Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Learn to enjoy mathematics.
Google
Web CTK
Best sites for teachers
Sites for teachers
Sites for parents
Terms of use
Awards

Interactive Activities
CTK Exchange
CTK Insights - a blog

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Best sites for teachers
Sites for teachers
Sites for parents

Education & Parenting

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

Reflections of a Line Through the Orthocenter: What is this about?
A Mathematical Droodle


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this applet
What if applet does not run?

Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The following result is due to L. Carnot (1801):

 

Reflect a line passing through the orthocenter H of ABC in the three sidelines. The three reflections concur on the circumcircle of the triangle.

The problem quite easily solved by "angle chasing" - counting angles which, due to the layout of the problem, come in just several related sizes. However, configurations differ in details and several cases ought to be considered. That said, the idea of the proof can be introduced with a particular case:

Assume a line through the orthocenter H of ABC crosses AC at Tb and BC at Ta. Denote angle CTbH as b and angle BTaH as a. Then

(1) a + b + ACB = 180o.

As we know, reflection Ka of the orthocenter H in BC lies on the circumcircle, as does its reflection Kb in AC. Next observe that the lines TaKa and TbKb are exactly the reflections of the given line in BC and AC, respectively. Assume the two meet in Q and let's show that Q lies on the circumcircle of the given triangle.

In QTaT angle Ta is obviously 180o - 2a, while angle Tb measures 180o - 2b. Thus

 
TaQTb = 180o - (180o - 2a) - (180o - 2b)
  = 2(a + b) - 180o
  = 2(180o - ACB) - 180o
  = 2(90o - ACB),

where we applied (1) on the last step. But this is exactly half the angular measure of arc KaCKb that subtends the angle KaQKb. The latter is therefore inscribed into the circumcircle, so that indeed this is where Q lies and TaKa and TbKb meet.

There are two more pairs of reflections to consider, but since each reflection participates in two pairs but intersects the circle only once (not counting the points K), all three pairs meet in the same point which is bound to be Q.

Copyright © 1996-2008 Alexander Bogomolny

29284644Page copy protected against web site content infringement by Copyscape


Search:
Keywords:


Latest on CTK Exchange
calculator suitable for high scho ...
Posted by albert1950
1 messages
10:42 AM, Jun-17-08

Constucting a triangle instructions
Posted by Gerald B.
3 messages
01:32 PM, May-20-08

Missing information
Posted by roboknight
2 messages
07:32 AM, Jun-22-08

An Interesting Formula And Algorithm
Posted by ddixonslc
1 messages
01:44 PM, Jun-19-08

Mistake on the page (an aside, Be ...
Posted by Max
4 messages
10:28 AM, Feb-28-08

Statistical estimation question
Posted by Ralph
2 messages
02:21 PM, Jul-01-08

fusc pseudocode
Posted by azi
1 messages
08:02 PM, Jun-29-08