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Morley's theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle.

In a talk some years ago, David Rusin made the provocative claim that Morley's theorem is a rare example of a striking theorem that defies generalization. The first ideas that come to everyone's mind—passing to higher dimensions or hyperbolic geometry for example—don't work.

The proof by Alain Connes yields a mild generalization of sorts, but not a very satisfying one in my opinion. Wikipedia claims that there are "various generalizations" of Morley's theorem, but by this it seems to mean extensions of Morley's theorem, i.e., further equilateral triangles that one can construct. This is not what I would, strictly speaking, call a "generalization."

So is David Rusin correct?

Are there no satisfactory generalizations of Morley's theorem?

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  • $\begingroup$ Perhaps Morley's Theorem is the limit of generalization. Are there interesting specializations of Morley's Theorem in plane geometry? Gerhard "Ask Me About System Design" Paseman, 2012.04.11 $\endgroup$ Apr 12 '12 at 6:10
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    $\begingroup$ I would guess that there's no reason to expect what you might call "poetry preserving" generalizations - ones that involve pretty analogues of trisections and equilateral triangles. But perhaps by first recasting Morley's theorem as some (big ugly) algebraic identity, one could then see the identity as a specialization of one even more complex. $\endgroup$ Apr 12 '12 at 7:35
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Please forgive me if you are aware of this result (as it is linked from the Wikipedia page, albeit in another context), but there is a paper by Richard K. Guy called "The lighthouse theorem, Morley & Malfatti—a budget of paradoxes" in the American Mathematical Monthly. The eponymous theorem could be considered a generalization of Morley's theorem:

Lighthouse Theorem. Two sets of $n$ lines at equal angular distances, one set through each of the points $B$, $C$, intersect in $n^2$ points that are the vertices of $n$ regular $n$-gons.

Naturally, it is not clear how this would qualify as a generalization, but the connecting observation is the following:

The Morley Miracle. The nine edges of the equilateral triangles of the Lighthouse Theorem for $n=3$ are the Morley lines of a triangle.

Properly, the Lighthouse Theorem should be enlarged to include enough observations to make this connection. For example, the $n^2$ lines of the $n$ regular $n$-gons form $n$ families of $\binom{n}{2}$ parallel lines; if $n$ is odd, then the $n$-gons are homothetic. Moreover, there is an angle duplication result that establishes the presence of the trisectors.

From Guy's point of view, the particularly pleasant appearance of Morley's theorem is due to the fact that $\binom{n}{2} = n$ for $n=3$. For comparison, the case $n=2$ is even simpler and may be regarded as the statement that the altitudes of a triangle concur. (The $n$ $n$-gons are an orthocentric system.) The case $n=4$ gives some properties of Malfatti circles. For all of these interpretations, Guy wrestles with the "paradox" that you recover theorems about a triangle even though you don't start with any triangles.

Again, my apologies if you're aware of all of this. I imagine you may be, in which case I justify my answer as simply too long for a comment!

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    $\begingroup$ Thanks for this...it is very nice, and I was not aware of it! $\endgroup$ Apr 22 '12 at 22:29
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The generalization I was hoping for would start with: "Given any simplex in R^n, ..."; the case n=2 of this theorem would then be Morley's theorem.

I recall starting with a random tetrahedron in R^3 and trying a bunch of constructions looking for something regular to appear: I believe the variations I tried included trisecting and quadrisecting the dihedral angles, and drawing a few sets of regularly-spaced rays out of each vertex. Any three planes, any ray-plane pair, and occasional pairs of rays provide points of intersection, but I don't recall finding even any isosceles triangles among those points of intersection. Perhaps I miscalculated (or am mis-remembering)?

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Morley originally found this theorem as a trivial case of much more complicated theorems. Anyone who says this theorem defies generalization is really just saying that they are unaware of its history.

See Oakley and Baker's 1978 paper for extensive discussion of Morley's theorem and over 100 references.

See also this question and answer, somewhat similar in spirit to the present question.

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    $\begingroup$ What were these more complicated theorems? $\endgroup$ Jun 29 '12 at 6:04
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Please see here is our work about extension of Morley's theorem

On some extensions of Morley's trisector theorem

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  • $\begingroup$ I look forward your comments and your interest. Thank you! $\endgroup$ May 26 '20 at 0:38
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    $\begingroup$ Very nice extensions. I work 5 years but can not give a nice generalization of the Morley theorem $\endgroup$ May 30 '20 at 3:14
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    $\begingroup$ Your generalization also is a generalization of Napoleon theorem $\endgroup$ May 30 '20 at 3:32
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    $\begingroup$ Thank you very much Mr Dao Thanh Oai for your interest. $\endgroup$ Jun 29 '20 at 15:38
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    $\begingroup$ Dear Mr Dao, we can't trisect an angle with ruler and compass so in general, we can't construct Morley triangle by ruler and compass, thus, general theorem is the same. But converse construction is possible. $\endgroup$ Jul 1 '20 at 0:57
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A recent "generalization":

Tran, Q. H. "Morley’s trisector Theorem for isosceles tetrahedron." Acta Mathematica Hungarica (2021): 1-8. DOI.

Abstract. We extend Morley’s trisector theorem in the plane to an isosceles tetrahedron in three-dimensional space. We will show that the Morley tetrahedron of an isosceles tetrahedron is also isosceles tetrahedron. Furthermore, by the formula for distance in barycentric coordinate, we introduce and prove a general theorem on an isosceles tetrahedron.

