# Morley's trisector theorem

Morley's trisector theorem If each vertex angle of the outer triangle is trisected, Morley's trisector theorem states that the purple triangle will be equilateral.

In plane geometry, Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the first Morley triangle or simply the Morley triangle. The theorem was discovered in 1899 by Anglo-American mathematician Frank Morley. It has various generalizations; in particolare, if all of the trisectors are intersected, one obtains four other equilateral triangles.

Contenuti 1 Prove 2 Side and area 3 Morley's triangles 4 Related triangle centers 5 Guarda anche 6 Appunti 7 Riferimenti 8 External links Proofs There are many proofs of Morley's theorem, some of which are very technical.[1] Several early proofs were based on delicate trigonometric calculations. Recent proofs include an algebraic proof by Alain Connes (1998, 2004) extending the theorem to general fields other than characteristic three, and John Conway's elementary geometry proof.[2][3] The latter starts with an equilateral triangle and shows that a triangle may be built around it which will be similar to any selected triangle. Morley's theorem does not hold in spherical[4] and hyperbolic geometry.

Fig 1. Elementary proof of Morley's trisector theorem One proof uses the trigonometric identity {displaystyle peccato(3teta )=4sin theta sin(60^{circ }+teta )peccato(120^{circ }+teta )} (1) quale, by using of the sum of two angles identity, can be shown to be equal to {displaystyle peccato(3teta )=-4sin ^{3}theta +3sin theta .} The last equation can be verified by applying the sum of two angles identity to the left side twice and eliminating the cosine.

Points {stile di visualizzazione D,e,F} are constructed on {stile di visualizzazione {sopra {AVANTI CRISTO}}} as shown. abbiamo {displaystyle 3alpha +3beta +3gamma =180^{circ }} , the sum of any triangle's angles, Così {displaystyle alpha +beta +gamma =60^{circ }.} Perciò, the angles of triangle {displaystyle XEF} sono {displaystyle alfa ,(60^{circ }+beta ),} e {stile di visualizzazione (60^{circ }+gamma ).} From the figure {displaystyle peccato(60^{circ }+beta )={frac {sopra {DX}}{sopra {XE}}}} (2) e {displaystyle peccato(60^{circ }+gamma )={frac {sopra {DX}}{sopra {XF}}}.} (3) Also from the figure {displaystyle angle {AYC}=180^{circ }-alpha -gamma =120^{circ }+beta } e {displaystyle angle {AZB}=120^{circ }+gamma .} (4) The law of sines applied to triangles {displaystyle AYC} e {displaystyle AZB} yields {displaystyle peccato(120^{circ }+beta )={frac {sopra {corrente alternata}}{sopra {AY}}}sin gamma } (5) e {displaystyle peccato(120^{circ }+gamma )={frac {sopra {AB}}{sopra {AZ}}}sin beta .} (6) Express the height of triangle {stile di visualizzazione ABC} in two ways {displaystyle h={sopra {AB}}peccato(3beta )={sopra {AB}}cdot 4sin beta sin(60^{circ }+beta )peccato(120^{circ }+beta )} e {displaystyle h={sopra {corrente alternata}}peccato(3gamma )={sopra {corrente alternata}}cdot 4sin gamma sin(60^{circ }+gamma )peccato(120^{circ }+gamma ).} where equation (1) was used to replace {displaystyle peccato(3beta )} e {displaystyle peccato(3gamma )} in these two equations. Substituting equations (2) e (5) nel {displaystyle beta } equation and equations (3) e (6) nel {gamma di stili di visualizzazione } equation gives {displaystyle h=4{sopra {AB}}sin beta cdot {frac {sopra {DX}}{sopra {XE}}}cdot {frac {sopra {corrente alternata}}{sopra {AY}}}sin gamma } e {displaystyle h=4{sopra {corrente alternata}}sin gamma cdot {frac {sopra {DX}}{sopra {XF}}}cdot {frac {sopra {AB}}{sopra {AZ}}}sin beta } Since the numerators are equal {stile di visualizzazione {sopra {XE}}cdot {sopra {AY}}={sopra {XF}}cdot {sopra {AZ}}} o {stile di visualizzazione {frac {sopra {XE}}{sopra {XF}}}={frac {sopra {AZ}}{sopra {AY}}}.} Since angle {displaystyle EXF} and angle {displaystyle ZAY} are equal and the sides forming these angles are in the same ratio, triangles {displaystyle XEF} e {displaystyle AZY} are similar.

