# Can you go backwards from a Morley triangle?

Definition 1. If T is a triangular, allow E (T) be the triangular created by the junctions of the surrounding trisectors of the (inside) angles of T. Synonymously, E (T) will certainly be called the Morley triangular for T.

Remark. By the theory called "Morley' s Miracle", E (T) is constantly equilateral.

Inspiration: I assume it is a rather all-natural and also ideal inquiry to ask in its very own right, yet actually for me it is a tipping rock (given that the solution remains in the affirmative) of an additional inquiry that is the one that I REALLY intend to ask. I do nt recognize whether it is ideal for a customer to "cascade" inquiries in a solitary blog post, yet I' ll proceed and also do so. If it is not ideal, please allow me recognize. Because instance the inquiry that adheres to can be overlooked as a main inquiry in the meantime, and also considered as merely component of the background/motivation for the here and now inquiry.

While including access to the "what stunned you" area wiki, I saw that of the access got on covering an equilateral triangular with squares. Below is the link:

The heads - up on this remarkable outcome ("One shock for me-- What is the optimum means to cover an equilateral triangular with 2 squares? It was nt addressed appropriately till 2009. http://www2.stetson.edu/~efriedma/squcotri/") results from Silas Pike.

I quickly asked yourself whether there was a means to connect this to Morley' s Miracle. So, below is what I thought of:

Definition 2. For each and every favorable integer n, allow f (n) be the topmost favorable number s such that an equilateral triangular of side of size s can be covered by n device squares.

Definition 3. For each and every favorable integer n, allow g (n) be the location of any kind of triangular T such that E (T) has side of size f (n). (Notice that for g to be well - specified calls for an affirmative response to the main inquiry of this blog post.)

Inquiry: What is the actions of g? (i.e., "What does its chart appearance like?") Is it logarithmic, or rapid, or what? THAT' S what I actually intended to ask.

Regards,

Mike Jones

22. May.2011 (Beijing time)

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2022-06-07 14:32:15
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The straightforward solution is no. Start with any kind of crooked triangular, and also locate its Morley triangular. There is an equilateral triangular which additionally offers the very same Morley triangular (simply place a triangular with angles $4\pi/9$, $4\pi/9$, and also $\pi/9$ on each side of the Morley triangular to locate its vertices), so the Morley triangular does not specify the form of the initial triangular.