CTK Exchange
Front Page
Movie shortcuts
Personal info
Awards
Reciprocal links
Terms of use
Privacy Policy

Interactive Activities

Cut The Knot!
MSET99 Talk
Games & Puzzles
Arithmetic/Algebra
Geometry
Probability
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
My Logo
Math Poll
Other Math sit's
Guest book
News sit's

Recommend this site

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Products to download and subscription Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "Morley's theorem. little error on comment on proof by Zsolt"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange Thoughts and Suggestions Topic #29
Reading Topic #29
Amic
guest
Jan-31-08, 10:57 AM (EST)
 
"Morley's theorem. little error on comment on proof by Zsolt"
 
   On this page :

https://www.cut-the-knot.org/triangle/Morley/MorleyZsolt.shtml

I read
"
There are two special cases that should be mentioned. There is the case a = b = c = 20, when the theorem is obvious. The other case is when two of a, b, c add up to 40 without being equal. In this case, two of the six points coincide but the other four are distinct so the triangle ABC can still be constructed from them.
"

There is a mistake : the other case is when a or b or c=30, and in fact we cannot define the line joining two points which are the same...

And I don't see why the case a=b=c=20 is a special case...

So in my opinion this proof is wrong for a triangle ABC with a right angle.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
alexbadmin
Charter Member
2167 posts
Jan-31-08, 12:14 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
1. "RE: Morley's theorem"
In response to message #0
 
   You are right on both accounts: the special case is when one of the angles is 30°, not when all of them are equal or two add up to 40°.

In the exceptional case, the proof needs an amendment: one of the lines should be drawn not through two points (which coincide and do not determine a line) but through a (double) point at computable angles to the bases of the two adjacent trapezoids.

I made a remark to that effect at the bottom of the page.

Many thanks for bringing this up.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

Copyright © 1996-2018 Alexander Bogomolny

Search:
Keywords:

Google
Web CTK