Similar angles {displaystyle AYZ} e {displaystyle XFE} equal {stile di visualizzazione (60^{circ }+gamma )} , and similar angles {displaystyle AZY} e {displaystyle XEF} equal {stile di visualizzazione (60^{circ }+beta ).} Similar arguments yield the base angles of triangles {displaystyle BXZ} e {displaystyle CYX.} In particular angle {displaystyle BZX} is found to be {stile di visualizzazione (60^{circ }+alfa )} and from the figure we see that {displaystyle angle {AZY}+angle {AZB}+angle {BZX}+angle {XZY}=360^{circ }.} Substituting yields {stile di visualizzazione (60^{circ }+beta )+(120^{circ }+gamma )+(60^{circ }+alfa )+angle {XZY}=360^{circ }} where equation (4) was used for angle {displaystyle AZB} e quindi {displaystyle angle {XZY}=60^{circ }.} Similarly the other angles of triangle {displaystyle XYZ} are found to be {displaystyle 60^{circ }.} Side and area The first Morley triangle has side lengths[5] {stile di visualizzazione a^{primo }=b^{primo }=c^{primo }=8Rsin(A/3)peccato(B/3)peccato(C/3),,} where R is the circumradius of the original triangle and A, B, and C are the angles of the original triangle. Since the area of an equilateral triangle is {stile di visualizzazione {tfrac {mq {3}}{4}}a'^{2},} the area of Morley's triangle can be expressed as {stile di visualizzazione {testo{La zona}}=16{mq {3}}R^{2}peccato ^{2}(A/3)peccato ^{2}(B/3)peccato ^{2}(C/3).} Morley's triangles Morley's theorem entails 18 equilateral triangles. The triangle described in the trisector theorem above, called the first Morley triangle, has vertices given in trilinear coordinates relative to a triangle ABC as follows: A-vertex = 1 : 2 cos(C/3) : 2 cos(B/3) B-vertex = 2 cos(C/3) : 1 : 2 cos(A/3) C-vertex = 2 cos(B/3) : 2 cos(A/3) : 1 Another of Morley's equilateral triangles that is also a central triangle is called the second Morley triangle and is given by these vertices: A-vertex = 1 : 2 cos(C/3 − 2π/3) : 2 cos(B/3 − 2π/3) B-vertex = 2 cos(C/3 − 2π/3) : 1 : 2 cos(A/3 − 2π/3) C-vertex = 2 cos(B/3 − 2π/3) : 2 cos(A/3 − 2π/3) : 1 The third of Morley's 18 equilateral triangles that is also a central triangle is called the third Morley triangle and is given by these vertices: A-vertex = 1 : 2 cos(C/3 − 4π/3) : 2 cos(B/3 − 4π/3) B-vertex = 2 cos(C/3 − 4π/3) : 1 : 2 cos(A/3 − 4π/3) C-vertex = 2 cos(B/3 − 4π/3) : 2 cos(A/3 − 4π/3) : 1 The first, secondo, and third Morley triangles are pairwise homothetic. Another homothetic triangle is formed by the three points X on the circumcircle of triangle ABC at which the line XX −1 is tangent to the circumcircle, where X −1 denotes the isogonal conjugate of X. This equilateral triangle, called the circumtangential triangle, has these vertices: A-vertex = csc(C/3 − B/3) : csc(B/3 + 2C/3) : −csc(C/3 + 2B/3) B-vertex = −csc(A/3 + 2C/3) : csc(A/3 − C/3) : csc(C/3 + 2A/3) C-vertex = csc(A/3 + 2B/3) : −csc(B/3 + 2A/3) : csc(B/3 − A/3) A fifth equilateral triangle, also homothetic to the others, is obtained by rotating the circumtangential triangle π/6 about its center. Called the circumnormal triangle, its vertices are as follows: A-vertex = sec(C/3 − B/3) : −sec(B/3 + 2C/3) : −sec(C/3 + 2B/3) B-vertex = −sec(A/3 + 2C/3) : sec(A/3 − C/3) : −sec(C/3 + 2A/3) C-vertex = −sec(A/3 + 2B/3) : −sec(B/3 + 2A/3) : sec(B/3 − A/3) An operation called "extraversion" can be used to obtain one of the 18 Morley triangles from another. Each triangle can be extraverted in three different ways; il 18 Morley triangles and 27 extravert pairs of triangles form the 18 vertices and 27 edges of the Pappus graph.[6] Related triangle centers The centroid of the first Morley triangle is given in trilinear coordinates by Morley center = X(356) = cos(A/3) + 2 cos(B/3)cos(C/3) : cos(B/3) + 2 cos(C/3)cos(A/3) : cos(C/3) + 2 cos(A/3)cos(B/3).

The first Morley triangle is perspective to triangle ABC:[7] the lines each connecting a vertex of the original triangle with the opposite vertex of the Morley triangle concur at the point 1st Morley–Taylor–Marr center = X(357) = sec(A/3) : sec(B/3) : sec(C/3). See also Angle trisection Hofstadter points Morley centers Notes ^ Bogomolny, Alessandro, Morley's Miracle, Cut-the-knot, recuperato 2010-01-02 ^ Bogomolny, Alessandro, J. Conway's proof, Cut-the-knot, recuperato 2021-12-03 ^ Conway, John (2006), "The Power of Mathematics", in Blackwell, Alan; Mackay, Davide (eds.), Power (PDF), Cambridge University Press, pp. 36–50, ISBN 978-0-521-82377-7, recuperato 2010-10-08 ^ Morley's Theorem in Spherical Geometry, Java applet. ^ Weisstein, Eric W. "First Morley Triangle". Math World. Recuperato 2021-12-03. ^ Guy (2007). ^ Fox, M. D.; and Goggins, J. R. "Morley's diagram generalised", Gazzetta matematica 87, novembre 2003, 453–467. References Connes, Alain (1998), "A new proof of Morley's theorem", Publications Mathématiques de l'IHÉS, S88: 43–46. Connes, Alain (Dicembre 2004), "Symmetries" (PDF), European Mathematical Society Newsletter, 54. Coxeter, H. S. M.; Greitzer, S. l. (1967), Geometria rivisitata, L'Associazione Matematica d'America, LCCN 67-20607 Francis, Richard L. (2002), "Modern Mathematical Milestones: Morley's Mystery" (PDF), Missouri Journal of Mathematical Sciences, 14 (1), doi:10.35834/2002/1401016. Tipo, Richard K. (2007), "The lighthouse theorem, Morley & Malfatti—a budget of paradoxes" (PDF), Mensile matematico americano, 114 (2): 97–141, doi:10.1080/00029890.2007.11920398, JSTOR 27642143, SIG 2290364, archiviato dall'originale (PDF) Su 2010-04-01. Oakley, C. O.; Baker, J. C. (1978), "The Morley trisector theorem", Mensile matematico americano, 85 (9): 737–745, doi:10.2307/2321680, JSTOR 2321680. Taylor, F. Glanville; Marr, w. l. (1913–14), "The six trisectors of each of the angles of a triangle", Proceedings of the Edinburgh Mathematical Society, 33: 119–131, doi:10.1017/S0013091500035100. External links Morleys Theorem at MathWorld Morley's Trisection Theorem at MathPages Morley's Theorem by Oleksandr Pavlyk, The Wolfram Demonstrations Project. Categorie: Theorems about triangles

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