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

Tangent Circles and an Isosceles Triangle II
Solution by Vladimir

The applet presents a solution to the following Sangaku problem: Given a circle S with center O and diameter AC and point B on AC. Form circle G with center P and diameter AB and an isosceles triangle BCE with E on the circle S. Circle W with center Q is inscribed in the curvilinear triangle formed by circles S and G and the line BE. Prove that QB is perpendicular to AC.

(Another solution appears elsewhere. The one below is remarkable in an absolute absence of calculations.)

 

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?

Reformulate the problem a little: (O), (P) are circles with diameters AC > AB tangent at A. Perpendicular bisector of BC cuts (O) at E, F, so that triangle BEC is isosceles (or BECF is a rhombus). Circle (Q) is internally / externally tangent to (O), (P) and to BE. Prove BQ is perpendicular to AC.

EB, FB cut (O) again at K, L. FE bisects angle ∠CFL = ∠CFB, implying EL = EC = EB so that B is the incenter of ΔCKL. Perpendicular to AC at B cuts (O) at M, N. By construction, (O) cuts circle (B) with center B and diameter MN in diametrically opposite points which means that the inversion in (B) with negative power -BM² takes S into itself. Line EBK through the inversion center also goes to itself. Power of B to (O) is -BM² = BM × BN = BA × BC, meaning that the inversion takes A to C. Circle (P) with diameter AB goes to line perpendicular to AC at C, the external bisector of the angle ∠KCL. Let Ik, Il be K-, L-excenters of ΔCKL. The inverse image (Q') of (Q) is tangent to IkIl, to BE = BIk, and (by Feuerbach's Theorem) externally tangent to the 9-point circle (O) of the triangle BIkIl, therefore (Q') is its excircle in the angle BIkIl, centered on the external bisector of the angle EBF, i.e., the angle IkBIl, perpendicular to the internal bisector BC of this angle, therefore BQQ' is perpendicular to AC.

This permits generalization to a non-isosceles triangle CKL with circumcircle (O) and incenter B and angle bisector CBA. (P) is circle with diameter AB, (Q) is circle internally tangent to (O), externally tangent to (P), and to the angle bisector KBE. No change in the proof that (O) is also tangent to the other angle bisector LBF.

Another solution makes use of the common inversion.

Inversion - Introduction

Copyright © 1996-2010 Alexander Bogomolny

35705121Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK