## A:

## ABC

The reference triangle

Equivalences:

• (ABC-X3 reflections)-of-(ABC-X3 reflections), (anticomplementary)-of-(medial), (circumorthic)-of-(2nd circumperp), (2nd Euler)-of-(hexyl), (1st excosine)-of-(2nd Zaniah), (outer-Garcia)-of-(outer-Garcia), (Johnson)-of-(Johnson), (medial)-of-(anticomplementary), (orthic)-of-(excentral), (tangential)-of-(intouch).

• Only for acute ABC: (1st circumperp)-of-(1st anti-circumperp), (2nd circumperp)-of-(circumorthic), (2nd Conway)-of-(2nd anti-Conway), (3rd Euler)-of-(3rd anti-Euler), (4th Euler)-of-(4th anti-Euler), (excentral)-of-(orthic), (hexyl)-of-(2nd Euler), (6th mixtilinear)-of-(6th anti-mixtilinear), (Wasat)-of-(anti-Wasat).A-vertex coordinates:

Trilinears1 : 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 theantialtimedial orthic axes triangle, orAAOA trianglefor 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-X

_{3}reflectionsThis triangle is obtained by reflecting A, B, C about the circumcenter X

_{3}. Reference: TCCT 6.12, p. 159.Equivalences:

• (1st anti-circumperp)-of-(2nd circumperp), (2nd anti-circumperp-tangential)-of-(anti-Mandart-incircle), (circumorthic)-of-(1st circumperp), (Euler)-of-(anticomplementary), (2nd Euler)-of-(excentral), (Mandart-incircle)-of-(2nd circumperp tangential), (medial)-of-(anti-Euler), (orthic)-of-(hexyl).

• Only for acute ABC: (1st circumperp)-of-(circumorthic), (2nd circumperp)-of-(1st anti-circumperp), (3rd Euler)-of-(4th anti-Euler), (4th Euler)-of-(3rd anti-Euler), (Hutson intouch)-of-(tangential).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 M

_{A}, M_{B}, M_{C}be the midpoints of sides BC, CA, AB, respectively, and H_{A}, H_{B}, H_{C}the feet of the altitudes from A, B, C, respectively. The triangles H_{A}M_{C}M_{B}, M_{C}H_{B}M_{A}, M_{B}M_{A}H_{C}are theA-, B- and C-altimedial triangles. (References: Hyacinthos 253 and preamble just before X15015)A-altimedial triangle trilinear vertex coordinates:

H_{A}0 : (-c^2+a^2+b^2)*c : (-b^2+c^2+a^2)*b

M_{C}b : a : 0

M_{B}c : 0 : aNote: Let P be a triangle center. Let P

_{A}be P-of-A-altimedial-triangle, and define P_{B}, P_{C}cyclically. Triangle P_{A}P_{B}P_{C}is theP-of-altimedial-triangles triangle, orP-altimedial trianglefor short.

## altimedial orthic axes (AOA)

Let L

_{A}be the orthic axis of theA-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 thealtimedial orthic axes triangle, orAOA trianglefor 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 A

_{1}, A_{2}and the Simson lines of A_{1}, A_{2}cut at A'. Define B', C' cyclically. Then A'B'C' is equilateral and it is the1st 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:

Trilinears2nd Altintas-isodynamic triangle:-((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

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

## adjunct anti-altimedial

Let A

_{A}B_{A}C_{A}, A_{B}B_{B}C_{B}, A_{C}B_{C}C_{C}be the A-, B- and C-anti-altimedialtriangles of ABC. The triangles A_{A}A_{B}A_{C}, B_{A}B_{B}B_{C}, C_{A}C_{B}C_{C}are theA-, B- and C- adjunct anti-altimedial triangles, respectively. (Reference: preamble just before X15015)A-adjunct anti-altimedial triangle trilinear vertex coordinates:

A_{A}-a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

A_{B}1/a : 1/(-a^2+b^2+c^2)*b : -1/c

A_{C}1/a : -1/b : 1/(-a^2+b^2+c^2)*cNote: 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"_{A}P"_{B}P"_{C}is theP-of-adjunct-anti-altimedial-triangles triangle, orP-adjunct-anti-altimedial trianglefor short.

## anticomplementary

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

Equivalences:

• (anti-Euler)-of-(ABC-X3 reflections), (3rd anti-Euler)-of-(1st circumperp), (4th anti-Euler)-of-(2nd circumperp), (anti-Wasat)-of-(excentral), (Aquila)-of-(outer-Garcia), (1st excosine)-of-(intouch), (Johnson)-of-(anti-Euler), (medial)-of-(Gemini 111), (orthic)-of-(2nd Conway), (tangential)-of-(inner-Conway), (X3-ABC reflections)-of-(Johnson).

• Only for acute ABC: (excentral)-of-(1st anti-circumperp), (hexyl)-of-(circumorthic), (6th mixtilinear)-of-(orthic), (Ursa-minor)-of-(tangential), (Wasat)-of-(3rd anti-Euler).A-vertex coordinates:

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

## anti-triangles

If T is a triangle of ABC then the

anti-T triangleis the triangle whose T-triangle is ABC.

## anti-altimedial

The triangle A

_{A}B_{A}C_{A}whose A-altimedial triangle is ABC. (Reference: preamble just before X15015)A-anti-altimedial triangle trilinear vertex coordinates:

A_{A}-a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

B_{A}a/(a^2-b^2+c^2) : 1/b : -1/c

C_{A}a/(a^2+b^2-c^2) : -1/b : 1/cNote: 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'_{A}P'_{B}P'_{C}is theP-of-anti-altimedial-triangles triangle, orP-anti-altimedial trianglefor short.

## anti-Aquila

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

Equivalences:

• (6th anti-mixtilinear)-of-(hexyl), (outer-Garcia)-of-(medial), (Kosnita)-of-(intouch), (orthic)-of-(2nd circumperp), (tangential)-of-(incircle-circles), (Trinh)-of-(Hutson intouch).A-vertex coordinates:

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

## anti-Ara

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

Equivalences:

• (Hutson intouch)-of-(anti-excenters-reflections).

• Only for acute ABC: (incircle-circles)-of-(circumorthic), (intouch)-of-(orthic).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.

Equivalences:

• (anticomplementary)-of-(2nd anti-extouch).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:

TrilinearsS^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:

Trilinearsanti-2nd Auriga:-(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

Trilinearswhere where D=sqrt(r*R+4*R^2).-(-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

## 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:

Trilinearsa*((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.

Equivalences:

• (ABC-X3 reflections)-of-(circumorthic), (6th anti-mixtilinear)-of-(Gemini 111), (circumorthic)-of-(ABC-X3 reflections), (Euler)-of-(4th anti-Euler), (2nd Euler)-of-(anti-Euler), (medial)-of-(3rd anti-Euler), (orthic)-of-(anticomplementary).

• Only for acute ABC: (2nd anti-circumperp-tangential)-of-(anti-Hutson intouch), (Mandart-incircle)-of-(tangential).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.

Equivalences:

• (ABC-X3 reflections)-of-(Mandart-incircle), (1st anti-circumperp)-of-(Hutson intouch), (circumorthic)-of-(intouch), (2nd Euler)-of-(Ursa-minor), (Mandart-incircle)-of-(5th mixtilinear).

• Only for acute ABC: (Hutson intouch)-of-(intangents).Notes:

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

2) Continuing with the construction 2) of the anti-tangential-midarc triangle, the line A_{b}A_{c}is tangent to the incircle of ABC in a point A'; B', C' are defined similarly. Then A'B'C' is the 2nd anti-circumperp-tangential triangle. (César Lozada, Dec. 8, 2021).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.

Equivalences:

• (medial)-of-(2nd anti-extouch).Note: The A-vertex of the anti-Conway triangle is the A' = {perpendicular to BC through X(578)} ∩ AX(6).

A-vertex coordinates:

Trilinearsa^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.

Equivalences:

• (medial)-of-(orthic).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.

Equivalences:

• (anti-Euler)-of-(Johnson), (Johnson)-of-(X3-ABC reflections).

• Only for acute ABC: (excentral)-of-(Ehrmann-side).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.

Equivalences:

• (anticomplementary)-of-(ABC-X3 reflections), (anti-Ehrmann-mid)-of-(Johnson), (3rd anti-Euler)-of-(2nd circumperp), (4th anti-Euler)-of-(1st circumperp), (anti-Wasat)-of-(hexyl), (anti-X3-ABC reflections)-of-(Gemini 111), (Johnson)-of-(anticomplementary).

• Only for acute ABC: (excentral)-of-(circumorthic), (hexyl)-of-(1st anti-circumperp), (Ursa-minor)-of-(anti-Hutson intouch), (Wasat)-of-(4th anti-Euler).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.

Equivalences:

• (anticomplementary)-of-(1st anti-circumperp), (anti-Euler)-of-(circumorthic), (4th anti-Euler)-of-(ABC-X3 reflections), (anti-Wasat)-of-(anticomplementary), (Johnson)-of-(4th anti-Euler).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.

Equivalences:

• (anticomplementary)-of-(circumorthic), (anti-Euler)-of-(1st anti-circumperp), (3rd anti-Euler)-of-(ABC-X3 reflections), (anti-Wasat)-of-(anti-Euler), (Johnson)-of-(3rd anti-Euler).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.

Equivalences:

• (anti-Hutson intouch)-of-(anti-Ara).

• Only for acute ABC: (5th mixtilinear)-of-(orthic).A-vertex coordinates:

Trilinears2*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.

Equivalences:

• (anticomplementary)-of-(anti-Conway), (medial)-of-(anti-Ascella).Note: Only when ABC is acute. A' = AX(3) ∩ perpendicular(X(1181), BC)

A-vertex coordinates:

TrilinearsS^2*a/SA : SB*b : SC*c

## anti-inner-Garcia

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

Note: If A"B"C" is the inner-Garcia triangle of ABC, then A' = A"X(3) ∩ perpendicular(X(6326), B"C").

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.

Equivalences:

• (anti-Ursa minor)-of-(anti-Euler), (tangential)-of-(ABC-X3 reflections), (Trinh)-of-(X3-ABC reflections).

• Only for acute ABC: (anti-Mandart-incircle)-of-(circumorthic), (Aquila)-of-(Trinh), (2nd circumperp tangential)-of-(1st anti-circumperp).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.

Equivalences:

• (anti-inverse-in-incircle)-of-(Johnson), (circumorthic)-of-(Ara), (tangential)-of-(X3-ABC reflections).

• Only for acute ABC: (Aquila)-of-(tangential).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.

Equivalences:

• (anti-incircle-circles)-of-(Johnson), (anti-Ursa minor)-of-(Gemini 111), (tangential)-of-(anticomplementary).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-centers

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

A-vertex coordinates:

anti-1st Kenmotu-centers:

Trilinearsanti-2nd Kenmotu-centers:-(b^2+c^2+2*S)/a : b : c

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

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

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

## anti-Lucas(-1) homothetic and anti-Lucas(+1) homothetic

The triangle A'B'C' whose Lucas(-1) homothetic/Lucas(+1) homothetic triangle is ABC.

A-vertex coordinates:

anti-Lucas(-1) homothetic:

Trilinearsanti-Lucas(+1) homothetic:S*(2*S^2-(SA+SW)*S+SA^2)/a : b*(2*S^2-(SB+SW)*S+SB^2) : c*(2*S^2-(SC+SW)*S+SC^2)

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

## anti-Mandart-incircle

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

Equivalences:

• (anti-Hutson intouch)-of-(2nd circumperp), (anti-Ursa minor)-of-(excentral), (2nd circumperp tangential)-of-(ABC-X3 reflections), (tangential)-of-(1st circumperp).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.

Equivalences:

• (1st anti-circumperp)-of-(Gemini 110), (2nd Euler)-of-(anti-X3-ABC reflections), (orthic)-of-(medial), (submedial)-of-(anticomplementary).

• Only for acute ABC: (anti-Aquila)-of-(2nd Euler).A-vertex coordinates:

Trilinears2*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.

Equivalences:

• 4th anti-Brocard-of-circumsymmedial.A-vertex coordinates:

Trilinearsa*((-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:

TrilinearsCoordinates not found yet

## 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.

Equivalences:

• (anti-Hutson intouch)-of-(Mandart-incircle), (tangential)-of-(2nd anti-circumperp-tangential).Notes:

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

2) Let a_{b}, a_{c}be the circles both tangent to AI at I=X(1) and passing the first through B and the second through C. Let A_{b}be the point, other than B, at which a_{b}cuts AB, and A_{c}the point, other than C, at which a_{c}cuts AC. Define {B_{c}, B_{a}}, {C_{a}, C_{b}} cyclically. Then the triangle bounded by lines A_{b}A_{c}, B_{c}B_{a}, C_{a}C_{b}is the anti-tangential-midarc triangle of ABC. (César Lozada, Dec. 8, 2021)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.

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

Build squares BCC_{a}B_{a}, CAA_{b}C_{b}and ABB_{c}A_{c}inwards ABC. A' = A_{b}B_{a}∩ A_{c}C_{a}.A-vertex coordinates:

Trilinears-(S^2-(SB+SC)*S-3*SB*SC)/(a*(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.

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

Build squares BCC_{a}B_{a}, CAA_{b}C_{b}and ABB_{c}A_{c}outwards ABC. A' = A_{b}B_{a}∩ A_{c}C_{a}.A-vertex coordinates:

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

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

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.

Equivalences:

• (anti-Hutson intouch)-of-(Euler), (anti-inverse-in-incircle)-of-(Gemini 110), (tangential)-of-(medial).

• Only for acute ABC: (anti-Mandart-incircle)-of-(orthic), (2nd circumperp tangential)-of-(2nd Euler).Construction: Let A

_{0}B_{0}C_{0}be the orthic triangle of ABC. Let A_{1}be the intersection of the parallel to A_{0}B_{0}through B and the parallel to A_{0}C_{0}through C. Let A_{2}be the midpoint of A and A_{1}. Build B_{2}, C_{2}cyclically. Then A_{2}B_{2}C_{2}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.

Equivalences:

• (anticomplementary)-of-(orthic), (anti-Euler)-of-(2nd Euler), (3rd anti-Euler)-of-(medial), (4th anti-Euler)-of-(Euler), (1st excosine)-of-(anti-Ara).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.

Equivalences:

• (anti-Euler)-of-(Gemini 110), (Ehrmann-mid)-of-(Johnson), (Johnson)-of-(medial).

• Only for acute ABC: (hexyl)-of-(6th anti-mixtilinear), (Hutson intouch)-of-(Trinh), (incircle-circles)-of-(tangential).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

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

## anti-inner/outer Yff

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

A-vertex coordinates:

anti-inner-Yff:

Trilinearsanti-outer-Yff:(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

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 triangleof 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:

Trilinearsa/(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.

Equivalences:

• (anticomplementary)-of-(outer-Garcia), (anti-incircle-circles)-of-(intouch), (circumorthic)-of-(excentral), (Ehrmann-side)-of-(hexyl), (2nd Euler)-of-(6th mixtilinear).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)Equivalences:

• Only for acute ABC: (excentral)-of-(tangential), (Wasat)-of-(1st excosine).A-vertex coordinates:

Trilinearsa*(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 O

_{A}be the circle {{B,C,B',C'}}, and define O_{B}and O_{C}cyclically. The three circles are othogonal to the incircle and their centers A",B",C" are the vertices of theAscella triangle. (Reference: Preamble just before X8726)A-vertex coordinates:

Trilinears2*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 v

^{2}= |UV|^{2}- u^{2}= power of V with respect to (U,u). Let U_{V,A}be the circle (V,w) obtained from (U,r) = A-excircle and V = incenter, and let L_{A}be the radical axis of U_{V,A}and the A-excircle. Define L_{B}and L_{C}cyclically. Let A' = L_{B}∩L_{C}, B' = L_{C}∩L_{A}, C' = L_{A}∩L_{B}. The triangle A'B'C' is theAtik 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, I

_{a}be the incenter and A-excenter of ABC. Let a' be the radical axis of the A-excircle and the circle with diameter II_{a}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:

Trilinearswhere D=sqrt(r*R+4*R^2)(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

## 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:

Trilinearswhere D=sqrt(r*R+4*R^2)(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

## Ayme

Let R

_{A}be the radical axis of the circumcircle and the A-excircle, and define R_{B}and R_{C}cyclically. Let T_{A}= R_{B}∩R_{C}, and define T_{B}and T_{C}cyclically. (T_{A}is also the radical center of the circumcircle and the B- and C- excircles.) TheAyme triangleis T_{A}T_{B}T_{C}. (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 A

_{b}be the rotation of A toward B by an angle |π/6| and let A_{c}be the rotation of A toward C by the same angle |π/6| (triangle OA_{b}A_{c}is an equilateral triangle). Build (B_{c}, B_{a}) and (C_{a}, C_{b}) cyclically and let A_{m}, B_{m}, C_{m}be the midpoints of B_{c}C_{b}, C_{a}A_{c}, A_{b}B_{a}, respectively. The triangle A_{m}B_{m}C_{m}is equilateral and is named theBankoff equilateral triangleof 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

## BCE

Let E

_{a}, E_{b}, E_{c}be the excenters of ABC and J_{a}, J_{b}, J_{c}the E_{a}-, E_{b}-, E_{c}- excenters of E_{a}BC, E_{b}CA and E_{c}AB, respectively. The triangle J_{a}J_{b}J_{c}is theBCE triangleof ABC.A-vertex coordinates:

Trilinears-1 : 1-2*sin(C/2) : 1-2*sin(B/2)

## BCE-incenters

Let E

_{a}, E_{b}, E_{c}be the excenters of ABC and I_{a}, I_{b}, I_{c}the incenters of E_{a}BC, E_{b}CA and E_{c}AB, respectively. The triangle I_{a}I_{b}I_{c}is theBCE-incenters triangleof ABC.A-vertex coordinates:

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

## BCI

Let I be the incenter of ABC and I

_{a}, I_{b}, I_{c}the incenters of IBC, ICA and IAB, respectively. The triangle I_{a}I_{b}I_{c}is theBCI triangleof ABC. (Reference: MathWorld -- BCI Triangle.)A-vertex coordinates:

Trilinears1 : 1+2*cos(C/2) : 1+2*cos(B/2)

## BCI-excenters

Let I be the incenter of ABC and E

_{a}, E_{b}, E_{c}the I- excenters of IBC, ICA and IAB, respectively. The triangle E_{a}E_{b}E_{c}is theBCI-excenters triangleof ABC.A-vertex coordinates:

Trilinears1 : 1-2*cos(C/2) : 1-2*cos(B/2)

## Bevan-antipodal

Let V be the Bevan point X(40) of ABC and V

_{a}, V_{b}, V_{c}the antipodes of V on the circumcircles of VBC, VCA and VAB, respectively. The triangle V_{a}V_{b}V_{c}is theBevan-antipodal triangleof 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 triangleof ABC. (Reference: MathWorld -- Brocard Triangles.)Equivalences:

• Only for acute ABC: (1st circumperp)-of-(7th Brocard), (2nd circumperp)-of-(10th Brocard).A-vertex coordinates:

Trilinearsa*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 triangleof 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 the2nd Brocard triangleof ABC. (Reference: MathWorld -- Brocard Triangles.)A-vertex coordinates:

Trilinearsb^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 triangleof ABC. (Reference: MathWorld -- Brocard Triangles.)A-vertex coordinates:

Trilinearsb^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 triangleof ABC. (Reference: MathWorld -- Brocard Triangles.)A-vertex coordinates:

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

## 5th Brocard

The

5th Brocard triangleis 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)Equivalences:

• (ABC-X3 reflections)-of-(10th Brocard), (1st anti-circumperp)-of-(1st Brocard).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 X39642)

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 X39642)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

## 10th Brocard

The reflection of 7th Brocard triangle in the center X(182) of Brocard circle is the

10th Brocard triangle. (Reference: Preamble just before X43803)Equivalences:

• (ABC-X3 reflections)-of-(7th Brocard), (circumorthic)-of-(1st Brocard).A-vertex coordinates:

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

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

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

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

• (ABC-X3 reflections)-of-(circumtangential), (anti-Aquila)-of-(Stammler), (2nd anti-circumperp-tangential)-of-(Stammler), (anti-X3-ABC reflections)-of-(Stammler), (circumorthic)-of-(circumtangential), (Euler)-of-(Stammler), (2nd Euler)-of-(Stammler), (4th Euler)-of-(Stammler), (outer-Garcia)-of-(circumtangential), (Hutson intouch)-of-(Stammler), (Johnson)-of-(circumtangential), (Kosnita)-of-(circumtangential).A-vertex coordinates:

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

## circumorthic

The circumcevian triangle of the orthocenter of ABC is its

circumorthic triangle.Equivalences:

• (ABC-X3 reflections)-of-(1st anti-circumperp), (1st anti-circumperp)-of-(ABC-X3 reflections), (anti-incircle-circles)-of-(anti-Ara), (Ehrmann-side)-of-(Johnson), (Euler)-of-(3rd anti-Euler), (2nd Euler)-of-(anticomplementary), (medial)-of-(4th anti-Euler), (orthic)-of-(anti-Euler).

• Only for acute ABC: (2nd anti-circumperp-tangential)-of-(tangential), (Aquila)-of-(orthic), (Mandart-incircle)-of-(anti-Hutson intouch).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).Equivalences:

• (ABC-X3 reflections)-of-(2nd circumperp), (2nd circumperp)-of-(ABC-X3 reflections), (Euler)-of-(hexyl), (3rd Euler)-of-(anticomplementary), (4th Euler)-of-(anti-Euler), (Hutson intouch)-of-(2nd circumperp tangential), (medial)-of-(excentral).A-vertex coordinates:

Trilinearsa : 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).Equivalences:

• (ABC-X3 reflections)-of-(1st circumperp), (Ara)-of-(incircle-circles), (1st circumperp)-of-(ABC-X3 reflections), (Euler)-of-(excentral), (3rd Euler)-of-(anti-Euler), (4th Euler)-of-(anticomplementary), (excentral)-of-(anti-Aquila), (Hutson intouch)-of-(anti-Mandart-incircle), (medial)-of-(hexyl).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).

Equivalences:

• (anti-Hutson intouch)-of-(1st circumperp), (anti-Mandart-incircle)-of-(ABC-X3 reflections), (anti-Ursa minor)-of-(hexyl), (tangential)-of-(2nd circumperp).A-vertex coordinates:

Trilinearsb*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 trianglehas 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.Equivalences:

• (ABC-X3 reflections)-of-(circumnormal), (circumorthic)-of-(circumnormal), (Ehrmann-mid)-of-(Stammler), (3rd Euler)-of-(Stammler), (outer-Garcia)-of-(circumnormal), (Johnson)-of-(circumnormal), (Kosnita)-of-(circumnormal), (Mandart-incircle)-of-(Stammler), (medial)-of-(Stammler).A-vertex coordinates:

Trilinears-1 : sin((B-C)/3)*csc((B+2*C)/3) : sin((C-B)/3)*csc((C+2*B)/3)

## contact

See

intouch triangle.

## Conway

Let a, b, c be the sidelengths BC, CA and AB, respectively, of ABC. Let A

_{b}be the point on CA such that AA_{b}= -(a/b)*AC and A_{c}the point on BA such that AA_{c}= -(a/c)*AB. Define B_{c}, B_{a}, C_{a}, C_{b}similarly. TheConway triangleis the triangle bounded by the lines A_{b}A_{c}, B_{c}B_{a}and C_{a}C_{b}. (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 A_{2}= A_{b}B_{a}∩A_{c}C_{a}, and define B_{2}and C_{2}cyclically. The triangle A_{2}B_{2}C_{2}is the2nd Conway triangleof ABC. (Reference: Preamble just before X9776)Equivalences:

• (Ara)-of-(inner-Conway), (excentral)-of-(anticomplementary), (Wasat)-of-(Gemini 111).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 A_{3}be the intersection of the tangents to the Conway circle at B_{a}and C_{a}, and define B_{3}, C_{3}cyclically. The triangle A_{3}B_{3}C_{3}is the3rd Conway triangle. (Reference: X10434)

Note: The six points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}lie on a circle named the Conway circle.Equivalences:

• (anticomplementary)-of-(inverse-in-Conway).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 A_{4}be the intersection of the tangents to the Conway circle at A_{b}and A_{c}, and define B_{4}, C_{4}cyclically. The triangle A_{4}B_{4}C_{4}is the4th Conway triangle. (Reference: X10434)

Note: The six points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}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 A_{5}be the intersection of the tangents to the Conway circle at B_{c}and C_{b}, and define B_{5}, C_{5}cyclically. Triangle A_{5}B_{5}C_{5}is named the5th Conway triangle. (Reference: X10434)

Note: The six points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}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 A

_{b}be the point on CA such that AA_{b}= (a/b)*AC and A_{c}the point on BA such that AA_{c}= (a/c)*AB. Define B_{c}, B_{a}, C_{a}, C_{b}similarly. Theinner-Conway triangleis the triangle bounded by the lines A_{b}A_{c}, B_{c}B_{a}and C_{a}C_{b}. (Reference: Preamble just before X11677)Equivalences:

• (anti-Ara)-of-(2nd Conway).Notes: See this construction at MathWorld -- Conway Circle.

Points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}and C_{b}do not lie on a conic.A-vertex coordinates:

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

## D:

## D-triangle

See

4th Brocard triangle.

## Dao

Let (a) denote the ellipse through A with foci B and C and define (b) and (c) cyclically. Let d

_{a}be the line joining the intersection points of of (b) and (c) and denote A'= d_{a}∩ BC. Build B' and C' cyclically. A'B'C' is theDao triangle. (Reference: Preamble just before X(45738))A-vertex coordinates:

Trilinears0 : c*(a+b)*(a-b+c)^2 : b*(a+c)*(a+b-c)^2

## 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 A

_{B}the the point of intersection of the circle {{P,B,C}} and the line AB. Define A_{C}symmetrically, and define B_{C}, B_{A}, C_{A}, C_{B}cyclically. These six points of intersection are on a circle if and only if P = X(6). The centers A_{1}, B_{1}, C_{1}of the circles {{X(6),B,C}}, {{X(6),C,A}}, {{X(6),A,B,}}, respectively, are the vertices of the1st 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 A_{2}= B_{C}B_{A}∩C_{A}C_{B}, and define B_{2}and C_{2}cyclically. The triangle A_{2}B_{2}C_{2}is the2nd Ehrmann triangle. (Reference: Preamble just before X8537)A-vertex coordinates:

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

## Ehrmann circumscribing

Let T

_{C1}=A_{1}B_{1}C_{1}be the triangle obtained by rotating ABC about the 1st Ehrmann pivot, P(5), by an angle of 2π/3, so that T_{C1}circumscribes ABC. Triangle T_{C1}is the1st Ehrmann circumscribing triangle. Let T_{C2}=A_{2}B_{2}C_{2}be the triangle obtained by rotating ABC about the 2nd Ehrmann pivot, U(5), by an angle of -2π/3, so that T_{C2}circumscribes ABC. Triangle T_{C2}is the2nd 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 T_{1}=A_{1}B_{1}C_{1}and T_{2}=A_{2}B_{2}C_{2}are two distinct triangles, the cross-triangle T*=A*B*C* of T_{1}and T_{2}is the triangle having vertices A*=B_{1}C_{2}∩ B_{2}C_{1}, B*=C_{1}A_{2}∩ C_{2}A_{1}and C*=A_{1}B_{2}∩ A_{2}B_{1}.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 T

_{I1}=A_{1}B_{1}C_{1}be the triangle obtained by rotating ABC about the 1st Ehrmann pivot, P(5), by an angle of -2π/3, so that T_{I1}is inscribed in ABC. Triangle T_{I1}is the1st Ehrmann inscribed triangle. Let T_{I2}=A_{2}B_{2}C_{2}be the triangle obtained by rotating ABC about the 2nd Ehrmann pivot, U(5), by an angle of 2π/3, so that T_{I2}is inscribed in ABC. Triangle T_{I2}is the2nd 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:Trilinears0 : (S-sqrt(3)*SA)*c : (S+sqrt(3)*SA)*b

2nd Ehrmann inscribed triangle:Trilinears0 : (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 T_{1}=A_{1}B_{1}C_{1}and T_{2}=A_{2}B_{2}C_{2}are two distinct triangles, the mid-triangle of T_{1}and T_{2}is the triangle having vertices the midpoints of {A_{1},A_{2}}, {B_{1},B_{2}} and {C_{1},C_{2}}.Equivalences:

• (anti-X3-ABC reflections)-of-(Johnson), (Johnson)-of-(Euler).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

Ehrmann-side triangle. (Reference: Preamble just before X18300)

Note: if T_{1}=A_{1}B_{1}C_{1}and T_{2}=A_{2}B_{2}C_{2}are two distinct triangles, the side-triangle T*=A*B*C* of T_{1}and T_{2}is the triangle having vertices A*=B_{1}B_{2}∩ C_{1}C_{2}, B*=C_{1}C_{2}∩ A_{1}A_{2}and C*=A_{1}A_{2}∩ B_{1}B_{2}.Equivalences:

• (circumorthic)-of-(Johnson), (2nd Euler)-of-(X3-ABC reflections), (orthic)-of-(anti-Ehrmann-mid).

• Only for acute ABC: (Aquila)-of-(2nd Euler).A-vertex coordinates:

Trilinearsa*(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 T_{1}=A_{1}B_{1}C_{1}and T_{2}=A_{2}B_{2}C_{2}are two distinct triangles, the vertex-triangle T*=A*B*C* of T_{1}and T_{2}is the triangle having vertices A*=B_{1}C_{1}∩ B_{2}C_{2}, B*=C_{1}A_{1}∩ C_{2}A_{2}and C*=A_{1}B_{1}∩ A_{2}B_{2}.Equivalences:

• (Kosnita)-of-(Johnson), (tangential)-of-(Ehrmann-mid).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 triangleof ABC. (Reference: MathWorld -- Euler Triangle.)

A*, B*, C* are also the reflections of the vertices of the medial triangle in X(5).Equivalences:

• (ABC-X3 reflections)-of-(medial), (1st anti-circumperp)-of-(4th Euler), (circumorthic)-of-(3rd Euler), (2nd Euler)-of-(Wasat), (Johnson)-of-(Ehrmann-mid), (medial)-of-(Johnson).

• Only for acute ABC: (1st circumperp)-of-(2nd Euler), (2nd circumperp)-of-(orthic), (4th Euler)-of-(anti-Wasat), (Hutson intouch)-of-(anti-Ursa minor).A-vertex coordinates:

Trilinears2((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)Equivalences:

• (ABC-X3 reflections)-of-(orthic), (1st anti-circumperp)-of-(Euler), (6th anti-mixtilinear)-of-(X3-ABC reflections), (circumorthic)-of-(medial), (Ehrmann-side)-of-(anti-X3-ABC reflections), (Euler)-of-(anti-Wasat), (orthic)-of-(Johnson).

• Only for acute ABC: (anti-Aquila)-of-(Ehrmann-side), (2nd anti-circumperp-tangential)-of-(anti-Ursa minor), (Aquila)-of-(6th anti-mixtilinear).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)Equivalences:

• (ABC-X3 reflections)-of-(4th Euler), (1st circumperp)-of-(medial), (2nd circumperp)-of-(Euler), (4th Euler)-of-(Johnson), (medial)-of-(Wasat).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)Equivalences:

• (ABC-X3 reflections)-of-(3rd Euler), (1st circumperp)-of-(Euler), (2nd circumperp)-of-(medial), (Euler)-of-(Wasat), (3rd Euler)-of-(Johnson).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 isT(-1,3). Reference: TCCT, p. 173.Equivalences:

• (Ara)-of-(Hutson intouch), (excentral)-of-(5th mixtilinear).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 isT(-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.

Equivalences:

• (anticomplementary)-of-(1st circumperp), (anti-Euler)-of-(2nd circumperp), (Ara)-of-(intouch), (2nd circumperp)-of-(Aquila), (2nd Conway)-of-(medial), (hexyl)-of-(ABC-X3 reflections), (Johnson)-of-(hexyl), (medial)-of-(6th mixtilinear), (Ursa-minor)-of-(anti-Mandart-incircle), (Wasat)-of-(anticomplementary).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 A

_{b}, A_{c}be the points where the A-excosine circle cuts again the sidelines AC and AB, respectively, and build {B_{c}, B_{a}}, {C_{a}, C_{b}} cyclically. The triangle A_{1}B_{1}C_{1}bounded by the lines A_{b}A_{c}, B_{c}B_{a}and C_{a}C_{b}is the1st excosine triangle of ABCand the triangle A_{2}B_{2}C_{2}bounded by the lines B_{c}C_{b}, C_{a}A_{c}and A_{b}B_{a}is the2nd 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 calledA-excosine circleof ABC. This circle cuts AB and AC again at two points which are the extremities of a diameter of it. (Reference: MathWorld -- Excosine Circle.)Equivalences:

• 1st excosine = (anticomplementary)-of-(tangential), (anti-Wasat)-of-(Ara).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 triangleof ABC.Equivalences:

• (Aries)-of-(intouch).A-vertex coordinates:

Trilinears0 : 1/(b*(a+b-c)) : 1/(c*(a-b+c))

## 2nd extouch

Let A

_{A},A_{B}, A_{C}be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define B_{B}, B_{C}, B_{A}and C_{C}, C_{A}, C_{B}cyclically.

Let A_{2}= B_{C}B_{A}∩C_{A}C_{B}, and define B_{2}and C_{2}cyclically. The2nd extouch triangleis A_{2}B_{2}C_{2}. (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 A

_{A},A_{B}, A_{C}be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define B_{B}, B_{C}, B_{A}and C_{C}, C_{A}, C_{B}cyclically.

Let A_{3}= C_{A}A_{C}∩A_{B}B_{A}, and define B_{3}and C_{3}cyclically. The3rd extouch triangleis A_{3}B_{3}C_{3}. (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 A

_{A},A_{B}, A_{C}be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define B_{B}, B_{C}, B_{A}and C_{C}, C_{A}, C_{B}cyclically.

Let A_{4}= B_{C}C_{A}∩B_{A}C_{B}, and define B_{4}and C_{4}cyclically. The4th extouch triangleis A_{4}B_{4}C_{4}. (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 A

_{A},A_{B}, A_{C}be the touchpoints of the A-excircle and the lines BC, CA, AB, respectively, and define B_{B}, B_{C}, B_{A}and C_{C}, C_{A}, C_{B}cyclically.

Let D_{1}= B_{B}B_{C}∩C_{C}C_{A}, and define D_{2}and D_{3}cyclically. Let E_{1}= B_{B}B_{A}∩C_{C}C_{B}, and define E_{2}and E_{3}cyclically. Define A_{5}= D_{2}E_{2}∩D_{3}E_{3}, and define B_{5}and C_{5}cyclically. The5th extouch triangleis A_{5}B_{5}C_{5}. (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 triangleof ABC.Equivalences:

• (anticomplementary)-of-(1st half-diamonds).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 triangleof ABC.Equivalences:

• (anticomplementary)-of-(2nd half-diamonds).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 L

_{a}the parallel to BC through F. Let A_{b}and A_{c}be the points, others than F, where L_{a}cuts the circumcircles of FAC and FAB, respectively. Construct B_{c}, B_{a}and C_{a}, C_{b}cyclically. Let A'=A_{b}C_{b}∩A_{c}B_{c}and similarly B' and C'. The triangle A'B'C' is equilateral and is named the1st 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 A

_{b}and A_{c}be points on AC and AB, respectively, such that FA_{b}A_{c}is an equilateral trianglewith the same orientation as ABC. Denote the centroid of FA_{b}A_{c}as A' and similarly build B' and C'. The triangle A'B'C' is equilateral and is named the3rd Fermat-Dao equilateral triangle. (Reference: X16267)Let F=X(14) be the 2nd Fermat point of ABC. Let A

_{b}and A_{c}be points on AC and AB, respectively, such that FA_{b}A_{c}is an equilateral trianglewith opposite orientation than ABC. Denote the centroid of FA_{b}A_{c}as A" and similarly build B" and C". The triangle A"B"C" is equilateral and is named the4th 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 B

_{a}and C_{a}be points on BC such that the triangle FB_{a}C_{a}is equilateral. Define the pairs {C_{b}, A_{b}} and {A_{b}, B_{c}}cyclically. Let A' be the point other than F that lies on the circles (FA_{c}B_{c}) and (FC_{b}A_{b}), and define B' and C' cyclically. Then the triangle A'B'C' is equilateral and is the5th 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 A

_{b}and A_{c}be points on AC and AB, respectively, such that FA_{b}A_{c}is an equilateral trianglewith the same orientation as ABC. Let A' be the midpoint of A_{b}and A_{c}and define B' and C' cyclically. The triangle A'B'C' is equilateral and is named the7th Fermat-Dao equilateral triangle. (Reference: X396)Let F=X(14) be the 2nd Fermat point of ABC. Let A

_{b}and A_{c}be points on AC and AB, respectively, such that FA_{b}A_{c}is an equilateral trianglewith opposite orientation than ABC. Let A" be the midpoint of A_{b}and A_{c}and define B" and C" cyclically. The triangle A"B"C" is equilateral and is named the8th 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 B

_{a}, C_{a}be the points on BC such that FB_{a}C_{a}is an equilateral triangle. Define C_{b}, A_{b}and A_{c}, B_{c}cyclically. Let A', B', C' be the midpoints of A_{b}A_{c}, B_{c}B_{a}, C_{a}C_{b}, respectively. The triangle A'B'C' is equilateral and is named here the9th 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:Trilinears2*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 A

_{b}and A_{c}be the points where the circle FBC cuts again AC and AB, respectively. Define B_{c}, B_{a}and C_{a}, C_{b}cyclically. Denote A', B', C' the circumcenters of AB_{c}C_{b}, BC_{a}A_{c}and CB_{a}A_{b}, respectively. Then the triangle A'B'C' is equilateral and is here named the11th 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 A

_{b}and A_{c}be the points where the circle FBC cuts again AC and AB, respectively. Define B_{c}, B_{a}and C_{a}, C_{b}cyclically. Let A', B', C' be centroids of AB_{c}C_{b}, BC_{a}A_{c}, CA_{b}B_{a}respectively. Then the triangle A'B'C' is equilateral and is the13th 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 A

_{o}B_{o}C_{o}be the outer-Fermat triangle of ABC. Denote p'_{a}the polar of A with respect to the circle with center A_{o}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 the15th Fermat-Dao equilateral triangle.Let A

_{i}B_{i}C_{i}be the inner-Fermat triangle of ABC. Denote p"_{a}the polar of A with respect to the circle with center A_{i}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 the16th Fermat-Dao equilateral triangle. (Reference: X16960)Equivalences:

• 15th Fermat-Dao = (anticomplementary)-of-(1st isodynamic-Dao), (anti-Ehrmann-mid)-of-(1st isodynamic-Dao), (3rd anti-Euler)-of-(1st isodynamic-Dao), (anti-inverse-in-incircle)-of-(3rd isodynamic-Dao), (anti-Mandart-incircle)-of-(1st isodynamic-Dao), (Ara)-of-(3rd isodynamic-Dao), (2nd Conway)-of-(3rd isodynamic-Dao), (excentral)-of-(1st isodynamic-Dao), (1st excosine)-of-(3rd isodynamic-Dao), (6th mixtilinear)-of-(3rd isodynamic-Dao), (reflection)-of-(1st isodynamic-Dao), (tangential)-of-(1st isodynamic-Dao).

• 16th Fermat-Dao = (anticomplementary)-of-(2nd isodynamic-Dao), (anti-Ehrmann-mid)-of-(2nd isodynamic-Dao), (3rd anti-Euler)-of-(2nd isodynamic-Dao), (anti-inverse-in-incircle)-of-(4th isodynamic-Dao), (anti-Mandart-incircle)-of-(2nd isodynamic-Dao), (Ara)-of-(4th isodynamic-Dao), (2nd Conway)-of-(4th isodynamic-Dao), (excentral)-of-(2nd isodynamic-Dao), (1st excosine)-of-(4th isodynamic-Dao), (6th mixtilinear)-of-(4th isodynamic-Dao), (reflection)-of-(2nd isodynamic-Dao), (tangential)-of-(2nd isodynamic-Dao).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 A

_{f}B_{f}C_{f}.Then triangles A

- Let A* be the reflection of G in A
_{f}and A_{1}the reflection of A* in A; build B_{1}and C_{1 }cyclically.- Let A* be the reflection of G in A and A
_{2}the reflection of A* in A_{f}; build B_{2}and C_{2}cyclically.- Let A* be the reflection of A in G and A
_{3}the reflection of A* in A_{f}; build B_{3}and C_{3}cyclically.- Let A* be the reflection of A
_{f}in G and A_{4}the reflection of A* in A; build B_{4}and C_{4}cyclically._{1}B_{1}C_{1}, A_{2}B_{2}C_{2}, A_{3}B_{3}C_{3}, A_{4}B_{4}C_{4}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 the1st, 2nd, 3rd and 4th outer-Fermat-Dao-Nhi triangles, respectively.Equivalences:

• 1st inner-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(2nd inner-Fermat-Dao-Nhi), (anticomplementary)-of-(3rd outer-Fermat-Dao-Nhi), (anti-Ehrmann-mid)-of-(3rd outer-Fermat-Dao-Nhi), (anti-Euler)-of-(4th outer-Fermat-Dao-Nhi), (3rd anti-Euler)-of-(3rd outer-Fermat-Dao-Nhi), (4th anti-Euler)-of-(4th outer-Fermat-Dao-Nhi), (anti-Hutson intouch)-of-(4th outer-Fermat-Dao-Nhi), (anti-Mandart-incircle)-of-(3rd outer-Fermat-Dao-Nhi), (Aquila)-of-(4th outer-Fermat-Dao-Nhi), (circumorthic)-of-(2nd inner-Fermat-Dao-Nhi), (2nd circumperp tangential)-of-(4th outer-Fermat-Dao-Nhi), (excentral)-of-(3rd outer-Fermat-Dao-Nhi), (outer-Garcia)-of-(2nd inner-Fermat-Dao-Nhi), (hexyl)-of-(4th outer-Fermat-Dao-Nhi), (Johnson)-of-(2nd inner-Fermat-Dao-Nhi), (Kosnita)-of-(2nd inner-Fermat-Dao-Nhi), (reflection)-of-(3rd outer-Fermat-Dao-Nhi), (tangential)-of-(3rd outer-Fermat-Dao-Nhi), (X3-ABC reflections)-of-(4th outer-Fermat-Dao-Nhi).

• 2nd inner-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(1st inner-Fermat-Dao-Nhi), (anticomplementary)-of-(4th outer-Fermat-Dao-Nhi), (anti-Ehrmann-mid)-of-(4th outer-Fermat-Dao-Nhi), (anti-Euler)-of-(3rd outer-Fermat-Dao-Nhi), (3rd anti-Euler)-of-(4th outer-Fermat-Dao-Nhi), (4th anti-Euler)-of-(3rd outer-Fermat-Dao-Nhi), (anti-Hutson intouch)-of-(3rd outer-Fermat-Dao-Nhi), (anti-Mandart-incircle)-of-(4th outer-Fermat-Dao-Nhi), (Aquila)-of-(3rd outer-Fermat-Dao-Nhi), (circumorthic)-of-(1st inner-Fermat-Dao-Nhi), (2nd circumperp tangential)-of-(3rd outer-Fermat-Dao-Nhi), (excentral)-of-(4th outer-Fermat-Dao-Nhi), (outer-Garcia)-of-(1st inner-Fermat-Dao-Nhi), (hexyl)-of-(3rd outer-Fermat-Dao-Nhi), (Johnson)-of-(1st inner-Fermat-Dao-Nhi), (Kosnita)-of-(1st inner-Fermat-Dao-Nhi), (reflection)-of-(4th outer-Fermat-Dao-Nhi), (tangential)-of-(4th outer-Fermat-Dao-Nhi), (X3-ABC reflections)-of-(3rd outer-Fermat-Dao-Nhi).

• 3rd inner-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(4th inner-Fermat-Dao-Nhi), (anti-Aquila)-of-(2nd outer-Fermat-Dao-Nhi), (2nd anti-circumperp-tangential)-of-(2nd outer-Fermat-Dao-Nhi), (anti-X3-ABC reflections)-of-(2nd outer-Fermat-Dao-Nhi), (circumorthic)-of-(4th inner-Fermat-Dao-Nhi), (Ehrmann-mid)-of-(1st outer-Fermat-Dao-Nhi), (Euler)-of-(2nd outer-Fermat-Dao-Nhi), (2nd Euler)-of-(2nd outer-Fermat-Dao-Nhi), (3rd Euler)-of-(1st outer-Fermat-Dao-Nhi), (4th Euler)-of-(2nd outer-Fermat-Dao-Nhi), (outer-Garcia)-of-(4th inner-Fermat-Dao-Nhi), (Hutson intouch)-of-(2nd outer-Fermat-Dao-Nhi), (Johnson)-of-(4th inner-Fermat-Dao-Nhi), (Kosnita)-of-(4th inner-Fermat-Dao-Nhi), (Mandart-incircle)-of-(1st outer-Fermat-Dao-Nhi).

• 4th inner-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(3rd inner-Fermat-Dao-Nhi), (anti-Aquila)-of-(1st outer-Fermat-Dao-Nhi), (2nd anti-circumperp-tangential)-of-(1st outer-Fermat-Dao-Nhi), (anti-X3-ABC reflections)-of-(1st outer-Fermat-Dao-Nhi), (circumorthic)-of-(3rd inner-Fermat-Dao-Nhi), (Ehrmann-mid)-of-(2nd outer-Fermat-Dao-Nhi), (Euler)-of-(1st outer-Fermat-Dao-Nhi), (2nd Euler)-of-(1st outer-Fermat-Dao-Nhi), (3rd Euler)-of-(2nd outer-Fermat-Dao-Nhi), (4th Euler)-of-(1st outer-Fermat-Dao-Nhi), (outer-Garcia)-of-(3rd inner-Fermat-Dao-Nhi), (Hutson intouch)-of-(1st outer-Fermat-Dao-Nhi), (Johnson)-of-(3rd inner-Fermat-Dao-Nhi), (Kosnita)-of-(3rd inner-Fermat-Dao-Nhi), (Mandart-incircle)-of-(2nd outer-Fermat-Dao-Nhi).

• 1st outer-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(2nd outer-Fermat-Dao-Nhi), (anticomplementary)-of-(3rd inner-Fermat-Dao-Nhi), (anti-Ehrmann-mid)-of-(3rd inner-Fermat-Dao-Nhi), (anti-Euler)-of-(4th inner-Fermat-Dao-Nhi), (3rd anti-Euler)-of-(3rd inner-Fermat-Dao-Nhi), (4th anti-Euler)-of-(4th inner-Fermat-Dao-Nhi), (anti-Hutson intouch)-of-(4th inner-Fermat-Dao-Nhi), (anti-Mandart-incircle)-of-(3rd inner-Fermat-Dao-Nhi), (Aquila)-of-(4th inner-Fermat-Dao-Nhi), (circumorthic)-of-(2nd outer-Fermat-Dao-Nhi), (2nd circumperp tangential)-of-(4th inner-Fermat-Dao-Nhi), (excentral)-of-(3rd inner-Fermat-Dao-Nhi), (outer-Garcia)-of-(2nd outer-Fermat-Dao-Nhi), (hexyl)-of-(4th inner-Fermat-Dao-Nhi), (Johnson)-of-(2nd outer-Fermat-Dao-Nhi), (Kosnita)-of-(2nd outer-Fermat-Dao-Nhi), (reflection)-of-(3rd inner-Fermat-Dao-Nhi), (tangential)-of-(3rd inner-Fermat-Dao-Nhi), (X3-ABC reflections)-of-(4th inner-Fermat-Dao-Nhi).

• 2nd outer-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(1st outer-Fermat-Dao-Nhi), (anticomplementary)-of-(4th inner-Fermat-Dao-Nhi), (anti-Ehrmann-mid)-of-(4th inner-Fermat-Dao-Nhi), (anti-Euler)-of-(3rd inner-Fermat-Dao-Nhi), (3rd anti-Euler)-of-(4th inner-Fermat-Dao-Nhi), (4th anti-Euler)-of-(3rd inner-Fermat-Dao-Nhi), (anti-Hutson intouch)-of-(3rd inner-Fermat-Dao-Nhi), (anti-Mandart-incircle)-of-(4th inner-Fermat-Dao-Nhi), (Aquila)-of-(3rd inner-Fermat-Dao-Nhi), (circumorthic)-of-(1st outer-Fermat-Dao-Nhi), (2nd circumperp tangential)-of-(3rd inner-Fermat-Dao-Nhi), (excentral)-of-(4th inner-Fermat-Dao-Nhi), (outer-Garcia)-of-(1st outer-Fermat-Dao-Nhi), (hexyl)-of-(3rd inner-Fermat-Dao-Nhi), (Johnson)-of-(1st outer-Fermat-Dao-Nhi), (Kosnita)-of-(1st outer-Fermat-Dao-Nhi), (reflection)-of-(4th inner-Fermat-Dao-Nhi), (tangential)-of-(4th inner-Fermat-Dao-Nhi), (X3-ABC reflections)-of-(3rd inner-Fermat-Dao-Nhi).

• 3rd outer-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(4th outer-Fermat-Dao-Nhi), (anti-Aquila)-of-(2nd inner-Fermat-Dao-Nhi), (2nd anti-circumperp-tangential)-of-(2nd inner-Fermat-Dao-Nhi), (anti-X3-ABC reflections)-of-(2nd inner-Fermat-Dao-Nhi), (circumorthic)-of-(4th outer-Fermat-Dao-Nhi), (Ehrmann-mid)-of-(1st inner-Fermat-Dao-Nhi), (Euler)-of-(2nd inner-Fermat-Dao-Nhi), (2nd Euler)-of-(2nd inner-Fermat-Dao-Nhi), (3rd Euler)-of-(1st inner-Fermat-Dao-Nhi), (4th Euler)-of-(2nd inner-Fermat-Dao-Nhi), (outer-Garcia)-of-(4th outer-Fermat-Dao-Nhi), (Hutson intouch)-of-(2nd inner-Fermat-Dao-Nhi), (Johnson)-of-(4th outer-Fermat-Dao-Nhi), (Kosnita)-of-(4th outer-Fermat-Dao-Nhi), (Mandart-incircle)-of-(1st inner-Fermat-Dao-Nhi).

• 4th outer-Fermat-Dao-Nhi = (ABC-X3 reflections)-of-(3rd outer-Fermat-Dao-Nhi), (anti-Aquila)-of-(1st inner-Fermat-Dao-Nhi), (2nd anti-circumperp-tangential)-of-(1st inner-Fermat-Dao-Nhi), (anti-X3-ABC reflections)-of-(1st inner-Fermat-Dao-Nhi), (circumorthic)-of-(3rd outer-Fermat-Dao-Nhi), (Ehrmann-mid)-of-(2nd inner-Fermat-Dao-Nhi), (Euler)-of-(1st inner-Fermat-Dao-Nhi), (2nd Euler)-of-(1st inner-Fermat-Dao-Nhi), (3rd Euler)-of-(2nd inner-Fermat-Dao-Nhi), (4th Euler)-of-(1st inner-Fermat-Dao-Nhi), (outer-Garcia)-of-(3rd outer-Fermat-Dao-Nhi), (Hutson intouch)-of-(1st inner-Fermat-Dao-Nhi), (Johnson)-of-(3rd outer-Fermat-Dao-Nhi), (Kosnita)-of-(3rd outer-Fermat-Dao-Nhi), (Mandart-incircle)-of-(2nd inner-Fermat-Dao-Nhi).A-vertices coordinates:

1th to 4th inner-Fermat-Dao-Nhi trianglesTrilinears1th to 4th outer-Fermat-Dao-Nhi trianglesA_{1}= (7*S-3*sqrt(3)*a^2)/a : (-2*S+3*sqrt(3)*SC)/b : (-2*S+3*sqrt(3)*SB)/c

TrilinearsA_{2}= (5*S-3*sqrt(3)*a^2)/a : (-4*S+3*sqrt(3)*SC)/b : (-4*S+3*sqrt(3)*SB)/c

TrilinearsA_{3}= (-4*S-3*sqrt(3)*a^2)/a : ( 5*S+3*sqrt(3)*SC)/b : ( 5*S+3*sqrt(3)*SB)/c

TrilinearsA_{4}= (-8*S-3*sqrt(3)*a^2)/a : ( S+3*sqrt(3)*SC)/b : ( S+3*sqrt(3)*SB)/c

TrilinearsA_{1}= (-7*S-3*sqrt(3)*a^2)/a : ( 2*S+3*sqrt(3)*SC)/b : ( 2*S+3*sqrt(3)*SB)/c

TrilinearsA_{2}= (-5*S-3*sqrt(3)*a^2)/a : ( 4*S+3*sqrt(3)*SC)/b : ( 4*S+3*sqrt(3)*SB)/c

TrilinearsA_{3}= ( 4*S-3*sqrt(3)*a^2)/a : (-5*S+3*sqrt(3)*SC)/b : (-5*S+3*sqrt(3)*SB)/c

TrilinearsA_{4}= ( 8*S-3*sqrt(3)*a^2)/a : ( -S+3*sqrt(3)*SC)/b : ( -S+3*sqrt(3)*SB)/c

## Feuerbach

The

Feuerbach triangleis 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 trianglehas vertices the reflections of the vertices of 2nd circumperp triangle on the sidelines of ABC.A-vertex coordinates:

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

## 2nd Fuhrmann

The

2nd Fuhrmann trianglehas 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 Brocardouter-Gallatly-Kiepert: See

1st Brocard reflected

## inner-Garcia

Let T

_{A}T_{B}T_{C}be the extouch triangle of a triangle ABC, and let L_{A}be the line perpendicular to line BC at T_{A}. Of the two points on L_{A}at distance r=inradius-of-ABC from T_{A}, 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 theinner Garcia triangle. (Reference: X5587)A-vertex coordinates:

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

## outer-Garcia

Let T

_{A}T_{B}T_{C}be the extouch triangle of a triangle ABC, and let L_{A}be the line perpendicular to line BC at T_{A}. Of the two points on L_{A}at distance r=inradius-of-ABC from T_{A}, 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 theouter Garcia triangleand A"B"C" is the inner Garcia triangle. (Reference: X5587)Equivalences:

• (anti-Aquila)-of-(anticomplementary), (Kosnita)-of-(inner-Conway), (medial)-of-(Aquila).A-vertex coordinates:

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

## Garcia-Moses

Let A'B'C' be the cevian triangle of X(1), M

_{a}the midpoint of AA', and define M_{b}and M_{c}cyclically. Let A" = BM_{c}∩ CM_{b}, and define B" and C" cyclicaly. The triangle A"B"C" is theGarcía-Moses triangle. Reference: X51504)A-vertex coordinates:

Trilinears1 : (a + c)/b : (a + b)/c

## Garcia-reflection

The

Garcia-reflection trianglehas vertices the reflections of the excenters on the midpoints of their corresponding sides. (Reference: X15995)A-vertex coordinates:

Trilinears1 : (c - a)/b : (b - a)/c

## Gemini triangles 1 to 111

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 BB

_{a}C_{a}C, CC_{b}A_{b}A, AA_{c}B_{c}B on the sides BC, CA, AB of triangle ABC (inwards). Then, the triangle A'B'C' enclosed by the lines B_{a}C_{a}, C_{b}A_{b}, A_{c}B_{c}is theinner-Grebe triangleof ABC.A-vertex coordinates:

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

## outer-Grebe

Erect squares BB

_{a}C_{a}C, CC_{b}A_{b}A, AA_{c}B_{c}B on the sides BC, CA, AB of triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the lines B_{a}C_{a}, C_{b}A_{b}, A_{c}B_{c}is theouter-Grebe triangleof 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 B

_{c}the reflection of B in the C-cevian of P and C_{b}the reflection of C in the B-cevian-of P. Denote as a' the circle {{A,B_{c},C_{b}}} 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.Equivalences:

• 1st half-diamonds triangle = (medial)-of-(inner-Fermat).

• 2nd half-diamonds triangle = (medial)-of-(outer-Fermat).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.)

Equivalences:

• 1st half-diamonds-central = (Ehrmann-mid)-of-(outer-Napoleon), (3rd Euler)-of-(outer-Napoleon), (Mandart-incircle)-of-(outer-Napoleon), (medial)-of-(outer-Napoleon).

• 2nd half-diamonds-central = (Ehrmann-mid)-of-(inner-Napoleon), (3rd Euler)-of-(inner-Napoleon), (Mandart-incircle)-of-(inner-Napoleon), (medial)-of-(inner-Napoleon).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.)Equivalences:

• 1st half-squares = (anticomplementary)-of-(outer-Vecten), (6th mixtilinear)-of-(3rd outer-Vecten).

• 2nd half-squares = (anticomplementary)-of-(inner-Vecten).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 C

_{A}be the point other than C' in which the perpendicular to BC from C' meets the incircle, let B_{A}be the point other than B' in which the perpendicular to BC from B' meets the incircle, and let A_{0}be the point of intersection of lines BC_{A}and CB_{A}. Define B_{0}and C_{0}cyclically. A_{0}B_{0}C_{0}is the1st 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 A

_{0}B_{0}C_{0}be the 1st Hatzipolakis triangle of ABC. Let A_{1}be the orthogonal projection of A_{0}onto line BC, and define B_{1}and C_{1}cyclically. The triangle A_{1}B_{1}C_{1}is the2nd Hatzipolakis triangle. (Reference: X1119)A-vertex coordinates:

Trilinears0 : 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 A

_{b}and A_{c}be the orthogonal projections of A' on AC and AB, respectively, and denote N_{a}the nine-point-center of AA_{b}A_{c}; build N_{b}and N_{c}cyclically. N_{a}N_{b}N_{c}is the3rd Hatzipolakis triangle. (Reference: X17817)A-vertex coordinates:

Trilinears2*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 A

_{B}be the orthogonal projection of A' onto line AB, and define B_{C}and C_{A}cyclically. Let A_{C}be the orthogonal projection of A' onto line AC, and define B_{A}and C_{B}cyclically. Let A'_{B}be the reflection of A in A_{B}, and let A'_{C}be the reflection of A in A_{C}. Let N_{A}be the nine-point center of the triangle AA'_{B}A'_{C}, and define N_{B}and N_{C}cyclically. The triangle N_{A}N_{B}N_{C}, is theHatzipolakis-Moses triangle. (Reference: X6145)A-vertex coordinates:

Trilinears2*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 J

_{a}, J_{b}, J_{c}, define A* as the point in which the perpendicular to AB through the excenter J_{b}meets the perpendicular to AC through the excenter J_{c}, and similarly define B* and C*. Then A*B*C* is thehexyl triangleof ABC. (Reference: MathWorld -- Hexyl Triangle.)Equivalences:

• (anticomplementary)-of-(2nd circumperp), (anti-Euler)-of-(1st circumperp), (anti-X3-ABC reflections)-of-(6th mixtilinear), (excentral)-of-(ABC-X3 reflections), (Johnson)-of-(excentral), (6th mixtilinear)-of-(anti-Aquila), (Ursa-minor)-of-(2nd circumperp tangential), (Wasat)-of-(anti-Euler).A-vertex coordinates:

Trilinearsa^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 L

_{A}be the line through X(7) parallel to B'C', and let A_{B}= AB∩L_{A}and A_{C}= AC∩L_{A}. Define B_{C}and C_{A}cyclically, and define B_{A}and C_{B}cyclically. (The six points A_{B}, B_{C}, C_{A}, A_{C}, B_{A}, C_{B}lie on the Adams circle). Let A* = A_{B}C_{B}∩A_{C}B_{C}, and define B* and C* cyclically. The triangle A*B*C* is theHonsberger triangle. (Reference: X7670)A-vertex coordinates:

Trilinears1/((b+c)*a-(b-c)^2) : -1/(b*(a-b+c)) : -1/(c*(a+b-c))

## Hung-Feuerbach

Let F

_{a}F_{b}F_{c}be the Feuerbach triangle of ABC and (K_{a}) the circle, other than the nine-point-circle, passing through F_{b}and F_{c}and tangent to the A-excircle (I_{a}). Let A' be the touchpoint of (K_{a}) and (I_{a}) and name as t_{a}the tangent to both circles in A'. Build t_{b}and t_{c}cyclically. The triangle bounded by t_{a}, t_{b}and t_{c}is named here theHung-Feuerbach triangleof 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 P

_{A}be the point closer to line BC and let Q_{A}be the other point. Define P_{B}and P_{C}cyclically, and define Q_{B}and Q_{C}cyclically. Let L_{A}be the line tangent to the A-excircle at P_{A}, and define L_{B}and L_{C}cyclically. Theinner-Hutson triangleis the triangle bounded by the lines L_{A}, L_{B}and L_{C}. (Reference: X363)A-vertex coordinates:

Trilinearswhere x = (-a+b+c)/2, y = (a-b+c)/2, z = (a+b-c)/2 and δ

-(2*δ^{2}*sin(A/2)-y^{2}*sin(B/2)-z^{2}*sin(C/2)) :

(2*(c*z-δ^{2})*b*sin(A/2)-(a-c)*y^{2}*sin(B/2)-b*z^{2}*sin(C/2))/b :

(2*(b*y-δ^{2})*c*sin(A/2)-(a-b)*z^{2}*sin(C/2)-c*y^{2}*sin(B/2))/c

^{2}= (4*R*r+r^{2})/2

## outer-Hutson

Continuing the construction of the inner-Hutson triangle, let M

_{A}be the line tangent to the A-excircle at Q_{A}, and define M_{B}and M_{C}cyclically. Theouter-Hutson triangleis the triangle bounded by the lines M_{A}, M_{B}and M_{C}. (Reference: X363)A-vertex coordinates:

Trilinearswhere x = (-a+b+c)/2, y = (a-b+c)/2, z = (a+b-c)/2 and δ

-(2*δ^{2}*sin(-A/2)-y^{2}*sin(B/2)-z^{2}*sin(C/2)) :

(2*(c*z-δ^{2})*b*sin(-A/2)-(a-c)*y^{2}*sin(B/2)-b*z^{2}*sin(C/2))/b :

(2*(b*y-δ^{2})*c*sin(-A/2)-(a-b)*z^{2}*sin(C/2)-c*y^{2}*sin(B/2))/c

^{2}= (4*R*r+r^{2})/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)Equivalences:

• (ABC-X3 reflections)-of-(intouch), (anti-Ara)-of-(excenters-reflections), (1st anti-circumperp)-of-(midarc), (2nd anti-circumperp-tangential)-of-(2nd tangential-midarc), (circumorthic)-of-(2nd midarc), (1st circumperp)-of-(2nd anti-circumperp-tangential), (2nd circumperp)-of-(Mandart-incircle), (Euler)-of-(Ursa-minor), (Mandart-incircle)-of-(tangential-midarc).A-vertex coordinates:

Trilinears4*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 the1st 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 triangleof ABC. (Reference: Hyacinthos #24020)Equivalences:

• (Aries)-of-(anti-Ara).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:

Trilinears0 : 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 triangleof ABC. (Reference: X11034)Equivalences:

• (anti-Ara)-of-(2nd circumperp), (anti-X3-ABC reflections)-of-(intouch), (Johnson)-of-(inverse-in-incircle).A-vertex coordinates:

Trilinears2*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 t

_{a}be the other than BC interior common tangent of the incircle and the A-excircle; define t_{b}and t_{c}cyclically. The triangle A'B'C' bounded by these lines is theintangents triangle.Equivalences:

• (anti-Hutson intouch)-of-(2nd anti-circumperp-tangential), (tangential)-of-(Mandart-incircle).A-vertex coordinates:

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

## intouch

The

intouch triangleof ABC, also called thecontact triangle, has as vertices the touchpoints of the incircle and the sidelines of ABC.Equivalences:

• (ABC-X3 reflections)-of-(Hutson intouch), (anti-Ara)-of-(excentral), (1st anti-circumperp)-of-(2nd midarc), (2nd anti-circumperp-tangential)-of-(tangential-midarc), (anticomplementary)-of-(inverse-in-incircle), (circumorthic)-of-(midarc), (1st circumperp)-of-(Mandart-incircle), (2nd circumperp)-of-(2nd anti-circumperp-tangential), (Hutson intouch)-of-(5th mixtilinear), (incircle-circles)-of-(Aquila), (Mandart-incircle)-of-(2nd tangential-midarc), (medial)-of-(Ursa-minor), (X3-ABC reflections)-of-(incircle-circles).A-vertex coordinates:

Trilinears0 : 1/(b*(a-b+c)) : 1/(c*(a+b-c))

## inverse-in-Conway

Let A', B', C' be the inverses of A, B, C in the Conway circle. A'B'C' is the

inverse-in-Conway triangle. (Reference: X10439)Equivalences:

• (medial)-of-(3rd Conway).A-vertex coordinates:

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

## inverse-in-excircles

The

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

• (Johnson)-of-(incircle-circles), (medial)-of-(intouch).A-vertex coordinates:

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

## 1st and 2nd isodynamic-Dao

Let P

_{1}= X(15) be the 1st isodynamic point of ABC. Circles (A, P_{1}), (B, P_{1}) and (C, P_{1}) intersect by pairs at P_{1}, A', B', C'. Then the triangle A'B'C' is equilateral and is named the1st isodynamic-Dao equilateral triangle.Let P

_{2}= X(16) be the 2nd isodynamic point of ABC. Circles (A, P_{2}), (B, P_{2}) and (C, P_{2}) intersect by pairs at P_{2}, A", B", C". Then the triangle A"B"C" is equilateral and is called the2nd isodynamic-Dao equilateral triangle. (Reference: X16802)Equivalences:

• 1st isodynamic-Dao = (anticomplementary)-of-(3rd isodynamic-Dao), (anti-Ehrmann-mid)-of-(3rd isodynamic-Dao), (3rd anti-Euler)-of-(3rd isodynamic-Dao), (anti-Mandart-incircle)-of-(3rd isodynamic-Dao), (Ehrmann-mid)-of-(15th Fermat-Dao), (3rd Euler)-of-(15th Fermat-Dao), (excentral)-of-(3rd isodynamic-Dao), (Mandart-incircle)-of-(15th Fermat-Dao), (reflection)-of-(3rd isodynamic-Dao), (tangential)-of-(3rd isodynamic-Dao).

• 2nd isodynamic-Dao = (anticomplementary)-of-(4th isodynamic-Dao), (anti-Ehrmann-mid)-of-(4th isodynamic-Dao), (3rd anti-Euler)-of-(4th isodynamic-Dao), (anti-Mandart-incircle)-of-(4th isodynamic-Dao), (Ehrmann-mid)-of-(16th Fermat-Dao), (3rd Euler)-of-(16th Fermat-Dao), (excentral)-of-(4th isodynamic-Dao), (Mandart-incircle)-of-(16th Fermat-Dao), (reflection)-of-(4th isodynamic-Dao), (tangential)-of-(4th isodynamic-Dao).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 A

_{o}B_{o}C_{o}be the orthic triangle of ABC. Denote by A', B', C' the 1st isodynamic points X(15) of AB_{o}C_{o}, BC_{o}A_{o}and CA_{o}B_{o}, respectively, and A", B", C" the 2nd isodynamic points X(16) of AB_{o}C_{o}, BC_{o}A_{o}and CA_{o}B_{o}, respectively. Then the triangles A'B'C' and A"B"C" are equilateral and are the3rd and 4th isodynamic-Dao of ABC, respectively. (Reference: preamble just before X31683)Equivalences:

• 3rd isodynamic-Dao = (anti-Ara)-of-(15th Fermat-Dao), (2nd anti-Conway)-of-(15th Fermat-Dao), (6th anti-mixtilinear)-of-(15th Fermat-Dao), (Ehrmann-mid)-of-(1st isodynamic-Dao), (3rd Euler)-of-(1st isodynamic-Dao), (inverse-in-incircle)-of-(15th Fermat-Dao), (Mandart-incircle)-of-(1st isodynamic-Dao), (medial)-of-(1st isodynamic-Dao).

• 4th isodynamic-Dao = (anti-Ara)-of-(16th Fermat-Dao), (2nd anti-Conway)-of-(16th Fermat-Dao), (6th anti-mixtilinear)-of-(16th Fermat-Dao), (Ehrmann-mid)-of-(2nd isodynamic-Dao), (3rd Euler)-of-(2nd isodynamic-Dao), (inverse-in-incircle)-of-(16th Fermat-Dao), (Mandart-incircle)-of-(2nd isodynamic-Dao), (medial)-of-(2nd isodynamic-Dao).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 theA-, 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 T

_{ab}and T_{ac}be the touchpoints of the A-Jenkins circle with the B- and C- excircles, respectively. Denote (T_{bc}, T_{ba}) and (T_{ca}, T_{cb}) cyclically. The triangle bounded by the lines T_{ab}T_{ac}, T_{bc}T_{ba}and T_{ca}T_{cb}is named the3rd Jenkins triangle of ABC.A-vertex coordinates:

1st Jenkins triangle:

Trilinears2nd Jenkins triangle:

-(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

Trilinears3rd Jenkins triangle:-(a+b+c)*((b+c)*a+b^2+c^2)/(a*(a-b+c)*(a+b-c)) : (a+c)/b : (a+b)/c

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 T

_{a}be the touchpoint of the A-Jenkins circle and the A-excircle and denote T_{b}, T_{c}cyclically. The triangle T_{a}T_{b}T_{c}is theJenkins-contact triangleof 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 τ

_{a}be the tangent to the A-Jenkins circle at T_{a}(see Jenkins-contact triangle) and define τ_{b}and τ_{c}cyclically. The triangle bounded by these tangents is theJenkins-tangential triangleof 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 triangleare the centers of the Johnson circles J_{a}, J_{b}and J_{c}. (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 J_{a}is the circle {{B,C,X(4)}} and J_{b}and J_{c}are defined cyclically, then each one has radius=circumradius and J is the circumcircircle of ABC.

Note: The Johnson triangle is also known asCarnot triangle.Equivalences:

• (anticomplementary)-of-(Euler), (anti-Ehrmann-mid)-of-(anti-X3-ABC reflections), (anti-Euler)-of-(medial), (3rd anti-Euler)-of-(4th Euler), (4th anti-Euler)-of-(3rd Euler), (anti-X3-ABC reflections)-of-(anticomplementary), (Ehrmann-mid)-of-(anti-Euler), (Euler)-of-(anti-Ehrmann-mid), (medial)-of-(X3-ABC reflections), (X3-ABC reflections)-of-(Ehrmann-mid).

• Only for acute ABC: (2nd circumperp)-of-(Ehrmann-side), (excentral)-of-(2nd Euler), (hexyl)-of-(orthic), (incircle-circles)-of-(anti-inverse-in-incircle), (inverse-in-incircle)-of-(anti-incircle-circles).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 J

_{b}and J_{c}Johnson circles of ABC (seeJohnson triangle), let L_{A}be the one on the opposite side of BC from A, and let M_{A}be the other. Define L_{B}, M_{B}, L_{C}, M_{C}cyclically. Theinner-Johnson triangleis the triangle enclosed by the lines L_{A}, L_{B}and L_{C}. (Reference: X10522)Equivalences:

• (anti-Ursa minor)-of-(Ursa-major).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 J

_{b}and J_{c}Johnson circles of ABC (seeJohnson triangle), let L_{A}be the one on the opposite side of BC from A, and let M_{A}be the other. Define L_{B}, M_{B}, L_{C}, M_{C}cyclically. Theouter-Johnson triangleis the triangle enclosed by the lines M_{A}, M_{B}and M_{C}. (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 the1st Johnson-Yff triangleare the other points of intersection of Y'_{a}, Y'_{b}and Y'_{c}. (Reference: X10523).Equivalences:

• (Johnson)-of-(inner-Yff).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:

Trilinearsa/(-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 the2nd Johnson-Yff triangleare the other points of intersection of Y"_{a}, Y"_{b}and Y"_{c}. (Reference: X10523).Equivalences:

• (Johnson)-of-(outer-Yff).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 A

_{e}, A_{i}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 A_{i}closer to A than A_{e}. Build B_{i}, B_{e}, C_{i}, C_{e}cyclically. A_{e}B_{e}C_{e}and A_{i}B_{i}C_{i}are theK798e & 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 and 2nd Kenmotu-centers

Let A', B', C' be the centers of the inner-Kenmotu squares, as showed in MathWorld's Kenmotu Point. Triangle A'B'C' is named here the

1st Kenmotu-centers triangleof ABC. As there exists an outer version of these squares with centers A", B", C", the triangle A"B"C" is refered here as the2nd Kenmotu-centers triangleof ABC.Equivalences:

• 1st Kenmotu-centers = (anti-X3-ABC reflections)-of-(1st Kenmotu-free-vertices).

• 2nd Kenmotu-centers = (anti-X3-ABC reflections)-of-(2nd Kenmotu-free-vertices).A-vertex coordinates:

1st triangle:

Trilinears2nd triangle:(a^2+2*S)/a : b : c

Trilinears(a^2-2*S)/a : b : 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.Equivalences:

• (Kosnita)-of-(1st Kenmotu-free-vertices), (tangential)-of-(1st Kenmotu-centers).A-vertex coordinates:

Trilinearsa*(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.Equivalences:

• (Kosnita)-of-(2nd Kenmotu-free-vertices), (tangential)-of-(2nd Kenmotu-centers).A-vertex coordinates:

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

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

Let K

_{a}, K_{b}, K_{c}be the free vertices of the 1st(2nd) Kenmotu squares. Triangle K_{a}K_{b}K_{c}is the1st(2nd) Kenmotu free vertices triangle.(Reference: X12375)Equivalences:

• 1st Kenmotu-free-vertices = (X3-ABC reflections)-of-(1st Kenmotu-centers).

• 2nd Kenmotu-free-vertices = (X3-ABC reflections)-of-(2nd Kenmotu-centers).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 triangleare the circumcenters of the triangles BOC, COA, AOB, where O is the circumcenter of ABC. (Reference: X1658)Equivalences:

• (Ehrmann-vertex)-of-(Johnson), (tangential)-of-(anti-X3-ABC reflections), (Trinh)-of-(ABC-X3 reflections).

• Only for acute ABC: (anti-Aquila)-of-(tangential).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 A

_{e}B_{e}C_{e}circumscribing ABC and such that A lies between B_{e}and C_{e}, B lies between C_{e}and A_{e}and C lies between A_{e}and B_{e}. The triangle defined here is the A_{e}B_{e}C_{e}with largest area. (Reference: Preamble just before X36761.)

This triangle is the antipedal triangle of X(13). The A-vertex 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 B

_{c}, C_{b}be the circumcenters of CC'B and BB'C, respectively, and build C_{a}, A_{c}, A_{b}and B_{a}cyclically. Denote the circumcenters of B_{c}C_{b}A, C_{a}A_{c}B, A_{b}B_{a}C as O_{a}, O_{b}, O_{c}, respectively. Then, in each case, the triangle O_{a}O_{b}O_{c}is equilateral. These triangles are named theinner- 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 triangleis the cevian triangle of X(598).A-vertex coordinates:

Trilinears0 : 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 A

_{b}, A_{c}be the points where the antiparallel to BC through X(6) cuts the sides AC and AB, respectively. Define B_{c}, B_{a}, C_{a}, C_{b}cyclically. Build equilateral triangles B_{c}C_{b}A', C_{a}A_{c}B' and A_{b}B_{a}C' allwith the same orientation as ABCand build equilateral triangles B_{c}C_{b}A", C_{a}A_{c}B" and A_{b}B_{a}C", allwith opposite orientation than ABC. Then A'B'C' and A"B"C" are both equilateral triangles and are named the1st Lemoine-Dao equilateral triangleand the2nd 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 BCC

_{A}B_{A}(see figure). Lines AB_{A}and AC_{A}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'_{B}A'_{C}A"_{B}A"_{C}is a square named theA-Lucas square(and also theA-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 theA-, B-, C- Lucas circlesand result to be pairwise tangents and simultaneously tangent internally to the circumcircle of ABC. The circle externally tangent to the Lucas circles is named theLucas inner circle.

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

Denote:

A_{0}= the center of the A-Lucas circle, and similarly B_{0}, C_{0};

A_{T}= the touchpoint of the B- and C- Lucas circles, and similarly B_{T}, C_{T};

A_{p}= the antipode of A in the A-Lucas circle, and similarly B_{p}, C_{p};

A_{i}= the touchpoint of the A- Lucas circle and the Lucas inner circle, and similarly B_{i}, C_{i};

A_{2}= the center of the A-Lucas secundary central circle, and similarly B_{2}, C_{2};

A_{2O}= the touchpoints of the A-Lucas secundary central circle and the circumcircle, and similarly B_{2O}, C_{2O};

A_{2B}, A_{2C}= the touchpoints of the A-Lucas secundary central circle and the B- and C- Lucas circles, respectively, and similarly B_{2C}, B_{2A}, C_{2A}, C_{2B}. These six point lie on theLucas 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 nameLucaswithLucas(-1); examples:A-Lucas(-1) squareandA-Lucas(-1) circleinstead ofA-Lucas squareandA-Lucas circle. Moreover, if the construction is generalized and BCC_{A}B_{A}is a rectangle whose sides are in the proportion length:width=t:1 (i.e, BB_{A}= CC_{A}= t*BC), then all the terms described can be reused just by replacingLucasbyLucas(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: A

_{p}, B_{p}and C_{p}. See Lucas triangles. (Reference: X6457)A-vertex coordinates:

Lucas antipodal:Trilinearscos(B)*cos(C)-sin(A) : -cos(B) : -cos(C)

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

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

The tangents at A

_{p}, B_{p}and C_{p}to the A-, B- and C-Lucas circles, respectively, bound theLucas 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 triangleis 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: A

_{0}, B_{0}and C_{0}. See Lucas triangles.Equivalences:

• Lucas central = (tangential)-of-(Lucas tangents).

• Lucas(-1) central = (tangential)-of-(Lucas(-1) tangents).A-vertex coordinates:

Lucas central:Trilinearscos(A)+2*sin(A) : cos(B) : cos(C)

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

## Lucas homothetic and Lucas(-1) homothetic

The

Lucas homothetic triangleis enclosed by the lines A"_{B}A"_{C}, B"_{C}B"_{A}and C"_{B}C"_{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: A

_{i}, B_{i}and C_{i}. See Lucas triangles.A-vertex coordinates:

Lucas inner:Trilinears2*cos(A)+3*sin(A)/2 : 2*cos(B)+sin(B) : 2*cos(C)+sin(C)

Lucas(-1) inner:Trilinears2*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 A

_{i}, B_{i}and C_{i}. See Lucas triangles. (Reference: Preamble just before X6395)Equivalences:

• Lucas inner tangential = (tangential)-of-(Lucas inner).

• Lucas(-1) inner tangential = (tangential)-of-(Lucas(-1) inner).A-vertex coordinates:

Lucas inner tangential:Trilinears4*cos(A)+sin(A) : 4*cos(B)+3*sin(B) : 4*cos(C)+3*sin(C)

Lucas(-1) inner tangential:Trilinears4*cos(A)-sin(A) : 4*cos(B)-3*sin(B) : 4*cos(C)-3*sin(C)

## Lucas reflection and Lucas(-1) reflection

Let L

_{a}be the reflection of the line B_{0}C_{0}in BC and define L_{b}, L_{c}cyclically (see Lucas triangles). TheLucas reflection triangleis the triangle limited by L_{a}, L_{b}, L_{c}. (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: A

_{2}, B_{2}and C_{2}. See Lucas triangles. (Reference: X6199)A-vertex coordinates:

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

Lucas(-1) secondary central:Trilinearscos(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 B

_{2A}C_{2A}, C_{2B}A_{2B}and A_{2C}B_{2C}(see Lucas triangles). (Reference: X6199)A-vertex coordinates:

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

Lucas(-1) 1st secondary tangents:Trilinearscos(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 A

_{2B}A_{2C}, B_{2C}B_{2A}and C_{2A}C_{2B}(see Lucas triangles). (Reference: X6199)A-vertex coordinates:

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

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

## Lucas tangents and Lucas(-1) tangents

Vertices: A

_{T}, B_{T}and C_{T}. See Lucas triangles.A-vertex coordinates:

Lucas tangents:Trilinearscos(A) : cos(B)+sin(B) : cos(C)+sin(C)

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

## M:

## MacBeath

The

MacBeathtriangle 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:

Trilinears0 : 1/(b^3*(a^2-b^2+c^2)) : 1/(c^3*(a^2+b^2-c^2))

## Malfatti

The

Malfatti triangle, or better named, theinner-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.)There is a second triad of circles each tangent to two sidelines of ABC and to the other two circles, but not packed inside ABC. These circles are named the

outer-Malfatti circles. The centers of these circles are vertices of theouter-Malfatti triangle.Equivalences:

• inner-Malfatti = (tangential)-of-(inner-Malfatti-touchpoints).

• outer-Malfatti = (tangential)-of-(outer-Malfatti-touchpoints).A-vertex coordinates:

inner-Malfatti: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

outer-Malfatti: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

## Malfatti-touchpoints

Let A' be the touchpoint of the inner-B- and inner-C-Malfatti circles and define B', C' cyclically. The triangle A'B'C' is the

inner-Malfatti-touchpoints triangleof ABC. (Reference: X46876)Let A" be the touchpoint of the outer-B- and outer-C-Malfatti circles and define B", C" cyclically. The triangle A"B"C" is the

outer-Malfatti-touchpoints triangleof ABC.A-vertex coordinates:

inner-Malfatti-touchpoints:Trilinears(1+cos(B/2))*(1+cos(C/2))/(1+cos(A/2)) : (1+cos(C/2))^2 : (1+cos(B/2))^2

outer-Malfatti-touchpoints:Trilinears(1-cos(B/2))*(1-cos(C/2))/(1-cos(A/2)) : (1-cos(C/2))^2 : (1-cos(B/2))^2

## 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 triangleof 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 triangleof ABC.Equivalences:

• (ABC-X3 reflections)-of-(2nd anti-circumperp-tangential), (1st anti-circumperp)-of-(intouch), (2nd anti-circumperp-tangential)-of-(5th mixtilinear), (circumorthic)-of-(Hutson intouch), (orthic)-of-(Ursa-minor).

• Only for acute ABC: (Hutson intouch)-of-(anti-tangential-midarc).A-vertex coordinates:

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

## McCay

The

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

Note: Let G be the centroid and A'B'C' the2nd Brocard triangleof 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:

Trilinearsa*(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 triangleis the cevian triangle of the centroid.Equivalences:

• (ABC-X3 reflections)-of-(Euler), (1st anti-circumperp)-of-(3rd Euler), (anticomplementary)-of-(Gemini 110), (2nd anti-Conway)-of-(excentral), (circumorthic)-of-(4th Euler), (Euler)-of-(Johnson), (outer-Garcia)-of-(anti-Aquila), (Johnson)-of-(anti-X3-ABC reflections), (orthic)-of-(Wasat), (tangential)-of-(2nd Zaniah).

• Only for acute ABC: (1st circumperp)-of-(orthic), (2nd circumperp)-of-(2nd Euler), (3rd Euler)-of-(anti-Wasat), (excentral)-of-(6th anti-mixtilinear), (inverse-in-incircle)-of-(tangential), (6th mixtilinear)-of-(submedial).A-vertex coordinates:

Trilinears0 : 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 triangleof ABC. (Reference: MathWorld -- Mid-Arc Triangle.)Equivalences:

• (ABC-X3 reflections)-of-(2nd midarc), (1st circumperp)-of-(Hutson intouch), (2nd circumperp)-of-(intouch), (4th Euler)-of-(Ursa-minor), (Hutson intouch)-of-(2nd tangential-midarc).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 triangleof ABC.Equivalences:

• (ABC-X3 reflections)-of-(midarc), (1st circumperp)-of-(intouch), (2nd circumperp)-of-(Hutson intouch), (3rd Euler)-of-(Ursa-minor), (Hutson intouch)-of-(tangential-midarc).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 triangleof ABC. (It is also known as thehalf-altitude triangle)A-vertex coordinates:

Trilinears2*a : (a^2+b^2-c^2)/b : (a^2-b^2+c^2)/c

## mixtilinear

The

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

Note: A circle that isinternallytangent to two sides of a triangle and to the circumcircle is called amixtilinear incircle. There are three mixtilinear incircles, one corresponding to each angle of the triangle (reference: MathWorld -- Mixtilinear Incircles.). Also, there are three circles that areexternallytangent to two sides of a triangle and to the circumcircle and are called amixtilinear 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 excirclesis the2nd mixtilinear triangle. It is alson known as theouter-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 trianglehas 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 trianglehas 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 I

_{ab}and I_{ac}be the touchpoints of the A-mixtilinear incircle and the sidelines AC and AB, respectively, and define I_{bc}, I_{ba}, I_{ca}, I_{cb}similarly. Denote A_{5}=I_{ab}I_{ba}∩I_{ac}I_{ca}, B_{5}=I_{bc}I_{cb}∩I_{ba}I_{ab}, C_{5}=I_{ca}I_{ac}∩I_{cb}I_{bc}. A_{5}B_{5}C_{5}is the5th mixtilinear triangle. (See note in mixtilinear triangle).

Note: This triangle is also namedCaelum triangle. (Reference: Preamble just before X7955)Equivalences:

• (anti-Hutson intouch)-of-(intouch), (anti-Mandart-incircle)-of-(2nd anti-circumperp-tangential), (2nd circumperp tangential)-of-(Mandart-incircle), (orthic)-of-(excenters-reflections), (tangential)-of-(Hutson intouch).A-vertex coordinates:

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

## 6th mixtilinear

Let E

_{ab}and E_{ac}be the touchpoints of the A-mixtilinear excircle and the sidelines AC and AB, respectively, and define E_{bc}, E_{ba}, E_{ca}, E_{cb}similarly. The6th mixtilinear triangleis the triangle bounded by the lines E_{ab}E_{ac}, E_{bb}E_{ba}and E_{ca}E_{cb}. (See note in mixtilinear triangle). (Reference: Preamble just before X7955)Equivalences:

• (anticomplementary)-of-(excentral), (hexyl)-of-(Aquila), (X3-ABC reflections)-of-(hexyl).A-vertex coordinates:

Trilinearsa^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 E

_{ab}and E_{ac}be the touchpoints of the A-mixtilinear excircle and the sidelines AC and AB, respectively, and define E_{bc}, E_{ba}, E_{ca}, E_{cb}similarly. Denote A_{7}=E_{ab}E_{ba}∩E_{ac}E_{ca}, B_{7}=E_{bc}E_{cb}∩E_{ba}E_{ab}, C_{7}=E_{ca}E_{ac}∩E_{cb}E_{bc}. A_{7}B_{7}C_{7}is the7th 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)

## 8th mixtilinear

Let ω'

_{a}, ω'_{b}and ω'_{c}be the A-, B- and C- mixtilinear incircles of ABC, and Ω_{i}the inner Apollonius circle of them. Denote as A', B', C' the touchpoints of Ω_{i}and ω'_{a}, ω'_{b}and ω'_{c}, respectively. The triangle A'B'C' is the8th-mixtilinear triangleof ABC. (Reference: Preamble just before X44840)A-vertex coordinates:

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

## 9th mixtilinear

Let ω"

_{a}, ω"_{b}and ω"_{c}be the A-, B- and C- mixtilinear excircles of ABC, and Ω_{o}the outer Apollonius circle of them. Denote as A", B", C" the touchpoints of Ω_{o}and ω"_{a}, ω"_{b}and ω"_{c}, respectively. The triangle A"B"C" is the9th-mixtilinear triangleof ABC. (Reference: Preamble just before X44840)A-vertex coordinates:

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

## inner-mixtilinear

See

mixtilinear triangle.

## outer-mixtilinear

See

2nd mixtilinear triangle.

## Montesdeoca-Hung

Let (Ap) be the Apollonius circle and (K

_{a}), (K_{b}), (K_{c}) the circles defined in theHung-Feuerbach triangle. Let L_{a}be the radical axis of (Ap) and (K_{a}) and define L_{b}and L_{c}similarly. The triangle enclosed by these lines is theMontesdeoca-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 b

_{a}and b_{c}be the two angle trisectors of B, and c_{a}and c_{b}the two angle trisectors of C, with b_{a}and c_{a}closer to BC than the others. Build a_{b}and a_{c}similarly. Define A_{1}= b_{a}∩c_{a}and similarly B_{1}and C_{1}. The triangle A_{1}B_{1}C_{1}is equilateral and is named the1st Morley triangle. (MathWorld -- First Morley Triangle.)A-vertex coordinates:

Trilinears1 : 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 b

_{a}around B an angle Pi/3 (counter-clockwise) and let b'_{a}be the resulting line. Next rotate c_{a}around C an angle Pi/3 (clockwise) and denote c'_{a}the resulting line. Define A_{2}= b'_{a}∩c'_{a}and similarly B_{2}and C_{2}. The triangle A_{2}B_{2}C_{2}is equilateral and is named the2nd 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 b

_{a}around B an angle Pi/3 (clockwise) and let b"_{a}be the resulting line. Next rotate c_{a}around C an angle Pi/3 (counter-clockwise) and denote c"_{a}the resulting line. Define A_{3}= b"_{a}∩c"_{a}and similarly B_{3}and C_{3}. The triangle A_{3}B_{3}C_{3}is equilateral and is named the3rd 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)

## 1st Morley-adjunct

With the same notation used in 1st Morley triangle, denote A'

_{1}= b_{c}∩c_{b}and similarly B'_{1}and C'_{1}. The triangle A'_{1}B'_{1}C'_{1}is the1st Morley-adjunct triangle. (MathWorld -- First Morley Adjunct Triangle.)

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

Trilinears2 : sec(C/3) : sec(B/3)

## 2nd Morley-adjunct

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate b

_{c}around B an angle Pi/3 (counter-clockwise) and let b'_{c}be the resulting line. Next rotate c_{b}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'_{2}B'_{2}C'_{2}is equilateral is the2nd 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)

## 3rd Morley-adjunct

Suppose that ABC is labeled counter-clockwise. With the same notation used in the 1st Morley triangle, rotate b

_{c}around B an angle Pi/3 (clockwise) and let b"_{a}be the resulting line. Next rotate c_{b}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'_{3}B'_{3}C'_{3}is the3rd 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 A

_{n}B_{n}C_{n}be the n^{th}Morley-adjunct triangle (n=1,2,3). Let B_{a}and C_{a}be points on BC such that A_{n}B_{a}C_{a}is an equilateral triangle having the same orientation as ABC; define C_{b}and A_{b}cyclically, and define A_{c}and B_{c}cyclically. Let A"_{n}be the midpoint of A_{b}and A_{c}, and define B"_{n}and C"_{n}cyclically. Then A"_{n}B"_{n}C"_{n}is named then. (Reference: X16839-to-X16841)^{th}Morley-adjunct-midpoint triangleA-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 A

_{n}B_{n}C_{n}be the n^{th}Morley triangle (n=1,2,3). Let B_{a}and C_{a}be points on BC such that A_{n}B_{a}C_{a}is an equilateral triangle having the same orientation as ABC; define C_{b}and A_{b}cyclically, and define A_{c}and B_{c}cyclically. Let A"_{n}be the midpoint of A_{b}and A_{c}, and define B"_{n}and C"_{n}cyclically. Then A"_{n}B"_{n}C"_{n}is an equilateral triangle and is named then. (Reference: X3602-to-X3604)^{th}Morley-midpoint equilateral triangleA-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 H

_{a}H_{b}H_{c}be the orthic triangle of ABC and (J_{a}) the circle, other than the nine-point-circle, passing through H_{b}and H_{c}and tangent to the A-excircle (I_{a}). Let A' be the touchpoint of (J_{a}) and (I_{a}) and define B' and C' cyclically. The triangle A'B'C' is named here theMoses-Hung triangleof 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-Miyamoto

Let γ be the incircle of ABC. Let γ

_{a}be the circle through B and C and internally tangent to γ. Outside ABC, let Γ_{a}be the circle externally tangent to AB, AC, γ_{a}, and let A' be the center of Γa. Define B', C' cyclically. The triangle A'B'C' is theMoses-Miyamoto triangle. (Reference: X52804)A-vertex coordinates:

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

## Moses-Miyamoto-Apollonius, 1st and 2nd

In a scalene acute triangle ABC, let M

_{a}M_{b}M_{c}be its medial triangle. Let Ω_{a}be the circle centered at M_{a}and passing through B and C. Inside ABC, let γ_{a}be the circle externally tangent to lines AB, AC and circle Ω_{a}. Define Ω_{b}, Ω_{c}γ_{b}and γ_{c}cyclically. Inside ABC, let γ be the circle internally tangent to Ω_{a}, Ω_{b}, Ω_{c}. Then there exists a circle Γ that is tangent to the four circles γ, γ_{a}, γ_{b}and γ_{c}.Let A' be the touchpoint of Γ and γ

_{a}, and define B' and C' cyclically. The triangle A'B'C' is the1st Moses-Miyamoto-Apollonius triangle. (Reference: X52804)

In a scalene acute triangle ABC, let M

_{a}M_{b}M_{c}be its medial triangle. Let Ω_{a}be the circle centered at M_{a}and passing through B and C. Inside ABC, let γ_{a}be the larger circle internally tangent to lines AB, AC and circle Ω_{a}. Define Ω_{b}, Ω_{c}γ_{b}and γ_{c}cyclically. Outside ABC, let γ be the circle internally tangent to Ω_{a}, Ω_{b}, Ω_{c}. Then there exists a circle Γ that is tangent to the four circles γ, γ_{a}, γ_{b}and γ_{c}.Let A" be the touchpoint of Γ and γ

_{a}, and define B" and C" cyclically. The triangle A"B"C" is the2nd Moses-Miyamoto-Apollonius triangle. (Reference: X52804)

A-vertex coordinates:

1st Moses-Miyamoto-Apollonius triangle:

Trilinears2nd Moses-Miyamoto-Apollonius triangle:-2/((a+b-c)*(a-b+c)-2*S) : 1/(b*(a-b+c)) : 1/(c*(a+b-c))

Trilinears-2/((a+b-c)*(a-b+c)+2*S) : 1/(b*(a-b+c)) : 1/(c*(a+b-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)Note: Moses-Soddy triangle is the anticevian-triangle of X(514).

