Three Points Casey's Theorem

Circle (S) is tangent to the circumcircle of points A, B, C. ATa, BTb, CTc are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·ATa + AC·BTb = AB·CTc.


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.


What if applet does not run?

Solution

References

  1. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 2, Moscow, 1952, #123(a).

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Circle (S) is tangent to the circumcircle of points A, B, C. ATa, BTb, CTc are tangent to (S). Assume that M - the point of tangency of the two circles - is in the arc AB. Prove that

BC·ATa + AC·BTb = AB·CTc.


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.


What if applet does not run?

Solution

The problem is just a particular instance of Casey's theorem, wherein three of the four circles degenerate into points. Still, I'll give a proof independent of Casey's theorem.

Point M serves the center of similitude of the two circles. If A', B', C' are the second points of intersection of AM, BM, CM, with (S) then

A'M/AM = B'M/BM = C'M/CM = ρ/r

where r and ρ are the radii of the circumcircle (ABC) and that of (S), respectively. It follows that

AA'/AM = BB'/BM = CC'/CM = (r ± ρ)/ρ.

(The sign is chosen depending on whether the circumcircle (ABC) is outside or inside (S).)

On the other hand, by the Power of a Point Theorem,

AM·AA' = (ATa)²,
BM·BB' = (BTb)²,
CM·CC' = (CTc)²,

such that

AM² / (ATa)² = AM / AA',
BM² / (BTb)² = BM / BB',
CM² / (CTc)² = CM / CC',

implying

AM / ATa = BM / BTb = CM / CTc = ρ / (r ± ρ),

which simply shows that the required identity

BC·ATa + AC·BTb = AB·CTc.

is a direct consequence of that of Ptolemy's applied to the quadrilateral ABCM:

BC·AM + AC·BM = AB·CM.

Ptolemy's Theorem

  1. Ptolemy's Theorem
  2. Sine, Cosine, and Ptolemy's Theorem
  3. Useful Identities Among Complex Numbers
  4. Ptolemy on Hinges
  5. Thébault's Problem III
  6. Van Schooten's and Pompeiu's Theorems
  7. Ptolemy by Inversion
  8. Brahmagupta-Mahavira Identities
  9. Casey's Theorem
  10. Three Points Casey's Theorem
  11. Ptolemy via Cross-Ratio
  12. Ptolemy Theorem - Proof Without Word
  13. Carnot's Theorem from Ptolemy's Theorem

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 61156620

Search by google: