On some extensions of Morley's trisector theorem
OON SOME EXTENSIONS OF MORLEY’S TRISECTORTHEOREM
NIKOS DERGIADES AND TRAN QUANG HUNG
Abstract.
We establish a simple generalization for the famous theorem ofMorley about trisectors in triangle with a purely synthetic proof using onlyangle chasing and similar triangles. Furthermore, based on the converse con-struction, another simple extension of Morley’s Theorem is created and proven. Introduction
Over one hundred years ago in 1899, Frank Morley introduced a geometric re-sult. This result was so classic that Alexander Bogomolny once said "it enteredmathematical folklore"; see [1]. Morley’s marvelous theorem states as follows:
Theorem 1 (Morley, 1899) . The three points of intersection of the adjacent tri-sectors of the angles of any triangle form an equilateral triangle.
Figure 1.
Morley’s marvelous theoremMany mathematicians consider Morley’s Theorem to be one of the most beautifultheorems in plane Euclidean geometry. Throughout history, numerous proofs havebeen proposed; see [1, 2, 3, 5, 7, 9, 10]. There was a generalization of Morley’sTheorem using projective geometry in [4]. Some extensions to this theorem has
Mathematics Subject Classification.
Key words and phrases.
Morley’s trisector theorem, Morley’s triangle, equilateral triangles,perspective triangles. a r X i v : . [ m a t h . HO ] M a y NIKOS DERGIADES AND HUNG QUANG TRAN been recently analyzed by Richard Kenneth Guy in [6]. Guy’s extensions were veryextensive and deep research on Morley’s Theorem.In the main part of this paper, we would like to offer and prove synthetically asimple generalization of Morley’s Theorem. Generalized theorem is introduced asfollows:
Theorem 2 (Generalized theorem) . Let
ABC be a triangle. Assume that threepoints X , Y , Z , and the intersection D = BZ ∩ CY , E = CX ∩ AZ , F = AY ∩ BX lie inside triangle ABC , they also satisfy the following conditions • ∠ BXC = 120 ◦ + ∠ Y AZ , ∠ CY A = 120 ◦ + ∠ ZBX , and ∠ AZB = 120 ◦ + ∠ XCY . • The points X , Y , and Z lie on the bisectors of angles ∠ BDC , ∠ CEA , and ∠ AF B , respectively.Then triangle
XY Z is an equilateral triangle.
When X , Y , and Z are the intersections of the adjacent trisectors of triangle ABC , it is easily seen that they satisfy two conditions of Theorem 2. Thus Theorem1 is a direct consequence of Theorem 2.An important property of the pair of triangles
ABC and
XY Z is introduced inthe following theorem:
Theorem 3.
The triangles
ABC and
XY Z of the Theorem 2 are perspective.
At the last section of this paper, we shall apply a converse construction to findanother extension for Morley’s Theorem. Some new equilateral triangles in a givenarbitrary triangle are also found. The family of these new equilateral trianglesclosely relate to the construction of the Morley’s equilateral triangle.2.
Proofs of the theorems
The solutions for Theorem 2 and Theorem 3 will be showed in this section.
Proof of Theorem 2.
The main idea of this proof comes from [2]. (See Figure 2).Set ∠ Y AZ = x , ∠ ZBX = y , and ∠ XCY = z . Since X lies inside triangle DBC (because X lies inside triangle ABC and it lies on bisector of angle ∠ BDC , too),we have ∠ BDC = ∠ BXC − y − z = 120 ◦ + x − y − z. Similarly, ∠ CEA = 120 ◦ + y − z − x and ∠ AF B = 120 ◦ + z − x − y .On the sides of an equilateral triangle X (cid:48) Y (cid:48) Z (cid:48) , the isosceles triangles D (cid:48) Z (cid:48) Y (cid:48) , E (cid:48) X (cid:48) Z (cid:48) , and F (cid:48) Y (cid:48) X (cid:48) are constructed outwardly such that ∠ Y (cid:48) D (cid:48) Z (cid:48) = ∠ Y DZ , ∠ Z (cid:48) E (cid:48) X (cid:48) = ∠ ZEX , and ∠ X (cid:48) F (cid:48) Y (cid:48) = ∠ XF Y .Take the intersections A (cid:48) = E (cid:48) Z (cid:48) ∩ F (cid:48) Y (cid:48) , B (cid:48) = F (cid:48) X (cid:48) ∩ DZ (cid:48) , C (cid:48) = D (cid:48) Y (cid:48) ∩ E (cid:48) X (cid:48) .From quadrilateral A (cid:48) E (cid:48) X (cid:48) F (cid:48) , it is deduced that ∠ A (cid:48) = 360 ◦ − ∠ Z (cid:48) E (cid:48) X (cid:48) − (cid:20)(cid:18) ◦ − ∠ Z (cid:48) E (cid:48) X (cid:48) (cid:19) + 60 ◦ + (cid:18) ◦ − ∠ X (cid:48) F (cid:48) Y (cid:48) (cid:19)(cid:21) − ∠ X (cid:48) F (cid:48) Y (cid:48) , this implies that ∠ A (cid:48) = 120 ◦ − ∠ Z (cid:48) E (cid:48) X (cid:48) + ∠ X (cid:48) F (cid:48) Y (cid:48) ◦ − ◦ − x x. N SOME EXTENSIONS OF MORLEY’S TRISECTOR THEOREM 3
11 1 xy zXZ YDE FAB C Z' Y'X'D'E' F'A'B' C'
Figure 2.
Proof of generalized theoremAn analogous argument shows that ∠ B (cid:48) = y and ∠ C (cid:48) = z. Since D (cid:48) X (cid:48) is bisector of ∠ B (cid:48) D (cid:48) C (cid:48) (from the constructions of isosceles triangle D (cid:48) Z (cid:48) Y (cid:48) and equilateral triangle X (cid:48) Y (cid:48) Z (cid:48) ), (cid:52) DBX ∼ (cid:52) D (cid:48) B (cid:48) X (cid:48) (because they havesame angles y , ∠ BDC ), and (cid:52) DCX ∼ (cid:52) D (cid:48) C (cid:48) X (cid:48) (because they have same angles z , ∠ BDC ), we obtain XBX (cid:48) B (cid:48) = DXD (cid:48) X (cid:48) = XCX (cid:48) C (cid:48) or XBXC = X (cid:48) B (cid:48) X (cid:48) C (cid:48) , and also ∠ B (cid:48) X (cid:48) C (cid:48) = ∠ B (cid:48) D (cid:48) C (cid:48) + ∠ B (cid:48) + ∠ C (cid:48) = ∠ BXC.
Two previous conditions point out that (cid:52)
XBC ∼ (cid:52) X (cid:48) B (cid:48) C (cid:48) (s.a.s).Analogously, (cid:52) Y CA ∼ (cid:52) Y (cid:48) C (cid:48) A (cid:48) , and (cid:52) ZAB ∼ (cid:52) Z (cid:48) A (cid:48) B (cid:48) .Finally, from these similar triangles, it follows that ∠ BAC = ∠ B (cid:48) A (cid:48) C (cid:48) , ∠ CBA = ∠ C (cid:48) B (cid:48) A (cid:48) , and ∠ ACB = ∠ A (cid:48) C (cid:48) B (cid:48) , so (cid:52) ABC ∼ (cid:52) A (cid:48) B (cid:48) C (cid:48) .This takes us to the conclusion that (cid:52) XY Z ∼ (cid:52) X (cid:48) Y (cid:48) Z (cid:48) , it might be worthpointing out that XY Z is equilateral, and completes the proof of generalized the-orem. (cid:3)
The above proof of generalized theorem also shows that Morley’s Theorem canbe proven simply using similar triangles and angle chasing in the same way.
NIKOS DERGIADES AND HUNG QUANG TRAN
The barycentric coordinates will be used in the next proof for Theorem 3, see[8].
XZ YDE FAB CP
Figure 3.
Perspective triangles
Proof of theorem 3.
Without loss of generality, assume that the sidelengths of theequilateral triangle
XY Z is . Therefore, in barycentric coordinates X = (1 : 0 : 0) , Y = (0 : 1 : 0) , Z = (0 : 0 : 1) . Because D , E , and F lie on perpendicular bisector of sides BC , CA , and AB ,respectively, assume that coordinates of D , E , and F as follows: D = ( − p : 1 : 1) , E = (1 : − q : 1) , F = (1 : 1 : − r ) . Now using the equation of lines [8], we find that A = ( − q : r ) , B = ( p : − r ) , C = ( p : q : − . Obviously, the lines AX , BY , and CZ concur at the point P ( p : q : r ) . Thisfinishes the proof. (cid:3) Converse construction
In this section, some newly discovered equilateral triangles based on a givenarbitrary triangle are found.Now coming back to Theorem 2, even though it is really a generalization ofTheorem 1 and has been proven, we only see one possible case that is Morley’sTheorem. That will be make less sense if we only see one application of generalizedtheorem which is Theorem 1. In order to exclude the objection that it can not finda triangle
XY Z , satisfying the conditions of the generalized theorem except onlythe case of Morley’s triangle, we now show a converse construction with giving apurely synthetic proof.