A-vertex coordinates:

Trilinears1/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 triangleof 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 triangleof 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 triangleis 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.)Equivalences:

• (anticomplementary)-of-(2nd half-diamonds-central), (anti-Ehrmann-mid)-of-(2nd half-diamonds-central), (3rd anti-Euler)-of-(2nd half-diamonds-central), (anti-Mandart-incircle)-of-(2nd half-diamonds-central), (excentral)-of-(2nd half-diamonds-central), (reflection)-of-(2nd half-diamonds-central), (tangential)-of-(2nd half-diamonds-central).A-vertex coordinates:

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

## outer-Napoleon

The

outer Napoleon triangleis 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.)Equivalences:

• (anticomplementary)-of-(1st half-diamonds-central), (anti-Ehrmann-mid)-of-(1st half-diamonds-central), (3rd anti-Euler)-of-(1st half-diamonds-central), (anti-Mandart-incircle)-of-(1st half-diamonds-central), (excentral)-of-(1st half-diamonds-central), (reflection)-of-(1st half-diamonds-central), (tangential)-of-(1st half-diamonds-central).A-vertex coordinates:

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

## 1st Neuberg

The triangle A

_{1}B_{1}C_{1}formed by joining the centers of the Neuberg circles is the1st 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 A

_{2}B_{2}C_{2}formed by joining the centers of the reflections of the Neuberg circles on the opposite sides of ABC is the2nd 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).

Equivalences:

• (ABC-X3 reflections)-of-(2nd Euler), (1st anti-circumperp)-of-(medial), (anticomplementary)-of-(2nd anti-Conway), (6th anti-mixtilinear)-of-(anticomplementary), (circumorthic)-of-(Euler), (Ehrmann-side)-of-(Ehrmann-mid), (2nd Euler)-of-(Johnson), (medial)-of-(anti-Wasat), (submedial)-of-(Gemini 111), (tangential)-of-(anti-Ara).

• Only for acute ABC: (anti-Aquila)-of-(circumorthic), (Mandart-incircle)-of-(anti-Ursa minor).A-vertex coordinates:

Trilinears0 : 1/(b*(a^2-b^2+c^2)) : 1/(c*(a^2+b^2-c^2))

## orthic axes

Let L

The orthic axes triangle is also the cevian triangle of X(4) with respect to the orthic triangle._{A}be the orthic axis of triangle BCX(4), and define L_{B}, L_{C}cyclically. The triangle bounded by these lines is theorthic axes triangle. (Reference: X2501)

A-vertex coordinates:

Trilinears2*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:

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

## orthocentroidal-isogonic

Let A'B'C' be the orthocentroidal triangle of ABC and A", B", C" the isogonal conjugates of A', B', C' with respect to ABC, respectively. A"B"C" is the

orthocentroidal-isogonic triangle of ABC.A-vertex coordinates:

Trilinears2*SB*SC/a : b*SB : c*SC

## 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:

Trilinears2*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 p

_{a}be the parabola with focus A and directrix BC. Let B_{a}be the point, closer to C, at which p_{a}cuts AC and let C_{a}be the point, closer to B, at which p_{a}cuts AB. Build C_{b}, A_{b}and A_{c}, B_{c}cyclically. Points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}lie on an ellipse named thePaasche ellipse.Let A

_{H}= A_{b}C_{b}∩A_{c}B_{c}, B_{H}= B_{c}A_{c}∩B_{a}C_{a}, C_{H}= C_{a}B_{a}∩C_{b}A_{b}. A_{H}B_{H}C_{H}is thePaasche-Hutson triangle. (Reference: X(37994))A-vertex coordinates:

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

## 1st Pamfilos-Zhou

The Pamfilos-Zhou A-rectangle R

_{A}= AA_{B}A_{A}A_{C}is the rectangle of maximal area such that A is a vertex of R_{A}, B lies on the line A_{A}A_{C}, and C lies on the line A_{A}A_{B}. The Pamfilos-Zhou B- and C-rectangles, R_{B}and R_{C}, are defined cyclically.

Let A' = B_{C}B_{A}∩C_{A}C_{B}, and define B' and C' cyclically. The1st Pamfilos-Zhoutriangle 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 R

_{A}= AA_{B}A_{A}A_{C}is the rectangle of maximal area such that A is a vertex of R_{A}, B lies on the line A_{A}A_{C}, and C lies on the line A_{A}A_{B}. The Pamfilos-Zhou B- and C-rectangles, R_{B}and R_{C}, are defined cyclically.

Let A" = C_{A}A_{C}∩A_{B}B_{A}, and define B" and C" cyclically. The2nd Pamfilos-Zhoutriangle 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 A

_{1}be the intersection, other than X(111), of the Parry circle and the line AX(111), and define B_{1}and C_{1}cyclically. The triangle A_{1}B_{1}C_{1}is the1st Parry triangle. (Reference: Preamble just before X9123)

Note: TheParry circleis the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.Equivalences:

• (ABC-X3 reflections)-of-(2nd Parry).A-vertex coordinates:

Trilinears3*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 A

_{2}be the intersection, other than X(110), of the Parry circle and line AX(110), and define B_{2}and C_{2}cyclically. The triangle A_{2}B_{2}C_{2}is the2nd Parry triangle. (Reference: Preamble just before X9123)

Note: TheParry circleis the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.Equivalences:

• (ABC-X3 reflections)-of-(1st Parry).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 A

_{3}be the intersection, other than X(2), of the Parry circle and the A-median, and define B_{3}and C_{3}cyclically. The triangle A_{3}B_{3}C_{3}, is the3rd Parry triangle. (Reference: Preamble just before X9123)

Note: TheParry circleis the circle passing through the isodynamic points X(15) and X(16) and the centroid X(2) of a triangle.A-vertex coordinates:

Trilinearsa*(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 triangleis the vertex triangle of ABC and theintangets triangle. Reference: TCCT, p. 162.

Note: Thevertex triangleA*B*C* of triangles T_{1}=A_{1}B_{1}C_{1}and T_{2}=A_{2}B_{2}C_{2}has vertex A*=B_{1}B_{2}∩C_{1}C_{2}, and B*, C* are defined cyclically.Note: The Pelletier triangle is the anticevian triangle of X(650) and, as an anticevian triangle, it is the tangential triangle of the circumconic with perspector X(650), i.e., it is the tangential triangle of the Feuerbach circum-hyperbola.

A-vertex coordinates:

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

## 1st Przybyłowski-Bollin

Let D

_{1}be the 1st isodynamic center X(15) of ABC and I'_{a}, I'_{b}and I'_{c}the incenters of BCD_{1}, CAD_{1}and ABD_{1}, respectively. The triangle I'_{a}I'_{b}I'_{c}is the1st Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted asAiBiCiin ETC.A-vertex coordinates:

Trilinearsa*(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 D

_{1}be the 1st isodynamic center X(15) of ABC and J'_{a}the excenter of BCD_{1}opposed to D_{1}. Define J'_{b}and J'_{c}similarly. The triangle J'_{a}J'_{b}J'_{c}is the2nd Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted asAaBbCbin ETC.A-vertex coordinates:

Trilinearsa*(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 D

_{2}be the 2nd isodynamic center X(16) of ABC and I"_{a}, I"_{b}and I"_{c}the incenters of BCD_{2}, CAD_{2}and ABD_{2}, respectively. The triangle I"_{a}I"_{b}I"_{c}is the3rd Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as(AiBiCi)*in ETC.A-vertex coordinates:

Trilinearsa*(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 D

_{2}be the 2nd isodynamic center X(16) of ABC and J"_{a}the excenter of BCD_{2}opposed to D_{2}. Define J"_{b}and J"_{c}similarly. The triangle J"_{a}J"_{b}J"_{c}is the4th Przybyłowski-Bollin triangle. (Reference: Preamble just before X11752)

Note: This triangle is denoted as(AaBbCb)*in ETC.A-vertex coordinates:

Trilinearsa*(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)

## Q:

## R:

## reflection

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

reflection triangleof ABC.A-vertex coordinates:

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

## Roussel (equilateral)

If the angular trisectors of a triangle are produced to the circumcircle, then the chords of adjacent trisectors bound 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

In a triangle ABC with incenter I, let A', B', C' be the orthogonal projections of I in BC, CA and AB, respectively. Then the line B'C', the perpendicular to BC through A' and the A-median of ABC concur in a point A

_{1}and points B_{1}and C_{1}are defined cyclically. The triangle A_{1}B_{1}C_{1}is the1st Savin triangleof ABC. (Reference: Preamble just before X44301)No geometric construction was given for 2nd Savin triangle.

A-vertex coordinates:

1st triangle:

Trilinears2nd triangle:2/(-a+b+c) : 1/b : 1/c

Trilinears2/(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 L

_{a}, L_{b}and L_{c}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 B_{c}the point of intersection, other than S, of L_{b}and the circle (C,|CS|) and C_{b}the point of intersection, other than S, of L_{c}and the circle (B,|BS|). Define O'_{a}as the circumcenter of SB_{c}C_{b}and similarly O'_{b}and O'_{c}. The triangle O'_{a}O'_{b}O'_{c}is the1st Schiffler triangle. (Reference: X6595)A-vertex coordinates:

Trilinears1/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 L

_{a}, L_{b}and L_{c}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 B_{c}the point of intersection, other than S, of L_{b}and the circle (C,|CS|) and C_{b}the point of intersection, other than S, of L_{c}and the circle (B,|BS|). Define O"_{a}as the circumcenter of SB_{c}C_{b}and similarly O"_{b}and O"_{c}. The triangle O"_{a}O"_{b}O"_{c}is the2nd 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 triangleof ABC.Note: The Schroeter triangle is the anticevian triangle of X(523).

A-vertex coordinates:

Trilinears1/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 A

_{o}B_{o}C_{o}, A_{i}B_{i}C_{i}be the outer- and inner-Soddy triangles of ABC. Then B_{o}C_{o}C_{i}B_{i}are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of theSoddy 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 triangleis 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 triangleis 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, A

_{o}B_{o}C_{o}the outer-Soddy triangle and A_{i}B_{i}C_{i}the inner-Soddy triangles of ABC. Then:

(Reference: Preamble before X(31528))

- B'C'C
_{i}B_{i}are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of the2nd inner-Soddy triangle.

- B'C'C
_{o}B_{o}are concyclic and similarly the other quartets of vertices. The centers of the circles circumscribing these quartets are the vertices of the2nd outer-Soddy triangle.A-vertex coordinates:

2nd inner-Soddy triangle:

Trilinears2nd outer-Soddy triangle:((-a+b+c)*a+2*S)/(a*(-a+b+c)) : 1 : 1

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

## inner-squares

Build outwards ABC the square BCC

_{A}B_{A}(see figure). Lines AB_{A}and AC_{A}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'_{B}A'_{C}A"_{B}A"_{C}is a square named theA-inner-inscribed-square(and also theA-Lucas square). The B- and C- inner-inscribed-squares can be built similarly. The centers of these squares are the vertices of theinner-squares triangles. (MathWorld -- Inner Inscribed Squares Triangle.)A-vertex coordinates:

Trilinears2*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 BCC

_{A}B_{A}(see figure). Lines AB_{A}and AC_{A}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'_{B}A'_{C}A"_{B}A"_{C}is a square named theA-outer-inscribed-square(and also theA-Lucas(-1) square). The B- and C- outer-inscribed-squares can be built similarly. The centers of these squares are the vertices of theouter-squares triangles. (MathWorld -- Outer Inscribed Squares Triangle.)A-vertex coordinates:

Trilinears2*a : (a^2+b^2-c^2-2*S)/b : (a^2+c^2-b^2-2*S)/c

## Stammler

The

Stammler triangleis 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.)Equivalences:

• (anticomplementary)-of-(circumtangential), (anti-Ehrmann-mid)-of-(circumtangential), (anti-Euler)-of-(circumnormal), (3rd anti-Euler)-of-(circumtangential), (4th anti-Euler)-of-(circumnormal), (anti-Hutson intouch)-of-(circumnormal), (anti-Mandart-incircle)-of-(circumtangential), (Aquila)-of-(circumnormal), (2nd circumperp tangential)-of-(circumnormal), (excentral)-of-(circumtangential), (hexyl)-of-(circumnormal), (reflection)-of-(circumtangential), (tangential)-of-(circumtangential), (X3-ABC reflections)-of-(circumnormal).A-vertex coordinates:

Trilinearscos(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 triangleis the cevian triangle of the Steiner point X(99). (MathWorld -- Steiner Triangle.)A-vertex coordinates:

Trilinears0 : (b^2-c^2)*(a^2-b^2)/b : (c^2-b^2)*(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 R

_{A}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 O_{A}be the center R_{A}, and define O_{B}and O_{C}cyclically. The central triangle O_{A}O_{B}O_{C}is named thesubmedial triangle. (Reference: X9813)Equivalences:

• (6th anti-mixtilinear)-of-(medial), (orthic)-of-(Gemini 110).A-vertex coordinates:

Trilinears2*a*b*c : (3*a^2+b^2-c^2)*c : (3*a^2+c^2-b^2)*b

## symmedial

The

symmedial triangleis the cevian triangle of the symmedian point X(6).A-vertex coordinates:

Trilinears0 : 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 triangleis 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.Equivalences:

• (anti-Hutson intouch)-of-(ABC-X3 reflections), (anti-incircle-circles)-of-(anti-X3-ABC reflections), (anti-inverse-in-incircle)-of-(medial), (anti-Ursa minor)-of-(anticomplementary), (Ehrmann-vertex)-of-(anti-Ehrmann-mid), (Kosnita)-of-(X3-ABC reflections), (medial)-of-(1st excosine), (orthic)-of-(Ara).

• Only for acute ABC: (anti-Aquila)-of-(anti-incircle-circles), (anti-Mandart-incircle)-of-(1st anti-circumperp), (Aquila)-of-(Kosnita), (2nd circumperp tangential)-of-(circumorthic).Note: The tangential triangle is the anticevian triangle of X(6).

A-vertex coordinates:

Trilinears-a : b : c

## (1st) tangential-midarc

The

tangential mid-arctriangle 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.)Equivalences:

• (anti-Hutson intouch)-of-(2nd midarc), (anti-Mandart-incircle)-of-(Hutson intouch), (2nd circumperp tangential)-of-(intouch), (tangential)-of-(midarc).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-arctriangle 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)Equivalences:

• (anti-Hutson intouch)-of-(midarc), (anti-Mandart-incircle)-of-(intouch), (2nd circumperp tangential)-of-(Hutson intouch), (tangential)-of-(2nd midarc).A-vertex coordinates:

Trilinears2*b*c*sin(A/2) : c*(2*sin(B/2)*a-(a+b-c)) : b*(2*sin(C/2)*a-(a+c-b))

## Thomson

The Thomson cubic K002 cuts the circumcircle of a triangle ABC in A, B, C and other three points A', B', C'. Let A' be the point on the branch of K002 passing through A and choose B' and C' similarly. A'B'C' is the

Thomson triangle.

Algebraic coordinates of vertices of Thomson triangle are very complicated to be written here.

Some other used triangles related to Thomson triangles are:Thomson-anticomplementary,Thomson-excentral,Thomson-medialandThomson-orthic. Of course, algebraic expressions for the vertices of these triangles are also very complicated.Numeric coordinates with respect to ETC's 6-9-13 triangle:

Thomson:

TrilinearsThomson-anticomplementary:{{3.87136295325922, -1.48498096032229, 2.8819453221583},

{-0.629263011432855, 2.55535829974244, 2.16199935659095},

{12.9519946390386, 12.4724370423599, -10.9719433039008}}

TrilinearsThomson-excentral:

{{8.45136867434929, 16.5127763024386, -11.6918892694805},

{17.452620603748, 8.43209778230062, -10.2519973383382},

{-9.70989469721579, -11.4020597029505, 16.0158879826651}}

TrilinearsThomson-medial:

{{5.3112525691825, 19.0782363806524, -12.0186926597053},

{20.7720702849278, 5.71159495994938, -9.90062600648572},

{-3.15806347765949, -4.47821668670003, 8.19853610085032}}

TrilinearsThomson-orthic:

{{6.16136581380771, 7.51389767105683, -4.40497197365852},

{8.41167879615739, 5.49372804102233, -4.04499899087295},

{1.62104997091644, 0.535188669709552, 2.52197233937788}}

Trilinears

{{1.67590632156687, 4.23860048384595, -0.0672464630167511},

{5.51552805619645, 1.04218899889407, 0.373520687505509},

{1.53291973629658, 0.614305674705425, 2.50787452112836}}

## inner tri-equilateral

In a triangle ABC, let's inscribe three congruent equilateral triangles PA

_{b}A_{c}, PB_{c}B_{a}and PC_{a}C_{b}, with B_{a}, C_{a}on BC, C_{b}, A_{b}on CA and A_{c}, B_{c}on AB. There are two points P making possible this construction: P = P_{i}=X(15) and P = P_{o}=X(16). The equilateral triangles obtained in each case are here named theA-, B-, C- inner/outer equilateral triangles, respectively.

For the inner-equilateral triangles, theinner tri-equilateral triangleA_{i}B_{i}C_{i}is defined as the triangle bounded by the lines A_{b}A_{c}, B_{c}B_{a}and C_{a}C_{b}. (Reference: Preamble above X10631)A-vertex coordinates:

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

## outer tri-equilateral

In a triangle ABC, let's inscribe three congruent equilateral triangles PA

_{b}A_{c}, PB_{c}B_{a}and PC_{a}C_{b}, with B_{a}, C_{a}on BC, C_{b}, A_{b}on CA and A_{c}, B_{c}on AB. There are two points P making possible this construction: P = P_{i}=X(15) and P = P_{o}=X(16). The equilateral triangles obtained in each case are here named theA-, B-, C- inner/outer equilateral triangles, respectively.

For the outer-equilateral triangles, theouter tri-equilateral triangleA_{o}B_{o}C_{o}is defined as the triangle bounded by the lines A_{b}A_{c}, B_{c}B_{a}and C_{a}C_{b}. (Reference: Preamble above X10631)A-vertex coordinates:

Trilinearsa*(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 thetri-squares trianglesof ABC.A-vertex coordinates:

1st tri-square triangle:

Trilinears2nd tri-square triangle:2*S/a : (a^2+3*b^2-c^2+2*S)/b : (a^2-b^2+3*c^2+2*S)/c

Trilinears3rd tri-square triangle:-2*S/a : (a^2+3*b^2-c^2-2*S)/b : (a^2-b^2+3*c^2-2*S)/c

Trilinears4th tri-square triangle:4*(a^2+S)/(a*(b^2+c^2-a^2+2*S)) : 1/b : 1/c

Trilinears4*(a^2-S)/(a*(b^2+c^2-a^2-2*S)) : 1/b : 1/c

## tri-squares-central

The

tri-squares-centraltriangles are the triangles whose vertices are the centers of the squares built in thetri-squares triangles. (Reference: Preamble above X13637).A-vertex coordinates:

1st tri-square-central triangle:

Trilinears2nd tri-square-central triangle: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

Trilinears3rd tri-square-central triangle: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

Trilinears4th tri-square-central triangle:(a^2+2*S)/a : (b^2+S)/b : (c^2+S)/c

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

## Trinh

There is a unique equilateral triangle AA

_{1}A_{2}inscribed in the circumcircle of triangle ABC, where, for concreteness, the labels are fixed so that AA_{1}A_{2}has the same orientation as ABC. Let BB_{1}B_{2}and CC_{1}C_{2}be the corresponding equilateral triangles. Let A' = B_{1}B_{2}∩C_{1}C_{2}, B' = C_{1}C_{2}∩A_{1}A_{2}and C' = A_{1}A_{2}∩B_{1}B_{2}. The triangle A'B'C' is named theTrinh triangleof ABC. (Reference: X7688)Equivalences:

• (anti-Hutson intouch)-of-(anti-X3-ABC reflections), (Kosnita)-of-(ABC-X3 reflections).

• Only for acute ABC: (anti-Aquila)-of-(anti-Hutson intouch).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 triangleof ABC. (Reference: Preamble just before X17603)Equivalences:

• (Ursa-minor)-of-(inner-Johnson).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 triangleof ABC. (Reference: Preamble just before X17603)Equivalences:

• (anticomplementary)-of-(intouch), (anti-Euler)-of-(Hutson intouch), (3rd anti-Euler)-of-(2nd midarc), (4th anti-Euler)-of-(midarc), (excentral)-of-(Mandart-incircle), (hexyl)-of-(2nd anti-circumperp-tangential).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.Equivalences:

• (medial)-of-(2nd half-squares).A-vertex coordinates:

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

## 2nd and 3rd inner-Vecten

Given a triangle ABC, build the square σ

_{a}= AA_{b}A_{a}A_{c}, with thesame orientationas ABC, and such that B and C lie on the lines A_{a}A_{b}and A_{a}A_{c}, respectively. Define σ_{b}= BB_{c}B_{b}B_{a}and σ_{c}= CC_{a}C_{c}C_{b}cyclically. Let A_{o}, B_{o}, C_{o}be the centers of these squares.

A_{a}B_{b}C_{c}and A_{o}B_{o}C_{o}are the2nd inner-Vectenand3rd inner-Vectenof ABC, respectively. (Reference: Preamble just before X32488).Equivalences:

• 3rd inner-Vecten = (6th anti-mixtilinear)-of-(2nd half-squares), (orthic)-of-(inner-Vecten).A-vertex coordinates:

2nd inner-Vecten triangle:

Trilinears3rd inner-Vecten:1/(a*SA) : 1/(b*(SB-S)) : 1/(c*(SC-S))

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.Equivalences:

• (excentral)-of-(3rd outer-Vecten), (medial)-of-(1st half-squares).A-vertex coordinates:

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

## 2nd and 3rd outer-Vecten

Given a triangle ABC, build the square σ

_{a}= AA_{b}A_{a}A_{c}, withopposite orientationthan ABC, and such that B and C lie on the lines A_{a}A_{b}and A_{a}A_{c}, respectively. Define σ_{b}= BB_{c}B_{b}B_{a}and σ_{c}= CC_{a}C_{c}C_{b}cyclically. Let A_{o}, B_{o}, C_{o}be the centers of these squares.

A_{a}B_{b}C_{c}and A_{o}B_{o}C_{o}are the2nd outer-Vectenand3rd outer-Vectenof ABC, respectively. (Reference: Preamble just before X32488).Equivalences:

• (6th anti-mixtilinear)-of-(1st half-squares), (orthic)-of-(outer-Vecten).A-vertex coordinates:

2nd outer-Vecten triangle:

Trilinears3rd outer-Vecten:1/(a*SA) : 1/(b*(SB+S)) : 1/(c*(SC+S))

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

## Vijay 1st to 7th

In the plane of a triangle ABC, let A'B'C' = medial triangle and A"B"C" = orthic triangle, and let E

_{a}denote the ellipse that passes through A and has foci B' and C'. Define ellipses E_{b}and E_{c}cyclically. Let A_{b}= the point, other than A, in E_{a}∩ AB, and define B_{c}and C_{a}cyclically and let A_{c}= the point, other than A, in E_{a}∩ AC, and define B_{a}and C_{b}cyclically.Triangles 1st to 7th Vijay (A

_{1}B_{1}C_{1}to A_{7}B_{7}C_{7}) are defined as follows: (Reference: Preamble just before X44733).

- A
_{1}= A_{b}C_{b}∩ A_{c}B_{c}, and define B_{1}and C_{1}cyclically;- A
_{2}= B_{a}B_{c}∩ C_{a}C_{b}, and define B_{2}and C_{2}cyclically;- A
_{3}= A_{b}B_{c}∩ A_{c}C_{b}, and define B_{3}and C_{3}cyclically;- A
_{4}= A_{b}C_{a}∩ A_{c}B_{a}, and define B_{4}and C_{4}cyclically;- A
_{5}= the point in E_{b}∩ E_{c}closer to A, and define B_{5}and C_{5}cyclically;- A
_{6}= the point in E_{b}∩ E_{c}farthest from A, and define B_{6}and C_{6}cyclically;- A
_{7}= BC ∩ A_{5}A_{6}, and define B_{7}and C_{7}cyclically.A-vertex coordinates:

1st triangle:

Trilinears2nd triangle:-(a+b+c)*(3*a^3+(b+c)*a^2-(b-c)^2*a+(b^2-c^2)*(b-c))/(2*(b+c)*a) : ((2*b+c)*a+b*c+c^2)*(a+b-c)/b : ((b+2*c)*a+b^2+b*c)*(a-b+c)/c

Trilinears3rd triangle:(a+b+c)*(3*a^3+(b+c)*a^2-(b-c)^2*a+(b^2-c^2)*(b-c))*b*c/(a^2*(-a^2+b^2+c^2)) :

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

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

Trilinears4th triangle:(a+b+c)*(3*a^3+(b+c)*a^2-(b-c)^2*a+(b^2-c^2)*(b-c))/(4*a^2*(b+c)) :

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

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

Trilinears5th triangle:-(a^7-(b+c)*a^6-(b-c)^2*a^5+(b^2-c^2)*(b-c)*a^4-((b^2-c^2)^2-4*b^2*c^2)*a^3+((b^2-c^2)^2-4*b^2*c^2)*(b+c)*a^2+

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

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

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

Trilinears6th triangle:(a+b+c)*(a^2-b^2+c^2)*(a^2+b^2-c^2)/a :

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

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

Trilinears7th triangle:-(a+b+c)*(a^2-b^2+c^2)*(a^2+b^2-c^2)/a :

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

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

Trilinears0 : (a^2+b^2-c^2)*(a+b)/b : (a^2-b^2+c^2)*(a+c)/c

## Vijay-Paasche-Hutson (without index)

See

Vijay-Paasche-polar.

## Vijay-Paasche-Hutson triangles 1 to 31

See here.

## Vijay-Paasche-midpoints

The medial triangle of the

1st Vijay-Paasche-Hutson triangleis theVijay-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 p

_{a}be the parabola with focus A and directrix BC. Let B_{a}be the point, closer to C, at which p_{a}cuts AC and let C_{a}be the point, closer to B, at which p_{a}cuts AB. Build C_{b}, A_{b}and A_{c}, B_{c}cyclically. Points A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}lie on an ellipse named thePaasche ellipse.Let A

_{P}be the pole of A with respect to the Paasche ellipse and build B_{P}, C_{P}cyclically. A_{P}B_{P}C_{P}is theVijay-Paasche-polar triangle. (Reference: X(37994))Note: In some ETC centers ETC, this triangle is refered as

Vijay-Paasche-Hutson (without index). In other centers it is namedVijay-Paasche-tangents.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

See

Vijay-Paasche-polar.

## Vu-Dao-isodynamic equilateral

Let A

_{0}B_{0}C_{0}be the orthic triangle of ABC. Let A_{15}= X(15)-of-AB_{0}C_{0}, and define B_{15}and C_{15}cyclically, so that A_{15}B_{15}C_{15}is the 3rd isodynamic-Dao equilateral triangle of ABC.Let A'

_{15}be the point, other than A, in which the line AA_{15}meets the circle {{A,B_{0},C_{0}}}, and define B'_{15}and C'_{15}cyclically. The triangle A'_{15}B'_{15}C'_{15}is theVu-Dao-X(15)-isodynamic equilateral triangle.If X(15) is replaced by X(16) in the above construction, the resulting triangle, A'

_{16}B'_{16}C'_{16}, is theVu-Dao-X(16)-isodynamic equilateral triangle.(Reference: X(5478))A-vertex coordinates:

Vu-Dao-X(15)-isodynamic equilateral triangleTrilinearsVu-Dao-X(16)-isodynamic equilateral triangle((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)/cTrilinears((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 (O

_{a}) the circle with diameter AA' and r_{a}the radical axis of (O_{a}) and (I); build (O_{b}), (O_{c}) and r_{b}, r_{c}cyclically. The triangle A*B*C* bounded by r_{a}, r_{b}, r_{c}is the Wasat triangle of ABC. (Reference: Preamble just before X21616)Equivalences:

• (anticomplementary)-of-(3rd Euler), (anti-Euler)-of-(4th Euler), (Ara)-of-(2nd Zaniah), (2nd Conway)-of-(Gemini 110), (excentral)-of-(medial), (hexyl)-of-(Euler).Note: A* is also the radical center of the incircle and the B- and C- excircles. (Nov 5, 2021)

A-vertex coordinates:

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

## X:

## X

_{3}-ABC reflectionsThis triangle is obtained by reflecting the circumcenter X

_{3}about A, B, C. Reference: TCCT 6.13, p. 160.Equivalences:

• (anticomplementary)-of-(Johnson), (Johnson)-of-(anti-Ehrmann-mid).

• Only for acute ABC: (hexyl)-of-(Ehrmann-side), (6th mixtilinear)-of-(2nd Euler).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 theX-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 I

_{AC}I_{AB}, I_{BA}I_{BC}, and I_{CA}I_{CB}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'I_{BC}I_{CB}, I_{AC}B'I_{CA}, I_{AB}I_{BA}C', and A'B'C' are congruent. The innermost triangle A'B'C' is called theYff central triangle(Kimberling 1998, pp. 94-95). (See this figure in MathWorld).A-vertex coordinates:

Trilinears2*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 triangleis the cevian triangle of X(190).A-vertex coordinates:

Trilinears0 : -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 r

_{1}=r*R/(R+r) and r_{2}=r*R/(R-r), respectively. The centers of the circles in the first triplet are the vertices of theinner-Yff triangle. (Reference: Preamble just before X10037)Equivalences:

• (Johnson)-of-(1st Johnson-Yff).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 r

_{1}=r*R/(R+r) and r_{2}=r*R/(R-r), respectively. The centers of the circles in the second triplet are the vertices of theouter-Yff triangle. (Reference: Preamble just before X10037)Equivalences:

• (Johnson)-of-(2nd Johnson-Yff).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 L

_{A}be the line, other than BC, tangent to the B- and C-inner Yff circles. Define L_{B}, L_{C}cyclically. Let A' = L_{B}∩L_{C}and define B', C' cyclically. Triangle A'B'C' is theinner-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 M

_{A}be the line, other than BC, tangent to the B- and C-outer Yff circles. Define M_{B}, M_{C}cyclically. Let A" = M_{B}∩M_{C}and define B", C" cyclically. Triangle A"B"C" is theouter-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 triangleis 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:

Trilinears2*S*(SA^2-R^2*SA-S^2)/R : c*(S^2+SA*SC) : b*(S^2+SA*SB)

## Yiu tangents

Let t

_{A}be the tangent at A to the A-Yiu circle, and define t_{B}and t_{C}cyclically. Let A' = t_{B}∩t_{C}, and define B' and C' cyclically. Triangle A'B'C' is theYiu 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 A

_{m}B_{m}C_{m}be the medial triangle, I the incenter and A' the touchpoint of the incircle with the side BC. Then the lines AA', IA_{m}and B_{m}C_{m}are concurrent at a point A_{1}. Denote B_{1}and C_{1}cyclically. The triangle A_{1}B_{1}C_{1}is the1st Zaniah triangle of ABC.Continuing with the previous construction, let J

_{a}be the A-excenter and A" the touchpoint of the A-excircle with the side BC. Then the lines AA", J_{a}A_{m}and B_{m}C_{m}are concurrent at a point A_{2}. Denote B_{2}and C_{2}cyclically. The triangle A_{2}B_{2}C_{2}is the2nd Zaniah triangle of ABC. (Reference: Preamble before X18214)Equivalences:

• 1st Zaniah = (Aries)-of-(2nd Zaniah).

• 2nd Zaniah = (anti-Ara)-of-(Wasat).A-vertex coordinates:

1st Zaniah triangle:

Trilinears2nd Zaniah triangle:2 : (a+b-c)/b : (a-b+c)/c

Trilinears2 : (a-b+c)/b : (a+b-c)/c

Last changes:

2023.09.04 - Corrected: 2nd Conway

2023.06.20 - Added: Moses-Miyamoto, Moses-Miyamoto-Apollonious

2023.01.15 - Added: Download coordinates link