynamic center X(16) of ABC and I"a, I"b and I"c the incenters of BCD2, CAD2 and ABD2, respectively. The triangle I"aI"bI"c is the 3rd Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as (AiBiCi)* in ETC.

A-vertex coordinates:

Trilinears    a*(sqrt(3)*SA-S) : sqrt(3)*SB*b-(b+V)*S : sqrt(3)*SC*c-(c+V)*S
where V = 2*sqrt(SW-sqrt(3)*S)

## 4th Przybyłowski-Bollin

Let D2 be the 2nd isodynamic center X(16) of ABC and J"a the excenter of BCD2 opposed to D2. Define J"b and J"c similarly. The triangle J"aJ"bJ"c is the 4th Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as (AaBbCb)* in ETC.

A-vertex coordinates:

Trilinears    a*(sqrt(3)*SA-S) : sqrt(3)*SB*b-(b-V)*S : sqrt(3)*SC*c-(c-V)*S
where V = 2*sqrt(SW-sqrt(3)*S)

# R:

## reflection

Let A' be the reflection of A in BC and define B', C' similarly. A'B'C' is the reflection triangle of ABC.

A-vertex coordinates:

Trilinears    -1 : (a^2+b^2-c^2)/(a*b) : (a^2-b^2+c^2)/(a*c)

## Roussel

If the angular trisectors of a triangle are produced to the circumcircle, then the chords of adjacent trisectors form an equilateral triangle named the Roussel triangle. (Reference)

A-vertex coordinates:

Trilinears
-a^2-2*(c*cos(B/3)+b*cos(C/3))*a+4*b*c*(4*cos(A/3)^2-1)*cos(B/3)*cos(C/3) :
(a+2*b*cos(C/3))*(2*cos(A/3)*c+b)+4*c*cos(A/3)*cos(B/3)*(2*cos(A/3)*b+c) :
(a+2*c*cos(B/3))*(2*cos(A/3)*b+c)+4*b*cos(A/3)*cos(C/3)*(2*cos(A/3)*c+b)

# S:

## 1st & 2nd Savin

No construction given. (Reference: Preamble just before X44301)

A-vertex coordinates:

1st triangle:
Trilinears    2/(-a+b+c) : 1/b : 1/c
2nd triangle:
Trilinears    2/(3*a+b+c) : 1/b : 1/c

## 1st Schiffler

It is well known that if P lies on the Neuberg cubic then the Euler lines La, Lb and Lc of PBC, PCA and PAB, respectively, are concurrent. In particular, for P=X(1) and P=X(3065), the point of concurrence is the Schiffler center S=X(21).

Let P=X(1) and Bc the point of intersection, other than S, of Lb and the circle (C,|CS|) and Cb the point of intersection, other than S, of Lc and the circle (B,|BS|). Define O'a as the circumcenter of SBcCb and similarly O'b and O'c. The triangle O'aO'bO'c is the 1st Schiffler triangle. (Reference: X6595)

A-vertex coordinates:

Trilinears    1/a : (a-c)/(a^2-(b+2*c)*a+c^2-b^2) : (a-b)/(a^2-(c+2*b)*a+b^2-c^2)

## 2nd Schiffler

It is well known that if P lies on the Neuberg cubic then the Euler lines La, Lb and Lc of PBC, PCA and PAB, respectively, are concurrent. In particular, for P=X(1) and P=X(3065), the point of concurrence is the Schiffler center S=X(21).

Let P=X(3065) and Bc the point of intersection, other than S, of Lb and the circle (C,|CS|) and Cb the point of intersection, other than S, of Lc and the circle (B,|BS|). Define O"a as the circumcenter of SBcCb and similarly O"b and O"c. The triangle O"aO"bO"c is the 2nd Schiffler triangle. (Reference: X6596)

A-vertex coordinates:

Trilinears    -1/a : (a-c)/(a^2+(b-2*c)*a+c^2-b^2) : (a-b)/(a^2+(c-2*b)*a+b^2-c^2)

## Schröeter

Let A'B'C' and A"B"C" be the medial and orthic triangles of ABC, respectively. Define A*=B'C'∩B"C" and similarly B* and C*. A*B*C* is the Schröeter triangle of ABC.

Note: The Schroeter triangle is the anticevian triangle of X(523).

A-vertex coordinates:

Trilinears    1/a : (a^2-c^2)/(b*(b^2-c^2)) : (a^2-b^2)/(c*(c^2-b^2))

## 1st Sharygin

Let A', B', C' be the feet of the internal angles bisectors. The perpendicular bisectors of AA', BB', CC' bound a triangle A*B*C* called the 1st Sharygin triangle. (Reference)

A-vertex coordinates:

Trilinears    -a^2+b*c : a*b+c^2 : a*c+b^2

## 2nd Sharygin

Let A", B", C" the feet of the external angles bisectors. The perpendicular bisectors of AA", BB", CC" bound a triangle A*B*C* called the 2nd Sharygin triangle. (Reference)

A-vertex coordinates:

Trilinears    -a^2+b*c : a*b-c^2 : a*c-b^2

## Soddy

Let AoBoCo, AiBiCi be the outer- and inner-Soddy triangles of ABC. Then BoCoCiBi are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of the Soddy triangle. (Reference: Preamble before X(31528))

Note: The Soddy triangle is the anticevian triangle of X(7).

A-vertex coordinates:

Trilinears    -1/(a*(-a+b+c)) : 1/(b*(a-b+c)) : 1/(c*(a+b-c))

## inner-Soddy

Given a triangle ABC with inner Soddy center S', the inner Soddy triangle is the triangle A'B'C' formed by the points of tangency of the inner Soddy circle with the three mutually tangent circles centered at each of the vertices of ABC. (MathWorld -- Soddy Triangles.)

Note: Given three noncollinear points, construct three tangent circles such that one is centered at each point and the circles are pairwise tangent to one another. Then there exist exactly two nonintersecting circles that are tangent to all three circles. These are called the inner and outer Soddy circles, and their centers are called the inner S and outer Soddy centers S^', respectively. (MathWorld -- Soddy Circles.)

A-vertex coordinates:

Trilinears    (a*(-a+b+c)+2*S)/(a*(-a+b+c)) : (b*(a+c-b)+S)/(b*(a-b+c)) : (c*(a+b-c)+S)/(c*(a-c+b))

## outer-Soddy

Given a triangle ABC with outer Soddy center S", the outer Soddy triangle is the triangle A"B"C" formed by the points of tangency of the outer Soddy circle with the three mutually tangent circles centered at each of the vertices of ABC. (MathWorld -- Soddy Triangles.)

Note: Given three noncollinear points, construct three tangent circles such that one is centered at each point and the circles are pairwise tangent to one another. Then there exist exactly two nonintersecting circles that are tangent to all three circles. These are called the inner and outer Soddy circles, and their centers are called the inner S and outer Soddy centers S^', respectively. (MathWorld -- Soddy Circles.)

A-vertex coordinates:

Trilinears    (a*(-a+b+c)-2*S)/(a*(-a+b+c)) : (b*(a+c-b)-S)/(b*(a-b+c)) : (c*(a+b-c)-S)/(c*(a-c+b))

## 2nd inner-Soddy and 2nd outer-Soddy

Let A'B'C' be the intouch triangle of ABC, AoBoCo the outer-Soddy triangle and AiBiCi the inner-Soddy triangles of ABC. Then:

1. B'C'CiBi are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of the 2nd inner-Soddy triangle.

2. B'C'CoBo are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of the 2nd outer-Soddy triangle.
(Reference: Preamble before X(31528))

A-vertex coordinates:

2nd inner-Soddy triangle:
Trilinears    ((-a+b+c)*a+2*S)/(a*(-a+b+c)) : 1 : 1
2nd outer-Soddy triangle:
Trilinears    ((-a+b+c)*a-2*S)/(a*(-a+b+c)) : 1 : 1

## inner-squares

Build outwards ABC the square BCCABA (see figure). Lines ABA and ACA cut BC at A'B and A'C, respectively, and the perpendiculars through these points cut AB and AC at A"C and A"B, respectively. The quadrilateral A'BA'CA"BA"C is a square named the A-inner-inscribed-square (and also the A-Lucas square). The B- and C- inner-inscribed-squares can be built similarly. The centers of these squares are the vertices of the inner-squares triangles. (MathWorld -- Inner Inscribed Squares Triangle.)

A-vertex coordinates:

Trilinears    2*a : (a^2+b^2-c^2+2*S)/b : (a^2+c^2-b^2+2*S)/c

## outer-squares

Build inwards ABC the square BCCABA (see figure). Lines ABA and ACA cut BC at A'B and A'C, respectively, and the perpendiculars through these points cut AB and AC at A"C and A"B, respectively. The quadrilateral A'BA'CA"BA"C is a square named the A-outer-inscribed-square (and also the A-Lucas(-1) square). The B- and C- outer-inscribed-squares can be built similarly. The centers of these squares are the vertices of the outer-squares triangles. (MathWorld -- Outer Inscribed Squares Triangle.)

A-vertex coordinates:

Trilinears    2*a : (a^2+b^2-c^2-2*S)/b : (a^2+c^2-b^2-2*S)/c

## Stammler

The Stammler triangle is the triangle formed by the centers of the Stammler circles. It is an equilateral triangle. (MathWorld -- MathWorld -- Stammler Triangle.)

Note: The Stammler circles are the three circles (apart from the circumcircle), that intercept the sidelines of a reference triangle ABC in chords of lengths equal to the corresponding side lengths a, b, and c. (MathWorld -- Stammler Circles.)

A-vertex coordinates:

Trilinears    cos(A)-2*cos((B-C)/3) : cos(B)+2*cos(B/3+2*C/3) : cos(C)+2*cos(2*B/3+C/3)

## Steiner

The Steiner triangle is the cevian triangle of the Steiner point X(99). (MathWorld -- Steiner Triangle.)

A-vertex coordinates:

Trilinears    0 : -c*(a^2-b^2) : b*(a^2-c^2)

## submedial

Let ABC be a triangle, and suppose that C' is a point on side AB and that B' is a point on side AC. Let r(B'C') be the rectangle whose vertices are B', C', and the orthogonal projections of B' and C' onto side BC. Let RA be the rectangle r(B'C') of maximal area, which is obtained by taking A'B'C' to be the medial triangle of ABC. Let OA be the center RA, and define OB and OC cyclically. The central triangle OAOBOC is named the submedial triangle. (Reference: X9813)

A-vertex coordinates:

Trilinears    2*a*b*c : (3*a^2+b^2-c^2)*c : (3*a^2+c^2-b^2)*b

## symmedial

The symmedial triangle is the cevian triangle of the symmedian point X(6).

A-vertex coordinates:

Trilinears    0 : 1/c : 1/b

# T:

## T(1,2)

See Aquila triangle.

## T(-1,3)

See excenters-incenter reflections triangle.

## T(-2,1)

See excenters-incenter midpoints triangle.

## tangential

The tangential triangle is the triangle enclosed by the lines tangent to the circumcircle of a given triangle ABC at its vertices. It is therefore the antipedal triangle of ABC with respect to the circumcenter O.

Note: The tangential triangle is the anticevian triangle of X(6).

A-vertex coordinates:

Trilinears    -a : b : c

## (1st) tangential-midarc

The tangential mid-arc triangle of a reference triangle ABC is the triangle A'B'C' whose sides are the tangents to the incircle at the intersections of the internal angle bisectors with the incircle, where the points of intersection nearest the vertices are chosen. (MathWorld -- Tangential Mid-Arc Triangle.)

A-vertex coordinates:

Trilinears    -2*b*c*sin(A/2) : c*(2*sin(B/2)*a+a+b-c) : b*(2*sin(C/2)*a+a+c-b)

## 2nd tangential-midarc

The 2nd tangential mid-arc triangle of a reference triangle ABC is the triangle A'B'C' whose sides are the tangents to the incircle at the intersections of the internal angle bisectors with the incircle, where the points of intersection furthest from the vertices are chosen. (Reference: Preamble above X8075)

A-vertex coordinates:

Trilinears    2*b*c*sin(A/2) : c*(2*sin(B/2)*a-(a+b-c)) : b*(2*sin(C/2)*a-(a+c-b))

## inner tri-equilateral

In a triangle ABC, let's inscribe three congruent equilateral triangles PAbAc, PBcBa and PCaCb, with Ba, Ca on BC, Cb, Ab on CA and Ac, Bc on AB. There are two points P making possible this construction: P = Pi=X(15) and P = Po=X(16). The equilateral triangles obtained in each case are here named the A-, B-, C- inner/outer equilateral triangles, respectively.

For the inner-equilateral triangles, the inner tri-equilateral triangle AiBiCi is defined as the triangle bounded by the lines AbAc, BcBa and CaCb. (Reference: Preamble above X10631)

A-vertex coordinates:

Trilinears    a*(SA-sqrt(3)*S)/(SA+sqrt(3)*S) : b : c

## outer tri-equilateral

In a triangle ABC, let's inscribe three congruent equilateral triangles PAbAc, PBcBa and PCaCb, with Ba, Ca on BC, Cb, Ab on CA and Ac, Bc on AB. There are two points P making possible this construction: P = Pi=X(15) and P = Po=X(16). The equilateral triangles obtained in each case are here named the A-, B-, C- inner/outer equilateral triangles, respectively.

For the outer-equilateral triangles, the outer tri-equilateral triangle AoBoCo is defined as the triangle bounded by the lines AbAc, BcBa and CaCb. (Reference: Preamble above X10631)

A-vertex coordinates:

Trilinears    a*(SA+sqrt(3)*S)/(SA-sqrt(3)*S) : b : c

## tri-squares

Inscribe three squares into a triangle ABC such that each square has two vertices on two distinct sides of ABC and the other vertices of the three squares coincide at the vertices of another triangle A'B'C'. See this figure. (Reference: Preamble above X13637).

There are four triangles A'B'C' for this construction. These are named the tri-squares triangles of ABC.

A-vertex coordinates:

1st tri-square triangle:
Trilinears     2*S/a : (a^2+3*b^2-c^2+2*S)/b : (a^2-b^2+3*c^2+2*S)/c
2nd tri-square triangle:
Trilinears    -2*S/a : (a^2+3*b^2-c^2-2*S)/b : (a^2-b^2+3*c^2-2*S)/c
3rd tri-square triangle:
Trilinears    4*(a^2+S)/(a*(b^2+c^2-a^2+2*S)) : 1/b : 1/c
4th tri-square triangle:
Trilinears    4*(a^2-S)/(a*(b^2+c^2-a^2-2*S)) : 1/b : 1/c

## tri-squares-central

The tri-squares-central triangles are the triangles whose vertices are the centers of the squares built in the tri-squares triangles. (Reference: Preamble above X13637).

A-vertex coordinates:

1st tri-square-central triangle:
Trilinears    2*(3*a^2+4*S)/a : (3*b^2+c^2-a^2+2*S)/b : (b^2+3*c^2-a^2+2*S)/c
2nd tri-square-central triangle:
Trilinears    2*(3*a^2-4*S)/a : (3*b^2+c^2-a^2-2*S)/b : (b^2+3*c^2-a^2-2*S)/c
3rd tri-square-central triangle:
Trilinears    (a^2+2*S)/a : (b^2+S)/b : (c^2+S)/c
4th tri-square-central triangle:
Trilinears    (a^2-2*S)/a : (b^2-S)/b : (c^2-S)/c

## Trinh

There is a unique equilateral triangle AA1A2 inscribed in the circumcircle of triangle ABC, where, for concreteness, the labels are fixed so that AA1A2 has the same orientation as ABC. Let BB1B2 and CC1C2 be the corresponding equilateral triangles. Let A' = B1B2∩C1C2, B' = C1C2∩A1A2 and C' = A1A2∩B1B2. The triangle A'B'C' is named the Trinh triangle of ABC. (Reference: X7688)

A-vertex coordinates:

Trilinears    -a*(S^2+3*SA^2) : b*(S^2-3*SA*SB) : c*(S^2-3*SA*SC)

# U:

## Ursa-major

In a triangle ABC, the common internal tangents of the incircle and the A-excircle touch them in four concyclic points. Let {a'} be the circle through these touchpoints and denote a" the radical axis of {a'} and the A-excircle. Build b" and c" cyclically. The triangle bounded by a", b" and c" is the Ursa-major triangle of ABC. (Reference: Preamble just before X17603)

A-vertex coordinates:

Trilinears    -(b+c)*a^2+2*(b^2+c^2)*a-(b+c)*(b^2+c^2) : ((a-b)^2+(2*a-c)*c)*(a-c) : ((a-c)^2+(2*a-b)*b)*(a-b)

## Ursa-minor

In a triangle ABC, the common internal tangents of the incircle and the A-excircle touch them in four concyclic points. Let {a'} be the circle through these touchpoints and denote a" the radical axis of {a'} and the incircle. Build b" and c" cyclically. The triangle bounded by a", b" and c" is the Ursa-minor triangle of ABC. (Reference: Preamble just before X17603)

A-vertex coordinates:

Trilinears    -(b+c)*a-(b-c)^2 : (a-b+c)*(a-c) : (a+b-c)*(a-b)

# V:

## inner-Vecten

Erect squares internally on each side of a triangle ABC. The centers of this squares are the vertices of the inner-Vecten triangle.

A-vertex coordinates:

Trilinears    -a : (SC-S)/b : (SB-S)/c

## 2nd and 3rd inner-Vecten

Given a triangle ABC, build the square σa = AAbAaAc, with the same orientation as ABC, and such that B and C lie on the lines AaAb and AaAc, respectively. Define σb = BBcBbBa and σc = CCaCcCb cyclically. Let Ao, Bo, Co be the centers of these squares.

AaBbCc and AoBoCo are the 2nd inner-Vecten and 3rd inner-Vecten of ABC, respectively. (Reference: Preamble just before X32488).

A-vertex coordinates:

2nd inner-Vecten triangle:
Trilinears    1/(a*SA) : 1/(b*(SB-S)) : 1/(c*(SC-S))
3rd inner-Vecten:
Trilinears    (3*S^2+SB*SC-2*S*SW)/(a*SA) : (SC-S)/b : (SB-S)/c

## outer-Vecten

Erect squares externally on each side of a triangle ABC. The centers of this squares are the vertices of the outer-Vecten triangle.

A-vertex coordinates:

Trilinears    -a : (SC+S)/b : (SB+S)/c

## 2nd and 3rd outer-Vecten

Given a triangle ABC, build the square σa = AAbAaAc, with opposite orientation than ABC, and such that B and C lie on the lines AaAb and AaAc, respectively. Define σb = BBcBbBa and σc = CCaCcCb cyclically. Let Ao, Bo, Co be the centers of these squares.

AaBbCc and AoBoCo are the 2nd outer-Vecten and 3rd outer-Vecten of ABC, respectively. (Reference: Preamble just before X32488).

A-vertex coordinates:

2nd outer-Vecten triangle:
Trilinears    1/(a*SA) : 1/(b*(SB+S)) : 1/(c*(SC+S))
3rd outer-Vecten:
Trilinears    (3*S^2+SB*SC+2*S*SW)/(a*SA) : (SC+S)/b : (SB+S)/c

See here.

## Vijay-Paasche-midpoints

The medial triangle of the 1st Vijay-Paasche-Hutson triangle is the Vijay-Paasche-midpoints triangle. (Reference: X(37881))

A-vertex coordinates:

Trilinears    (2*R+b)*(2*R+c)/R : (4*R*(R+b)+b*c+(b-c)*a)/b : (4*R*(R+c)+c*b+(c-b)*a)/c

## Vijay-Paasche-polar

Let pa be the parabola with focus A and directrix BC. Let Ba be the point, closer to C, at which pa cuts AC and let Ca be the point, closer to B, at which pa cuts AB. Build Cb, Ab and Ac, Bc cyclically. Points Ab, Ac, Bc, Ba, Ca, Cb lie on an ellipse named the Paasche ellipse.

Let AP be the pole of A with respect to the Paasche ellipse and build BP, CP cyclically. APBPCP is the Vijay-Paasche-polar triangle. (Reference: X(37994))

A-vertex coordinates:

Trilinears    -(-2*a*R+S)^2/a : (S^2+2*R*(a+b+4*R)*S+2*a*b*(2*R^2+S))/b : (S^2+2*R*(a+c+4*R)*S+2*a*c*(2*R^2+S))/c

## Vijay-Paasche-tangents

Let pa be the parabola with focus A and directrix BC. Let Ba be the point, closer to C, at which pa cuts AC and let Ca be the point, closer to B, at which pa cuts AB. Build Cb, Ab and Ac, Bc cyclically. Points Ab, Ac, Bc, Ba, Ca, Cb lie on an ellipse named the Paasche ellipse.

Let AT be the intersection of the tangents to the Paasche ellipse at Ab and Ac and build BT, CT cyclically. ATBTCT is the Vijay-Paasche-tangents triangle. (Reference: X(37884))

A-vertex coordinates:

Trilinears    -(-2*a*R+S)^2/a : (S^2+2*R*(a+b+4*R)*S+2*a*b*(2*R^2+S))/b : (S^2+2*R*(a+c+4*R)*S+2*a*c*(2*R^2+S))/c

## Vu-Dao-isodynamic equilateral

Let A0B0C0 be the orthic triangle of ABC. Let A15 = X(15)-of-AB0C0, and define B15 and C15 cyclically, so that A15B15C15 is the 3rd isodynamic-Dao equilateral triangle of ABC.

Let A'15 be the point, other than A, in which the line AA15 meets the circle {{A,B0,C0}}, and define B'15 and C'15 cyclically. The triangle A'15B'15C'15 is the Vu-Dao-X(15)-isodynamic equilateral triangle.

If X(15) is replaced by X(16) in the above construction, the resulting triangle, A'16B'16C'16, is the Vu-Dao-X(16)-isodynamic equilateral triangle. (Reference: X(5478))

A-vertex coordinates:

Vu-Dao-X(15)-isodynamic equilateral triangle
Trilinears    ((S^2+3*SB*SC)*a^2-4*sqrt(3)*S*SB*SC)/(4*a*SA*(sqrt(3)*a^2-2*S)) : (sqrt(3)*SC-S)/b : (sqrt(3)*SB-S)/c
Vu-Dao-X(16)-isodynamic equilateral triangle
Trilinears    ((S^2+3*SB*SC)*a^2+4*sqrt(3)*S*SB*SC)/(4*a*SA*(sqrt(3)*a^2+2*S)) : (sqrt(3)*SC+S)/b : (sqrt(3)*SB+S)/c

# W:

## Walsmith

The Walsmith triangle has A-vertex the intersection of the A-symmedian of ABC and the perpendicular at X(125) to BC. (Reference: Bernard Gibert cubic K1091)

A-vertex coordinates:

Trilinears    -(2*a^6-(3*b^4-4*b^2*c^2+3*c^4)*a^2+(b^4-c^4)*(b^2-c^2))/(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)/a : b : c

## Wasat

Let ABC be a triangle, A'B'C' its incentral triangle and (I) its incircle. Denote (Oa) the circle with diameter AA' and ra the radical axis of (Oa) and (I); build (Ob), (Oc) and rb, rc cyclically. The triangle A*B*C* bounded by ra, rb, rc is the Wasat triangle of ABC. (Reference: Preamble just before X21616)

A-vertex coordinates:

Trilinears    (b+c)/a : (c-a)/b : (b-a)/c

# X:

## X3-ABC reflections

This triangle is obtained by reflecting the circumcenter X3 about A, B, C. Reference: TCCT 6.13, p. 160.

A-vertex coordinates:

Trilinears    -(3*S^2+SB*SC)/a : SB*b : SC*c

## X-parabola-tangential

Let A*B*C* be the vertex triangle of the medial and orthic triangles of ABC and A'B'C' the medial triangle of A*B*C*. Then A, B, C, A', B', C' lie on a parabola here named the X-parabola of ABC. The tangents to this parabola at A,B,C bound the X-parabola-tangential triangle. (Reference: X12064)

Note: The X-parabola-tangential triangle is the anticevian triangle of X(115).

A-vertex coordinates:

Trilinears    -(b^2-c^2)^2/a : (a^2-c^2)^2/b : (a^2-b^2)^2/c

# Y:

## Yff central

Let three isoscelizers IACIAB, IBAIBC, and ICAICB be constructed on a triangle ABC, one for each side. This makes all of the inner triangles similar to each other. However, there is a unique set of three isoscelizers for which the four interior triangles A'IBCICB, IACB'ICA, IABIBAC', and A'B'C' are congruent. The innermost triangle A'B'C' is called the Yff central triangle (Kimberling 1998, pp. 94-95). (See this figure in MathWorld).

A-vertex coordinates:

Trilinears    2*sin(A/2) : (2*a*sin(B/2)+a+b-c)/b : (2*a*sin(C/2)+a-b+c)/c

## Yff contact

The Yff contact triangle is the cevian triangle of X(190).

A-vertex coordinates:

Trilinears    0 : -1/b/(a-c) : 1/c/(a-b)

## inner-Yff

The Yff circles are the two triplets of congruent circles in which each circle is tangent to two sides of a reference triangle (see MathWorld -- Yff Circles.). The circles in each triplet have radius r1=r*R/(R+r) and r2=r*R/(R-r), respectively. The centers of the circles in the first triplet are the vertices of the inner-Yff triangle. (Reference: Preamble just before X10037)

A-vertex coordinates:

Trilinears    -(a^4-2*(b^2+b*c+c^2)*a^2+(b^2-c^2)^2)/(2*a^2*b*c) : 1 : 1

## outer-Yff

The Yff circles are the two triplets of congruent circles in which each circle is tangent to two sides of a reference triangle (see MathWorld -- Yff Circles.). The circles in each triplet have radius r1=r*R/(R+r) and r2=r*R/(R-r), respectively. The centers of the circles in the second triplet are the vertices of the outer-Yff triangle. (Reference: Preamble just before X10037)

A-vertex coordinates:

Trilinears    (a^4-2*(b^2-b*c+c^2)*a^2+(b^2-c^2)^2)/(2*a^2*b*c) : 1 : 1

## inner-Yff tangents

Let LA be the line, other than BC, tangent to the B- and C-inner Yff circles. Define LB, LC cyclically. Let A' = LB∩LC and define B', C' cyclically. Triangle A'B'C' is the inner-Yff tangents triangle. (Reference: X10527)

A-vertex coordinates:

Trilinears    (b+c-a)*(a^3+(b+c)*a^2-(b^2+c^2)*a-(b^2-c^2)*(b-c))/(4*a^2*b*c) : 1 : 1

## outer-Yff tangents

Let MA be the line, other than BC, tangent to the B- and C-outer Yff circles. Define MB, MC cyclically. Let A" = MB∩MC and define B", C" cyclically. Triangle A"B"C" is the outer-Yff tangents triangle. (Reference: X10527)

A-vertex coordinates:

Trilinears    -(b+c-a)*(a^3+(b+c)*a^2-(b^2-4*b*c+c^2)*a-(b^2-c^2)*(b-c))/(4*a^2*b*c) : 1 : 1

## Yiu

The Yiu triangle is the triangle formed by the centers of the Yiu circles. (Reference)

Note: The A- Yiu circle of a triangle ABC is the circle through A, the reflection of B in AC and the reflection of C in AB. (MathWorld -- Yiu Triangle.)

A-vertex coordinates:

Trilinears    2*S*(SA^2-R^2*SA-S^2)/R : c*(S^2+SA*SC) : b*(S^2+SA*SB)

## Yiu tangents

Let tA be the tangent at A to the A-Yiu circle, and define tB and tC cyclically. Let A' = tB∩tC, and define B' and C' cyclically. Triangle A'B'C' is the Yiu tangents triangle. (Reference: X7495)

Note: The A- Yiu circle of a triangle ABC is the circle through A, the reflection of B in AC and the reflection of C in AB. (MathWorld -- Yiu Triangle.)

A-vertex coordinates:

Trilinears    -(SB+3*SC)*(SC+3*SB)/a : (SA+3*SC)*(SC+3*SB)/b : (SA+3*SB)*(SB+3*SC)/c

# Z:

## 1st and 2nd Zaniah

In a triangle ABC, let AmBmCm be the medial triangle, I the incenter and A' the touchpoint of the incircle with the side BC. Then the lines AA', IAm and BmCm are concurrent at a point A1. Denote B1 and C1 cyclically. The triangle A1B1C1 is the 1st Zaniah triangle of ABC.

Continuing with the previous construction, let Ja be the A-excenter and A" the touchpoint of the A-excircle with the side BC. Then the lines AA", JaAm and BmCm are concurrent at a point A2. Denote B2 and C2 cyclically. The triangle A2B2C2 is the 2nd Zaniah triangle of ABC. (Reference: Preamble before X18214)

A-vertex coordinates:

1st Zaniah triangle:
Trilinears    2 : (a+b-c)/b : (a-b+c)/c
2nd Zaniah triangle:
Trilinears    2 : (a-b+c)/b : (a+b-c)/c

Last changes:

2021.08.21 - Added: Vijay-Paasche-Hutson 1 to 31, 1st- & 2nd- Savin, Moses-Soddy
2021.05.06 - Added: anti-1st Parry, anti-2nd Parry, anti-3rd tri-squares-central, anti-4th tri-squares-central, anti-X3-ABC reflections, anti-inner-Yff, anti-outer-Yff
2021.05.06 - Added: anti-1st Auriga, anti-2nd Auriga, anti-Ehrmann-mid, anti-inner-Garcia, anti-1st Kenmotu-free-vertices, anti-2nd Kenmotu-free-vertices
2021.04.24 - Added: note to 4th mixtilinear