N SOME EXTENSIONS OF MORLEY’S TRISECTOR THEOREM 5
Theorem 4 (Converse construction) . Arbitrary isosceles triangles
DY Z , EZX ,and
F XY are constructed outwardly of an equilateral triangle
XY Z with bases thesides of
XY Z , such that the pairs of lines ( EZ, F Y ) , ( F X, DZ ) , and ( DY, EX ) meet at A , B , and C in the same place with D , E , and F , respectively, relative tothe sides of XY Z . Then ∠ BXC = 120 ◦ + ∠ ZAY, ∠ CY A = 120 ◦ + ∠ XBZ, ∠ AZB = 120 ◦ + ∠ Y CX. x xy yz zZ YXD FE AB C
Figure 4.
Proof of converse construction
Proof.
Since X lies on the bisector of ∠ D , and the sides XY , XZ of the equilateraltriangle XY Z are equally inclined to the sides DB , DC and so we have the equalityof angles designated as x . Analogously, we have the equality of angles designated y and z . So ∠ BXC = 360 ◦ − y − ◦ − z = 120 ◦ + (180 ◦ − y − z ) = 120 ◦ + ∠ ZAY.
Similarly, ∠ CY A = 120 ◦ + ∠ XBZ , ∠ AZB = 120 ◦ + ∠ Y CX , this would finishthe proof. (cid:3)
On the configuration of Theorem 4, let P , Q , and R be the circumcenters oftriangles AY Z , BZX , and
CXY , respectively. We use angle chasing ∠ CXR = ∠ BXQ = 90 ◦ − x, so ∠ BXR = ∠ QXC = 360 ◦ − y − z − ◦ + 90 ◦ − x = 390 ◦ − x − y − z. This means that there are six equal angles ∠ BXR = ∠ CXQ = ∠ CY P = ∠ AY R = ∠ AZQ = ∠ BZP.
At this point, using above conditions of angles as hypothesis, we propose anotherextension of Morley’s Theorem as follows:
NIKOS DERGIADES AND HUNG QUANG TRAN
Z YXD FE AB CP RQ
Figure 5.
On the configuration of Theorem 4
Theorem 5 (Extension of Morley’s Theorem) . Locate the points X , Y , and Z lying inside a given triangle ABC such that ∠ BXR = ∠ CXQ = ∠ CY P = ∠ AY R = ∠ AZQ = ∠ BZP, where P , Q , and R are circumcenters of triangles AY Z , BZX , and
CXY , re-spectively, and lying inside that respective triangles. Then triangle
XY Z is anequilateral triangle. x xz zy yZ YXAB CP RQ
Figure 6.
Extension of Morley’s Theorem
Proof.
From the hypothesis we conclude that ∠ CXR = ∠ QXB , this leads to ◦ − ∠ XY C = 90 ◦ − ∠ BZX,
N SOME EXTENSIONS OF MORLEY’S TRISECTOR THEOREM 7 so ∠ XY C = ∠ BZX = x. Similarly, we conclude the designation of angles y and angles z . From these ∠ BXR = ∠ CXX = 360 ◦ − y − z − ∠ X + 90 ◦ − x = 450 ◦ − x − y − z − ∠ X. Hence, we get the same equalities ∠ CY P = ∠ AY R = 450 ◦ − x − y − z − ∠ Y, and ∠ AZQ = ∠ BZP = 450 ◦ − x − y − z − ∠ Z. Thus from six equal angles of the hypothesis, it is easy to show that ∠ X = ∠ Y = ∠ Z. Therefore, the triangle
XY Z is equilateral. The theorem is proven. (cid:3)
Note that Theorem 5 will become Morley’s Theorem if adding more the condi-tions ∠ BXR = ∠ CXQ = ∠ CY P = ∠ AY R = ∠ AZQ = ∠ BZP = 150 ◦ . Finally, we conclude the article with an interesting consequence of Theorem 5where all six equal angles (in Theorem 5) are ◦ (see Figure 7). Z YXDE CFAB
Figure 7.
Consequence of Theorem 5
Theorem 6 (Consequence of Theorem 5) . Select three points X , Y , and Z lyinginside a given triangle ABC and satisfying the following conditions
NIKOS DERGIADES AND HUNG QUANG TRAN • BZ and CY meet at circumcenter of triangle AY Z . • CX and AZ meet at circumcenter of triangle BZX . • AY and BX meet at circumcenter of triangle CXY .Then triangle
XY Z is an equilateral triangle.
Acknowledgment.
The authors would like to express their sincere gratitude anddevote the most respect to two deceased mathematicians Alexander Bogomolny andRichard Kenneth Guy who devoted their love and appreciation to the recreationalmathematics, and they have also made a great contribution to the developmentprocess and introducing the famous theorem of Morley.
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