The point is to foster discussion. (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. The triangle A1B1C1 is the 1st Vijay-Paasche-Hutson triangle. For many mathematicians between the 16th and 18th century, their belief was a reason for studying mathematics. Note: If the tangents at B and C to the circumcircle of a triangle ABC intersect at A', then the circle with center A' passing through B and C is called A-excosine circle of ABC. (Reference: X17807) The Conway triangle is the triangle bounded by the lines AbAc, BcBa and CaCb. (Reference: X(5478)), The Walsmith triangle has A-vertex the intersection of the A-symmedian of ABC and the perpendicular at X(125) to BC. Notes: For any ABC. (Reference), The triangle A2B2C2 formed by joining the centers of the reflections of the Neuberg circles on the opposite sides of ABC is the 2nd Neuberg triangle. 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). According to one website (condensed from E T Bell’s Men of Mathematics): Euler remained a Christian all of his life and often read to his family from the Bible. Let A' be the insimilicenter of the circumcircle and A-excircle, and define B' and C' cyclically. Let P=X(1) and Bc the point of intersection, other than S, of Lb and the circle (C,|CS|) and Cb the point of intersection, other than S, of Lc and the circle (B,|BS|). The triangle A"B"C" is equilateral and is named the 8th Fermat-Dao equilateral triangle. Given a triangle ABC, build the square σa = AAbAaAc, with the same orientation as ABC, and such that B and C lie on the lines AaAb and AaAc, respectively. (TCCT, p. 163). Of the two points on LA at distance r=inradius-of-ABC from TA, let A' be the one farther from A and let A" be the closer. (Reference: preamble just before X15015). His name is associated with angles, approximation, circles, cycle, criterion, graphs, operator’s, polynomials, pseudo primes, … and Euler’s identity: said to be the most beautiful formula in the world. his wife, he had 13 children of which 5 survived. Note: The anti-Mandart-incircle triangle is (the tangential triangle)-of-(the 1st circumperp triangle). The Pelletier triangle is the vertex triangle of ABC and the intangets triangle. (Reference: X1118), Let A0B0C0 be the 1st Hatzipolakis triangle of ABC. The triangle A'B'C' whose Atik triangle is ABC. Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of triangle ABC (inwards). The circumcevian triangle of the centroid of ABC is its circummedial triangle. Notes: For any ABC. See Lucas triangles. Define A5 = D2E2∩D3E3, and define B5 and C5 cyclically. A female clinical psychologist who is skeptical about religion … Press J to jump to the feed. (Reference). (Reference: Preamble just before X36761.) ( Log Out /  The triangle A'B'C' whose 5th Brocard triangle is ABC. Euler's inequality, in the form stating that, for all triangles inscribed in a given circle, the maximum of the radius of the inscribed circle is reached for the equilateral triangle and only for it, is valid in absolute geometry. A*B*C* is the midheight triangle of ABC. (Reference: X7606) Build (Bc, Ba) and (Ca, Cb) cyclically and let Am, Bm, Cm be the midpoints of BcCb, CaAc, AbBa, respectively. Le cercle des neuf points d'Euler est homothétique du cercle circonscrit au triangle dans deux homothéties : Notons I1 le milieu de [BC], I2 le milieu de [AC] et I3 le milieu de [AB]. Puzzle / Euler’s Triangle. A-vertex coordinates: The triangle whose vertices are the excenters of ABC. Note: The 4th anti-Euler triangle is (the anticomplementary triangle)-of-(the circumorthic triangle). (Reference), Let A", B", C" the feet of the external angles bisectors. Define B', B", C', C" cyclically. The mixtilinear triangle is the triangle connecting the centers of the mixtilinear incircles. Let TI1=A1B1C1 be the triangle obtained by rotating ABC about the 1st Ehrmann pivot, P(5), by an angle of -2π/3, so that TI1 is inscribed in ABC. A-vertex coordinates: In particular, for P=X(1) and P=X(3065), the point of concurrence is the Schiffler center S=X(21). Diderot was interested, and, according to De Morgan, Euler advanced toward Diderot, and said gravely, and in a tone of perfect conviction: “Sir, (a + bn) / n = x , hence God exists; reply! This triangle is not equilateral.  1st half-squares triangle: Let A'B'C' be the intouch triangle of ABC. Dan Graves in his Scientists of Faith has this to say of Euler: Euler retained his firm Calvinist beliefs throughout life, holding daily prayer and worship in his home and sometimes preaching. The anti-Wasat triangle of ABC is (the 3rd anti-Euler triangle)-of-(the medial triangle). Note: Neither of both Ehrmann inscribed triangles is a central triangle. Let A2 be the intersection, other than X(110), of the Parry circle and line AX(110), and define B2 and C2 cyclically. (compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL) See here for a comparison of all solutions. (See note in mixtilinear triangle). Then A*B*C* is the hexyl triangle of ABC. The triangle A'B'C' is equilateral and is named here the 9th Dao-Fermat equilateral triangle. 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*. From MathWorld). (Reference: X7411) Guest articles are sought after for the purpose of bringing more diverse viewpoints to the topics of mathematics and theology. Next rotate cb around C an angle Pi/3 (clockwise) and denote c'b the resulting line. Let Ab, Ac be the points where the A-excosine circle cuts again the sidelines AC and AB, respectively, and build {Bc, Ba}, {Ca, Cb} cyclically. (Reference), The triangles whose vertices are the reflections of the vertices of the 1st Brocard triangle in the sidelines of ABC is the 1st Brocard-reflected triangle of ABC. A-vertex coordinates: There are three mixtilinear incircles, one corresponding to each angle of the triangle (Reference). Extend OI so that it intersects the circumcircle at P and Q; then PI × QI = AI × IL = 2Rr, so (R + d)(R − d) = 2Rr, i.e.