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 Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "Property of the line through incenter and circumcenter"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange High school Topic #348
Reading Topic #348
alexbadmin
Charter Member
1897 posts
Aug-04-06, 08:43 AM (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  
"Property of the line through incenter and circumcenter"
 
   R. Honsberger in From Erdos to Kiev, pp 199-201 treats a problem from Crux Mathematicorum, 1987, 160:

"As usual, let I and O be the incenter and circumcenter, respectively, of ÄABC. Suppose angle C is 30°, and that the side AB is laid off along each of the other two sides to give points D and E so that

EA = AB = BD.

Prove that the segment DE is both equal and perpendicular to IO."

Now, I think that the problem is a mixture of two:


  1. DE is perpendicular to IO,
  2. DE = IO.

The first one is independent of the magnitude of angle C: DE is perpendicular to IO anyway. The second one is the consequence of the angle requirement.

As a preliminary, the construction of points D and E can be repeated for the other two sides AC and BC, giving three lines. Perhaps, curiously, the three are parallel, see

https://www.cut-the-knot.org/Curriculum/Geometry/ThreeParallelsInTriangle.shtml


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
alexbadmin
Charter Member
1897 posts
Aug-05-06, 08:29 AM (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: Property of the line through incenter and circumcenter"
In response to message #0
 
   See solutions at

https://www.cut-the-knot.org/Curriculum/Geometry/InAndCircumcenter.shtml


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
me
guest
Oct-13-06, 11:12 AM (EST)
 
2. "RE: Property of the line through incenter and circumcenter"
In response to message #0
 
   If K, L, M in BC, CA, AB are feet of the external bisectors of the angles A, B, C, then KB/KC*LC/LA*MA/MB = b/c*c/a*a/b = 1 and K, L, M are collinear by Menelaus theorem. AL = bc/(c - a), AM = bc/(b - a) and AM/AL = (c - a)/(b - a) = AD/AE, triangles ADE, AML are centrally similar and DE parallel to LM, same as KLM. If I_a, I_b, I_c are excenters of the triangle ABC opposite to A, B, C, then I_aA, I_bB, I_cC are altitudes of the excentral triangle I_aI_bI_c, I its orthocenter, O its 9-point circle center, and the reflection O' of I in O its circumcenter, O'OI its Euler line. Quadrilaterals I_bI_cBC, I_cI_aCA, I_aI_bAB are all cyclic because of the right angles, KB*KC = KI_b*KI_c, LC*LA = LI_c*LI_a, MA*MB = MI_a*MI_b. Therefore KLM is radical axis of the circumcircles (O), (O') of the triangles ABC, I_aI_bI_c, perpendicular to the center line O'O, same as O'OI. In conclusion, DE is parallel to KLM, which is perpendicular to O'OI.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
alexbadmin
Charter Member
1897 posts
Oct-15-06, 07:50 AM (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  
3. "RE: Property of the line through incenter and circumcenter"
In response to message #2
 
   Beautiful. Thank you.


  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