A "Morley tetrahedron" is determined by planes trisecting each dihedral angle.

Tran, Q. H. is likely MO user TranQuangHung.

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  • $\begingroup$ Thanks to @MartinSleziak for sending me the paper (which is behind a paywall). $\endgroup$ Oct 31 at 12:43
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Hello friends here is my Generalisation of Napoleon Configuration and Morley Configuration

  1. Morey Generalisation

  2. Pure Generalisation of Napoleon Configuration


Sorry Sir Glorfindel, as I was new on this channel so I am not much experience about how to use it. I will re write the generalisation here and important line that will be helpful.


Napoleon configuration generalisation statement : Let there be six arbitrary point A1,A2,....,A6 and let the apex of Equilateral triangle made on base {A1,A2},....{A5,A6} be {B1,B2,....,B6} then G(B1,B2,B3) ; G(B3,B4,B5);G(B5,B6,B1) makes equilateral triangle and when the points A3 and A4 will coincide and cyclically, point (A5,A6) and (A1,A2) coincides then we will get Napoleon configuration . So we can say that Napoleon Configuration is very special case of this configuration & This will valid also when all the six equilateral triangle are drawn inwardly .(You can see more regarding in above link ).


Angle Trisector theorem or Generalisation of Morley configuration: Link

Description___________ THE ANGLE TRISECTOR and Bisector have a very long history . The angle bisector theorem appears as Proposition 03 of Book VI of Euclid elements . Before talking about Angle trisection ,we first look for Angle Bisector (See Figure 01). Take two line segment A1A3 , A2A4 intersect at O . Let the other two lines B1B3 & B2B4 be it’s angle bisector such that ∠A1OB2=∠B2OA2& ∠A2OB3= ∠B3OA3 and when { A1,A2,.....,B1,B2,.....,B4} lies on circle then { A1A2A3A4} and {B1B2B3B4} makes Rectangle . This result may not be new and can be derived but we explain this because we are doing similar thing on Angle trisector. Figure 1.0 Now Angle Trisector is a classical problem and both famous Mathematician Euclid and Archimedes have try to work on them . The Angle Trisection is used in Morley & it is one of the mysterious topic . Here is the description of our observations (see figure 2.0) . Figure 2.0 Let ( A1A3), (A2A4), (B1B5), (B2B6) , (B3B7) ;(B4B8) be line segment such that {A1,A2,A3,A4,B1,.....,B8) lies on the circle such that ∠A1OB1=x ∠B1OB5=x ∠B2OA2=x ∠A4OB6=x ∠B6OB5=x ∠B5OB3=x ∠A2OB3=60-x ∠B3OB4=60-x ∠B4OA3=60-x ∠A4OB7=60-x ∠B7OB8=60-x ∠B8OB1=60-x Or in simple words we can say that line B1B5,B2B6,B3B7 , B4B8 acts as a angle Trisector of line A1A3& A2A4 which intersect at O. Then following Results always holds true which is independent of the variable X. [1]m(B5,O) ; m( B4,B6) ; (B3,B6) makes equilateral triangle [2] m(B3,B6); m( B1,B4);m( B3,B4) makes equilateral triangle. [3] m(B4,B1); m(B3,B1) ; m( B2,O) makes equilateral triangle. [4] m(B6,O) ; m(B5,B7);m( B5,B8) makes equilateral triangle. [5] m(B5,B8) ; m( B7,B8);m( B2,B7) makes equilateral triangle. [6] m(B2,B7) ; m(B2,B8); m( B1,O) makes equilateral triangle & m( B4,B7) ;m(O,B5); m(O,B6) makes equilateral triangle Similarly we can observe m(B3,B8); m(O,B1); m( O,B2)makes equilateral triangle. [7] m{B6,m(B5,O)} ; m{B5,m(B6,O)} ;m{m(B4,B6);m(B5,B7)} makes equilateral triangle. [8] (A1A4),(A2A3),(B7B8), (B3B4),are parallel & Similarly (B1,B2);(A1,A2); (A3,A4) ; (B5,B6) are parallel. [9] m(B4,B7); m( B2,B5) ;m(B3,B6) &m( B4,B7); m( B1,B6); m(B5,B8) makes equilateral triangle. [10] m(B2,B5); m(B3,B8); m(B1,B4) &m(B3,B8) ; m( B1,B6); m(B2,B7) makes 2 equilateral triangle (see Figure 2.0).

Best regards Jayendra Jha and Sankalp Savarn

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  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Glorfindel
    Oct 21 at 7:01
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    $\begingroup$ Hi Jayendra and welcome to the MathOverflow. I embedded the links to your work in the text: however, your post is still quite difficult to read. Pehaps the following Math Jax quick tutorial may be useful to you. $\endgroup$ Oct 21 at 9:58
  • $\begingroup$ Thanks Danielle Tampieri for your help. Yes you are right that my paragraph are difficult to read but if anyone reads it from link then they will find easy as in those link I have also give the images that will be helpful. And thanks for Math Jax quick tutorial as I am just start read and understand it . $\endgroup$ Oct 21 at 16:24
  • $\begingroup$ You should write by Latex and add Picture. I can not understand what you wrote. $\endgroup$ Oct 25 at 9:23

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