Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Learning Math Online
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic
Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
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

Sites for parents

Education & Parenting

Manifesto  |  Bookstore  |  Contents  |  Amazon store  |  Term index  |  What changed?  |  Contact  |  Recommend
RSS Feed: Recent changes at CTK

Cherchez le quadrilatère cyclique II: 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 © 1996-2010 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

The applet provides an illustration to a generalization of the following problem from an outstanding collection by T. Andreescu and R. Gelca:

  Consider a semicircle of center O and diameter AB. A line intersects AB at M and the semicircle at C and D in such a way that BM < AM and CM < DM. The circumcircles of triangles AOD and BOC intersect a second time at K. Show that KM and KO are perpendicular.

(This is one of the problems from the 1996 Balkan Mathematical Olympiad.)

 

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?

The generalization is twofold:

  • First, AB need not be a diameter of the given circle. Suffice it for the quadrilateral ABCD to be cyclic.

  • Second, the inequalities BM < AM and CM < DM can be done away with, although the proof then may have to be split into several cases.

The outline of the proof is as follows:

  • First we show that quadrilaterals ABLK and CDKL are cyclic. By this means we introduce two circles meeting at K and L and thus having KL as their radical axis. Adjoining the given circle gives a set of three circles with AB, CD, and KL pairwise radical axes implying M is a radical center of the three circles. Thus L lies on KM.

  • Due to the first step, we need now prove that KO is perpendicular to KL (instead of KM.)

First step: ABLK and CDKL are cyclic quadrilaterals. Indeed,

 
∠AKB= ∠AKO + ∠BKO
 = ∠ADO + ∠BCO
 = (180o - ∠AOD)/2 + (180o - ∠BOC)/2
 = 180o - ∠AOD/2 - ∠BOC/2
 = 180o - arcAD/2 - arcBC/2
 = 180o - ∠ABD - ∠BAC
 = 180o - ∠ABL - ∠BAL
 = ∠ALB,

where (and also later on) the arcs are those of the given circle. This says that quadrilateral ABLK is cyclic. Using the cyclicity of ABLK we proceed to establish the cyclicity of CDKL:

 
∠DKL= 360o - ∠AKD - ∠AKL
 = 360o - ∠AOD - (180o - ∠ABL)
 = 180o - ∠AOD + ∠ABL
 = 180o - ∠AOD + ∠ABD
 = 180o - arcAD + arcAD/2
 = 180o - arcAD/2
 = 180o - ∠ACD
 = 180o - ∠LCD.

Wherefrom we obtain the identity

  ∠DKL + ∠DCL = 180o,

so that quadrilateral CDKL is also cyclic. Thus L does lie on KM.

Second step: To prove the orthogonality of KM and KO note that

  ∠AKL + ∠ABL = 180o

implying that orthogonality of KL and KO is equivalent to

(1) ∠AKO + ∠ABL = 90o.

But quadrilateral ADKO is cyclic: ∠AKO = ∠ADO. Now from (1):

 
∠AKO + ∠ABL= ∠ADO + ∠ABD
 = (180o - ∠AOD)/2 + ∠ABD
 = 90o - arcAD/2 + arcAD/2
 = 90o,

as required. KO KL.

References

  1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.3.12 (p. 13)

Radical Axis and Radical Center

Copyright © 1996-2010 Alexander Bogomolny

35697415Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK