play and relax: games for kids games
  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

Friendly Kiepert's Perspectors


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?

Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C. Bottema's theorem asserts that the midpoint M of the segment AbBa is independent of C. The point M also serves as the apex of two right-angled (at M) isosceles triangles: AMB and AcMBc. In the proof of Bottema's theorem we have established an identity

(1)(Ab + Ba)/2 = (a + b)/2 + (b - a)·i/2,

a and b are complex number representations of vertices A and B, respectively.

We have the following generalization due to Floor van Lamoen.

Fix a real number t and construct four points:

(tl)
Xl = Xl(t) = B + t·(Ab - B),
Yl = Yl(t) = A + t·(Ba - A)

and

(tu)
Xu = Xu(t) = Ac + 1/t·(Ab - Ac),
Yu = Yu(t) = Bc + 1/t·(Ba - Bc).

Define Ml and Mu as the midpoints of XlYl and XuYu. Then the three points C, Ml and Mu are collinear.

The proof generalizes that of Bottema's theorem. Following the latter it is using complex numbers. a and b have been already introduced as complex numbers corresponding to points A and B. Let g correspond to C, and ml, mu, xl, xu, yl, yu correspond to Ml, Mu, Xl, Xu, Yl, and Yu.

(tl) and (tu) could be rewritten as

(t'l)
xl = b - i·t·(g - b),
yl = a + i·t·(g - a)

and

(t'u)
xu = (g + i·(g - a)) + 1/t·(g - a),
yu = (g - i·(g - b)) + 1/t·(g - b).

To prove that Ml, Mu, and C are collinear, we only have to show that the ratio

 
K = (ml - g) / (mu - g)

is real. However from (t'l)

(2l) ml - g = (a + b)/2 + i·t·(b - a)/2 - g.

On the other hand, from (t'u)

(2u) mu - g = g - i·(b - a)/2 - g/t + (a + b)/2t - g
= 1/t·((a + b)/2 + i·t·(b - a)/2 - g)

It is thus clear that K = t, a real number.

(The applet suggests a different proof based on the observations that XlYl||XuYu and angles MlBA and XlCB are equal. This proof elucidates the meaning of t: t = tan(XlCB) = tan(MlAB), whereas 1/t = tan(XuCAc) = tan(MuAcBc).)

It follows that the Kiepert perspectors K(f) and K(p/2 - f) are related by friendship.

References

  1. F. van Lamoen, Friendship Among Triangle Centers, Forum Geometricorum 1 (2001), pp. 1-6.

Copyright © 1996-2008 Alexander Bogomolny

29397579Page copy protected against web site content infringement by Copyscape


Search:
Keywords:


Latest on CTK Exchange
try this puzzle ?/?? + ?/?? + ?/? ...
Posted by albert1950
0 messages
07:05 AM, Jul-10-08

Monty Hall Problem
Posted by linkdon
71 messages
03:06 PM, Jul-18-08

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

Golden Number linked to 666
Posted by Titus
4 messages
02:53 PM, Jul-19-08

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

trisection of an angle
Posted by Pytheus
2 messages
06:39 AM, Jul-14-08

You can drill a square hole
Posted by Giorgis
1 messages
10:15 PM, Jul-12-08